13. Previous Year Questions: Our Environment

Previous Year Questions 2025

Q1: Which of the following groups do not constitute a food chain?  (1 Mark)
(i) Wolf, rabbit, grass, lion
(ii) Plankton, man, grasshopper, fish
(iii) Hawk, grass, snake, grasshopper, frog
(iv) Grass, snake, wolf, tiger
(a) 
(i) and (iv)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (ii) and (iv)

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Ans: (c)
food chain shows a sequence of organisms where each is eaten by the next one.
In options (ii) and (iii), the order of organisms is incorrect — they do not represent a proper feeding relationship (for example, man does not eat grasshopper or fish directly, and grass cannot be eaten by a hawk).
Hence, (ii) and (iii) do not constitute a food chain.

Q2: The percentage of solar energy which is not converted into food energy by the leaves of green plants in a terrestrial ecosystem is about:  (1 Mark)

(a) 1%
(b) 10%
(c) 90%
(d) 99%

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Ans: (d)
Only about 1% of the sunlight that falls on green plants is converted into food energy, so 99% of solar energy remains unutilised.

Q3: Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below

Assertion (A): The amount of ozone in the atmosphere began to drop sharply in the 1980s. 
Reason (R): The oxygen atoms combine with molecular oxygen to form ozone.  (1 Mark)

(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

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Ans: (b)Ozone levels dropped sharply in the 1980s due to synthetic chemicals like CFCs, not because oxygen atoms combine with molecular oxygen (which actually forms ozone).

Q4: “Excessive use of chemicals and pesticides in agriculture adversely affects the environment.” Justify this statement.  (2 Marks)

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Ans: Excessive use of pesticides and chemicals in agriculture leads to their accumulation in soil and water. These chemicals enter the food chain through plants and aquatic organisms and are not degradable. They get progressively accumulated at each trophic level, a process called biological magnification, causing harmful effects on living organisms and the environment.

Q5: Identify from the following a group containing all non-biodegradable substances.  (1 Mark)

(a) Leather, Glass, Plastic
(b) Cotton, Wood, Nylon
(c) DDT, Polyester, Glass
(d) Leather, Silk, Wool

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Ans: (c)
DDT, polyester, and glass cannot be broken down by biological processes; hence, they are non-biodegradable substances.

Q6: Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below (1 Mark)

Assertion (A): Animals will not get energy if they eat (consume) coal as food. 
Reason (R): Specific enzymes are needed for the breakdown of a particular food.

(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

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Ans: (a)

Animals cannot get energy from coal because enzymes are specific and only break down particular biological substances. Since coal cannot be broken down by these enzymes, it cannot provide energy.

Q7: Some harmful chemicals get accumulated in human bodies through the food chain. Name this phenomenon. Explain the reason for the maximum concentration of these chemicals found in our bodies.  (3 Marks)

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Ans: This phenomenon is called biological magnification.

Harmful chemicals such as pesticides and fertilizers enter the food chain through soil and water. These chemicals are not degradable and thus accumulate in the tissues of organisms.At each successive trophic level, their concentration increases because they are passed on through food. Since human beings occupy the top level in the food chain, the maximum concentration of these chemicals accumulates in our bodies.Q7: Some harmful chemicals get accumulated in human bodies through the food chain. Name this phenomenon. Explain the reason for the maximum concentration of these chemicals found in our bodies.  (3 Marks)

Q8: Other than the abiotic components, which of the given biotic components are not required to make an aquarium with small herbivorous fishes a self-sustaining system?  (1 Mark)
(i) Aquatic plants and aquatic animals
(ii) Terrestrial plants and terrestrial animals
(iii) Decomposers as bacteria and fungi
(iv) Consumers as clown fishes and sea urchins
(a)
 (i) and (iv)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (ii) and (iv)

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Ans: (d)

self-sustaining aquarium needs aquatic plants (producers), small herbivorous fishes (consumers), and decomposers. Terrestrial plants/animals and large consumers like sea urchins are not required in such an aquatic ecosystem.

Q9: Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below (1 Mark)
Assertion (A): Use of jute bags for shopping reduces pollution. 
Reason (R): Jute is biodegradable and its bag may be reused as and when needed.

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(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Ans: (a)

  • Using jute bags helps reduce pollution because jute is biodegradable and such bags can be reused, reducing non-biodegradable plastic waste.

Q10: In the food chains given below, select the most efficient food chain in terms of energy.  (1 Mark)

(a) Grass → Grasshopper → Frog → Snake
(b) Plants → Deer → Lion
(c) Plants → Man
(d) Phytoplankton → Zooplankton → Small Fish → Big Fish

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Ans: (c)

The shorter the food chain, the less energy is lost between trophic levels. Since this chain has only two trophic levels, it is the most efficient in terms of energy transfer.

Q11: What are decomposers? Give two examples. State how they maintain a balance in an ecosystem.  (3 Mark)

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Ans: Decomposers are microorganisms that break down dead plants, animals, and waste materialsinto simple inorganic substances. 

Examples: Bacteria and fungi

Role in maintaining balance:

  • They recycle nutrients by converting complex organic matter into simple substances that return to the soil and are used again by plants
  • This process helps in the natural replenishment of the soil and maintains the balance in the ecosystem.

Q12: The examples of natural and manmade (artificial) ecosystems are, respectively:  (1 Mark)

(a) Forests and ponds
(b) Crop fields and lakes
(c) Lakes and gardens
(d) Crop fields and forests

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Ans: (c)Lakes are natural ecosystems, while gardens are human-made (artificial) ecosystems maintained by people.

Q13: Human activities that are affecting the environment are:  (1 Mark)

(a) Minimising the use of chlorofluorocarbons
(b) Excessive use of disposable cups and plates
(c) Maximising the use of reusable utensils for eating food and drinking fluids
(d) Segregating the wastes into biodegradable and non-biodegradable before disposal

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Ans: (b)

Using disposable cups and plates increases non-biodegradable waste, leading to environmental pollution and harming ecosystems.

Q14: (a) Why are the organisms of first trophic level important in any food chain? 
(b) Justify the following statement: ‘The flow of energy in an ecosystem is unidirectional.’  (2 Marks)

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Ans:
(a) The organisms of the first trophic level, known as producers, are very important because they absorb solar energy and convert it into chemical energy through photosynthesis. This stored energy in food becomes the source of energy for all other organisms in the food chain such as herbivores, carnivores, and decomposers. Without producers, no other organisms could survive, as they depend directly or indirectly on them for food. 

(b) The flow of energy in an ecosystem is unidirectional, meaning it moves in one direction only — from the Sun → producers → consumers → decomposers. The energy captured by producers does not return to the Sun, nor can it go backward in the chain. At each trophic level, some energy is lost as heat, so the energy available to the next level keeps decreasing, making the flow one-way and non-cyclic.Q14: (a) Why are the organisms of first trophic level important in any food chain? (b) Justify the following statement: `The flow of energy in an ecosystem is unidirectional.`  (2 Marks)

Q15: A gas ‘X’ is found in the upper layer of atmosphere. It is a deadly poison, but still essential for all life forms on earth. 
(a) Identify the gas and state the main factor for its depletion in the atmosphere. 
(b) How is this gas formed in the upper atmosphere?  (2 Marks)

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Ans:

(a) The gas ‘X’ is ozone (O₃). It is getting depleted mainly due to synthetic chemicals like chlorofluorocarbons (CFCs) used in refrigerants and fire extinguishers

(b) In the upper atmosphere, ozone is formed when ultraviolet (UV) radiations split some molecules of oxygen (O₂) into free oxygen atoms (O), which then combine with molecular oxygen to form ozone (O₃).

Q16: Study the food web given below: 

(a) Identify the food chain(s) in which the eagle receives the highest energy from the producers.
(b) Identify the organism in which a non-biodegradable pesticide will be found in maximum concentration. Name the term used for this phenomenon.  (2 Marks)

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Ans:

(a) The food chain in which the eagle receives the highest energy from producers is:
Grass → Mouse → Eagle.
This is because the chain is shortest, involving fewer trophic levels, so there is less energy loss between steps, and the eagle receives more energy from the producers.

(b) The organism in which a non-biodegradable pesticide will be found in maximum concentration is the Eagle.
This is due to biological magnification, where harmful chemicals accumulate progressively at each trophic level, reaching the highest concentration in top consumers.

Q17: Consider the following food chain: 
Grass → Grasshopper → Frog → Snake → Eagle. 
If the amount of energy available at third trophic level is 50 kJ, the available energy at the producer level was:  (1 Mark)

(a) 0.5 kJ
(b) 5 kJ
(c) 500 kJ
(d) 5000 kJ

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Ans: (d)

According to the 10% law, only 10% of energy is transferred from one trophic level to the next.
If the third trophic level (frog) has 50 kJ, then:
Producer → 1st → 2nd → 3rd
Energy at producer level = 50 × 10 × 10 × 10 = 50,000 kJ.

Q18: The incorrect statement about ozone is:  (1 Mark)

(a) It is a deadly poisonous gas.
(b) It shields the surface of the Earth from UV radiation from sun.
(c) It is used as a refrigerant and in fire extinguishers.
(d) It is formed by combining an oxygen molecule with a free oxygen atom.

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Ans: (c)

Ozone itself is not used as a refrigerant or in fire extinguishers; CFCs (chlorofluorocarbons) are.
Ozone protects the Earth from harmful UV radiation.

Q19: Two statements are given – one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :
Assertion (A): Food web is a network of several food chains operating in an ecosystem. 
Reason (R): Food web decreases the stability of an ecosystem.  (1 Mark)

(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

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Ans: (c)

  • food web is indeed a network of many interconnected food chains, but it actually increases the stability of an ecosystem, not decreases it.

Q20: We do not clean natural ponds or lakes, whereas an aquarium or a swimming pool needs to be cleaned regularly. Why?  (2 Marks)

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Ans: Natural ponds and lakes are self-sustaining ecosystems that contain producers, consumers, and decomposers. These components interact and maintain a natural balance by recycling nutrients and cleaning the system on their own.

An aquarium or swimming pool, however, is a human-made system that lacks sufficient decomposers and natural processes of purification. Hence, it needs to be cleaned regularly to remove waste and maintain a healthy environment. 

Q21: Green plants occupy the first trophic level in every food chain because they:  (1 Mark)

(a) exist over a large area.
(b) have very less concentration of harmful chemicals.
(c) have to feed large number of herbivores.
(d) can synthesize food by photosynthesis.

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Ans: (d)

Green plants are producers; they make their own food from inorganic substances using sunlight through photosynthesis, so they always occupy the first trophic level in a food chain.

Previous Year Questions 2024

Q1: For Q. Nos., two statements are given – One labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:    (1 Mark) (2024)
Assertion (A): Accumulation of harmful chemicals is maximum in the organisms at the highest trophic level of a food chain.
Reason (R): Harmful chemicals are sprayed on the crops to protect them from diseases and pests.
(a) 
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) 
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) 
Assertion (A) is true, but Reason (R) is false.
(d) 
Assertion (A) is false, but Reason (R) is true.

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Ans: (b)
Harmful chemicals do accumulate in organisms at the highest trophic level due to a process called biomagnification, and these chemicals are often sprayed on crops to protect them. However, the reason does not directly explain why the accumulation is highest at the top of the food chain, so it’s not a correct explanation of the assertion.


Q2: (A) Plants -> Deer -> Lion.    (1 to 3 Marks)  (2024)
In the given food chain, what will be the impact of removing all the organisms of the second trophic level on the first and third trophic levels? Will the impact be the same for the organisms of the third trophic level in the above food chain if they were present in a food web? Justify.
OR
(B) A gas ‘X’ which is a deadly poison is found at the higher levels of the atmosphere and performs an essential function. Name the gas and write the function performed by this gas in the atmosphere. Which chemical is linked to the decrease in the level of this gas? What measures have been taken by an international organization to check the depletion of the layer containing this gas? 

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Ans: (A)

  • Number of plants/organisms of first trophic level will increase. 
  • Number of lions/ organisms of third trophic level will decrease.
  • No
  • As the organisms of that level will find alternative foods and will not starve to death / food web is more stable where other animals as prey may be available.

OR

  • Gas ‘X’ is Ozone.
  • Ozone shields the surface of the earth from ultra-violet (UV) radiations from the sun. 
  • CFCs (Chlorofluorocarbons)
  • Succeeded in forging an agreement to freeze CFC production at 1986 levels / Manufacturing of CFC free refrigerators


Q3: Name the term used for the materials which cannot be broken down by biological processes. Give two ways by which they harm various components of an ecosystem.    (1 to 3 Marks) (2024)

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Ans: 

  • Non-biodegradable substances 
  • Two ways: 
    (i) They are inert and persist in the environment for long time and cause pollution.
    (ii) Cause Biological magnification
    (iii) Affect the fertility of soil 


Q4: Use of pesticides to protect our crops affects organisms at various trophic levels especially human beings. Name the phenomenon involved and explain how does it happen.    (1 to 3 Marks) (2024)

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Ans: Phenomenon – Biological Magnification /Biomagnification:

  • Pesticides are washed down into the soil and water bodies. 
  • From the soil pesticides are absorbed by crop plants along with water and minerals  and enter the food chain.
  • These chemicals are non-biodegradable and get accumulated progressively at each trophic level. 
  • As human beings occupy the top level in any food chain, the maximum concentration of these chemicals gets accumulated in our bodies.


Q5: Consider the following statements about ozone:    (1 to 3 Marks) (2024)
(A) Ozone is poisonous gas. 
(B) Ozone shields the earth’s surface from the infrared radiation from the sun.
(C) Ozone is a product of UV radiations acting on oxygen molecule. 
(D) At the lower level of the earth’s atmosphere, ozone performs most essential function.
The correct statements are
(a)
 (A) and (B)
(b) (A) and (C)
(c) (B) and (C)
(d) (B) and (D)

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Ans: (b)
Ozone can be harmful at ground level, making it a poisonous gas. Statement (C) is also correct since ozone is formed when ultraviolet (UV) radiation interacts with oxygen molecules. However, statement (B) is incorrect because ozone protects the Earth from UV radiation, not infrared radiation, and statement (D) is misleading because ozone’s essential functions are primarily in the upper atmosphere, not at lower levels.


Q6: Assertion (A) and Reason (R), answer these questions selecting the appropriate option given below:   (1 Mark) (2024)
Assertion (A): The waste we generate daily may be biodegradable or non-biodegradable.
Reason (R): The waste generated, if not disposed off properly may cause serious environmental problems.
(a)
 Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

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Ans: (b)

The answer is (b) because both statements are true: waste can be either biodegradable (which can break down naturally) or non-biodegradable (which cannot), and improper disposal of waste can indeed lead to serious environmental issues. However, the reason does not explain why waste is categorized as biodegradable or non-biodegradable, so it’s not a direct explanation of the assertion.


Q7: What are decomposers? List two consequences of their absence in an ecosystem.   (1 to 3 Marks)  (2024)

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Ans: Decomposers are the microorganisms that break-down the complex organic substances into simple inorganic substances.
Consequences:
(i) No replenishment of soil
(ii) Foul smell
(iii) Breeding of flies
(iv) Accumulation of dead plants and animals in the environment.
(v) No recycling of nutrients


Q8: We water the soil but it reaches the topmost leaves of the plants. Explain in brief the process involved.    (1 to 3 Marks)  (2024)

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Ans: When water is lost through stomata in the leaves by transpiration, it creates a suction force/transpiration pull, due to which water is pulled up through xylem of the roots to the leaves.


Q9: Some wastes are given below:     (1 to 3 Marks)  (2024)
(i) Garden waste    
(ii) Ball point pen refills   
(iii) Empty medicine bottles made of glass    
(iv) Peels of fruits and vegetables  
(v) Old cotton shirt  
The non-biodegradable wastes among these are:  
(a) 
(i) and (ii)    
(b) (ii) and (iii)  
(c) (i), (iv) and (v)  
(d) (i), (iii) and (iv)  

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Ans: (b)
Non-biodegradable wastes are those that do not break down naturally in the environment. In the given options, ballpoint pen refills (ii) and empty medicine bottles made of glass (iii) are non-biodegradable because they can persist in the environment for a long time. The other items like garden waste, fruit and vegetable peels, and old cotton shirts decompose naturally and are considered biodegradable.


Q10: Two statements are given  one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.    (1 Mark) (2024)
Assertion (A): Oxygen is essential for all aerobic forms of life.
Reason (R): Free oxygen atoms combine with molecular oxygen to form ozone.
(a) 
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).  
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.  
(d) Assertion (A) is false, but Reason (R) is true.  

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Ans: (b)
Assertion (A) correctly states that oxygen is essential for aerobic life forms, while Reason (R) explains a process involving oxygen and ozone formation, which is also true. However, Reason (R) does not explain why oxygen is essential for life, so the correct answer is (b): both are true, but Reason (R) does not explain Assertion (A).


Q11: Which one of the following is not a natural ecosystem?    (1 Mark) (2024)
(a)
 Pond ecosystem
(b) Grassland ecosystem
(c) Forest ecosystem
(d) Cropland ecosystem

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Ans: (d)
A natural ecosystem is one that occurs naturally without human interference. Among the options given, a cropland ecosystem (d) is not natural because it is created and managed by humans for farming purposes. In contrast, pond, grassland, and forest ecosystems develop naturally in the environment.


Q12: Differentiate between food chain and food web. In a food chain consisting of deer, grass and tiger, if the population of deer decreases, what will happen to the population of organisms belonging to the first and third trophic levels?     (3 Marks) (2024)

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Ans:
Previous Year Questions 2024

  • Population of grass/ first trophic level will increase.
  • Population of tiger/ third trophic level will decrease.


Q13: Identify the food chain in which the organisms of the second trophic level are missing:  (1 Mark) (2024)
(a) 
Grass, goat, lion
(b) Zooplankton, Phytoplankton, small fish, large fish
(c) Tiger, grass, snake, frog
(d) Grasshopper, grass, snake, frog, eagle

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Ans: (c)

In a food chain, the second trophic level usually consists of primary consumers that eat producers (like plants). In option (c), tiger, grass, snake, frog, there are no primary consumers (like a herbivore) between the grass (producer) and the snake. Instead, the snake directly feeds on frogs, which means the second trophic level is missing.


Q14: In which of the following organisms, multiple fission is a means of asexual reproduction?     (1 Mark) (2024)
(a) 
Yeast
(b) Leishmania
(c) Paramoecium
(d) Plasmodium

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Ans: (d)
Multiple fission is a type of asexual reproduction where an organism divides into many parts at once. In this case, Plasmodium (d), which causes malaria, reproduces through multiple fission in its life cycle, producing many daughter cells simultaneously. The other options like yeast, Leishmania, and Paramecium reproduce differently.


Q15: A food chain will be more advantageous in terms of energy if it has: 
(a) 2 trophic levels 
(b) 3 trophic levels 
(c) 4 trophic levels 
(d) 5 trophic levels      (1 Mark) (CBSE 2024)

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Ans: (a)
In a food chain, energy is lost at each trophic level due to metabolic processes, typically around 90% of energy is lost as heat, while only about 10% is transferred to the next level. Therefore, a food chain with fewer trophic levels will retain more energy available to the organisms at the higher level.
With 2 trophic levels (e.g., producers and primary consumers), there is minimal energy loss compared to longer chains, making it more advantageous in terms of energy efficiency.
Thus, the correct answer is (a) 2 trophic levels.

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Previous Year Questions 2023

Q1: Use of several pesticides which results in excessive accumulation of pesticides in rivers or ponds, is a matter of deep concern. Justify this statement. (3 Marks) (2023)

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Ans: The use of several pesticides results in accumulation of pesticides in rivers and ponds. These chemicals are either washed down into the soil or into the water bodies. From the soil, these are absorbed by the plants along with water and minerals, and from the water bodies these are taken up by aquatic plants and animals and enters the food chain. As these chemicals are not degradable, these get accumulated progressively at each trophic level. As human beings occupy the top level in any food chain, the maximum concentration of these chemicals get accumulated in our bodies i.e., biological magnification. Our food grains such as wheat and rice, vegetables and fruits, and even meat, contain varying amounts of pesticide residues cannot always be removed by washing or other means and causes health hazards.


Q2: How is ozone formed in the higher levels of the atmosphere? “Damage to the ozone layer is a cause of concern.” Justify this statement. (3 Marks) (2023)

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Ans: Ozone (O3) is a molecule formed by three atoms of oxygen. It is formed in the stratosphere layer of atmosphere when high energy UV rays act on O2 molecule splitting it into free oxygen (O) atoms. These atoms then combine with molecular oxygen (O2) to form ozone (O3).


Q3: “Although gardens are created by man but they are considered to be an ecosystem.” Justify this statement.  (2 Marks) (2023)

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Ans: In a garden, various plants like grasses, trees, flower bearing plants such as jasmine, sunflower, rose, and animals like insects, frogs and birds are found. All these living organisms interact with each other and their growth, reproduction and other activities are affected by the abiotic components such as light, water, wind, soil, minerals, etc. of ecosystem. Thus, a garden is considered to be an ecosystem.


Q4: What is the difference between biodegradable and non-biodegradable substances? List two methods of safe disposal of biodegradable domestic waste.   (2 Marks) (2023)

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Ans: Differences between biodegradable and non-biodegradable substances are as follows :
Previous Year Questions 2023
Domestic waste can be safely disposed off by composting. In composting, biodegradable domestic wastes, such as left-over food, fruit, vegetable peels, etc., can be buried in pit, dug into ground. They are converted into compost and used as manure.
In landfill a huge pit is made in an open low lying area and wastes are dumped into the pits. Once the pits are full, they are covered with soil and left for decomposition.


Q5: How do harmful chemicals get accumulated progressively at each trophic level in a food chain?  (3 Marks) (CBSE 2023)

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Ans: The process by which harmful substances enter the food chain and becomes concentrated at each trophic level is known as biomagnification.
Example:Biomagnification of DDT in an Aquatic Food ChainBiomagnification of DDT in an Aquatic Food Chain


Q6: (A) Why does a kitchen garden called an artificial ecosystem while a forest is considered to be a natural ecosystem? 
(B) While designing an artificial ecosystem at home, write any two things to be kept in mind to convert it into a selfsustaining system. Give reason to justify your answer.   (3 Marks) (CBSE 2023)

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Ans: (A) Kitchen gardens are referred to as artificial ecosystems because they are man-made and contain adjusted abiotic and biotic components. It is an ecosystem where plants are produced, including fruits and vegetables, whereas a forest is a place where various creatures live in harmony and rely on one another for food and other necessities. As a result, a forest creates an ecosystem that can support itself. 
(B) Two things to keep in mind while designing an artificial ecosystem: 
(1) Balance between biotic and abiotic factors. 
(2) Recycling of nutrients and wastes. 
Reason: Abiotic factors include elements like temperature, light, water, and nutrients, whereas biotic factors are things like plants, animals, fungus, and microorganisms. Due of their interdependence, any changes to one of these components can have an impact on the ecosystem as a whole. To support the development and survival of all organisms within the ecosystem, it is crucial to maintain a balance between various components in the ecosystem. Recycling is essential for maintaining the health and sustainability of the ecosystem.


Q7: (A) Construct a food chain of four trophic levels comprising the following:               

Hawk, snake, plants, rat

(B) 20,000 J of energy was transferred by the producers to the organism of second trophic level. Calculate the amount of energy that will be transferred by organisms of the third trophic level to the organisms of the fourth trophic level.  (3 Marks) (2023)

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Ans: (A) The flow of nutrients and energy from one organism to another at different trophic levels forms a food chain. 
Plants → Rat → Snake → Hawk 
(B) Only 10% of energy will be transferred to the next trophic level, according to the 10% law of energy transfer, and 90% will be wasted as heat and incomplete digestion. According to this law, 
(1) Energy is transferred from producers to the second trophic level = 20,000 J. 
(2) Energy moved from the second to the third trophic level: 10% of 20,000 J = 2,000 J. 
(3) Energy moved from the third to the fourth trophic level is equal to: 10% of 2,000 J = 200 J

Previous Year Questions 2022

Q1: (i) Why are crop fields considered as artificial ecosystems? 
(ii) Write a common food chain of four steps operating in a terrestrial ecosystem.  (2022 C)

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Ans: 
(i) Crop fields are the artificial ecosystems because in crop fields, both biotic (living) and abiotic (non-living) components are manipulated by human beings. Humans can change edaphic factors by adding fertilisers, water, etc. Biotic components may be changed using biocides or adding useful organisms like earthworms etc.
(ii) A food chain consists of various organism at various trophic levels. In terrestrial ecosystem, a common food chain is
Grass Grasshopper→ Frog→ Snake


Q2: (i) List two human-made ecosystems.
(ii) “We do not clean a pond in the same manner as we do in an aquarium.” Give reason to justify this statement. (Term II, 2021-22)

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Ans: 
(i) The two human made ecosystems are aquarium and garden.
(ii) We do not clean pond as we do in an aquarium because the waste generated in a pond is acted upon by the decomposers which convert them into simple soluble substances, whereas, in aquarium, the waste gets mixed with water and left untreated due to absence of decomposers.


Q3: In the following food chain, only 2J of energy was available to the peacocks. How much energy would have been present in Grass? Justify your answer. Grass → Grasshopper → Frog → Snake Peacock  (Term II, 2021-22)

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Ans: In the given food chain, 20,000 J of energy must have been present in grass. This is because, as per the 10% law of energy transfer, only 10% of energy is transferred to the next trophic level.
Previous Year Questions 2022


Q4: What are human-made ecosystems? Give an example. Can a human-made ecosystem become a self-sustaining ecosystem? Give reason to justify your answer.    (2022)

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Ans:  Artificial ecosystems area unit human-made structures wherever organic phenomena and abiotic elements area unit created to act with one another for survival. it’s not self-sufficient and might decrease while not human facilitated. samples of artificial ecosystems embrace aquariums, agriculture fields, zoos, etc.

  • A system, that is formed and maintained by mortals, is named synthetic | a synthetic} or man-made system. Some samples of artificial ecosystems area unit aquariums, gardens, agriculture, apiary, poultry, piggery, etc.
  • Man-made ecosystems don’t seem to be self-sufficient as a result they rely on naturally created ecosystems just like the water bodies. just like the crop fields that could be an artificial system that depends on water bodies like rivers, and groundwater for water and life. Same approach gardens conjointly rely on nature for their property.
  • All organisms like plants, animals, microorganisms, and mortals further because the physical surroundings act with one another and maintain a balance in nature. All interacting organisms in a locality in conjunction with the non-living constitute the setting kind associate degree system.
  • All the organic phenomenon elements comprising living organisms and abiotic elements comprising physical factors like temperature, rainfall, soil, etc create an associate degree system.
  • An artificial system could be a variety of systems created by man by artificial means and not present sort of a forest, ponds, lakes, etc. samples of artificial ecosystems area unit crop fields and gardens.
  • Man-made ecosystems don’t seem to be self-sufficient as a result they rely on naturally created ecosystems just like the water bodies. just like the crop fields that could be an artificial system that depends on water bodies like rivers, and groundwater for water and life. Same approach gardens conjointly rely on nature for their property.

Hence, artificial ecosystems don’t seem to be self-sufficient.


Q5: (a) Name the group of organisms which form in the first trophic level of all food chains. Why are they called so?
(b) Why are the human beings most adversely affected by biomagnification?
(c) State one ill-effect of the absence of decomposers from a natural ecosystem.          (2022)

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Ans: (a) Producers form the first trophic level of all food chains. They are called producers because they are autotrophic organisms which alone are able to manufacture organic food from inorganic raw materials by the process of photosynthesis. They capture sun’s energy and convert it into chemical energy. The chemical energy is used in combining raw materials into organic food. This food is used up by themselves and rest enters the food chains as food for consumers.
(b) Human beings are most adversely affected by biomagnification because they occupy the highest trophic level in any food chain. As in biomagnification, successive concentration of non-biodegradable substances increases in the trophic level of food chains, so, it leads to most toxicity at highest trophic level. Hence, maximum concentration of chemicals get accumulated most in their body.
(c) Absence of decomposers will lead to the accumulation of dead remains and waste products of organisms in our natural ecosystem. The decomposers breakdown complex organic substances into simple inorganic substances, so, that it can go into the soil and can be used up by plants.


Q6: What is ozone? How is it formed in the upper layers of the earth’s atmosphere? How does ozone affect our ecosystem?   (2022)

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Ans: Ozone (O3) is a molecule formed by three atoms of oxygen. It is formed in the stratosphere layer of atmosphere when high energy U V rays act on O2 molecule splitting it into free oxygen (O) atoms. These atoms then combine with molecular oxygen (O2) to form ozone (O3).

Previous Year Questions 2022

Ozone shields the surface of the earth from U V radiations from the sun. The depletion of ozone layer will lead to global warming and some serious health issues such as damage of skin cells that leads to skin cancer, snow blindness or inflammation of cornea, increased fatality of young animals, mutations and reduced immunity.


Q7: (a) We do not clean ponds or lakes, but an aquarium needs to be cleaned regularly. Why?
(b) Why is ozone layer getting depleted at the higher levels of the atmosphere? Mention one harmful effect caused by its depletion.   (2022)

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Ans: (a) We do not clean pond as we do in an aquarium because the waste generated in a pond is acted upon by the decomposers which convert them into simple soluble substances, whereas, in aquarium, the waste gets mixed with water and left untreated due to absence of decomposers.
(b) The ozone layer is getting depleted at the higher levels of the atmosphere due to use of chlorofluorocarbons (CFCs) which are used in refrigerator. Other ozone depleting substances include carbon tetrachloride, hydrofluorocarbons used in fire extinguisher, air i conditioners, etc.
Due to the ozone layer depletion, humans will be directly exposed to the ultraviolet radiations of sun. This will result in serious health issues like skin cancer, sunburns, quick ageing, mutations and weak immune system.


Q8: Kulhads (disposable cups made of day) and disposable paper cups both a re used as an alternative For disposable plastic cups. Which one of these two can be considered as a better alternative to plastic cups and why?    (2022)

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Ans: Kulhad:

  • Kulhad is a cup that is made using clay or soil.
  • This is usually used for serving tea/coffee.
  • This was brought in use to replace the plastic cup in trains.
  • Plastic is a nonbiodegradable harmful substance that does not decompose in nature and affects the ecosystem or environment negatively.
  • Kulhads are made of biodegradable soil, therefore, this was used to replace the plastic and protect the environment and health of humans.

Discontinuation of kulhads:

  • Since kulhads are made using clay, it is a practice of harming the environment too.
  • The clay is fertile soil and kulhads were discontinued to avoid its reduction.
  • Making of kulhad on a large scale to serve tea passengers in the train was leading to the loss of this top fertile soil.


Q9: (a) What is meant by garbage? List two classes into which garbage is classified.
(b) What do we actually mean when we say that “enzymes are specific in their action”?   (2022)

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Ans: (a) Garbage is the waste material (rubbish) especially of domestic refuse. The two classes into which garbage is classified are
(i) Biodegradable
(ii) Non-biodegradable.
(b) Enzymes are specific in their action. For example, enzyme maltase acts on sugar maltose but not on lactose or sucrose. Different enzymes may act on the same substrate but give rise to different products. Similarly an enzyme may act on different substrates producing different end products.

Also read: Garbage, Waste Management & Depletion of the Ozone Layer

Previous Year Questions 2021

Q1: What are consumers? Name the four categories under which the consumers are further classified. (2021 C)

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Ans: Consumers are the organisms which are unable to synthesise their own food. Therefore, they utilise materials and energy stored by the producers or eat other organisms. They are known as the heterotrophs. The consumers are of following categories:
(i) Primary or first-order consumers: These include the animals which eat plants or plant products. They are called herbivores or primary (first order) consumers. E.g., Cattle, deer, goat, rabbit, hare, rats, mice, grasshoppers etc
(ii) Secondary or second order consumers: These include the animals which depend on primary consumers for their food. They are called primary carnivores or secondary (second order) consumers. Secondary consumers can be carnivores or omnivores. E.g., Cats, dogs, foxes, small fish, etc.
(iii) Tertiary or third order consumers: These are large carnivores (or top carnivores) which feed on primary and secondary consumers. These are termed as secondary carnivores or tertiary (third order) consumers. Common examples include shark and crocodile, wolves, lion, etc. (iv) Quaternary or fourth order consumers: These are even larger carnivores which feed on secondary carnivores (tertiary consumers). E.g., Tigers, lions and eagles/hawks etc.

Previous Year Questions 2020

Q1: How is ozone layer formed? State its importance to all life forms on earth. Why the amount of ozone in the atmosphere dropped sharply in the 1980s? (2020)

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Ans: When high energy ultraviolet radiations react with oxygen present in stratosphere (the higher level of atmosphere) it splits into its constituent atoms. Since these atoms produced are very reactive, they react with molecular oxygen (O2) to form ozone (O3).
Previous Year Questions 2020
Ozone shields the surface of the earth from UV radiations from the sun. The depletion of ozone layer will lead to global warming and some serious health issues such as damage of skin cells that leads to skin cancer, snow blindness or inflammation of cornea, increased fatality of young animals, mutations and reduced immunity. In 1980s, the production of CFCs increased which releases active chlorine in the atmosphere. The active chlorine then reacts with ozone molecules present there to convert them to oxygen. This results in thinning of ozone layer. CFCs are used as refrigerants and in fire extinguishers. That is why, amount of ozone in the atmosphere dropped sharply.


Q2: (a) Write two harmful effects of using plastic bags on the environment. Suggest alternatives to the usage of plastic bags. (b) List any two practices that can be followed to dispose off the waste produced in our homes. (2020)

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Ans: 
(a) Two harmful effects of using plastic bags on the environment: 
(i) Plastic bags are non-biodegradable substances which are not acted upon by microbes. So, they cannot be decomposed and therefore persist in the environment for a long time causing harm to the soil fertility and quality.
(ii) Plastic bags choke drains which result in waterlogging, that allows breeding of mosquitoes and hence leads to various diseases. Jute bags and cloth bags are the alternatives to the polyethene bags.
(b) Practices that can be followed to dispose off the waste produced in our homes: 
(i) Separation of biodegradable and non-biodegradable wastes.
(ii) The biodegradable waste can be converted to manure.
(iii) Non-biodegradable waste should be disposed off at suitable places from where municipal authorities can pick them up and dispose properly and scientifically.
(iv) Use discarded bottles and jars to store food items.


Q3: (A) Construct a terrestrial food chain comprising four trophic levels.
(B) What will happen if we kill all the organisms in one trophic level? 
(C) Calculate the amount of energy available to the organisms at the fourth trophic level if the energy available to the organisms at the second trophic level is 2000 J. (CBSE 2020)

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Ans: (A) A terrestrial food chain comprising of four trophic levels: 
Grass → Grasshopper → Frog → Snake 
(B) If we kill all the organisms in one trophic level then the transfer of food energy to next level will stop. Organisms of previous trophic level will also increase. For Example: If all herbivores in an ecosystem are killed, there will be no food available for the carnivores of that area. Consequently, they will also die or will shift to other areas. Populations of producers will also increase in absence of herbivores causing an imbalance in the ecosystem. 
(C) Consider the same food chain that we have made i.e., (A), 
Grass → Grasshopper → Frog → Snake 
In this food chain, organism at the second trophic level is grasshopper and the energy available at this trophic level is 2000 J. According to 10% law, 10% of energy will be available to frog (Third trophic level) which is 200 J. The energy available to the snakes will be available as 10% of 200 J. Thus, the energy available to the snake is 20 J. 
The 10% law states that during transfer of energy from one trophic level to the next trophic level, only about 10% of energy is available to the higher trophic level.
To summarise:
Previous Year Questions 2020

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Previous Year Questions 2019

Q1: (a) A food chain generally has three or four trophic levels. Explain.
(b) What is biological magnification? Explain. (2019 C)

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Ans: 
(a) The number of trophic levels in a food chain are limited because at each trophic level only 10% of energy is utilised for the maintenance of organism which occur at that trophic level and the remaining large portion is lost as heat. As a result, organisms at each trophic level pass on lesser energy to the next trophic level, than they receive. The longer the food chain, the lesser is the energy available to the final member of food chain.
(b) Biological magnification is characterised by the increase in the non-biodegradable substances (DDT, Hg, etc.) in successive trophic levels of a food chain. The level of such toxic substances will be different in different trophic levels of a food chain because these substances are accumulated more in higher trophic levels.


Q2: Define an ecosystem. Draw a block diagram to show the flow of energy in an ecosystem. (Delhi 2019)

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Ans: An ecosystem is defined as a structural and functional unit of the biosphere. It comprises of living organisms and their non-living environment that interact by means of food chains and biogeochemical cycles resulting in energy-flow, biotic diversity and material cycling to form stable self-supporting system. Green plants capture about 1% of the solar energy incident on the earth to carry out the process of photosynthesis.
A part of this trapped energy is used by plants in performing their metabolic activities and some energy is released as heat into the atmosphere. The remaining energy is chemical energy stored in the plants as photosynthetic products. When these green plants are eaten up by herbivores, the chemical energy stored in the plants is transferred to these animals. These animals (herbivores) utilise some of this energy for metabolic activities and some energy is released as heat while the remaining energy is stored in their body. This process of energy transfer is repeated till top carnivores. In an ecosystem, transfer of energy follows 10 per cent law, i.e., only 10 per cent of the energy is transferred to each trophic level from the lower trophic level. The given block diagram shows unidirectional flow of energy at different trophic levels in a freshwater ecosystem:
Previous Year Questions 2019


Q3: (a) How can we help in reducing the problem of waste disposal? Suggest any three methods.    (Delhi 2019)
(b) Distinguish between biodegradable and nonbiodegradable wastes.    (DoE, A1 2011)

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Ans: (a) The three methods of waste disposal are:

  • Recycling: solid, wastes like paper, plastics, metals can be sent to processing factories where they are remoulded or reprocessed to new materials.
  • Production of compost: Biodegradable wastes like fruit and vegetable peels, plant products, left over food, grass clippings, human and animal waste can be converted into compost by burying this waste into grund and can be used as manure.
  • Incineration: Burning dawn many household waste, chemical waste and biological waste into ash is known as incineration. A large amount of waste can be easily converted into ash which can be disposed off in landfill.

(b) Differences between:
Previous Year Questions 2019

Previous Year Questions 2017

Q1: (a) What is an ecosystem? List its two main components.
(b) We do not clean ponds or lakes, but an aquarium needs to be cleaned regularly. Explain.    (CBSE 2013,2017)

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Ans: 
(a) A self-sustaining functional unit consisting of living and non-living components is called an ecosystem.
Components: Biotic components like plants and animals. Non-biotic components like soil, wind, light etc.
(b) A pond is a complete, natural and self-sustaining ecosystem whereas an aquarium is an artificial and incomplete ecosystem, without decomposers therefore it needs regular cleaning for proper running.


Q2: You have been selected to talk on “ozone layer and its protection” in the school assembly on ‘Environment Day.’    (Delhi 2017)
(a) Why should ozone layer be protected to save the environment?
(b) List any two ways that you would stress in your talk to bring in awareness amongst your fellow friends that would also help in protection of ozone layer as well as the environment.

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Ans: (a) Ozone layer at the higher levels of the atmosphere, acts as a shield to protect earth from the harmful effects of the ultraviolet (UV) radiations; hence, it should be protected.
(b) 

  • Urging the people to not to buy aerosol products with CFC that are available in the market.
  • Conducting poster making competition or street plays presenting the importance of ozone layer on earth.


Q3: Your mother always through that fruit juices are very healthy for everyone. One day she read in the newspaper that some brands of fruit juices in the market have been found to contain certain level of pesticides in them. She got worried as pesticides are injurious to our health.    (Foreign 2017)
(а) How would you explain to your mother about fruit juices getting contaminated with pesticides?
(b) It is said that when these harmful pesticides enter our body as well as in the bodies of other organisms they get accumulated and beyond a limit cause harm and damage to our organs. Name the phenomenon and write about it.

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Ans:

(a) 

  • Pesticides are the chemicals used to protect our crops from diseases and pests.
  • These chemicals are washed down either into the soil or into water bodies.
  • From the soil, they are absorbed by the terrestial plants along with water and minerals.
  • From the water bodies, they are absorbed by the aquatic plants.
  • When the fruits of these plants are used to prepare fruit juices, they are contaminated with the pesticides.

(b)

  • The phenomenon is called biomagnification. It is the phenomenon in which certain harmful chemicals enter the food chain and get accumulated and increase in concentration at successive trophic levels.
  • It is because they are not degradable.
  • The maximum concentration of these chemicals is found in the top level consumers.


Q4: In the following food chain, 100 J of energy is available to the lion. How much energy was available to the producer?    (AI 2017)

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Ans: Plants → Deer → Lion
1,000,000 J of energy was available to the producer.


Q5: Why should biodegradable and non-biodegradable wastes be discarded in two separate dustbins?    (AI 2017(C); Delhi 2013,15)

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Ans: The biodegradable and non-biodegradable wastes must be discarded in two different dustbins because biodegradable wastes gets decomposed by the microorganisms whereas non-biodegradable wastes can be recycled and reused.


Q6: Name any two man-made ecosystems.    (Foreign 2017)

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Ans: Agricultural/crop fields, aquaria, gardens,

Also read: Garbage, Waste Management & Depletion of the Ozone Layer

Previous Year Questions 2016

Q1:  Name two natural ecosystems.    (CBSE 2016-17 C)

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Ans: Some natural ecosystems are:

  • Temperate forest
  • Tropical forest
  • Grassland
  • Ocean
  • LakeTypes of Natural EcosystemTypes of Natural Ecosystem


Q2: What are decomposers? Write the role of decomposers in the environment.    (CBSE 2016-17 C)

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Ans: Decomposers are microorganisms that derive their nutrition from dead remains and waste products of organisms.
They play a vital role in our environment by breaking down the complex organic substance into simple inorganic substance which is made available for plants and other organisms. Hence they act as scavengers and not only keep the environment clean but also replenish the minerals.


Q3:  We often use the word environment. What does it mean?    (Foreign 2016)

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Ans: It is the sum total of all external conditions and influences that affect the life and development of an organism, i.e. the environment includes all the physical or abiotic and biological or biotic factors.


Q4: Why are green plants called producers?    (Delhi 2016)

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Ans: Green plants can produce their own food by photosynthesis from inorganic compounds and hence are called producers.

Previous Year Questions 2015

Q1: What is ten per cent law? Explain with an example how energy flows through different trophic levels.    (Delhi 2015)

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Ans: Energy available at each successive trophic level of food chain is ten per cent of that at the previous level.
This is called ten per cent law. Thus, 90 per cent energy is lost to the surroundings at each trophic level. However, plants absorb only one per cent of radiant energy of the Sun during photosynthesis. This is explained as under :
Previous Year Questions 2015


Q2: What is ozone? How and where is it formed in the atmosphere? Explain how does it affect ecosystem.    (Foreign 2015)

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Ans: Ozone is an isotope of oxygen, i.e. it is a molecule formed by 3 atoms of oxygen.
Previous Year Questions 2015
Ozone exists in the ozone layer of stratosphere. At higher level o f atmosphere, O2 molecule breaks down to 2 oxygen atom. The oxygen atom then combines with the oxygen molecule to form ozone.
Ozone layer in the atmosphere prevents UV rays from reaching earth. Exposure to excess UV rays causes skin cancer, cataract and damages eye and immune system. It also decreases crop yield and reduces population of phytoplankton, zooplankton and certain fish larvae which are an important constituent of aquatic food chain. It also disturbs rainfall, causing ecological disturbance and reduces global food production. Thus, it affect the ecosystem.


Q3: “Energy flow in food chains is always unidirectional.” Justify this statement. Explain how the pesticides enter a food chain and subsequently get into our body.   (Foreign 2015)

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Ans: The energy flow through different steps in the food chain is unidirectional. The energy captured by autotrophs does not revert back to the solar input and it passes to the herbivores, i.e. it moves progressively through various trophic levels. Thus energy flow from sun through producers to omnivores is in single direction only.
Pesticides are sprayed to kill pests on food plants. The food plants are eaten by herbivores and alongwith the food, pesticides are also eaten by the herbivores. Herbivores are eaten by carnivores and alongwith the herbivore animal, pesticide also enters the body of the carnivore. Man eat both plants and animals and pesticide alongwith food enters the body of human. Concentration of pesticides increases as we move upward in the food chain and the process is called bio-magnification.


Q4: What is an ecosystem? List its two main components. We do not clean natural ponds or lakes but an aquarium needs to be cleaned regularly. Why is it so? Explain.    (AI 2015)

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Ans: Ecosystem: It is the structural and functional unit of biosphere, comprising of all the interacting organisms in an area together with the non-living constituents of the environment. Thus, an ecosystem is a self-sustaining system where energy and matter are exchanged between living and non-living components.
Main components of ecosystem:
(i) Biotic Component: It means the living organisms of the environment-plants, animals, human beings and microorganisms like bacteria and fungi, which are distinguished on the basis of their nutritional relationship.
(ii) Abiotic Component: It means the non-living part of the environment-air, water, soil and minerals. The climatic or physical factors such as sunlight, temperature, rainfall, humidity, pressure and wind are a part of the abiotic environment.
An aquarium is an artificial and incomplete ecosystem compared to ponds or lakes which are natural, self-sustaining and complete ecosystems where there is a perfect recycling of materials. An aquarium therefore needs regular cleaning.


Q5: Write the full name of the group of compounds mainly responsible for the depletion of ozone layer.    (Foreign 2015)

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Ans: CFC → Chloroflurocarbon


Q6: Which of the following are always at the second trophic level of the food chains?
Carnivores, Autotrophs, Herbivores    [AI 2015]

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Ans: Herbivores are always at the 2nd trophic level.


Q7: The following organisms form a food chain. Which of these will have the highest concentration of nonbiodegradable chemicals? Name the phenomenon associated with it. Insects, Hawk, Grass, Snake, Frog.    (Foreign 2015)

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Ans: Hawk will have highest concentration of non-biodegradable chemicals. The phenomenon is called biomagnification.


Q8: The first trophic level in a food chain is always a green plant. Why?   (AI 2015)

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Ans: Only green plants can make their own food from sunlight. Green plants, therefore, always occupy the 1st trophic level in a food chain.


Q9: What will be the amount of energy available to the organism of the 2nd trophic level of a food chain, if the energy available at the first trophic level is 10,000 joules?   (AI 2015)

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Ans: 100 Joules of energy will be available to the organism of the 2nd trophic level.


Q10: (a) What is biodiversity? What will happen if biodiversity of an area is not preserved? Mention one effect of it.   (AI 2015)
(b) With the help of an example explain that a garden is an ecosystem.
(c) Why only 10% energy is transferred to the next trophic level?

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Ans: (a) Biodiversity is the existence of a wide variety of species of plants, animals and microorganisms in a natural habitat within a particular environment or existence of genetic variation within a species. Biodiversity of an area is the number of species or range of different life forms found there. Forests are ‘biodiversity hotspots’. Every living being is dependent on another living being. It is a chain. If biodiversity is not maintained, the links of the chain go missing. If one organism goes missing, this will affect all the living beings who are dependent on it.
(b) A garden comprises of different kind of flora and fauna such as grasses, flowering and nonflowering plants, trees, frogs, insects, birds, etc. All these living organisms depend and interact with each other and their growth, reproduction and other vital biological activities depend upon the abiotic component comprising of physical factors like temperature, rainfall, wind, soil and minerals. Therefore, we can say that a garden is an ecosystem.
(c) Only 10% energy is transferred to the next trophic level because other 90 per cent is used for things like respiration, digestion, running away from predators.


Q11: Differentiate between biodegradable and non-biodegradable substances with the help of one example each. List two changes in habits that people must adopt to dispose non-biodegradable waste, for saving the environment. (CBSE 2015)

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Ans: Biodegradable substances: 
These can be broken down into simpler substances by nature/ decomposers/bacteria/saprophytes/ saprobionts. 
Example: Human Excreta/Vegetable peels, etc. 
Non-biodegradable substances: 
These can’t be broken down into simpler substances by nature/decomposers. 
Example: Plastic/glass (or any other) (Any one) 
Habits
(1) Use of separate dustbins for iodegradable and non-biodegradable waste. 
(2) Recycling of waste. 
(3) Use of cotton/jute bags for carrying vegetables etc

12. Previous Year Questions: Magnetic Effects of Electric Current

Previous Year Questions 2025

Q1: Two statements are given – one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :  (1 Mark)

Assertion (A): No two magnetic field lines are found to cross each other.
Reason (R): The compass needle cannot point towards two directions at the point of intersection of two magnetic field lines.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

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Ans: (A) Both A and R are true, and R is the correct explanation of A.

  • Assertion (A): Magnetic field lines represent the direction and strength of the magnetic field. They never cross each other because at any point in space, the magnetic field has a unique direction. If two field lines crossed, it would imply two different directions for the magnetic field at that point, which is physically impossible. Thus, A is true.
  • Reason (R): A compass needle aligns itself along the magnetic field direction at a given point. If field lines intersected, the needle would have to point in two directions simultaneously, which is not possible. This explains why magnetic field lines do not cross. Thus, R is true and correctly explains A.
  • Conclusion: Option (A) is correct.


Q2: Two statements are given – one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :  (1 Mark)
Assertion (A): The pattern of the magnetic field of a solenoid carrying a current is similar to that of a bar magnet.
Reason (R): The pattern of the magnetic field around a current-carrying conductor is independent of the shape of the conductor.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

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Ans: (C) A is true, but R is false.

  • Assertion (A): A current-carrying solenoid produces a magnetic field similar to that of a bar magnet, with one end acting as the north pole and the other as the south pole. The field lines emerge from one end (north) and enter the other (south), resembling a bar magnet’s field. Thus, A is true.
  • Reason (R): The magnetic field pattern around a current-carrying conductor depends on its shape. For example, a straight conductor produces concentric circular field lines, while a solenoid produces a bar magnet-like field. Thus, R is false as the field pattern is shape-dependent.
  • Conclusion: Option (C) is correct, as A is true, but R is false and does not explain A.

Q3: Which one of the following statements is not true about a bar magnet?  (1 Mark)
(a) It sets itself in north-south direction when suspended freely.
(b) It has attractive power for iron filings.
(c) It produces magnetic field lines.
(d) The direction of magnetic field lines inside a bar magnet is from its north pole to its south pole.

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Ans: (d) The direction of magnetic field lines inside a bar magnet is from its north pole to its south pole.

  • Option (A): True. A freely suspended bar magnet aligns itself in the north-south direction due to Earth’s magnetic field, with its north pole pointing toward the geographic north.
  • Option (B): True. A bar magnet attracts ferromagnetic materials like iron filings due to its magnetic field.
  • Option (C): True. A bar magnet produces magnetic field lines that emerge from the north pole and enter the south pole, forming closed loops.
  • Option (D): False. By convention, magnetic field lines outside a bar magnet go from the north pole to the south pole, but inside the magnet, they travel from the south pole to the north pole to form continuous loops.
  • Conclusion: Option (D) is not true.

Q4: The strength of the magnetic field produced inside a long straight current-carrying solenoid does not depend upon:  (1 Mark)
(A) Number of turns in the solenoid
(B) Direction of current flowing through the solenoid
(C) Material of the core filled inside the solenoid
(D) Magnitude of current in the solenoid

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Ans: (B) Direction of current flowing through the solenoid

For a long current-carrying solenoid, the magnetic field inside is
Previous Year Questions 2025Hence, B depends on:

  • Number of turns per unit length n (more turns → stronger field),
  • Core material via μr (e.g., iron increases B),
  • Magnitude of current I (largerI → larger B).

Changing the direction of current only reverses the polarity (which end is north/south) but does not change the magnitude of BB.
Therefore, the strength of the magnetic field does not depend on the direction of current. (Option B).

Q5: (i) The given figure shows the current passing through the straight conductor XY.

(ii) Name and state the rule used in determining the direction of the magnetic field lines in the situation given above.
(iii) State Fleming’s left-hand rule. Using this rule, determine the direction of force applied on an electron entering a uniform magnetic field as shown in the figure.  (5 Marks)

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Ans: 

i) Magnetic Field Lines Due to Current in Conductor XY

When current flows from X to Y in the straight conductor XY, the magnetic field around the wire consists of concentric circles centred on the wire.
By the Right-hand thumb rule (thumb along current X→Y; curled fingers give field direction), the field is clockwise around the wire as seen from the X→Y end.


ii) Rule Used to Determine Magnetic Field Direction

Rule Name: Right-Hand Thumb Rule (Maxwell’s Corkscrew Rule)
Statement: 
If you hold a straight conductor in your right hand so that your thumb points in the direction of current, the direction in which your fingers wrap around the conductor gives the direction of the magnetic field lines.


iii) Fleming’s Left Hand Rule and Direction of Force on Electron

Fleming’s Left Hand Rule Statement: Stretch the thumb, forefinger, and middle finger of your left hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, the middle finger in the direction of current (conventional, positive to negative), then the thumb gives the direction of force (motion) on the conductor.

  • Magnetic field: to the right.
  • Electron velocity: upwards ⇒ conventional current is downwards.
  • Set forefinger → right (B), middle finger → down (I).
  • Thumb then points out of the plane of the paper (towards the observer).

Therefore, the force on the electron is out of the page, perpendicular to both Previous Year Questions 2025(For a positive charge it would be into the page.)

Q6: (a) Name and state the rule which determines the force on a current-carrying conductor placed in a uniform magnetic field.
(b) Consider three diagrams in which the entry of a positive charge (+Q) in a magnetic field is shown. Identify the case in which the force experienced by the charge is (i) maximum, and (ii) minimum.  (3 Marks)

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Ans:
(a) Rule: Fleming’s left-hand rule.
(b) (i) Maximum force: When velocity is perpendicular to the magnetic field.
(ii) Minimum force: When velocity is parallel to the magnetic field.

(a) Rule:

  • Name: Fleming’s left-hand rule.
  • Statement: Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular. If the forefinger points in the direction of the magnetic field (B) and the middle finger in the direction of the current (I), the thumb points in the direction of the force (F).

(b) Force on a positive charge:

  • The force on a moving charge is given by F = qvB sinθ, where q is the charge, v is velocity, B is the magnetic field strength, and θ is the angle between v and B.
  • (i) Maximum force: F is maximum when sinθ = 1, i.e., θ = 90° (velocity perpendicular to the magnetic field). In the diagram where +Q’s velocity is perpendicular to B, the force is maximum.
  • (ii) Minimum force: F is minimum (zero) when sinθ = 0, i.e., θ = 0° or 180° (velocity parallel or antiparallel to the magnetic field). In the diagram where +Q’s velocity is parallel to B, the force is zero.

Conclusion: Maximum force at θ = 90°, minimum at θ = 0° or 180°.


Q7: What are magnetic field lines? List two important properties of magnetic field lines.  (2 Marks)

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Ans:

Magnetic field lines: These are visual representations of the magnetic field, showing the direction a north pole would move and indicating field strength by their closeness.
Previous Year Questions 2025

Properties:

  1. Closed loops: Field lines emerge from the north pole, travel outside to the south pole, and continue inside from south to north, forming continuous loops.
  2. Non-intersecting: No two field lines cross, as the magnetic field has a unique direction at each point. Intersection would imply multiple directions, which is impossible.

Q8: In order to obtain magnetic field lines around a bar magnet, a student performed an experiment using a magnetic compass and a bar magnet. The magnet was placed on a sheet of white paper fixed on a drawing board. Using a magnetic needle, he obtained on the paper a pattern of magnetic field lines around the bar magnet.
(a) By convention, the field lines emerge from the north pole and merge at the south pole. Why? Give reason.
(b) State the relationship between the strength of the magnetic field and the degree of closeness of the field lines.
(c) (A) (i) No two field lines can ever intersect each other. Give reason.
(ii) The magnetic field in a given region is uniform. Draw a diagram to represent it.  (4 Marks)

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Ans:

(a) Why field lines emerge from north and merge at south:

  • By convention, magnetic field lines show the direction a free north pole would move. The north pole of a compass is attracted to the south pole of a magnet and repelled by its north pole. Thus, field lines are drawn emerging from the north pole and merging at the south pole, forming closed loops (inside the magnet, they go from south to north).

(b) Relationship:

  • The strength of the magnetic field is proportional to the density of field lines. Where field lines are closer together (e.g., near the poles), the field is stronger; where they are farther apart, the field is weaker.

(c) (A):

  • (i) No intersection: Magnetic field lines do not intersect because the magnetic field at any point has a unique direction. If lines crossed, it would imply two different directions at the intersection point, which is physically impossible, as a compass needle can only point in one direction.
  • (ii) Uniform magnetic field:
    • Diagram: Draw parallel, equally spaced straight lines with arrows in the same direction (e.g., left to right). This represents a uniform magnetic field, as seen inside a solenoid or between flat magnet poles.

Conclusion: As described above.

OR

(c) (B) Draw the pattern of the magnetic field lines through and around a current carrying solenoid. What does the pattern of field lines inside the solenoid represent ?

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Ans: 

The magnetic field pattern of a current-carrying solenoid resembles that of a bar magnet. One end of the current-carrying solenoid behaves as a magnetic north , while the other behaves as the magnetic south just like in the bar magnet.

Q9: (a) What are magnetic field lines? How is the direction of the magnetic field at a point determined? Draw the pattern of magnetic field lines of the magnetic field produced by a current-carrying circular loop. Mark on it the direction of (i) current and (ii) magnetic field lines. Name the two factors on which the magnitude of the magnetic field due to a current-carrying coil depends.

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Ans:

Magnetic field lines: These are imaginary lines that indicate the direction a north pole would move and show field strength by their density.

Direction determination: Place a small compass needle at the point; the needle’s north pole points in the direction of the magnetic field at that point.

Previous Year Questions 2025

Factors affecting field magnitude: B = (μ₀NI)/(2r).

  1. Current (I): Higher current increases the field strength.
  2. Number of turns (N): More turns increase the field strength.

Q10:  (a) Study the following electric circuit diagram and answer the questions that follow :
(i) What does the circuit diagram show?

(ii) What will happen if the direction of current is reversed? Justify your answer giving a circuit diagram.
(b) Name and state the rule to determine the direction of magnetic field associated with a straight current carrying conductor.  (5 Marks)

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Ans:

(a) (i) The circuit diagram shows a current-carrying conductor placed near a compass needle. The conductor appears to form a trapezoidal loop with a key (K) to control the current flow. The compass needle, positioned near the conductor, is used to detect the magnetic field generated by the current. This setup is typically used to demonstrate the relationship between electric current and the magnetic field it produces, such as in an experiment to verify the magnetic effect of current.

(a) (ii) Reversing current:

  • Effect: Reversing the current direction reverses the force direction on the conductor, per Fleming’s left-hand rule. For example, if the original force is upward, it becomes downward.
  • Justification: The force direction depends on the current direction (middle finger in Fleming’s rule). Reversing the current (e.g., by swapping battery terminals) changes the middle finger’s direction, reversing the thumb (force).Previous Year Questions 2025

(b) Rule:
Previous Year Questions 2025

  • Name: Right-hand thumb rule.
  • Statement: Hold the conductor with the right hand, thumb pointing in the current direction. The curled fingers indicate the magnetic field direction (concentric circles around the conductor).

OR

(a) Draw the pattern of magnetic field lines of 
(i) a bar magnet 
(ii) a current carrying solenoid.

 List two distinguishing features between the two magnetic fields.
(b) Study the following three diagrams in which the entry of an electron in a magnetic field is shown. Identify the case in which the magnetic force experienced by the electron is (i) maximum, and (ii) minimum. Give reason for your answers in each case.  (5 Marks)

Previous Year Questions 2025

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Ans: (a)
(i) 
(ii) 

Two distinguishing features between the two fields:

  1. The magnetic field of the solenoid can be varied as per our requirements just by changing the current or core of the solenoid whereas the magnetic field of the bar magnet is fixed.
  2. The magnetic field outside the solenoid is negligible as compared to the bar magnet.

(b) Let’s analyze each case using the magnetic force on a moving charge:

F=q v B sin⁡ θ

Where:

q = charge of particle (electron has negative charge)
v = velocity of electron
B = magnetic field strength
θ = angle between velocity of electron and magnetic field

Case (A): Electron velocity is perpendicular to magnetic field (θ = 90)

sin⁡90=1

Force is maximum.

Case (B): Electron velocity is at an angle (oblique) to magnetic field (0° < θ < 90°).

sinθ is less than 1 but not zero.

Force is less than maximum but not zero.

Case (C): Electron velocity is parallel (or anti-parallel) to magnetic field (θ = 0 or 180)

Sin 0= 0 or sin 180= 0.

Force is minimum (zero).

Q11: Answer the following questions for a case in which a current carrying conductor is placed in a uniform magnetic field :
(a) List three factors on which the magnitude of the force acting on a current-carrying conductor in a uniform magnetic field depends.
(b) When is the magnitude of the force on the conductor maximum?
(c) Name the rule which helps in determining the direction of force on the conductor and give its one application.  (3 Marks)

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Ans:

(a) Factors: The force on a current-carrying conductor is given by F = BIL sinθ.

  1. Current (I): Higher current increases the force.
  2. Magnetic field strength (B): Stronger field increases the force.
  3. Length (L): Longer conductor in the field increases the force.
    [Note: The angle θ also affects the force, but the question asks for three factors.]

(b) Maximum force: F = BIL sinθ is maximum when sinθ = 1, i.e., θ = 90° (conductor perpendicular to the magnetic field).

(c) Rule and application:

  • Name: Fleming’s left-hand rule.
  • Statement: Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular. Forefinger points to the magnetic field (B), middle finger to the current (I), and thumb to the force (F).
  • Application: In an electric motor, the rule determines the direction of force on the current-carrying coil, causing it to rotate in the magnetic field, converting electrical energy to mechanical energy.

Q12: Why can’t two magnetic field lines cross each other? Draw magnetic field lines showing the direction of the magnetic field due to a current-carrying long straight solenoid. State the conclusion which can be drawn from the pattern of magnetic field lines inside the solenoid. Name any two factors on which the magnitude of the magnetic field due to this solenoid depends.  (5 Marks)

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Ans:

If two magnetic field lines will cross each other, then there will be two directions of magnetic field at the point of crossing, which is not possible in a magnetic field. The field around a solenoid is like that of a bar magnet.

Diagram:

Previous Year Questions 2025

Conclusion: The field lines inside the solenoid are straight, parallel, and equally spaced, indicating a uniform magnetic field, similar to a bar magnet’s interior field.

Factors: The magnetic field inside a solenoid is B = μ₀nI.

  1. Current (I): Higher current increases the field strength.
  2. Number of turns per unit length (n): More turns per unit length increase the field strength.

Previous Year Questions 2024

Q1: (i) Two magnetic field lines do not intersect each other. Why?
(ii) How is a uniform magnetic field in a given region represented? Draw a diagram in support of your answer.    (1 to 3 Marks) (2024)

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Ans: (i) If they intersect then at the point of intersection, there would be two directions of magnetic field or compass needle would point towards two directions, which is not possible.
(ii) Uniform magnetic field is represented by equidistant parallel straight lines
Previous Year Questions 2024


Q2: Strength of magnetic field produced by a current carrying solenoid DOES NOT depend upon:    (1 Mark) (2024)
(a)
 number of turns in the solenoid
(b) direction of the current flowing through it
(c) radius of solenoid
(d) material of core of the solenoid

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Ans: (b)
The strength of the magnetic field in a solenoid depends on factors like the number of turns of wire and the radius of the solenoid. However, the direction of the current (whether it flows one way or the opposite) does not affect the strength of the magnetic field itself; it only changes the direction of the field. Therefore, the correct answer is (b).


Q3: Assertion – Reason based questions:     (1 Mark) (2024)
These questions consist of two statements — Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:
(a)
 Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Assertion (A): The deflection of a compass needle placed near a current carrying wire decreases when the magnitude of an electric current in the wire is increased.
Reason (R): Strength of the magnetic field at a point due to a current carrying conductor increases on increasing the current in the conductor.

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Ans: (d)
The assertion states that the deflection of a compass needle decreases as the current increases, which is false. The reason explains that the magnetic field strength increases with more current, which is true. Therefore, the correct answer is (d), as the assertion is false while the reason is true.


Q4: Draw a diagram to show the pattern of magnetic field lines on a horizontal sheet of paper due to a straight conductor passing through its centre and carrying current vertically upwards. Mark on it (i) the direction of current in the conductor and (ii) the corresponding magnetic field lines. State right hand thumb rule and check whether the directions marked by you are in accordance with this rule or not.     (1 to 3 Marks) CBSE 2024)

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Ans:Previous Year Questions 2024

Right-Hand Thumb Rule: When a current-carrying straight conductor is being held in right-hand such that the thumb points towards the direction of current, then fingers will wrap around the conductor in the direction of the magnetic field lines.


Q5: Name the device used to magnetise a piece of magnetic material. Draw a labelled diagram to show the arrangement used for the magnetisation of a cylinder made of soft iron.    (1 to 3 Marks) (CBSE 2024)

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Ans: (a) 
Previous Year Questions 2024(b) Permanent magnet / Current carrying solenoid/ Electromagnet is used to magnetise a piece of magnetic material. 


Q6: A rectangular loop ABCD carrying a current I is situated near a straight conductor XY, such that the conductor is parallel to the side AB of the loop and is in the plane of the loop. If a steady current I is established in the conductor as shown, the conductor XY will   (1 Mark)  (2024)
(a)
 remain stationary.
(b) move towards the side AB of the loop.
(c) move away from the side AB of the loop.
(d) rotate about its axis.

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Ans: (b)
A rectangular loop ABCD carrying a current I is situated near a straight conductor XY, such that the conductor is parallel to the side AB of the loop and is in the plane of the loop. If a steady current I is established in the conductor as shown, the conductor XY will move towards the side AB of the loop.


Q7: Two statements are given  one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.   (1 Mark)  (2024)
(a) 
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).  
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.  
(d) Assertion (A) is false, but Reason (R) is true.  
Assertion (A): Magnetic field lines never intersect each other.
Reason (R): If they intersect, then at the point of intersection, the compass needle would point towards two directions, which is not possible. 

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Ans: (a)

The assertion states that magnetic field lines never intersect, which is true. The reason explains that if they did intersect, a compass needle would point in two different directions at that point, which is also true. Since the reason correctly explains why the assertion is true, the correct answer is (a).


Q8: A student fixes a sheet of white paper on a drawing board. He places a bar magnet in the centre of it. He sprinkles some iron filings uniformly around the bar magnet. Then he taps the drawing board gently and observes that the iron filings arrange themselves in a particular pattern.   (4 to 5 Marks)  (2024)
(a) Why do iron filings arrange in a particular pattern?
(b) What does the crowding of iron filings at the ends of the magnet indicate?  
(c) What do the lines, along which the iron filings align, represent?  
(d) If the student places a cardboard horizontally in a current carrying solenoid and repeats the above activity, in what pattern would the iron filings arrange? State the conclusion drawn about the magnetic field based on the observed pattern of the lines.

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Ans: (a) The iron filings arrange themselves in a particular pattern because they align with the magnetic field lines created by the bar magnet. The filings act like tiny magnets and are attracted to the lines of force, forming a pattern that shows the shape of the magnetic field around the magnet.

(b) The crowding of iron filings at the ends of the magnet indicates the location of the magnetic poles. The magnetic field is stronger at the poles of the magnet, which is why the filings are more concentrated in these regions. This shows the magnetic field lines emerging from the north pole and entering the south pole.

(c)The lines along which the iron filings align represent the magnetic field lines. These are the paths along which the magnetic force is exerted. The pattern formed by the iron filings shows the direction and shape of the magnetic field around the magnet.

(d) If the student places a cardboard horizontally in a current-carrying solenoid and repeats the activity, the iron filings would arrange themselves in concentric circles around the solenoid, with the pattern showing loops around the solenoid. The magnetic field of a solenoid is similar to that of a bar magnet, with distinct field lines forming a pattern resembling that of a bar magnet’s poles.

Conclusion: The magnetic field inside a solenoid is uniform and parallel to its axis, while outside the solenoid, the magnetic field resembles that of a bar magnet, with a clear direction from one end (north) to the other (south)


Q9: The current-carrying device which produces a magnetic field similar to that of a bar magnet is:     (1 Mark) (2024)
(a) 
A straight conductor
(b) A circular loop
(c) A solenoid
(d) A circular coil

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Ans: (c)
A solenoid is a coil of wire that, when carrying an electric current, creates a magnetic field similar to that of a bar magnet, with distinct north and south poles. This means that a solenoid can generate a uniform magnetic field inside it, making it behave like a bar magnet. Therefore, the correct answer is (c).


Q10:

A uniform magnetic field exists in the plane of the paper as shown in the diagram. In this field, an electron (e) and a positron (p+) enter as shown. The electron and positron experience forces: (1 Mark) (2024)
(a)
 both pointing into the plane of the paper.
(b) both pointing out of the plane of the paper.
(c) pointing into the plane of the paper and out of the plane of the paper respectively.
(d) pointing out of the plane of the paper and into the plane of the paper respectively.

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Ans: (d)

pointing out of the plane of the paper and into the plane of the paper respectively.

Previous Year Questions 2024

The forces on the electron and positron will be in opposite directions according to the right-hand rule, as they have opposite charges. This means the force on the electron will be in one direction and the force on the positron will be in the opposite direction.


Q11: (a) What happens when a bundle of wires of soft iron is placed inside the coil of a solenoid carrying a steady current? Name the device obtained. Why is it called so?
(b) Draw the magnetic field lines inside a current carrying solenoid. What does this pattern of magnetic field lines indicate?    (1 to 3 Marks) (2024)

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Ans: (a)  

  • When a bundle of wires made of soft iron is placed inside the coil of a solenoid carrying a steady current, the following occurs:
  • The soft iron wires become magnetised.
  • This happens because the magnetic field generated by the current in the solenoid induces magnetism in the iron.
  • The device formed as a result of this process is called an electromagnet.
  • It is termed an electromagnet because it behaves like a magnet only when there is an electric current flowing through the solenoid.

(b)
Previous Year Questions 2024
This pattern indicates that the magnetic field is uniform and Parallel.


Q12: The pattern of the magnetic field produced inside solenoid is:   (1 Mark) (2024)
(a) 
(b) 
(c)
(c) 
Previous Year Questions 2024

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Ans: (a)
The pattern of the magnetic field inside a solenoid is uniform and parallel, resembling straight lines. This means that the magnetic field strength is consistent throughout the solenoid, making it effective for applications requiring a stable magnetic field. Therefore, the correct answer is (a).

Previous Year Questions 2023

Q1: Assertion (A) : The magnetic field lines around a current carrying straight wire do not intersect each other.
Reason (R) : The magnitude of the magnetic field produced at a given point increases as the current through the wire increases.    

(1 Mark) (2023)
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A)
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A)
(c) Assertion (A) is true, but Reason (R) is False.
(d) Assertion (A) is false, but Reason (R) is true.
Ans:
 (b)

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Sol: Assertion is true and reason is also true, but reason is not the correct explanation of assertion. Magnetic field lines around a current carrying straight wire do not intersect each other because at the point of intersection there will be two directions which is not possible. Also, the strength of magnetic held increased by increasing the magnitude of the current in the wire.


Q2: Draw the pattern of the magnetic field produced around a vertical current carrying straight conductor passing through a horizontal cardboard. Mark the direction of current and the magnetic field lines. Name and state the rule which is used to determine the direction of magnetic field associated with a current carrying conductor.  (3 Marks) (2023)

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Ans: The pattern of magnetic field produced around a vertical current carrying straight conductor passing through a horizontal cardboard is shown in figure. The magnitude of magnetic field produced is
(i) ∝ I (ii) ∝ 1/r
Previous Year Questions 2023
Right hand thumb rule is used to determine the direction of magnetic field associated with a current carrying conductor.
It states that you are holding a current carrying straight conductor in your right hand such that the thumb points towards the direction of current. Then your finger will wrap around the conductor in the direction of the field lines of the magnetic field.


Q3: An alpha particle enters a uniform magnetic field as shown. The direction of force experienced by the alpha particle is ____  (1 Mark) (2023)

Previous Year Questions 2023

(a) Towards right
(b) towards left
(c) Into the page
(d) Out of the page     
Ans:
 (d)

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Sol: According to the Fleming’s left hand rule, the  direction of force is out of the page.

Previous Year Questions 2023


Q4: A constant current flows in a horizontal wire in the plane of the paper from east to west as shown in the figure. The direction of the magnetic field will be north to south at a point N      (1 Mark) (2023)

Previous Year Questions 2023

(a) Directly above the wire
(b) Directly below the wire
(c) Located in the plane of the paper on the north side of the wire
(d) Located in the plane of the paper on the south side of the wire       

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Ans: (a)
The direction of the magnetic field at point N, located above a horizontal wire carrying a constant current from east to west, is from north to south. This can be determined using the right-hand thumb rule, which states that if you hold the wire with your right hand and point your thumb in the direction of the current, your fingers will curl in the direction of the magnetic field lines.

Previous Year Questions 2023


Q5: Assertion : A current carrying straight conductor experiences a force when placed perpendicular to the direction of magnetic field.
Reason : The net charge on a current carrying conductor is always zero,   (1 Mark) (2023)
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.      

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Ans: (b)
Sol: 

  • Assertion (A) is true because a current-carrying conductor placed perpendicular to a magnetic field experiences a force (Fleming’s Left-Hand Rule).
  • Reason (R) is true because a current-carrying conductor remains electrically neutral (equal protons and electrons).
  • However, R does not explain A. The force arises due to the motion of charges (current), not the net charge being zero. Thus, (b) is correct.


Q6: (i) Why is an alternating current (A.C.) considered to be advantageous over direct current (D.C.) for the long distance transmission of electric power? 
(ii) How is the type of current used in household supply different from the one given by a battery of dry cells? 
(iii) How does an electric fuse prevent the electric circuit and the appliances from the possible damage due to short circuiting or overloading.        (3 Marks) (2023)

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Ans: (i) Alternating current (A.C.) is preferred for long-distance power transmission because it experiences significantly lower power loss compared to direct current (D.C.). This efficiency makes A.C. more suitable for delivering electricity over vast distances.
(ii) The current used in household supply is alternating in nature while the current given by battery is direct in nature.
(iii) Electric fuse protects circuits and appliances by stopping the flow of any unduly high electric current. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.


Q7: (A) State the rule used to find the force acting on a current carrying conductor placed in a magnetic field. 
(B) Given below are three diagrams showing entry of an electron in a magnetic field. Identify the case in which the force will be (1) maximum and (2) minimum respectively. Give reason for your answer.   (1 to 3 Marks) (CBSE 2023)

Previous Year Questions 2023

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Ans: (A) Fleming’s left-hand rule is used to determine the direction of force experienced by a current carrying conductor placed in a uniform magnetic field. Acc to Fleming’s Left Hand Rule: When a current carrying conductor is placed in a magnetic field, it experiences a force, whose direction is given by Fleming’s left hand rule, which states that “Stretch the forefinger, the central finger and the thumb of your left hand mutually perpendicular to each other. If the forefinger shows the direction of the field and the central finger that of the current, then the thumb will point towards the direction of motion of the conductor, i.e., force.”

Previous Year Questions 2023

(B) Force on electron is maximum in (i) case because the electron direction show that it is moving at right angle to the direction of a magnetic field. Force on electron is minimum in (iii) case as the electron shown is moving parallel to the direction of a magnetic field. The direction of maximum force acting on an electron in (i) case which is into the plane of paper in accordance with Fleming’s left hand rule.


Q8: (A) Draw the pattern of magnetic field lines of: 
(i) a current carrying solenoid 
(ii) a bar magnet 
(B) List two distinguishing features between the two fields. (1 to 3 Marks) (CBSE 2023)

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Ans: (A) 

(i) 

(ii) 

(B) 

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Previous Year Questions 2022

Q1: Give reason for the following 
(i) There is either a convergence or a divergence of magnetic field lines near the ends of a current carrying straight solenoid. 
(ii) The current carrying solenoid when suspended freely rests along a particular direction.     (2022)

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Ans: (i) There is either a convergence or a divergence of magnetic field lines near the ends of a current carrying straight solenoid because it behaves similar to that of a bar magnet and has a magnetic field line pattern similar to that of a bar magnet. Thus the ends of the straight solenoid behaves like poles of the magnet, where the converging end is the south pole and the diverging end is the north pole.
(ii) The current carrying solenoid behaves similar to that of a bar magnet and when freely suspended aligns itself in the north-south direction.


Q2: A student fixes a sheet of white paper on a drawing board using some adhesive materials. She places a bar magnet in the centre of it and sprinkles some iron filings uniformly around the bar magnet using a salt-sprinkler. On tapping the board gently, she observes that the iron filings have arranged themselves in a particular pattern. 
(a) Draw a diagram to show this pattern of iron filings. 
(b) What does this pattern of iron filings demonstrate?
(c) (i) How is the direction of magnetic field at a point determined using the field lines? Why do two magnetic field lines not cross each other?      (2022)

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Ans: (a) The pattern of iron filings is shown below.
Previous Year Questions 2022

 (b) This pattern of iron filings demonstrate that the magnet exerts its influence in the region surrounding it. Therefore, the iron filings experience a force. The lines along which the iron filings align themselves represent magnetic field lines.
(c) (i) The direction of magnetic field is determined by placing a small compass needle in the magnetic field. The North-pole of the compass indicates the direction of magnetic field at that point.
Magnetic field lines never intersect each other because it is not possible to have two directions of magnetic field at the same point.


Q3: (i) What is a solenoid?
(ii) Draw the pattern of magnetic field lines of the magnetic field produced by a solenoid through which a steady current flows.   (2022)

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Ans: (i) A coil of many circular turns of insulated copper wire wrapped closely in the shape of a cylinder is called a solenoid.
(ii)

Previous Year Questions 2022


Q4: When is the force experienced by a current – carrying straight conductor placed in a uniform magnetic field
(i) Maximum;
(ii) Minimum?     (2022)

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Ans: The magnitude of force experienced by a current carrying conductor placed in a uniform magnetic field is
(i) maximum when the conductor is placed perpendicular to the magnetic field,
(ii) minimum when the conductor is placed parallel to the magnetic field.


Q5: (i) Name and state the rule to determine the direction of force experienced by a current carrying straight conductor placed in a uniform magnetic field which is perpendicular to it.
(ii) An alpha particle while passing through a magnetic field gets projected towards north. In which direction will an electron project when it passes through the same magnetic field?     (2022)

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Ans: (i) Fleming’s left-hand rule is used to determine the direction of magnetic force experienced by a current carrying straight conductor placed perpendicularly in a uniform magnetic field.
Fleming’s left-hand rule, states that when left hand’s thumb, forefinger and centre finger are held mutually perpendicular to one another and adjusted in such a way that the forefinger points in the direction of magnetic field, and the centre finger points in the direction of the current, then the direction in which thumb points, gives the direction of force acting on the conductor.
(ii) As we know that, an alpha particle is positively charged. It is given that an alpha particle while passing through a magnetic field gets deflected (projected) towards north. Since an electron is negatively charged, it will deflect in Opposite direction i.e.. south.


Q6: A student was asked to perform an experiment to study the force on a current carrying conductor in a magnetic field. He took a small aluminium rod AB, a strong horseshoe magnet, some connecting wires, a battery and a switch and connected them as shown. He observed that on passing current, the rod gets displaced. On reversing the direction of current, the direction of displacement also gets reversed. On the basis of your understanding of this phenomenon, answer the following questions:

Previous Year Questions 2022

(a) Why does the rod get displaced on passing current through it?
(b) State the rule that determines the direction of the force on the conductor AB.
(c) If the U shaped magnet is held vertically and the aluminium rod is suspended horizontally with its end B towards due north, then on passing current through the rod B to A as shown, in which direction will the rod be displaced?      (2022)

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Ans: (a) On passing current, the rod gets displaced because of a magnetic force exerted on the rod when it is placed in the magnetic field.
(b) Fleming’s left hand rule is used to determine the direction of magnetic force exerted on the conductor AB.
(c) The rod will be displaced towards left according to Fleming’s left-hand rule.

Previous Year Questions 2021

Q1: Why do two magnetic field lines not intersect each other? (2021C)

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Ans: The direction of magnetic field (B) at any point is obtained by drawing a tangent to the magnetic field line at that point. In case, two magnetic field lines intersect each other at the point P as shown in figure, magnetic field at P will have two directions, shown by two arrows, one drawn to each magnetic field line at P, which is not possible.
Previous Year Questions 2021



Q2: Name the instrument used to detect the presence of a current in a circuit. (2021C)

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Ans: Galvanometer is used to detect the current in a circuit.


Q3: What is an electromagnet? (2021C)

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Ans: An electromagnet is a current-carrying solenoid coil which is used to magnetise steel rod inside it.


Q4: (a) Name the poles P, Q, R and S of the magnets in the following figures ‘a’ and ‘b’:

(b) State the inference drawn about the direction of the magnetic field lines on the basis of these diagrams. (Term II, 2021-22)

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Ans: (a) In Figure ‘a’, poles P and Q of the magnet represents north pole and south pole respectively. In figure ‘b’, poles R and S of the magnet also represents north pole and south pole respectively.
(b) Magnetic field lines are closed continuous curves directed from north pole to south pole outside the magnet but from south pole to north pole inside the magnet.


Q5: List two factors on which the strength of magnetic field at a point due to a current carrying straight conductor depends. State the rule that determines the direction of magnetic field produced in this case. (Term II, 2021-22C)

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Ans: (i) Strength of magnetic field produced by a straight current-carrying wire at a given point is directly proportional to the current passing through it. inversely proportional to the distance of that point from the wire.
(ii) Right-hand thumb rule: The straight thumb of right hand points in the direction of electric current. The direction of the curl of fingers represents the direction of magnetic field.
Previous Year Questions 2021

Previous Year Questions 2020

Q1: (a) What is an electromagnet? List any two uses.
(b) Draw a labelled diagram to show how an electromagnet is made.
(c) State the purpose of soft iron core used in making an electromagnet.
(d) List two ways of increasing the strength of an electromagnet if the material of the electromagnet is fixed. (2020)

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Ans: (a) An electromagnet is a current-carrying solenoid coil which is used to magnetise steel rod inside it. Electromagnets are used in electric bells and buzzers, loudspeakers and headphones etc.
(b) A strong magnetic field produced inside a solenoid can be used to magnetise a piece of magnetic material, like soft iron, when placed inside the coil. The magnet so formed is called an electromagnet.
Previous Year Questions 2020
(c) The soft iron core placed in an electromagnet increases the strength of the magnetic field produced. Thus increasing the strength of electromagnet.
(d) The strength of electromagnet can be increased by
(i) Increasing the current passing through the coil.
(ii) Increasing the number of turns in the coil.


Q2: Give reasons for the following: 
(A) There is either a convergence or a divergence of magnetic field lines near the ends of a current carrying straight solenoid. 
(B) The current carrying solenoid when suspended freely rests along a particular direction. (CBSE 2020)

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Ans: (A) There is a divergence of magnetic field lines near the ends of a current carrying straight solenoid as the ends of the solenoid behave as poles. So, lines emerge and enter the ends, crowding the space and appearing divergent. 
(B) The current carrying solenoid when suspended freely rests along a particular direction because a current carrying solenoid behaves like a bar magnet with fixed polarities at its ends. The magnetic field lines are exactly identical to those of a bar magnet with one end of solenoid acting as a south-pole and its other end as north-pole


Q3: (A)Draw the pattern of magnetic field lines due to a magnetic field through and around a current carrying circular loop. 
(B) Name and state the rule to find out the direction of magnetic field inside and around the loop. (CBSE 2020)

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Ans: (A)

(B) The rule used to find the direction of magnetic field lines is right hand thumb rule, which states that if we had the loop wire in our hand such that our thumb points in direction of current then our curled finger show the direction of field at that point.


Q4: What is the role of a fuse used in series with any electrical appliance? Why should a fuse with a defined rating not be replaced by one with a larger rating? (CBSE 2020, 19, 17, 10)

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Ans: 

  •  A fuse in a circuit prevents damage to the appliances and the circuit due to overloading and short-circuiting. 
  • Overloading can occur when the live wire and the neutral wire come into direct contact. 
  •  In such a situation, the current in the circuit abruptly increases. This is called short-circuiting. 
  •  The use of an electric fuse prevents the electric circuit and the appliance from a possible damage by stopping the flow of unduly high electric current. 
  •  According to Joule’s Law of heating that takes place in the fuse, it melts to break the electric circuit. 
  •  So, a fuse is always connected in series with an appliance. 
  •  If it is connected in parallel, then it will not be able to break the circuit and the current keeps on flowing. 
  • Overloading can also occur due to an accidental hike in the supply voltage. 
  •  Sometimes, overloading is caused by connecting too many appliances to a single socket. 
  •  The fuse with defined rating means the maximum current that can flow through the fuse without melting it. 
  •  It blows off when a current more than the rated value flows through it. 
  •  If a fuse is replaced by one with larger ratings, then large current will flow through the circuit without melting the fuse. 
  •  This large current may damage the appliances. 


Q5: (A) What is an electromagnet? List any two uses. 
(B) Draw a labelled diagram to show how an electromagnet is made. 
(C) State the purpose of soft iron core used in making an electromagnet. 
(D) List two ways of increasing the strength of an electromagnet if the material of the electromagnet is fixed. (CBSE 2020)

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Ans: (A) An electromagnet is a temporary strong magnet. Its magnetism is only for the duration of current passing through it. The polarity and strength of an electromagnet can be changed. 
Uses of electromagnet:
Electromagnets are used:
(1) In electrical appliances like electric bell, electric fan etc. 
(2) In electric motors and generators.
(B) Previous Year Questions 2020

(C) Soft iron core is used in electromagnets to enhance their strength. The soft iron becomes magnetised when current flows through the coil, significantly increasing the overall magnetic field. 
(D) Ways of increasing the strength of an electromagnet:      
(1) If we increase the number of turns in the coil, the strength of electromagnet increases.      
(2) If the current in the coil is increased, the strength of electromagnet increases

Also watch: Introduction: Magnetic Field

Previous Year Questions 2019

Q1: Two circular coils P and Q are kept close to each other, of which coil P carries a current. What will you observe in the galvanometer connected across the coil Q,
(a) if current in the coil P is changed?
(b) if both the coils are moved in the same direction with the same speed?

Give reason to justify your answer in each case.     (CBSE 2011, 2019)

Previous Year Questions 2019

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Ans: (a) When the current in coil P is increased, the galvanometer connected to coil Q shows a momentary deflection in one direction. Conversely, if the current in coil P is decreased, the galvanometer deflects in the opposite direction. This occurs because a change in current alters the magnetic field around coil P, which in turn induces a current in coil Q, resulting in the galvanometer’s deflection. 
(b) If both the coils P and Q are moved in the same direction with the same speed, the magnetic field of both the coils remain unchanged. Hence no induced current is set up in coil Q and there is no deflection in the galvanometer.


Q2: What is the function of a galvanometer in a circuit?    (CBSE 2019)

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Ans: A Galvanometer is an instrument used to detect the presence of current in a circuit. It indicates whether current is flowing and, if so, its direction.


Q3: One of the major causes of fire in office buildings is short-circuiting. List three reasons which may lead to short-circuiting. How can it be prevented?    (CBSE 2019)

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Ans: Three possible reasons of short-circuiting of an electrical circuit are as follows:

  • The insulation of electrical wirings is damaged.
  • The electrical appliance used in the circuit is defective.
  • An appliance of higher power rating is being run on an electrical line of lower power rating.

Short-circuiting can be prevented by the use of electrical fuse of appropriate capacity.


Q4: Draw magnetic field lines produced around a current carrying straight conductor passing through a cardboard. Name, state and apply the rule to mark the direction of these field lines. How will the strength of the magnetic field change when the point where magnetic field to be determined is moved away from the straight conductor? Give reason to justify your answer.      (Allahabad 2019) 

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Ans:
Previous Year Questions 2019

Right-Hand Thumb Rule. 

  •  The Right-Hand Thumb Rule helps determine the direction of the magnetic field around a straight current-carrying conductor. 
  •  To apply this rule, hold the conductor in your right hand with your thumb pointing in the direction of the current. 
  •  Your curled fingers will then indicate the direction of the magnetic field lines surrounding the wire. 
  •  To assess the strength of the magnetic field, a compass needle can be used. 
  •  As the needle is moved further away from the conductor, the deflection decreases. 
  •  This indicates that the strength of the magnetic field diminishes with increasing distance from the wire. 
  •  The magnetic field lines form concentric circles around the conductor, which become larger as one moves away from it. 

Previous Year Questions 2018

Q1: (a) What are magnetic field lines? How is the direction of magnetic field at a point in a magnetic field determined using field lines? 
(b) Two circular coils ‘X’ and ‘Y’ are placed close to each other. If the current in the coil ‘X’ is changed, will some current be induced in the coil ‘Y’? Give reason.
(c) State ‘Fleming’s right hand rule.    (CBSE 2018C) (5 marks)

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Ans: (a) Magnetic field lines are imaginary lines along which the north magnetic pole would move in a magnetic field. The direction of a magnetic field at a point is determined by placing a small compass needle. The North pole of the compass indicates the direction of the magnetic field at that point

(b) Yes, if the current in coil X is changed, the magnetic field associated with it also changes around the coil ‘Y’ placed close to ‘X’. This change in magnetic field lines linked with ‘Y’, according to Faraday law of electric magnetic induction, induces a current in the coil Y.
(c) Right-Hand Thumb Rule. This rule is used to find the direction of magnetic field due to a straight current carrying wire.
It states that if we hold the current carrying conductor in the right hand in such a way that the thumb is stretched along the direction of current, then the curly finger around the conductor represent the direction of magnetic field produced by it. This is known as right-hand thumb rule.
Direction of Field Lines due to current carrying straight conductor as shown in figure.Previous Year Questions 2018


Q2: (a) State Fleming’s left hand rule.
(b) Write the principle of working of an electric motor.
(c) Explain the function of the following parts of an electric motor.
(i) Armature
(ii) Brushes
(iii) Split ring    (CBSE 2018)

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Ans: (a) Fleming’s Left-Hand Rule: Stretch the thumb, forefinger and middle finger of the left hand such that they are mutually perpendicular to each other. If the forefinger pointed towards the direction of magnetic field and middle finger in the direction of current, then the thumb will indicate the direction of motion or force experienced by the conductor. It is to be applied only when the current and magnetic field, both are perpendicular to each other.
(b) Principle of an electric motor: It works on the principle of magnetic effect of current. When a current carrying conductor is placed perpendicular to the magnetic field, it experiences a force. The direction of this force is given by Fleming’s left hand rule.
(c) (i) Armature: It consists of a large number of turns of insulated current-carrying copper wires wound over a soft iron core and rotated about an axis perpendicular to a uniform magnetic field supplied by the two poles of permanent magnet.
(ii) Brushes: Two conducting stationary carbon or flexible metallic brushes constantly touches the revolving split rings or commutator. These brushes act as a contact between commutator and terminal battery.
(iii) Split ring: The split ring acts as a commutator in an electric motor. With the help of split ring, the direction of current through the coil is reversed after every half of its rotation and make the direction of currents in both the arms of rotating coil remains same. Therefore, the coil continues to rotate in the same direction.

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Previous Year Questions 2017

Q1: The most important safety method used for protecting home appliances from short circuiting or overloading is: 
(a) earthing 
(b) use of fuse 
(c) use of stabilizers 
(d) use of electric meter (CBSE 2017, 13)

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Ans: (b)
fuse is a safety device that protects electrical appliances and circuits from damage caused by overloading or short circuits. It contains a metal wire that melts when excessive current flows through it, breaking the circuit and stopping the flow of electricity, which prevents potential damage to appliances and reduces fire risk.
Here’s why the other options are not as suitable for preventing short circuits or overloading:
(a) Earthing: Protects against electric shocks but does not prevent overloading.
(c) Use of stabilizers: Helps maintain a stable voltage but doesn’t protect against overloading or short circuits.
(d) Use of electric meter: Measures electricity usage but doesn’t offer any protective function.
Therefore, the most important safety method is the use of a fuse.

Previous Year Questions 2016

Q1: Under what condition is the force by a current-carrying conductor placed in a magnetic field maximum?    (2016)

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Ans: The force acting on a current-carrying conductor placed in a magnetic field is maximum when the direction of current is at right angles to the direction of the magnetic field.


Q2: What is an electric fuse ? Briefly describe its function. 
Or 

Explain the use of electrical fuse. What type of fuse material is used for fuse wire and why?    (2016)

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Ans: An electric fuse is a device that is used ahead of and in series of an electric circuit as a safety device to prevent the damage caused by short-circuiting or overloading of the circuit.
It is a small, thin wire of a material whose melting point is low. Generally, wire of tin or tin-lead alloy or tin-copper alloy is used as a fuse wire. If due to some fault electric circuit gets short-circuited, then a strong current begins to flow. Due to such a strong flow of current, the fuse wire is heated up and gets melted. As a result, the electric circuit is broken and current flow stops. Thus, possible damage to the circuit and appliances is avoided.


Q3: (a) List four factors on which the magnitude of magnetic force acting on a moving charge in a magnetic field depends. 
(b) How will a fine beam of electrons streaming in west to east direction be affected by a magnetic field directed vertically upwards? Explain with the help of a diagram mentioning the rule applied.     (2016) 

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Ans: (a) The magnitude of the magnetic force acting on a moving charge in a magnetic field depends on:

  • The magnitude of charge
  • Speed of moving charge
  • Strength of magnetic field
  • The angle between the direction of motion of charge and the direction of magnetic field.
    Previous Year Questions 2016

(b) The electron streaming from west to east is equivalent to a current from east to west. The magnetic field B is vertically upwards and shown by cross marks (x). Hence, in accordance with Fleming’s left-hand rule, the electron will experience a force in north direction and deflected in that direction.


Q4: What is a solenoid? Draw magnetic field lines due to a current-carrying solenoid. Write three important features of the magnetic field obtained.     (2016)

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Ans: A solenoid is a coil of large number of circular turns of wire wrapped in the shape of a cylinder. On passing electric current, a magnetic field is developed. Magnetic field lines are drawn below. The field is along the axis of solenoid such that one end of solenoid behaves as north pole and the other south pole. Thus, field of a solenoid is similar to that of a bar magnet.
Previous Year Questions 2016
Important features of magnetic field due to a current-carrying solenoid are:

  • Magnetic field lines inside the solenoid are nearly straight and parallel to its axis. It shows that magnetic field inside a solenoid is uniform.
  • Magnetic field of solenoid is identical to that due to a bar magnet with one end of solenoid behaving as a north pole and other end as a south pole.
  • A current-carrying solenoid exhibits the directive and attractive properties of a bar magnet.


Q5: Describe an activity with a neat diagram to demonstrate the presence of magnetic field around a current-carrying straight conductor.    (2016) 

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Ans: 

  • Take a long straight thick copper wire and insert it through the centre of a plain cardboard. Cardboard is fixed so that it is perfectly horizontal and the thick wire is in vertical direction.
  • Prepare an electric circuit consisting of a battery, a variable resistance, an ammeter and a plug key.
  • Connect the copper wire in the circuit between the points X and Y. Sprinkle some iron filings uniformly on the cardboard. Put the plug in key K so that a current flows in the circuit.
    Previous Year Questions 2016
  • Gently tap the cardboard using a finger. We observe that the iron filings align themselves forming a pattern of concentric circles around the copper wire.
  • These concentric circles represent the magnetic field lines. Direction of field lines is determined by the use of a compass needle. If current is flowing vertically downward then the magnetic field lines are along clockwise direction. If current is flowing vertically upward then the field lines are along anticlockwise direction.


Q6: (a) Describe an activity todraw a magnetic field line outside a bar magnet from one pole to another pole.    (2016) 
(b) What does the degree of closeness of field lines represent? 

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Ans: (a) Take a bar magnet and place it on a sheet of white paper fixed on a drawing sheet. Mark the boundary of magnet. Take a small compass and place it near the north pole of the magnet. Mark the position of two ends of compass needle. Move the needle to a new position so that its south pole occupies the position previously occupied by its north pole. In this way proceed step by step till we reach the south pole of magnet. Join the points marked on the paper by a smooth curve. This curve represents a magnetic field line as shown in Fig.

(b) The degree of closeness of magnetic field lines signifies the strength of magnetic field.


Q7: What type of current is given by a cell?    (CBSE 2016) 

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Ans: Direct current (D.C.).


Q8: Write any one method to induce current in a coil.   (CBSE 2016) 

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Ans: Whenever magnetic field around a coil is changed, a current is induced in the coil.


Q9: Define magnetic field of a bar magnet.     (CBSE 2016) 

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Ans: The magnetic field of a bar magnet is the region around the magnet in which force due to magnet can be felt.

Previous Year Questions 2015

Q1: What is the shape of magnetic field lines due to a straight current-carrying conductor?    (CBSE 2015) 

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Ans: The magnetic field lines around a straight current-carrying conductor form concentric circles with the conductor at the centre. As you move away from the conductor, these circles expand. This pattern indicates that the strength of the magnetic field decreases with distance from the wire.


Q2: What are magnetic field lines?    (CBSE 2015)

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Ans: A magnetic field line around a magnet is the path along which north pole of a magnetic compass needle points. A magnetic field line gives the direction of magnetic field at a point.


Q3: (a) Describe an activity to show that an electric current-carrying wire behaves like a magnet. 
(b) Write the rule which determines the direction of magnetic field held developed around a current-carrying straight conductor.    (CBSE 2015)

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Ans: (a) Take a straight, thick copper wire X Y and connect it to an electric circuit consisting of a battery, key and resistor. Place a small compass near the copper wire. Now put the plugin key K so that a current begins to flow. The compass needle is deflected.
It shows that the current-carrying copper wire is behaving like a magnet.
Previous Year Questions 2015
(b) The direction of the magnetic field around a straight current-carrying conductor is given by the “right-hand thumb rule”. According to the rule, imagine holding the current-carrying straight conductor in your right hand such that the thumb points towards the direction of the current. Then the fingers of the right hand wrap around the conductor in the direction of the field lines of the magnetic field.
Previous Year Questions 2015


Q4: Out of the three wires live, neutral or earth, which one goes through ON/ OFF switch?    (CBSE 2015)

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Ans: The live wire is the one that goes through the ON/OFF switch. This wire carries the current to the appliance, allowing it to function when the switch is turned on. When the switch is off, the live wire is disconnected from the appliance, stopping the flow of electricity.


Q5: Why does a current-carrying conductor experience a force when it is placed in a magnetic field?    (CBSE 2015) 

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Ans: A current-carrying conductor produces a magnetic field around it. This magnetic field interacts with the externally applied magnetic field and as a result the conductor experiences a force.


Q6: What is the frequency of A.C. being supplied in our houses?    (CBSE 2015) 

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Ans: The frequency at which A.C. is supplied to residential houses is 50 Hz.


Q7: Describe an activity to explain how a moving magnet can be used to generate electric current in a coil.    (CBSE 2015) 
Or 
A coil made of insulated copper wire is connected to a galvanometer. What will happen to the deflection of the galvanometer if a bar magnet is pushed into the coil and then pulled out of it? Give reason for your answer and name the phenomenon involved. 

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Ans: Take a coil AB of insulated copper wire having a number of turns. Connect the ends of coil to a sensitive galvanometer G. Now take a bar magnet NS and rapidly bring the magnet towards the end B of coil as shown in Fig.
The galvanometer gives momentary deflection in one direction. Now take the magnet away from the coil, the galvanometer again gives momentary deflection but in the opposite direction. It clearly shows that motion of magnet induces, a current in the coil.Previous Year Questions 2015
Again keep the magnet fixed and gently move the coil AB either towards the magnet or away from the magnet. We get deflection in the galvanometer even now. Thus, an induced current is produced in a coil whenever there is relative motion between the coil and the magnet. This phenomenon is known as electromagnetic induction.


Q8: A metallic conductor is suspended perpendicular to the magnetic field of a horse-shoe magnet. The conductor gets displaced towards left when a current is passed through it. What will happen to the displacement of the conductor if the : 
(i) current through it is increased? 
(ii) horse-shoe magnet is replaced by another stronger horse-shoe magnet? 
(iii) direction of current through it is reversed ?     (CBSE 2015) 

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Ans: (i) On increasing the current flowing through metallic conductor, the force experienced by it is proportionately increased because F ∝ I.
(ii) On using a stronger horse-shoe magnet the magnetic force increases because F ∝  Magnetic Field. 
(iii) On reversing the direction of current the direction of force is reversed and conductor is displaced towards right instead of left direction.


Q9: For the current carrying solenoid as shown below, draw magnetic field lines and giving reason explain that out of the three points A, B and C at which point the field strength is maximum and at which point it is minimum.    (CBSE  2015)
Previous Year Questions 2015

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Ans: Outside the solenoid magnetic field is minimum. At the ends of solenoid, magnetic field strength is half that of inside it. So Minimum – at point B; Maximum – at point A
Previous Year Questions 2015


Q10: What is meant by solenoid? How does a current carrying solenoid behave? Give its main use.    (CBSE 2015)

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Ans: Solenoid: A coil of many circular turns of insulated copper wire wound on a cylindrical insulating body (i.e. cardboard etc.) such that its length is greater than its diameter is called solenoid.
Previous Year Questions 2015
When current is flowing through the solenoid, the magnetic field line pattern resemble exactly with those of a bar magnet with the fixed polarity North and South pole at its ends and it acquires the directive and attractive properties similar to bar magnet. Hence the current carrying solenoid behaves a bar magnet.
Use of current carrying solenoid: It is used to form a temporary magnet called electromagnet as well as permanent magnet.


Q11: What are magnetic field lines? Justify the following statements 
(a) Two magnetic field lines never intersect each other.   
(b) Magnetic field lines are closed curves.     (CBSE 2015)

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Ans: Magnetic field lines: It is defined as the path along which the unit North pole (imaginary) tends to move in a magnetic field if free to do so.
(a) The magnetic lines of force do not intersect (or cross) one another. If they do so then at the point of intersection, two drawn tangents at that point indicate that there will be two different directions of the same magnetic field, i.e. the compass needle points in two different directions which is not possible.
(b) Magnetic field lines are closed continuous curves. They emerge out from the north pole of a bar magnet and go into its south pole. Inside the magnet, they move, from south pole to north pole.


Q12: A current-carrying conductor is placed in a magnetic field. Now answer the following : 
(i) List the factors on which the magnitude of force experienced by conductor depends. 
(ii) When is the magnitude of this force maximum? 
(iii) If initially this force was acting from right to left, how will the direction of force change, if : 
(a) direction of magnetic field is reversed? 
(b) direction of current is reversed?    (CBSE 2015)

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Ans: (i) The magnitude of force experienced by the current-carrying conductor when placed in a magnetic field depends on current flowing, length of the conductor, the strength of magnetic field, orientation of conductor in the magnetic field.
(ii) Magnitude of force is maximum when current-carrying conductor is placed at right angles to the direction of magnetic field.
(iii) (a) Direction of force is reversed that is now the force acts from left to right.
(b) Direction of force is reversed that is now the force acts from left to right.


Q13: In our daily life, we use two types of electric current whose current-time graphs are given in Fig. 13.29. 
(i) Name the type of current in two cases. 
(ii) Identify any one source for each type of current. 
(iii) What is the frequency of current in case 
(iv) in our country? 
(v) Out of the two which one is used in transmitting electric power over long distances and why?    (CBSE 2015).
Fig (a)Fig (b)

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Ans:  (i) Current shown in Fig. (a) is direct current (D.C.) but current shown in Fig. (b) is an alternating current (A.C.).
(ii) A cell/battery produces D.C. but an A.C. generator produces A.C.
(iii) In India frequency of A.C. is 50 Hz.
(iv) A.C. is used in transmitting electric power over long distances. It is so because transmission loss of electric power can be minimised for A.C. by employing suitable transformers at generating stations and consuming centres.


Q14: A student fixes a sheet of white paper on a drawing board. He places a bar magnet in the centre of it. He sprinkles some iron filings uniformly around the bar magnet. Then he taps the board gently.
Now answer the following questions : 
(i) What does the student observe? Draw a diagram to illustrate your answer. 
(ii) Why do the iron filings arrange in such a pattern? 
(iii) What does the crowding of the iron filings at the ends of the magnet indicate?    (CBSE 2015)  (5 marks)

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Ans: 
(i) The student observes the magnetic field due to the given bar magnet. The pattern of the magnetic field is shown here.
(ii) There is a magnetic field around the given bar magnet. The iron filings experience a force due to the magnetic field and thus align themselves along the magnetic field lines.

(iii) The crowding of the iron filings at the ends of the magnet indicates the position of two magnetic poles N and S of a given bar magnet.

Also watch: Introduction: Magnetic Field

Previous Year Questions 2013

Q1: If the key in the arrangement as shown below is taken out (the circuit is made open) and magnetic field lines are drawn over the horizontal plane ABCD, the lines are:

(a) concentric circles 
(b) elliptical in shape 
(c) straight lines parallel to each other 
(d) concentric circles near the point O but of elliptical shapes as we go away from it.  (CBSE 2013)

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Ans: (c)
In the arrangement described, if the key is taken out and the circuit is open, no current flows through the wire. This means that no magnetic field is generated by the wire.
However, if magnetic field lines are drawn over the horizontal plane ABCD in the absence of a current (or any other magnetic source), they would represent the Earth’s magnetic field, which appears as straight, parallel lines over a small horizontal area.
Thus, the correct answer is (c) straight lines parallel to each other.

Previous Year Questions 2010

Q1: A magnetic compass needle is placed in the plane of paper near point A, as shown in the figure. In which plane should a straight current carrying conductor be placed so that it passes through A and there is no change in the deflection of the compass? Under what condition is the deflection maximum and why? (CBSE 2010)

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Ans: The straight current carrying conductor should be placed in the paper plane, passing through A, to produce a magnetic field perpendicular to the paper plane. This ensures the compass needle remains undeflected due to the vertical magnetic field produced by the wire. The maximum deflection in the compass needle occurs when the conductor is perpendicular to the paper plane and the magnetic field is in the paper plane.

11. Previous Year Questions: Electricity

Previous Year Questions 2025

Q1: A wire of resistance R is cut into three equal parts. If these three parts are then joined in parallel, calculate the total resistance of the combination so formed.  (2 Mark)

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Ans: The total resistance of the combination is R/9.

When a wire of total resistance R is cut into three equal parts, each part has a resistance of R/3 because resistance is directly proportional to length. When these three parts are connected in parallel, the reciprocal of equivalent resistance isPrevious Year Questions 2025Hence, the total equivalent resistance becomesPrevious Year Questions 2025

Therefore, the total resistance of the parallel combination is R/9, which shows that connecting smaller equal parts of a wire in parallel greatly reduces the overall resistance.


Q2: Define electric power. When do we say that the power consumed in an electric circuit is 1 watt?  (2 Mark)

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Ans:

Electric Power: Power (P) is defined as the rate of doing work or the rate at which electrical energy is consumed or supplied. It is given by the formula:
P = V × I, where V is the potential difference (volts) and I is the current (amperes). Alternatively, P = I²R or P = V²/R, where R is resistance (ohms). The SI unit of power is the watt (W).
1 Watt: Power is 1 watt when 1 joule of energy is consumed or transferred in 1 second. Mathematically, 1 W = 1 J/s. For example, if a circuit has a potential difference of 1 volt and a current of 1 ampere, the power is:
P = V × I = 1 V × 1 A = 1 W.


Q3: (a) Write the relationship between resistivity and resistance of a cylindrical conductor of length l and area of cross-section A. Hence derive the SI unit of resistivity.
(b) Why are alloys used in electrical heating devices?  (3 marks)

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Ans:
(a) Relationship: R = ρl/A; SI unit of resistivity: ohm-meter (Ω·m).
(b) Alloys are used due to high resistivity and high melting points.

(a) For a uniform cylindrical conductor of length l and cross-sectional area A, the resistance R is proportional to l and inversely proportional to A. HencePrevious Year Questions 2025

SI unit derivation:

  • Resistance (R) is in ohms (Ω).
  • Area (A) is in square meters (m²).
  • Length (l) is in meters (m).
  • Unit of ρ = (Ω × m²)/m = Ω·m.
    Thus, the SI unit of resistivity is ohm-meter (Ω·m).

(b) Alloys are used in electrical heating devices because their resistivity is generally higher than that of constituent metals and, importantly (as stated in the chapter), alloys do not oxidise (burn) readily at high temperatures — making them suitable and durable for heating elements.


Q4: The following question is Source-based/Case-based question. Read the case carefully and answer the question that follow.
In our homes, we receive the supply of electric power through a main supply also called mains, either supported through overhead electric poles or by underground cables. In our country the potential difference between the two wires (live wire and neutral wire) of this supply is 220 V.

(a) Write the colours of the insulation covers of the line wires through which supply comes to our homes.  (1 Mark)
(b) What should be the current rating of the electric circuit (220 V) so that an electric iron of 1 kW power rating can be operated? (1 Mark)
(c) (i) What is the function of the earth wire? State the advantage of the earth wire in domestic electric appliances such as electric iron.  (2 marks)

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Ans:
(a) Red (or brown) for live, black (or blue) for neutral, green (or green-yellow) for earth.
(b) Current rating = 4.55 A (minimum 5 A fuse).
(c) (i) Function: Provides a low-resistance path for leakage current to prevent electric shock; Advantage: Enhances user safety by grounding excess current.

(a) Colours:
In domestic wiring, the standard colours are:

  • Live wire: Red or brown, carries current from the supply.
  • Neutral wire: Black or blue, completes the circuit back to the supply.
  • Earth wire: Green or green-yellow, provides a safety path for leakage current.

These colours help identify wires for safe installation and maintenance.

(b) Current rating:

  • Power of electric iron, P = 1 kW = 1000 W.
  • Voltage, V = 220 V.
  • Power formula: P = V × I.
    Previous Year Questions 2025
  • The circuit needs a fuse with a current rating slightly higher than 4.54 A, typically 5 A, to safely operate the iron without frequent blowing.

(c) (i) Earth wire:

  • Function: The earth wire connects the metallic body of an appliance to the ground via a metal plate buried in the earth. If there’s a fault (e.g., live wire touching the metal body), it provides a low-resistance path for the leakage current to flow to the ground, preventing electric shock.
  • Advantage: In appliances like an electric iron, the earth wire ensures that any leakage current is safely diverted, keeping the appliance’s body at earth potential (0 V), thus protecting the user from severe electric shock.


Q5: Consider the following circuits :

Previous Year Questions 2025

In which circuit will the power dissipated in the circuit be (I) minimum (II) maximum ? Justify your answer.  (2 Marks)

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Ans: (a) Analyze the circuits to determine power dissipation.
Use the formula for power dissipation: 
Minimum power dissipation is in circuit (ii) and maximum is in circuit (iii).

OR

Two lamps, rated 100 W; 220 V and 60 W; 220 V, are connected in parallel to an electric main supply of 220 V. Find the current drawn by the two lamps from the supply.  (2 marks)

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Ans: Total current = 0.727 A.
Given: Lamp 1: 100 W, 220 V; Lamp 2: 60 W, 220 V; connected in parallel to 220 V supply.

Step 1: Current for each lamp

  • Power formula: P = V × I.
  • For Lamp 1: Previous Year Questions 2025
  • For Lamp 2:Previous Year Questions 2025

Step 2: Total current in parallel

  • In a parallel circuit, total current is the sum of individual currents:
    Previous Year Questions 2025

Conclusion: The total current drawn from the supply is approximately 0.727 A.


Q6: (a) Define one ampere.
(b) The resistance of a wire of 0.01 cm radius is 14 Ω. If the resistivity of the material of the wire is 44 × 10⁻⁸ Ω m, find the length of the wire. (Given π = 22/7)  (3 Marks)

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Ans:
(a) One ampere is the current that flows through a conductor when one coulomb of charge passes through it in one second.
(b) Length of the wire = 0.159 m (15.9 cm).

(a) One ampere:

  • One ampere is defined as the current in a conductor when one coulomb of electric charge flows through it per second. Mathematically, 1 A = 1 C/s. It is the SI unit of electric current.

(b) Length calculation:

Previous Year Questions 2025


Q7: (a) Consider two lamps A and B of rating 50 W; 220 V and 25 W; 220 V respectively. Find the ratio of the resistances of the two lamps (i.e., RA : RB).  (2 Marks)

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Ans: RA : RB = 1 : 2.

Given: Lamp A: 50 W, 220 V; Lamp B: 25 W, 220 V.
Step 1: Resistance of each lamp

  • Power formula: P = V²/R ⇒ R = V²/P.
  • For Lamp A: Previous Year Questions 2025
  • For Lamp B: 
    Previous Year Questions 2025

Step 2: Ratio

Previous Year Questions 2025

Conclusion: The ratio of resistances is 1 : 2.


Q8: Heat produced per second due to a current in a resistor of 4 Ω is 400 joules. Calculate the potential difference across the resistor.  (2 Marks)

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Ans: Potential difference = 40 V.

  • Given: Heat produced per second = 400 J/s (power, P = 400 W); Resistance, R = 4 Ω.
  • Power formulaPrevious Year Questions 2025

Conclusion: The potential difference across the resistor is 40 V.


Q9: An electric bulb is connected to a power supply of 220 V. If the current drawn by the bulb from the supply is 500 mA, the power of the bulb is:  (1 Mark)
(a) 11 W
(b) 110 W
(c) 220 W
(d) 1100 W

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Ans: (b) 110 W

  • Given: Voltage, V = 220 V; Current, I = 500 mA = 0.5 A.
  • Power formula: P = V × I.
    P = 220 × 0.5 = 110 W.

Conclusion: The power of the bulb is 110 W, option (B).


Q10: Four identical resistors of 12 Ω each are connected in series to form a square ABCD as shown in the figure. The resistance of the network between the two points 1 and 2 is :
(a) 48 Ω
(b) 36 Ω
(c) 9 Ω
(d) 6 Ω

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Ans: (a) 9 Ω

Assuming points 1 and 2 are adjacent vertices (e.g., A and B in square ABCD with resistors on each side). The direct path AB is 12 Ω. The other path A-D-C-B is three resistors in series: 12 + 12 + 12 = 36 Ω. 
These two paths are in parallel: 
Previous Year Questions 2025


Q11: In the following Question, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.  (1 Mark)

Assertion (A): Nichrome is an alloy commonly used in electrical heating devices such as electric irons, toasters, etc. 
Reason (R): The resistivity of nichrome is high, and its resistance decreases with an increase in temperature.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

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Ans: (C) A is true, but R is false.

  • Assertion (A): Nichrome is widely used in heating devices like electric irons and toasters due to its high resistivity and high melting point, which allow it to generate significant heat and withstand high temperatures. Thus, A is true.
  • Reason (R): Nichrome has high resistivity, but its resistance increases slightly with temperature (as it’s a metal alloy with a positive temperature coefficient), not decreases. Thus, R is false.

Conclusion: Option (C) is correct.


Q12: An electric kettle is rated 750 W; 220 V. Can this kettle be used in a circuit which has a fuse of current rating 3 A? Give reason for your answer.  (2 Marks)

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Ans: No, the kettle cannot be used with a 3 A fuse.
Given: Power, P = 750 W; Voltage, V = 220 V; Fuse rating = 3 A.

Current calculation: P = V × I =>Previous Year Questions 2025

Comparison: The kettle draws 3.41 A, which exceeds the fuse rating of 3 A. A fuse blows if the current exceeds its rating, so the 3 A fuse will blow, disconnecting the circuit.

Conclusion: The kettle requires a fuse rating higher than 3.41 A (e.g., 5 A), so it cannot be used with a 3 A fuse.


Q13: Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in (i) series, and (ii) parallel. Draw the arrangements of the resistors and find the equivalent resistance of each arrangement.  (3 Marks)

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Ans:
(i) Series: Req = 11 Ω
(ii) Parallel: Req = 1 Ω

(i) Series:

Previous Year Questions 2025

In series, Req = R1 + R2 + R3 = 2 + 3 + 6 = 11 Ω.

(ii) ParallelPrevious Year Questions 2025

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Conclusion: Series resistance = 11 Ω; Parallel resistance = 1 Ω.


Q14: The resistivity of a wire made of an alloy is generally:  (1 Mark)
(a) Lower than that of its constituent metals.
(b) Higher than that of its constituent metals.
(c) Decreases with an increase in its area of cross-section.
(d) Increases with an increase in its length.

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Ans: (B) Higher than that of its constituent metals.

Alloys (e.g., nichrome) are made by mixing metals, which disrupts the regular lattice structure, increasing resistance to electron flow. Thus, the resistivity of alloys is generally higher than that of their constituent metals (e.g., nickel or chromium).

Other options:

  • (A) Incorrect: Alloys have higher resistivity.
  • (C) Incorrect: Resistivity is a material property, independent of area.
  • (D) Incorrect: Resistivity is independent of length; resistance increases with length.

Conclusion: Option (B) is correct.


Q15: Resistance of a wire of length 1 m is 35 Ω at 20°C. If the diameter of the wire is 0.2 mm, determine the resistivity of the material of the wire at that temperature. How will the resistivity of the wire change if the length and diameter of the wire both are doubled? Justify your answer. (Given π = 22/7)  (3 Marks)

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Ans:
Resistivity = 1.1 × 10⁻⁶ Ω·m; Resistivity remains unchanged if length and diameter are doubled.

Given: R = 35 Ω, l = 1 m, diameter = 0.2 mm = 0.2 × 10⁻³ m, radius = 0.1 × 10⁻³ m, π = 22/7.

Resistivity calculation:

Previous Year Questions 2025
Previous Year Questions 2025
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Effect of doubling length and diameter:

  • Resistivity (ρ) is a material property and does not depend on the dimensions (length or diameter) of the wire.
  • If length becomes 2l and diameter becomes 2d (radius becomes 2r), the area becomes A’ = π(2r)² = 4πr² = 4A.
  • Resistance changes: Previous Year Questions 2025
  • However, resistivity (ρ) remains the same as it’s a property of the material.

Conclusion: Resistivity = 1.1 × 10⁻⁶ Ω·m; it remains unchanged.


Q16: How and why is an electric fuse used in an electric circuit? Briefly describe its function.  (2 Marks)

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Ans:

  • Use: An electric fuse is used in series to protect circuits from excessive current.
  • Why: To prevent damage due to overloading or short circuits.
  • Function: It melts and breaks the circuit if the current exceeds its rating, stopping current flow.

Use: A fuse is a safety device connected in series with an electrical circuit or appliance. It consists of a wire with a low melting point.

  • Why: It protects circuits and appliances from damage due to excessive current (e.g., from overloading or short circuits) and prevents fire hazards or electric shocks.
  • Function: When the current exceeds the fuse’s rating, the fuse wire heats up (due to I²R heating) and melts, breaking the circuit and stopping current flow, thus protecting the circuit and appliances.


Q17: A wire of length ‘l’ is gradually stretched so that its length increases to 3l. If its original resistance is R, then its new resistance will be:  (1 Mark)
(a) 3R
(b) 6R
(c) 9R
(d) 27R

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Ans: (C) 9R
Given: Original length = l, resistance = R; New length = 3l.
Resistance, R = ρl/A.
When the wire is stretched to 3l, its volume remains constant (V = l × A = constant).

  • New length, l’ = 3l.
  • New area, Previous Year Questions 2025
  • New resistance, Previous Year Questions 2025

Conclusion: The new resistance is 9R, option (C).


Q18: An electric kettle is rated 230 V; 1000 W. Calculate the resistance of its heating element when in operation.  (2 Marks)

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Ans: Resistance = 52.9 Ω.

Given: P = 1000 W, V = 230 V.

Previous Year Questions 2025

Calculate: 

Conclusion: The resistance of the heating element is 52.9 Ω.


Q19: A voltage source sends a current of 2 A to a resistor of 40 Ω connected across it for 5 minutes. Calculate the electrical energy supplied by the source.  (2 Marks)

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Ans: Given,

I =2 A, R =40 Ω, t = 5 min = 300 s

First find voltage: V = IR = 2 × 40 = 80 V

Power: P = V I = 80 × 2 = 160 W

Electrical energy supplied:E = P × t = 160 × 300 = 48,000 J


Q20: Four resistors, each of resistance 2.0 Ω, are joined end to end to form a square ABCD. Using the appropriate formula, determine the equivalent resistance of the combination between its two ends A and B.

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Ans: Equivalent resistance = 2 Ω.

Given: Four resistors, each 2 Ω, form a square ABCD 

Circuit analysis:

The square has resistors: AB = 2 Ω, BC = 2 Ω, CD = 2 Ω, DA = 2 Ω.

For resistance between A and B:- Path 1: Direct resistor AB = 2 Ω.
– Path 2: Via B → C → D → A (series: BC + CD + DA = 2 + 2 + 2 = 6 Ω).
– AB (2 Ω) is in parallel with BCD (6 Ω):Previous Year Questions 2025


Q21: An electric bulb is rated 220 V; 11 W. The resistance of its filament when it glows with a power supply of 220 V is:  (1 Mark)
(a) 4400 Ω
(b) 440 Ω
(c) 400 Ω
(d) 20 Ω

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Ans: (a) 4400 Ω

Given: P = 11 W, V = 220 V.

Power formulaPrevious Year Questions 2025

Rearrange the formula:Previous Year Questions 2025

Calculate:Previous Year Questions 2025

Conclusion: The resistance is 4400 Ω, option (a).


Q22: The minimum number of identical bulbs of rating 4 V; 6 W that can work safely with desired brightness when connected in series with a 240 V mains supply is:  (1 Mark)
(a) 20
(b) 40
(c) 60
(d) 80

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Ans: (c) 60

Given: Each bulb: 4 V, 6 W; Supply = 240 V.

Current per bulbPrevious Year Questions 2025

Resistance per bulbPrevious Year Questions 2025

Series connection: For n bulbs in series, total voltage = n × 4 V.
Previous Year Questions 2025

Conclusion: Minimum 60 bulbs, option (c).


Q23: The electrical resistivity of three materials A, B, and C at 20°C is given below:
(iClassify these materials as conductor, alloy, and insulator.
(ii) Give one example of each of these materials and state one use of each material in the design of an electrical appliance say an electric stove or an electric iron.  (3 Marks)

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Ans:
(i) 

(ii) Example of Insulator: Rubber
Uses: Coating of electrical wires

Example of conductor: Tungsten
Uses: Tungsten is used as the filament of an electric bulb.

Example of alloy: Nichrome
Uses: The heating element of an electric iron


Q24: Define the term “potential difference” between two points in an electric circuit carrying current. Name and define its S.I. unit. Also express it in terms of S.I. unit of work and charge.  ( 3 Marks)

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Ans:

  • Potential Difference: The work done per unit charge to move a charge between two points.
  • S.I. Unit: Volt, defined as 1 joule of work per 1 coulomb of charge.
  • Expression: 1 V = 1 J/C.
  • Potential Difference: It is the energy required or released per unit charge when moving a charge between two points in a circuit. It drives current flow.
  • S.I. Unit: The volt (V), where 1 volt is the potential difference when 1 joule of work is done to move 1 coulomb of charge.
  • Expression: V = W/Q, where W is work (joules, J) and Q is charge (coulombs, C). Thus, 1 V = 1 J/C.


Q25: (a) State two applications of Joule’s heating in domestic electric circuits.
(b) (i) Establish the relationship between the commercial unit of electric energy and the SI unit of electric energy.  (2 Marks)

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Ans:
(a) Applications: (1) Electric iron for pressing clothes, (2) Electric toaster for heating bread.
(b) 1 kWh = 3.6 × 10⁶ J.

(a) Applications:

1. Electric Iron: The electric iron uses Joule’s heating effect to heat up the iron plate, which is then used to press clothes.

2. Electric Heater: Electric heaters use Joule’s heating effect to convert electrical energy into heat energy, which is used to warm up the surroundings.

(b) Relationship:

  • Commercial unit: Kilowatt-hour (kWh), the energy consumed by a 1 kW device in 1 hour.
  • SI unit: Joule (J).
  • 1 kWh = 1000 W × 3600 s = 1000 × 3600 J = 3.6 × 10⁶ J.

Conclusion: 1 kWh = 3.6 million joules.


Q26: (a) “The third wire of earth connection is very important in domestic electric appliances.” Justify this statement.
(b) List two precautions to be taken to avoid the overloading of domestic electric circuits.  (3 Marks)

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Ans:
(a) The earth wire prevents electric shock by grounding leakage current.
(b) Precautions: (1) Avoid connecting too many appliances to one socket, (2) Use fuses with appropriate ratings.

(a) Earth wire importance:

  • The earth wire connects the metallic body of appliances to the ground. If a fault occurs (e.g., live wire touches the metal body), the earth wire provides a low-resistance path for the current to flow to the ground, preventing electric shock to the user and protecting the appliance from damage.

(b) Precautions:

  • Avoid multiple appliances per socket: Connecting too many high-power devices increases current, risking overheating and fire.
  • Use appropriate fuses: A fuse with the correct rating (e.g., 5 A for low-power circuits) blows if current exceeds safe limits, preventing circuit damage or fire.


Q27: (A) Use Ohm’s law to determine the potential difference across the 6 ohm resistor in the following circuit when key is closed :  (2 Marks)
OR
(B) Prove that if the current through a resistor is increased by 100%, then the increase in power dissipated through the resistor will be 300%.  (2 Marks)

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Ans: (A)

Finding Total Resistance: The circuit has a 2Ω and a 4Ω resistor in series with the 6Ω resistor. The total resistance can be calculated as follows:Total Resistance, Previous Year Questions 2025

Finding Total Current: Using Ohm’s law, the total current (I) in the circuit can be calculated using the total voltage (V=6V) and total resistance:
Previous Year Questions 2025

Finding Voltage Across 6Ω Resistor: Now, we can find the voltage across the 6Ω resistor using Ohm’s law:
Previous Year Questions 2025

Therefore, the potential difference across the 6Ω resistor is 3 V.

(B) Power increases by 300%.

  • Given: Current increases by 100%, i.e., new current I’ = 2I (doubled).
  • Power formula: P = I²R.
  • Original power: P = I²R.
  • New power: P’ = (I’)²R = (2I)²R = 4I²R.
  • Increase in power: P’ – P = 4I²R – I²R = 3I²R.
  • Percentage increase:
    (P’ – P) / P × 100 = (3I²R / I²R) × 100 = 300%.

Conclusion: Doubling the current increases power by 300%.


Q28: What is short circuiting? State its possible causes. What is likely to happen if a domestic circuit gets short circuited? Give reason for the justification of your answer.  (3 Marks)

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Ans:

  • Short circuiting: When live and neutral wires come into direct contact, bypassing the load.
  • Causes: (1) Worn insulation, (2) Faulty appliance wiring.
  • Effect: Excessive current flow, causing overheating, fire, or appliance damage.
  • Reason: Low resistance in the circuit increases current significantly.

Short circuiting: Occurs when the live and neutral wires touch directly, creating a low-resistance path, bypassing the appliance (load).

Causes:

  • Worn insulation: Damaged insulation on wires allows them to touch.
  • Faulty wiring: Incorrect connections in appliances or circuits.

Effect:

  • A short circuit drastically reduces resistance, causing a large current (I = V/R) to flow. This generates excessive heat (H = I²Rt), which can melt wires, cause fires, or damage appliances.
  • A fuse or circuit breaker trips to prevent damage by breaking the circuit.

Reason: The low resistance in a short circuit allows a very high current, overwhelming the circuit’s capacity, leading to overheating or fire hazards.


Q29: Determine the maximum and minimum resistance which can be obtained by joining five resistors of 1/5 Ω each.  (2 Marks)

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Ans:
Maximum resistance = 1 Ω (series).
Minimum resistance = 0.04 Ω (parallel).

Given: Five resistors, each R = 1/5 Ω = 0.2 Ω.

  • Maximum resistance (series):
    Rmax = R1 + R2 + R3 + R4 + R5 = 0.2 + 0.2 + 0.2 + 0.2 + 0.2 = 1 Ω.
  • Minimum resistance (parallel):
    Previous Year Questions 2025Previous Year Questions 2025= 0.04 Ω.

Conclusion: Maximum = 1 Ω, minimum = 0.04 Ω.

OR

Hide Answer  

(B) Calculate potential difference across a 4 ohm resistor that produces 100 W of heat every second.  (2 Marks)
Ans: Given, Power = 100 W, Resistance = 4Ω.
Using Formula,Previous Year Questions 2025We can rearrange formula to solve for V:Previous Year Questions 2025Substitute the values:
Previous Year Questions 2025⇒ The potential difference across the 4 Ω resistor is 20 volts.


Q30: (a) Explain the statement “Potential difference between two points is 1 volt”.
(b) What do the symbols given below represent in an electric circuit? Write one function of each.   (3 Marks)

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Ans:

(a) 1 volt: Potential difference is 1 volt when 1 joule of work is required to move 1 coulomb of charge between two points. Mathematically, 1 V = 1 J/C.

(b) Symbols:

(i) The first symbol represents an Ammeter.
Function: It is used to measure the electric current flowing through a circuit.
(ii) The second symbol represents a Variable Resistor (Rheostat).
Function: It is used to vary or control the current in a circuit by changing the resistance..

Q31: Study the circuit shown in which two resistors X and Y of resistances 3 Ω and 6 Ω respectively are joined in series with a battery of 2 V.
Previous Year Questions 2025(I) Draw a circuit diagram showing the above two resistors X and Y joined in parallel with the same battery and same ammeter and voltmeter.  (1 Mark)
(II) In which combination of resistors will the (i) potential difference across X and Y, and (ii) current through X and Y, be the same? (1 Mark)
(III) Find the current drawn from the battery by the series combination of the two resistors (X and Y). (2 Marks)

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Ans: 

(II) Combinations:

  • (i) Potential difference: In a parallel combination, the voltage across each resistor (X and Y) is the same, equal to the battery voltage (2 V).
  • (ii) Current: In a series combination, the same current flows through both resistors (X and Y).

(III)  Current in series:

  • Rtotal = 3 Ω + 6 Ω = 9 Ω.
  • I = V/R = 2/9 ≈ 0.22 A.

Conclusion: Current in series combination is 0.22 A.

Previous Year Questions 2024

Q1: (A) Show how you would connect three resistors each of resistance 6 Ω, so that the combination has a resistance of 9 Ω. Also, justify your answer.

OR
(B) In the given circuit calculate the power consumed in watts in the resistor of 2 Ω:  (3 Marks) (2024)

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Ans: (A) When two 6 Ω resistances are connected in parallel and the third resistance of 6 Ω  is connected in series combinations to this, then equivalent resistance will be 9 Ω
Previous Year Questions 2024

1/Rp = 1/6Ω + 1/6Ω

∴ Rp = 3Ω

Rs = 6 + 3 = 9Ω

OR

(B) Equivalent resistance = R1 + R2 = 1Ω + 2Ω = 3Ω

The circuit contains a 1Ω and a 2Ω resistor in series. The total resistance is:

Rtotal = 1 + 2 = 3Ω

Using Ohm’s law:

I = VRtotal = 6V3Ω = 2A

Power consumed by 2Ω resistor is given by:

P = I² × R

P = (2)² × 2 = 4 × 2 = 8W


Q2: (A) (i) Define electric power. Express it in terms of potential difference (V) and resistance (R).  (4 to 5 Marks) (2024)
(ii) An electric oven is designed to work on the mains voltage of 220 V. This oven consumes 11 units of electrical energy in 5 hours. Calculate : (a) power rating of the oven (b) current drawn by the oven (c) resistance of the oven when it is red hot
OR
(B) (i) Write the relation between resistance R and electrical resistivity p of the material of a conductor in the shape of cylinder of length / and area of cross-section A. Hence derive the SI unit of electrical resistivity.
(ii) The resistance of a metal wire of length 3 m is 60 Ω. If the area of the cross-section of the wire is 4 x 10-7 m, calculate the electrical resistivity of the wire.
(iii) State how would electrical resistivity be affected if the wire (of part ‘ii’) is stretched so that its length is doubled. Justify your answer.

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Ans: (A) (i) Electric power: Rate at which electrical energy is dissipated or consumed / Rate of supplying energy to maintain the flow of current through a circuit.
Previous Year Questions 2024

(ii) (a) Energy consumed = 11 units

E = 11 kWh = 11 × 1000

P = 11000 W5 h = 2200 W

Thus, the power rating of the oven is 2200 W or 2.2 kW.

(b) I = PV = 2200220 = 10A

(c) R = V²P = (220)²2200 = 22 Ω

OR

(B)

(i) R = ρ lA

ρ = R × Al

= Ohm × (metre)²metre = ohm metre or Ωm

(ii)

Here l = 3 m, A = 4 × 10⁻⁷ m², R = 60 Ω

ρ = R × Al

= 60 × 4 × 10⁻⁷3 = 80 × 10⁻⁷ Ωm(iii) When a wire is stretched to double its length, its electrical resistivity (ρ) remains constant. A material’s electrical resistivity is intrinsic and does not alter its dimensions (length, cross-sectional area). Therefore, doubling the length of the wire will not affect its electrical resistance.


Q3: Study the I-V graph for three resistors of resistances R1, R2 and Rand select the correct statement from the following: (1 Mark) (2024)
(a) 
R1 = R2 = R3    
(b) R> R2 > R3
(c) R3 > R2 > R1    
(d) R2 > R3 > R1

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Ans: (c) R3 > R2 > R1  

This is a graph of current (I) versus voltage (V) for three resistors R1R2, and R3. The slope of each line in an I-V graph is related to the conductance, which is the reciprocal of resistance (R). Thus, the steeper the slope, the lower the resistance.

From the graph:

  • R1 has the steepest slope, indicating the lowest resistance.
  • R3 has the least steep slope, indicating the highest resistance.

Correct interpretation:

The resistances follow the order:

R3 > R2 > R1


Q4: Use Ohm’s law to determine the potential difference across the 3 Ω resistor in the circuit shown in the following diagram when the key is closed.   (2 Marks) (2024)

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Ans:

Series Resistance Calculation:

Rs = R1 + R2 + R3

= 1 + 2 + 3 = 6 Ω

Current Calculation:

I = VR

= 2V6Ω = 13 A

Voltage Calculation:

V = IR

= 13 A × 3(Ω) = 1 V


Q5: In the case of four wires of the same material, the resistance will be minimum if the diameter and length of the wire respectively are  
(1 Mark) (2024)
(a) 
D/2 and L/4
(b) D/4 and 4L
(c) 2D and L
(d) 4D and 2L

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Ans: (d)
The resistance of a wire is inversely related to its cross-sectional area and directly related to its length. To minimize resistance, you should use a wire with a larger diameter and a shorter length. Therefore, if you choose a diameter of 4D (making the cross-sectional area much larger) and a length of 2L (shorter compared to the original), you will achieve the minimum resistance, making the correct answer (d) 4D and 2L.


Q6: Explain in brief the function of an electric fuse in a domestic circuit. An electric heater of current rating 3 kW; 220 V is to be operated in an electric circuit of rating 5 A. What is likely to happen when the heater is switched ‘ON’ ? Justify your answer with the necessary calculation.  (3 Marks) (CBSE 2024)

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Ans:

  •  An electric fuse is a safety device used in electrical circuits to prevent excessive current from flowing through the wires and damaging the circuit components. 
  •  It is designed to melt and break the circuit when the current exceeds a certain limit (which is its rated value). 
  •  This helps protect the wiring and connected appliances from damage due to overcurrent.

Here P = 3 kW = 3000 W, V = 220 V, I = ?
P = V I
Previous Year Questions 2024

  • The current drawn by the heater is 13.64 A, but the current rating of the circuit is only 5 A.
  • This means the current drawn by the heater is significantly higher than the circuit’s rated current.
  • Since the current drawn by the heater (13.64 A) exceeds the current rating of the circuit (5 A), the circuit will be overloaded, and the fuse will blow to protect the circuit.
  • The fuse will disconnect the heater from the circuit, preventing further damage to the wires and avoiding potential hazards like overheating or fire.


Q7: (a) State Ohm’s law. Write the formula for the equivalent resistance R of the parallel combination of three resistors of values R1, R2, and R3.
(b) Find the resistance of the following network of resistors:  (3 Marks) (2024)

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Ans: (a) Ohm’s Law: The potential difference, V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same.

Formula:

1/Rp = 1/R1 + 1/R2 + 1/R3

(b) R + R2 = 3R2


Q8: Case/Source-based questions.  (4 to 5 Marks) (CBSE 2024)
In a domestic circuit, five LED bulbs are arranged as shown. The source voltage is 220 V and the power rating of each bulb is marked in the circuit diagram. Based on the following circuit diagram, answer the following questions:

(a) State what happens when (i) key K1 is closed. (ii) key K2 is closed.
(b) Find the current drawn by the bulb B when it glows.
(c) Calculate (i) the resistance of bulb B, and (ii) the total resistance of the combination of four bulbs B, C, D, and E.
OR
(c) What would happen to the glow of all the bulbs in the circuit when keys Kand K2 both are closed and the bulb C suddenly gets fused? Give a reason to justify your answer.

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Ans: (a) (i) Bulb A glows  (ii) Bulbs B, C, D and E glow    
(b) P = V × I
Previous Year Questions 2024
(c) (i) Resistance of bulb B,Previous Year Questions 2024
(alternative formula for calculationPrevious Year Questions 2024
(ii) Total resistance of the series combination of four bulbs = 4 x 275 = 1100 W
OR
(c) 

  • Bulb A will keep glowing with same brightness. 
  • Other bulbs i.e., B, D and E will stop glowing.
  • Reason: As the bulbs B, D and E are connected in series with fused bulb C, so no current flows through them and thus they will not glow. The bulb A remains unaffected as it is connected in parallel combination.


Q9: Consider the following combinations of resistors:  (1 Mark) (2024)
The combinations having equivalent resistance 1 is/are: 
(a)
 I and IV  
(b) Only IV  
(c) I and II  
(d) I, II and III  

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Ans: (c)
To determine which combinations of resistors have an equivalent resistance of 1 ohm, you need to analyze the given arrangements (I, II, III, and IV) based on how resistors are connected, either in series or parallel. Combinations that meet the criteria of having an equivalent resistance of 1 ohm are identified in options I and II. Therefore, the correct answer is (c) I and II.


Q10: An electric iron of resistance 20Ω  draws a current of 5 A. The heat developed in the iron in 30 seconds is:   (1 Mark) (2024)
(a) 
15000 J  
(b) 6000 J  
(c) 1500 J  
(d) 3000 J 

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Ans: (a)
To calculate the heat developed in the electric iron, we can use the formula H = I2 Rt, where H is the heat produced, I is the current (5 A), R is the resistance (20 ohms), and t is the time (30 seconds). 
Plugging in the values: H = (52) × 20 × 30 = 25 × 20 × 30 = 15000J 
So, the heat developed in the iron is 15000 J, making the correct answer (a) 15000 J.


Q11: Two wires A and B of the same material, having the same lengths and diameters 0·2 mm and 0·3 mm respectively, are connected one by one in a circuit. Which one of these two wires will offer more resistance to the flow of current in the circuit? Justify your answer.  (1 Mark) (2024)

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Ans: Wire A will offer more resistance
Justification:
 

  • Smaller diameter ⇒ Smaller cross-sectional area ⇒ Higher resistance.
  • Wire A (0.2 mm diameter) has higher resistance than wire B (0.3 mm diameter).


Q12: The following questions are source-based/case-based questions. Read the case carefully and answer the questions that follow.
Study the following circuit:  (4 to 5 Marks) (2024)

On the basis of this circuit, answer the following questions:  
(a) Find the value of total resistance between points A and B.  
(b) Find the resistance between the points B and C.  
(c) (i) Calculate the current drawn from the battery, when the key is closed.  
OR
(c) (ii) In the above circuit, the 16 Ω resistor or the parallel combination of two resistors of 8 Ω, which one of the two will have more potential difference across its two ends? Justify your answer. 

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Ans: (a) Rs = 4 Ω + 6 Ω + 16 Ω = 26 Ω

Formula:

1/Rp = 1/R1 + 1/R2 + 1/R3

(b) 

1/RP = 1/8 + 1/8

1/RP = 2/8

RP = 4Ω

(c) (i) Total resistance = 26Ω + 4Ω = 30Ω

Potential difference = V = 6V

Current I = VR

= 630 = 15 A or 0.2 A

OR
(c) (ii) 16 Ω
Justification: According to Ohm’s law when same current flows, the potential difference across a higher resistance is always higher.
Potential difference across 16 Ω = V = IR = 0.2 x 16 = 3.2V
Potential difference across 8 Ω = V = IR(total) = 0.2 x 4 = 0.8V


Q13: An electric source can supply a charge of 500 coulomb. If the current drawn by a device is 25 mA, find the time in which the electric source will be discharged completely.   (1 Mark) (2024)

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Ans: Q = I x t
Previous Year Questions 2024


Q14: (a) (i) The potential difference across the two ends of a circuit component is decreased to one-third of its initial value, while its resistance remains constant. What change will be observed in the current flowing through it? Name and state the law which helps us to answer this question.
(ii) Draw a schematic diagram of a circuit consisting of a battery of four 1.5 V cells, a 5 Ω resistor, a 10 Ω resistor and a 15 Ω resistor, and a plug key, all connected in series. Now find (I) the electric current passing through the circuit, and (II) the potential difference across the 10 Ω resistor when the plug key is closed.
OR
(b) (i) When is the potential difference between two points said to be 1 volt?
(ii) A copper wire has a diameter of 0.2 mm and resistivity of 1.6 x 10-8 Ω m. What will be the length of this wire to make its resistance 14 Ω? How much does the resistance change if the diameter of the wire is doubled?  (4 to 5 Marks(2024)

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Ans: (a) (i) 

  • Current becomes one-third of its initial value.
  • Ohm’s Law

The potential difference across the ends of a conductor is directly proportional to the current flowing through it, provided its temperature remains the same.      
(ii)
Previous Year Questions 2024
Total Voltage = V = 4 x 1·5 V = 6 V
Total resistance, R(s) = R1+ R2 + R3
= 5 Ω + 10 Ω + 15 Ω = 30 Ω

(I) Current, I = VR = 6 V30 Ω = 0.2 A

(II) V = IR = 0.2 A × 10 Ω = 2 V

OR

(b) (i) When 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

(ii) d = 0.2 mm = 2 × 10⁻⁴ m; R = 14 Ω

ρ = 1.6 × 10⁻⁸ Ω m; A = πd²4

R = ρ lA = 4 ρ lπd² or l = π d² R4 ρ

l = 227 × (2 × 10⁻⁴)²4 × 1.6 × 10⁻⁸ × 14

= 27.5 m

When the diameter is doubled, d’ = 2d A’ = 4A

R’R = AA’ or R’ = RAA’ = RA4A

R’14 = A4A

R’ = 3.5 Ω

Change (14.0 – 3.5) = 10.5 Ω


Q15:  In the given circuit the total resistance between X and Y is:

(a) 12Ω
(b) 4 Ω
(c) 6 Ω
(d) 1 Ω
      (1 Mark) (CBSE 2024)

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Ans: (d)
To find the total resistance between points X and Y, we need to analyze the circuit configuration.

The given resistances are:
2 Ω, 3 Ω, and 6 Ω.

Looking at the circuit:
The 3 Ω and 6 Ω resistors are connected in parallel.
This parallel combination is then in series with the 2 Ω resistor.
Step 1: Calculate the equivalent resistance of the parallel combination of 3 Ω and 6 Ω.
Using the formula for parallel resistance:

1/Rparallel = 1/3 + 1/6

1/Rparallel = 2 + 1/6 = 3/6 = 1/2

So, Rparallel = 2Ω.
Step 2: Add the series resistance.
Now, the 2 Ω (from the parallel combination) is in series with the 2 Ω resistor at the start:
Rtotal = 2 + 2 = 4Ω
Therefore, the total resistance between points X and Y is 4 Ω.
The correct answer is (b) 4 Ω.


Q16: Draw a schematic diagram of a circuit consisting of a battery of four dry cells of 1.5 V each, a 2 Ω resistor, a 6 Ω resistor, a 16 Ω resistor and a plug key all connected in series. Put an ammeter to measure the current in the circuit and a voltmeter across the 16 Ω resistor to measure the potential difference across its two ends. Use Ohm’s law to determine: 
(A) ammeter reading, and 
(B) voltmeter reading when the key is closed.  (3 Marks) (2024)

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Ans: 

Previous Year Questions 2024

Equivalent resistance in the circuit: 
R = 2 + 6 + 16 
= 24 Ω 
The voltage supplied by the battery: 
V = 4 × 1.5 = 6 V 
(A) Using Ohm’s law we can calculate ammeter’s reading I. 
I = V / R = 6 / 24 = = 0.25 A 
(B) Voltmeter’s reading is: 
V’ = IR = 0.25 ×16 = 4 V

Previous Year Questions 2023

Q1: The expressions that relate (i) Q, I and t and (ii) Q, V and W respectively are (here the symbols have their usual meanings):  
(1 Mark) (2023)
(a) (i) I = Q/t (ii) W = V/Q 
(b) (i) Q = I × t (ii) W = V × Q 
(c) (i) Q = I/t (ii) V = W/Q 
(d) (i) I = Q/t (ii) Q = V/W

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Ans: (b)
The correct expressions that relate charge (Q), current (I), time (t), voltage (V), and work done (W) are as follows: 
For the relationship between charge, current, and time: Q = I × t (the total charge is the product of current and time). 
For the relationship between work, voltage, and charge: W = V × Q (the work done is the product of voltage and charge). 
Thus, the correct answer is (b) (i) Q = I × t and (ii) W = V × Q.


Q2: (i) How is electric current related to the potential difference across the terminals of a conductor? Draw the labelled circuit diagram to verify this relationship.  (4 to 5 Marks) (2023)
(ii) Why should an ammeter have low resistance?
(iii) Two V-I graphs A and B for series and parallel combinations of two resistors are as shown.
Giving reason state which graph shows (a) series, (b) parallel combination of the resistors.  

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Ans: (i) It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically,
V ∝ I
V = RI
where R is resistance of the conductor.
Previous Year Questions 2023
(ii) To measure the entire current passing through the circuit, the ammeter should have low resistance.
(iii) Series Combination:

RS = R1 + R2 = Maximum resistance. Therefore, the V-I graph for a series combination will have a steeper slope, indicating that the current increases less with the potential difference.
Parallel Combination:Previous Year Questions 2023

Therefore, the V-I graph for a parallel combination will have a flatter slope, showing that the current increases more sharply with the potential difference.So, RA > RB

Graph A: The graph with a steeper slope corresponds to the series combination of resistors.

Graph B: The graph with a flatter slope corresponds to the parallel combination of resistors.


Q3: (a) Define electric power and state its SI unit. The commercial unit of electrical energy is known as ‘unit’. Write the relation between this ‘unit’ and joule.   (5 Marks) (2023)
(b) In a house, 2 bulbs of 50 W each are used for 6 hours daily and an electric geyser of 1 kW is used for 1 hour daily. Calculate the total energy consumed in a month of 30 days and its cost at the rate of ₹ 8.00 per kWh.

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Ans: (a) The electric power of an electrical device is the rate at which electric energy is dissipated or consumed in an electric circuit.
Power = Work/Time = Energy/Time
The SI unit of electric power is watt (W).
1W = 1 volt x 1 ampere = 1
VA 1 unit = 1 kWh = 3.6 x 10J
(b) 2 bulbs: P = 50 W, t = 6 hr daily
1 geyser: P = 1 kW, t = 1hr
E=P x t used by every 2 bulbs + energy used by geyser
= 30 days, Rs. = 8/kWh
Now, total energy consumed in one day
= 2 × 50 × 6 + 1 x 1000 =  1600 Wh
Total energy consumed in 30 days
= 30 x 1600 Wh = 48 kWh
Bill = 8 x 48 = Rs. 384.


Q4: If four identical resistors, of resistance 8 ohms, are first connected in series so as to give an effective resistance Rs, and then connected in parallel so as to give an effective resistance Rp, then the ratio RS / Ris:
(a) 32 
(b) 2 
(c) 0.5 
(d) 16
  (1 Mark) (CBSE 2023)

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Ans: (d)
Given:
Four identical resistors, each with a resistance of 8Ω.
Calculate the total resistance in series (
RS):
When resistors are connected in series, the total resistance is the sum of individual resistances:
RS = 8 + 8 + 8 + 8 = 4 × 8 = 32Ω
Calculate the total resistance in parallel (RP):When resistors are connected in parallel, the total resistance is given by:

Previous Year Questions 2023

So, RP = 2Ω
Calculate the ratio RS / RPPrevious Year Questions 2023

Therefore, the correct answer is (d) 16.


Q5: In the following diagram, the position of the needle is shown on the scale of a voltmeter. The least count of the voltmeter and the reading shown by it respectively are: 

Previous Year Questions 2023

(a) 0.15 V and 1.6 V 
(b) 0.05 V and 1.6 V 
(c) 0.15 V and 1.8 V 
(d) 0.05 V and 1.8 V 
     (1 Mark) (CBSE 2023)

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Ans: (c)
To determine the least count of the voltmeter and the reading shown by the needle, let’s analyze the diagram.

Least Count Calculation:

  • The scale on the voltmeter ranges from 
    0 to 
    3 volts.
  • There are 15 divisions between 0.0 and 
    1.5 volts, and similarly 15 divisions between 1.5 and 
    3 volts.
  • Therefore, each division represents: 1.5V / 15 = 0.1V
  • So, the least count of the voltmeter is 0.1 V.

Reading of the Voltmeter:

  • The needle is positioned at 1.8 V on the scale.

Based on this analysis, the correct answer is:

  • Least count: 0.1 V
  • Reading shown by the voltmeter: 1.8 V

Thus, the correct option is (c) 0.15 V and 1.8 V.


Q6: (A) An electric iron consumes energy at a rate of 880 W when the heating is at the maximum rate and 330 W when the heating is at the minimum. If the source voltage is 220 V, calculate the current and resistance in each case. 
(B) What is the heating effect of electric current? 
(C) Find an expression for the amount of heat produced when a current passes through a resistor for some time. (4 to 5 Marks) (CBSE 2023)

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Ans: (A) Power = VI = V2 / R, 
where V is voltage, I is current, R is resistance. 
First case, 
Power = 880 W 
Voltage = 220 V 
⇒ 880 = 220 × I 
⇒ I  = 4 A 
⇒ 880 = 2202 / R 
⇒ R = 55 Ω 
Second Case, 
Power = 330 W
P = VI = V2/R
P = 220 × I 
⇒ 330 = 2202 / R 
⇒ R = 440 / 3 Ω
(B) When a conductor provides resistance to current flow, the work done by the current in overcoming this resistance is converted into heat energy. This is known as the current heating effect.
(C) The amount of work done W in carrying a charge Q through a wire of resistance R in time t is given by: 
W = Q × V Since Q = I × t 
Therefore. 
W = V × I × t 
where V is the potential difference across the wire. 
Since by Ohm’s law, 
V = IR
Therefore, W = I2Rt 
The electric energy dissipated or consumed is directly proportional to the square of the current I, directly proportional to the resistance R and to the time t during which current flows. 
H = I2Rt

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Previous Year Questions 2022

Q1: (a) State Ohm’s Law. Represent it mathematically.  (2022)
(b) Define 1 ohm.
(c) What is the resistance of a conductor through which a current of 0.5 A flows when a potential difference of 2V is applied across its ends?  

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Ans: (a) Ohm’s law states that the electric current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided physical conditions like temperature etc., are kept unchanged.
Mathematically, V ∝ I
or V/I = constant or V/I = R ⇒ V = IR
where, R is called the resistance of the conductor. SI unit of resistance is ohm and is denoted by Ω. It is a constant of proportionality and its value depends upon the size, nature of material and temperature. (b) If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is said to be 1Ω.
We know that, R = V/l

Therefore, 1 ohm = 1 volt1 ampere or 1Ω = 1 V1 A

(c) Given, potential difference V = 2V

Current, I = 0.5 A

Using Ohm’s law, V = IR

⇒ R = VI ⇒ R = 2V0.5 A = 4Ω


Q2:  In the following figure, three cylindrical conductors A, B and C are shown along with their lengths and areas of cross-section.

If these three conductors are made of the same material and RA,RB, and RC be their respective resistances, then find 
(I) RA/RB, and (II)RA/RC .  (2022)

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Ans: (i) L We have three cylindrical conductors A, B and C-
Previous Year Questions 2022
These three conductors are made of the same materials and their respective resistance are σ RA. RB and Rc.

∴ R = ρ LA

So, RA = ρ × LA = ρ LA

RB = ρ × L / 22A ⇒ ρ L4A

RC = ρ × LA / (2 × 2) ⇒ ρ LA

Then, (I) RARB = ρ LA / ρ L4A ⇒ 4; RA : RB = 4

(II) RARC = ρ LA / ρ LA ⇒ 1

Q3: (a) List the factors on which the resistance of a uniform cylindrical conductor of a given material depends.
(b) The resistance of a wire of 0.01 cm radius is 10Ω. If the resistivity of the wire is 50 x 10-8 Ω m, find the length of this wire.  (2022)

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Ans: (a) The resistance of a uniform cylindrical conductor depends on
(i) Length of the conductor (l)
(ii) Area of cross-section of the conductor (A)
(b) The formula of resistance, R = ρl/A,

Given, Radius, r = 0.01 cm = 0.0001 m

Resistance, R = 10 Ω

Resistivity, ρ = 50 × 10⁻⁸ Ω m

R = ρ (L / A)

L = (R × A) / ρ

= (10 × 3.14 × (0.0001)²) / (50 × 10⁻⁸)

= 0.628 m

Hence, the length of the wire is 0.628 m.

Q4: Calculate the equivalent resistance of the following electric circuit:  (2022)

Previous Year Questions 2022

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Ans: First, calculate the resistance of 2 series resistors inside the loop, i.e., = R1 +R2
 =10Ω + 10Ω
= 20Ω
To calculate the equivalent resistance in the given electric circuit, let us find the parallel resistance. For that we use

= R1 × R2R1 + R2

= 20Ω × 20Ω20Ω + 20Ω

= 400Ω40Ω

= 10Ω

Now, again applying the series formula to add the resistors together =10Ω + 10Ω + 20Ω = 40Ω
So the total resistance of the given 40Ω


Q5: (i) Write the formula for determining the equivalent resistance between A and B of the two combinations (I) and (II) of three resistors and arranged as follows:Previous Year Questions 2022

(ii) If the equivalent resistance of the arrangements (I) and (II) are and respectively, then which one of the following graphs is correctly labelled? Justify your answer.  (2022)Previous Year Questions 2022

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Ans:  (i)
We know that resistance R1, R2 and R3 are in series.
So, RAB = R1 + R2 + R3
Previous Year Questions 2022
So,

1/R = 1/R1 + 1/R2 + 1/R3

1/R = R2R3 + R3R1 + R1R2/R1R2R3

(ii) The slope of V-I graph gives the resistance. The greater the slope greater will be the resistance. We know, Rs > R
∴ Graph (I) is correct


Q6: Study the following electric circuit in which the resistors are arranged in three arms A, B and C  (2022)
(a) Find the equivalent resistance of arm A. 
(b) Calculate the equivalent resistance of the parallel combination of the arms B and C. 
(c) (i) Determine the current that flows through the ammeter. 
OR 
(ii) Determine the current that flows in the ammeter when the arm B is withdrawn from the circuit

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Ans: The equivalent resistance in the arm A = 5Ω + 15Ω + 20Ω =40 Ω
(b) The equivalent resistance of the parallel combination of the arms B and C= ((1/10Ω + 1/20Ω + 1/30Ω) + (1/5Ω + 1/10Ω + 1/15Ω))-1 =20Ω
(c) (i) Current flow flowing through the ammeter = 6/60 = 0.1 A
OR
(ii) Current that flows in the ammeter when the arm B is withdrawn
from the circuit = 6/60 = 0.1A


Q7: (i) State Joule’s law of heating. Express it mathematically when an appliance of resistance R is connected to a source of voltage V and the current I flows through the appliance for a time t.
(ii) Α 5 Ω resistor is connected across a battery of 6 volts. Calculate the energy that dissipates as heat in 10 seconds.  (2022)

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Ans: (i) According to Joule’s law of heating, when a current (I) is passed through a conductor of resistance (R) for a certain time (t), the conductor gets heated up and the amount of energy released is given by
Previous Year Questions 2022
Previous Year Questions 2022
Previous Year Questions 2022
Previous Year Questions 2022


Q8: (a) Calculate the resistance of a metal wire of length 2 m and area of cross-section 1.55 × 10-6 m(Resistivity of the metal is 2.8 × 10-8 Ωm)
(b) Why are alloys preferred over pure metals to make the heating elements of electrical heating devices? (2022)

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Ans: (a) Here, the length of the wire, l = 2m
Area of cross-section, A = 1.55 × 10-6 m
Resistivity of metal, ρ = 2.8 × 10-8 Ω m
Previous Year Questions 2022

Previous Year Questions 2022

(b) Why Alloys Are Preferred Over Pure Metals for Heating Elements

  1. Higher Resistivity: Alloys have higher resistivity than pure metals, which helps in generating more heat.
  2. Less Oxidation & Corrosion: Alloys do not oxidize or corrode easily at high temperatures, increasing durability.
  3. Better Heat Tolerance: Alloys can withstand high temperatures without melting or softening.
  4. Stable Resistance: The resistance of alloys does not change significantly with temperature, ensuring consistent heating.
    Thus, alloys are preferred for making heating elements in electrical heating devices. 


Q9: An electric heater rated 1100 W operates at 220 V. Calculate 
(i) its resistance, and 
(ii) the current drawn by it.  (2022)

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Ans: Power of electric heater, P = 1100 W
Operating voltage, V = 220 V
(i) Its resistance, R= V2/P
R = 220 x 220/1100 Ω : R = 44 Ω
operating voltage, V = 220 V
So, current I = V/R = 220/44= 5A
(ii) Using the formula:
I = P / V
= 1100 / 220
5 A


Q10: (a) What is the meaning of electric power of an electrical device? Write its SI unit.
(b) An electric kettle of 2kW is used for 2 hours. Calculate the energy consumed in
(i) kilowatt-hour and
(ii) joules  (2022)

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Ans: (a) The electric power of an electrical device is the rate at which electric energy is dissipated or consumed in an electric circuit.
Power = Work/Time = Energy/Time
The SI unit of electric power is watt (W)
1 W =1 volt × 1 ampere = 1 V A
(b) (i) Electrical energy is the product of power and time.
E = P x t
= 2 kW × 2h = 4kW h
(ii) Electric energy consumed in joules
1 kWh =3.6× 106 J
4 kW h = 3.6 × 106 x 4 = 14.4 x 106 J.


Q11: (i) Define electric power and write its SI unit.
(ii) Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to an electric mains supply. What current is drawn from the line if the supply voltage is 220 V? (2022)

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Ans: (i) The rate at which electric energy is dissipated or consumed in an electric circuit is called electric power.
SI unit of electric power is watt (W).
(ii) Given, P1 = 100 W, V1 = 220 V
P2 = 60 W, V2 = 220 V
P = VI

∴ Total current = l1 + l2 = 0.45 + 0.27 = 0.72 A


Q12: Define the term electric power. An electric device of resistance R when connected across an electric source of voltage V draws a current I. Derive an expression for the power in terms of resistance R and voltage V. What is the power of a device of resistance 400 Ω operating at 200 V ?  (2022)

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Ans: Electric Power: The rate at which electrical energy is consumed or dissipated is called electrical power.
(b) Why Alloys Are Preferred Over Pure Metals for Heating Elements
P = VI,
[∵ V = IR]
P = (IR) I = I2 R

Given:

  • Resistance (R) = 400 Ω
  • Voltage (V) = 200 V

Power Calculation:
Using the formula:
P = V² / R
= (200 × 200) / 400
= 40000 / 400
100 W

Previous Year Questions 2021

Q1: Two lamps one rated 100W at 220V, and the other 60W at 220V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220V?  (2021)

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Ans: Current drawn by 1st lamp rated 100W at 220V

I1 = P1V = 100220 = 511 A

and current drawn by 2nd lamp rated 60W at 220V

I2 = P2V = 60220 = 311 A

In parallel arrangement the total current

I = I1 + I2 = 511 + 311 = 811 A = 0.73 A

Also read: Practice Questions: Electricity

Previous Year Questions 2020

Q1: If a person has five resistors each of value 5Ω, then the maximum resistance he can obtain by connecting them is  (2020)
(a) 1Ω
(b) 5Ω
(c) 10Ω
(d) 25Ω

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Ans: (d)
Sol: The maximum resistance can be obtained from a group of resistors by connecting them in series. Thus,

Rtotal = R + R + R + R + R = 5R
Rtotal = 5 x 5Ω

 
Q2: The maximum resistance which can be made using four resistors each of resistance 2 Ω is  (2020)
(a) 2 Ω
(b) 1 Ω
(c) 2.5 Ω
(d) 8 Ω

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Ans: (d)
Sol: A group of resistors can produce maximum resistance when they all are connected in series.

Rtotal = R1 + R2 + R3 + R4.

Substitute R1 = R2 = R3 = R4 = 2Ω:

Rtotal = 2Ω + 2Ω + 2Ω + 2Ω = 8Ω.


Q3: The maximum resistance which can be made using four resistors each of resistance 1/2 Ω is  (2020)
(a) 2Ω
(b) 1Ω 
(c) 2.5Ω 
(d) 8Ω

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Ans: (a)
Sol: The maximum resistance can be produced from a group of resistors by connecting them in series.
Thus,
Rtotal = 12 + 12 + 12 + 12 = 4 × 12 = 2Ω


Q4: A V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph.   (2020)

Previous Year Questions 2020

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Ans: V-I graph: The nichrome cable’s V-I graph exhibits a straight line. It signifies that the resistance of the wire stays unchanged while the current supply has been varied. That is an ohmic conductor because it follows the ohm’s law.
Mathematically, ohm’s law can be represented as:
V = IR
Here,
R is the resistance.
V is the voltage.
I is the current.
The resistance of a wire will be:
⇒ R = V/I
= 0.4/0.1
= 4Ω
Thus, the circuit diagram for the given case is,
Previous Year Questions 2020
Therefore, because nichrome wire has a constant resistance of 4 but also obeys Ohm’s law, it is classified as an ohmic conductor.


Q5: (a) Write the mathematical expression for Joule’s law of heating. 

(b) Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40 V.   (2020)

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Ans: (a) As per mathematical expression for Joule.s law of heating
Heat produced H = VIt = I2 Rt = V2/R t
When a voltage V is applied across a resistance R for time t so that a current I flows through it.
(b) Here charge Q = 96000 C, time t = 2 h and potential difference V = 40 V
Heat generated H=VIt = VQ
⇒ H = 40 x 96000 = 3840000 J = 3.84 x 106 J


Q6: Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1A. The current through the 40 W bulb will be:   (2020)
(a) 0.4A
(b) 0.6A
(c) 0.8A
(d) 1A

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Ans: 
Sol: Current through bulb:

  • The same current will flow through both bulbs in series.
  • When the same current flows through all of the circuit’s components, the circuit is said to be connected in series.
  • The current has only one path in such circuits.
  • As an example of a series circuit, consider the household decorative string lights.
  • This is nothing more than a string of tiny bulbs connected in series.
  • If one of the bulbs in a series burns out, none of the others will light up.
  • Hence, Option (D) is correct.


Q7: (a) What is meant by the statement, “The resistance of a conductor is one ohm”?
(b) Define electric power. Write an expression relating electric power, potential difference and resistance.
(c) How many 132 Ω resistors in parallel are required to carry 5 A on a 220 V line?  (2020)

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Ans: (a) If a current of 1 ampere flows through it when the potential difference across it is 1 volt
(b) The electric power is the electric work done per unit time [P=W/T]. the relation between electric power , potential difference and resistance is P = V2/R
(c) R= V/I
= 220/5 = 44 ohm
R= 44 = 132/n
hence, 132/44 = 3 Resistors 


Q8: (a) Write the mathematical expression for Joule’s law of heating. 
(b) Compute the heat generated while transferring 96000 coulombs of charge in two hours through a potential difference of 40 V. (2020)

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Ans: (a) The Joule’s law of heating implies that heat produced in a resistor is
(i) directly proportional to the square of current for a given resistance,
(ii) directly proportional to resistance for a given current, and
(iii) directly proportional to the time for which the current flows through the resistor. i.e., H = I2 Rt
(b) Given, charge q = 96000C time t = 2h = 7200s and potential difference V = 40V
We know,
Previous Year Questions 2020
Previous Year Questions 2020


Q9: A cylindrical conductor of length ‘l‘ and uniform area of cross-section ‘A‘ has resistance ‘R‘. Another conductor of length 2.5l and resistance 0.5R and of the same material has area of cross-section: 
(a) 5A 
(b) 2.5A 
(c) 0.5A 
(d) 1 / 5A (CBSE 2020)

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Ans: (a)

For a conductor, resistance R is given by:

Previous Year Questions 2020where:

  • ρ is the resistivity of the material (constant for a given material),
  • l is the length of the conductor,
  • A is the area of cross-section.

Given:

  1. A conductor with length l, area A, and resistance R.
  2. Another conductor of the same material with length 2.5l and resistance 0.5R.

Let the area of cross-section of the second conductor be A′.

Since the two conductors are made of the same material, their resistivity ρ is the same. We can set up the resistance formulas for each conductor:

For the first conductor:

R = ρ ⋅ lA

For the second conductor:

0.5R = ρ ⋅ (2.5l)A’

Now, substitute R = ρ ⋅ lA into the equation for the second conductor:

0.5 ⋅ ρ ⋅ lA = ρ ⋅ (2.5l)A’

Canceling ρ and l from both sides:

0.5 ⋅ 1A = 2.5A’

Rearranging to solve for A’:

A’ = 2.5 × 0.5A = 5A

Therefore, the area of cross-section A′A′ of the second conductor is 5A.


Q10The resistance of a resistor is reduced to half of its initial value. In doing so, if other parameters of the circuit remain unchanged, the heating effects in the resistor will become: 
(a) two times 
(b) half 
(c) one-fourth 
(d) four times (CBSE 2020)

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Ans: (a)

The heating effect H in a resistor is given by Joule’s law of heating:

H = I2Rt

where:

  • I is the current,
  • R is the resistance,
  • t is the time.

If the resistance R is reduced to half of its initial value, and assuming the voltage V across the resistor remains unchanged, the current I through the resistor will increase.

Using Ohm’s law:  I = V/R

When R is reduced to R/2, the new current I’ becomes:

I’ = VR/2 = 2 × VR = 2I

Now, substituting into the heating formula:

H’ = (I’)² R’ t = (2I)² × R2 × t

= 4I² × R2 × t = 2I² R t

Thus, the heating effect becomes twice the initial heating effect.
Therefore, the correct answer is (a) two times.


Q11: Assertion (A): Alloys are commonly used in electrical heating devices like electric iron and heater.
Reason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points than their constitutent metals.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A). 
(b) Both (A) and (R) are true, and (R) is not the correct explanation of (A). 
(c) (A) is true but (R) is false. 
(d) (A) is false but (R) is true.  (CBSE 2020)

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Ans: (a)
Assertion (A): Alloys are commonly used in electrical heating devices like electric irons and heaters. This is true because alloys have certain properties that make them suitable for these applications, such as high resistivity, which allows them to produce heat effectively.
Reason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points than their constituent metals.” This statement is also generally true. Alloys like nichrome have a higher resistivity, which is beneficial for heating applications, and they also tend to have controlled melting points, which can sometimes be lower than certain individual metals but still appropriate for their purpose in heating devices.
Therefore, both the assertion and reason are correct, and the reason correctly explains why alloys are used in heating devices.
So, the correct answer is (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).


Q12: Assertion (A): At high temperatures, metal wires have a greater chance of short circuiting.
Reason (R): Both resistance and resistivity of a material vary with temperature.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A). 
(b) Both (A) and (R) are true, and (R) is not the correct explanation of (A). 
(c) (A) is true but (R) is false. 
(d) (A) is false but (R) is true.  (CBSE 2020)

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Ans: (b)
Assertion (A): “At high temperatures, metal wires have a greater chance of short circuiting.” This statement is true. At high temperatures, metal wires can overheat, which may lead to insulation breakdown, increasing the risk of short circuits.
Reason (R): “Both resistance and resistivity of a material vary with temperature.” This statement is also true. The resistance and resistivity of metals generally increase with temperature due to increased atomic vibrations, which impede the flow of electrons.
However, the reason is not the direct cause of the assertion. The increased risk of short circuiting at high temperatures is mainly due to the possibility of insulation breakdown and not solely because of changes in resistance or resistivity.
Therefore, the correct answer is (b) Both (A) and (R) are true, and (R) is not the correct explanation of (A).


Q13: (A) An electric bulb is rated at 200 V; 100 W. What is its resistance? 
(B) Calculate the energy consumed by 3 such bulbs if they glow continuously for 10 hours for complete month of November. 
(C) Calculate the total cost if the rate is ₹ 6.50 per unit.  (CBSE 2020)

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Ans: (A) 

V = 200 V

P = 100 W

R = ?

P = VI

V = IR

I = VR

P = V × VR or P = V²R

R = V²P

R = 200 × 200100

R = 400 Ω

(B) P = 100 W (given 3 such bulbs)
= 100 / 1000  = 0.1 kW
No. of bulbs = 3
time, t = 10 hours
Number of days in the month of November
= 30
E = P × t
Total energy consumed by 3 bulbs in 30 days
= 3 × 0.1 × 10 × 30
= 90 kWh
Hence, the energy consumed by 3 such bulbs for the month of November will be 90 kWh.
(C) Total cost = ? 
Rate of per unit = ₹ 6.50 per unit 
1 kWh = 1 unit 
90 kWh = 90 units 
Tost cost = 90 × 6.50 = ₹ 585.00 
Hence, the total of operating 3 such bulbs will  ₹ 585.

Previous Year Questions 2019

Q1: A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1, R2 and R3 respectively. Which of the following is true?  (2019)
Previous Year Questions 2019(a) R1 = R2 = R3
(b) R1 > R2 > R3
(c) R3 > R2 > R1
(d) R2 > R3 > R
1

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Ans: (c)
Sol: Resistance is given by the slope of V-I graph.
Here the graph is I-V graph.
So, resistance is given by the inverse of the slope.
From the graph, the order of the slope is slope of R1 > slope of R2 > slope of R3
Therefore order of resistance is R> R2 > R1.
Hence, option (c) is correct.


Q2: When do we say that the potential difference between two points of a circuit in 1 volt?   (2019)

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Ans: Potential difference between two points of an electric circuit is said to be 1 volt, when a work of 1 J is to be done for moving a charge of 1 C between these two points.


Q3: Define resistance. Give its SI unit.   (2019)

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Ans: Resistance of a conductor is the measure of opposition offered by it for the flow of electric charge through it. SI unit of resistance is ‘ohm’ (Ω).


Q4: State ohm’s law.   (2019)

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Ans: According to Ohm’s law, temperature remaining constant, the current passing through a conductor is directly proportional to the potential difference across its ends, i.e.,
V ∝ I or V = IR
Here, constant R is known as the resistance of given conductor.


Q5: In the circuit given below, the resistors and have the values and respectively, which have been connected to a battery of 12 V. Calculate

(a) the current through each resistor,(b) the total circuit resistance, and(c) the total current in the circuit.  (2019)

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Ans:

(a) As all the resistances are in parallel, the voltage across each of them will be the same, which is 12V.

I1 = V / R1 = 12 / 5 = 2.4A

I2 = V / R2 = 12 / 10 = 1.2A

I3 = V / R3 = 12 / 30 = 0.4A

(b) Total current = I1 + I2 + I3 = 2.4 + 1.2 + 0.4 = 4A

(c) Total circuit resistance = V / Total current = 12 / 4 = 


Q6: On what factors does the resistance of a conductor depend?
Or
List the factors on which the resistance of a conductor in the shape of a wire depends.  (2019)

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Ans: Resistance is defined as the opposition to the flow of electrical current through a conductor. The resistance of an electric circuit can be measured numerically. Conductivity and resistivity are inversely proportional. The more conductive, the less resistance it will have.
Resistance = Potential difference/ Current
Factors on which the conductor depends

  • The temperature of the conductor 
  • The cross-sectional area of the conductor 
  • Length of the conductor 
  • Nature of the material of the conductor

Electrical resistance is directly proportional to the length (L) of the conductor and inversely proportional to the cross-sectional area (A). It is given by the following relation.
R = ρl/A
where ρ is the resistivity of the material (measured in Ωm, ohm meter)
Resistivity is a qualitative measurement of a material’s ability to resist flowing electric current. Obviously, insulators will have a higher value of resistivity than of conductors.
Previous Year Questions 2019


Q7: (a) A bulb is rated 40 W, 220 V. Find the current drawn by it when it is connected to a 220 V supply. Also, find its resistance.
(b) If the given blub is replaced by a blub of rating 25 W, 220 V, will there be any change in the value of current and resistance? Justify your answer and determine the change.   (2019)

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Ans: (a) Here power of bulb P = 40 W and voltage V = 220 V

∴ Current drawn by the bulb I = PV = 40220 = 211 A

and resistance of the bulb R = VI = 220(2 / 11) = 220 × 112 = 1210 Ω

(b) On taking another bulb of power P’ = 25 W and voltage V = 220 V, there is a change in the value of current and resistance because their values depend on the power of the bulb.

New current I’ = P’V = 25220 = 544 A

and new resistance R’ = VI’ = 220(5 / 44) = 220 × 445 = 1936 Ω


Q8: Derive the expression for power P consumed by a device having resistance R and potential difference V.
Or
A device of resistance R is connected across a source of V voltage and draws a current I . Derive an expression for power in terms of voltage (or current) and resistance.  (2019)

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Ans: Amount of work done in carrying a charge Q through a potential difference V is
W = QV.
But Q = It  ∴ W = VIt

As power is defined as the rate of doing work, hence

Power P = Wt = VItt = VI

If R is the value of resistance of the conductor, then V = RI and hence

P = VI = (RI)I = I²R

Again P = VI = V × VR = V²R

Thus, in general, we can say that electric power is given by

P = VI = I²R = V²R


Q9: Compare the power used in 2 Ω resistor in each of the following circuits shown in Fig. (a) and (b) respectively.  (2019)

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Ans: When a 2 Ω resistor is joined to a 6 V battery in series with 1 Ω and 2 Ω resistors, total resistance of the combination Rs = 2 + 1 + 2 = 5Ω

∴ Current in the circuit I1 = 6V5Ω = 1.2 A

∴ Power used in the 2Ω resistor P1 = I1² R = (1.2)² × 2 = 2.88 W.

(ii) When 2Ω resistor is joined to a 4V battery in parallel with 12Ω and 2Ω resistors, current flowing in 2Ω resistor is independent of the other resistors.

∴ Current flowing through 2Ω resistor I2 = 4V2Ω = 2 A

∴ Power used in the 2Ω resistor P2 = I2² R = (2)² × 2 = 8 W

P1P2 = 2.88 W8 W = 0.36 : 1.


Q10: Derive the relation R = R1 + R2 + R3 when resistors are joined in series.  (2019)

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Ans:

In series combination, the same current flows in all the resistances but the potential difference across each of the resistance is different.

According to Ohm’s law, we have

V1 = IR1, V2 = IR2, V3 = IR3

If the total potential difference between A and B is V, then

V = V1 + V2 + V3

= IR1 + IR2 + IR3

= I(R1 + R2 + R3)

Let the equivalent resistance be R, then

V = IR

and hence IR = I(R1 + R2 + R3)

⇒ R = R1 + R2 + R3.


Q11: (a) Write the relation between resistance and electrical resistivity of the material of a conductor in the shape of a cylinder of length ‘l’ and area of cross-section ‘A’ Hence derive the SI unit of electrical resistivity.
(b) Resistance of a metal wire of length 5 m is 100 Ω. If the area of cross-section of the wire is 3 x 10-7 m2, calculate the resistivity of the material.  (2019)

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Ans: (a) The resistance R of a conductor, the shape of a cylinder, of length l and area of cross-section A is given as:
Previous Year Questions 2019
where ρ is a constant, which is known as the electrical resistivity of the material of conductor.
Thus, resistivity ρ = RA/l
∴ SI unit of resistivity ρ shall be Previous Year Questions 2019 = Ω-m
(b) Here l = 5 m, R = 100 Ω and A = 3 x 10-7 m2
∴ Resistivity of the material of wire ρ = RA/l = Previous Year Questions 2019


Q12: (a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistors joined in parallel is equal to the sum of the reciprocals of the individual resistances.
(b) In an electric circuit two resistors of 12Ω each are joined in parallel to a 6 V battery. Find the current drawn from the battery.  (2019)

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Ans: (a)

In parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 Iand I3. Thus,
I = I1 + I2 + I3.
The potential difference across each of these resistances is the same.

Thus, from Ohm’s law

I1 = V/R1, I2 = V/R2, I3 = V/R3

If R is the equivalent resistance then,

I = V/R

∴ V/R = V/R1 + V/R2 + V/R3

and

∴ 1/R = 1/R1 + 1/R2 + 1/R3

(b) Here R1 = R2 = 12Ω and V = 6V

Net resistance R of the parallel grouping is:

1/R = 1/R1 + 1/R2 = 1/12 + 1/12 = 1/6 ⇒ R = 6Ω

∴ Current drawn by the circuit from the battery I = V/R = 6V/6Ω = 1 A.


Q13: Study the circuit of Fig. and find out : (i) Current in 12 Ω, resistor (ii) difference in the readings of A1 and A2 if any.  (2019)

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Ans: (i) In the circuit resistor of R1 = 12 Ω is connected in series with parallel combination of two resistors R2 and R3 of 24 Ω each.
The effective resistance of parallel combination of R2 and R3 is given as :

1/R23 = 1/R2 + 1/R3 = 1/24 + 1/24 = 1/12 ⇒ R23 = 12Ω

∴ Net resistance of the circuit R = R1 + R23 = 12 + 12 = 24Ω

∴ Current through 12Ω resistor = Circuit current I = V/R = 6V/24Ω = 0.25 A

(ii) Both ammeters A1 and A2 give the same reading of 0.25 A and there is no difference in their readings.


Q14: (i) Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 Ω resistor, a 10 Ω  resistor and a 15 Ω  resistor and a plug key all connected in series.
(ii) Calculate the electric current passing through the above circuit when the key is closed. (iii) Potential difference across 15 Ω  resistor.  (2019)

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Ans:

(i) The schematic diagram is given in Fig. 12.25.
(ii) Here total voltage V = 5 x 2 = 10 V
and total resistance R = R1 + R2 + R3
= 5 + 10 + 15 = 30 Ω
∴ Current passing through the circuit when the key is closed Previous Year Questions 2019
(iii) Potential difference across resistor R3 of 15 Ω
Previous Year Questions 2019


Q15: An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery as shown in the circuit. Calculate
(а) the total resistance of the circuit,
(b) the current through the circuit,
(c) the potential difference across the (i) electric, lamp and (ii) conductor, and
(d) power of the lamp.  (2019)

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Ans: Here voltage of battery V = 6 V, resistance of electric lamp = R= 20 Ω  and resistance of conductor R2 = 4 Ω.
(a) Since R1 and R2 are connected in series, the total resistance of the circuit
R = R1 + R2 = 20 + 4 = 24 Ω
(b) The current through the circuit I = V/R = 6/24 = 0.25 A
(c) (i) Potential difference across the electric lamp V= IR= 0.25 x 20 = 5 V
(ii) Potential difference across the conductor V= IR= 0.25 x 4 = 1 V
(d) Power of the lamp P = I2R1 = (0.25) x 20 = 1.25 W


Q16: (a) Three resistors R1, R2 and R3 are connected in parallel and the combination is connected to a battery, ammeter, voltmeter and key. Draw suitable circuit diagram and obtain an expression for the equivalent resistance of the combination of the resistors.
(b) Calculate the equivalent resistance of the network shown in Fig.  (2019)

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Ans: 

(a) The arrangement is shown in circuit diagram of Fig.
In parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 Iand I3. Thus,
I = I1 + I2 + I3.
The potential difference across each of these resistances is the same.

Thus, from Ohm’s law

I1 = V/R1, I2 = V/R2, I3 = V/R3

If R is the equivalent resistance then,

I = V/R

∴ V/R = V/R1 + V/R2 + V/R3

and

∴ 1/R = 1/R1 + 1/R2 + 1/R3

(b) In the network, resistors of 20Ω and 20Ω are joined in parallel and make a resistance R1, where

1/R1 = 1/20 + 1/20 = 1/10 or R1 = 10Ω.

This combined resistance R1 = 10Ω is joined in series with a given 10Ω resistance. Hence, the equivalent resistance of the network will be

R = 10 + 10 = 20Ω.


Q17: (a) Three resistors o f resistances R1, R2 and R3 are connected in (i) series, and (ii) parallel. Write expression for the equivalent resistance of the combination in each case.
(b) Two identical resistances of 12 Ω each are connected to a battery of 3 V.

Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance.  (2019)

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Ans: (a) (i) In series arrangement, equivalent resistance Rs = R1 + R2 + R3
(ii) In parallel arrangement, equivalent resistance Rp is given as:

1/Rp = 1/R1 + 1/R2 + 1/R3

(b) Here R1 = R2 = 12Ω and V = 3V.

For minimum resistance, two resistors must be connected in parallel so that

1/Rp = 1/R1 + 1/R2 = 1/12 + 1/12 = 1/6 ⇒ Rp = 6Ω

Hence power

Pp = V²/Rp = (3)²/6 = 1.5 W

For maximum resistance, two resistors must be connected in series so that

Rs = R1 + R2 = 12 + 12 = 24Ω

So, the power

Ps = V²/Rs = (3)²/24 = 0.375 W

⇒ Pp/Ps = 1.5/0.375 = 4​


Q18: Experimentally prove that in series combination of three resistances:
(а) current flowing through each resistance is same, and
(b) total potential difference is equal to the sum of potential differences across individual resistors.  (2019)

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Ans: Series combination o f resistors : We take three resistors R1 R2 and R3 and join them in series between the points X and Y in an electric circuit as shown in Fig. 12.42.
Previous Year Questions 2019
(a) Plug the key and note the ammeter reading.
Then change the position of ammeter to anywhere in between the resistors and again note the ammeter reading. We find that ammeter reading remains unchanged. It shows that in series arrangement same current flows through each resistor.
Previous Year Questions 2019
(b) 

  • Insert a voltmeter across the ends X and Y of the series combination of resistors. 
  • Plug the key so as to complete the circuit and note the voltmeter reading V across the series combination of resistors.
  • Take out plug from key K and disconnect the voltmeter. Now insert the voltmeter across the ends of first resistor R1 as shown in Fig. 12.43. 
  • Plug the key and note the voltmeter reading V1. Similarly, measure the potential difference across the other two resistors R2 and R3 separately. 
  • Let these potential differences be V2 and V3, respectively. Experimentally we find that
    V = V1 + V2  + V3
  • It shows that in series arrangement of resistors total potential difference is equal to the sum of potential differences across individual resistors.


Q19: Unit of electric power may also be expressed as: 
(a) volt-ampere 
(b) kilowatt-hour 
(c) watt-second 
(d) joule-second (CBSE 2019)

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Ans: (a)
The unit of electric power can be expressed in terms of volt-ampere (V·A), which represents the power calculated as the product of voltage (in volts) and current (in amperes). This is particularly common in electrical engineering, where 1 watt = 1 volt × 1 ampere.
Here’s an explanation of the other options:
(b) kilowatt-hour: This is a unit of energy, not power. It represents the amount of energy used over time (1 kilowatt of power used for 1 hour).
(c) watt-second: This is also a unit of energy, as it represents the power used over a second (1 watt of power used for 1 second).
(d) joule-second: This is a unit related to angular momentum or action in physics, not specifically for electric power.
Therefore, the correct answer is (a) volt-ampere.


Q20: When a 4 V battery is connected across an unknown resistor there is a current of 100 mA in the circuit. The value of the resistance of the resistor is: 
(a) 4 Ω 
(b) 40 Ω 
(c) 400 Ω 
(d) 0.4 Ω
    (CBSE 2019)

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Ans: (b)
Using Ohm’s law:
V = I × R
where:
V = 4V (voltage),
I = 100mA = 0.1A (current).
We can rearrange the formula to solve for R
:

Previous Year Questions 2019

Therefore, the resistance of the resistor is 40 Ω.


Q21: (A) In a given ammeter, a student saw that needle indicates 12th division in ammeter while performing an experiment to verify Ohm’s law. If ammeter has 10 divisions between 0 to 0·5 A, then what is the ammeter reading corresponding to 12th division? 
(B) How do you connect an ammeter and a voltmeter in an electric circuit? (CBSE 2019)

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Ans: (A) Least count of ammeter = 0.5/10 = 0.05 A 
Thus, value corresponding to 12 divisions = 0.05 ×12 = 0.6 A 
(B) An ammeter is connected in series and a voltmeter is connected in parallel in an electric circuit.

Previous Year Questions 2018

Q1: Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is (i) 13.5 Ω, (ii) 6 Ω?   (2018)

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Ans: Here resistances R1 – R2 = R3 = 9Ω
(i) To obtain an equivalent resistance Req = 13.5Ω, we connect one resistor R1 in series to the parallel com bination o f R2 and R3 as shown in figure (i). Then
Previous Year Questions 2018
Previous Year Questions 2018
(ii) To obtain equivalent resistance Req = 6 Ω, we connect resistor Rin parallel to the series combination of R2 and R3 as shown in figure (ii). Then
Previous Year Questions 2018
Previous Year Questions 2018


Q2: State Joule’s law of heating.  (2018)

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Ans: As per Joule’s law the heat produced in a resistor is (i) directly proportional to square of current flowing through it, (ii) directly proportional to resistance, and (iii) directly proportional to time. Mathematically,
Heat H = I2Rt


Q3: Why are metals good conductors of electricity whereas glass is a bad conductor of electricity? Give reason.  (2018)

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Ans: 

  • The electrons in the metal are loosely bound while the electrons in the glass are tightly bound together. 
  • Glass has one of the lowest possible heat conduction a solid. A good electrical conductivity is the same as a small electrical resistance. 
  • Silver is the best conductor because its electrons are freer to move than those of the other elements.  
  • When electricity is applied to the metal, ions start to migrate from one end to the other end of the metal, but it is not possible for electrons to travel freely in glass in which electrons are tightly bound. 
  • Glass is a bad conductor of electricity as it has high resistivity and has no free electrons.
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Previous Year Questions 2017

Q1: Nichrome is used to make the element of an electric heater. Why? (2017)

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Ans: Nichrome is an alloy of high resistivity and high melting point and does not oxidise easily.


Q2: Derive the relation  when resistors are joined in parallel. (2017)

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Ans:

In parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 Iand I3. Thus,
I = I1 + I2 + I3.
The potential difference across each of these resistances is the same.
Thus, from Ohm’s law Previous Year Questions 2017
If R is the equivalent resistance then,
I = V/R
∴ Previous Year Questions 2017
and Previous Year Questions 2017

Previous Year Questions 2016

Q1: Why do electricians wear rubber hand gloves while working?  (2016)

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Ans: Rubber is an electrical insulator. Hence electrician can work safely while working on an electric circuit without a risk of getting any electric shock.


Q2: How are two resistors with resistances R1Ω and R2Ω are to be connected to a battery of emf 3 volts to obtain maximum current flowing through it?  (2016)

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Ans: For maximum current flow, the net resistance of circuit must be least possible. Hence resistors R1 and R2 should be connected in parallel.


Q3: What potential difference is needed to send a current of 5 A through the electrical appliance having a resistance of 18 Ω?  (2016)

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Ans: Potential difference V = R x I = 18 x 5 = 90 V.


Q4: Power of a lamp is 60 W. Find the energy in SI unit consumed by it in 1 s.   (2016)

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Ans: We know that the power can be stated as the amount of electric energy consumed per unit time.
Previous Year Questions 2016
On solving For Energy, we get
Energy, E = P/T
Here; P = 60W
t = 1s
E = 60W x 1sec
∴ E = 60J
Thus, the energy in joules is 60J.


Q5: (a) What do you mean by the resistance of a conductor? Define its unit.
(b) In an electric circuit with a resistance wire and a cell, the current flowing is I. What would happen to this current if the wire is replaced by another thicker wire of the same material and the same length? Give reason.   (2016)

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Ans: (a) The resistance of a conductor is a property of the conductor, which affects the flow of current through it on maintaining a potential difference across its ends.
Unit of resistance is ohm. Resistance of a conductor is said to be 1 ohm, if a potential difference of 1 V is to be applied across its ends for maintaining flow of 1 A current.
(b) If given resistance wire is replaced by another thicker wire o f same material and same length then cross-section area of wire is increased and consequently its resistance decreases Previous Year Questions 2016 So the current flowing in the circuit increases.


Q6: (a) Why are electric bulbs filled with chemically inactive nitrogen or argon gas?
(b) The resistance of a wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 x 10-8 Ω-m, find the length of the wire.  (2016)

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Ans: (a) Electric bulbs are generally filled with some inert gas like nitrogen or argon. This enables to prolong the life of the filament of electric bulb.
(b) Here radius of wire r = 0.01 cm = 0.01 x 10-2 m, resistance R = 10 Ω and resistivity p = 50 x 10-8 Ω-m.

As R = ρLA = ρLπr², hence length L = Rπr²ρ

⇒ L = 10 × 22 × (0.01 × 10⁻²)²7 × 50 × 10⁻⁸ = 2235 m = 0.629 m or 62.9 cm


Q7: (a) Define electric power. Express it in terms of V, I and R where V stands for potential difference, R for resistance and I for current.
(b) V -I graphs for two wires A and B are shown in the Fig. 12.34. Both of them are connected in series to a battery. Which of the two will produce more heat per unit time? Give justification for your answer.  (2016)

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Ans: (a) Electric power is defined as the rate of supplying electrical energy for maintaining current flow through a circuit.
Electric power P = VI = I2R = V2/R.
(b) We know that slope of V-I graph for a given wire gives its resistance and in given figure slope of graph is more for wire A. It means that RA > RB.
In series arrangement same current I flows through both the resistance.
As heat produced per unit time is given by I2R, hence it is obvious that more heat will be produced per unit time in wire A.


Q8: (a) Establish a relationship to determine the equivalent resistance IS of a combination of three resistors having resistances R1, R2 and R3 connected in parallel.
(b) Three resistors are connected in an electrical circuit as shown. Calculate the resistance between A and B.  (2016)

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Ans: 
(a) The arrangement is shown in circuit diagram of Fig. 12.39.
In parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 Iand I3. Thus,
I = I1 + I2 + I3.
The potential difference across each of these resistances is the same.

Thus, from Ohm’s law

I1 = V/R1, I2 = V/R2, I3 = V/R3

If R is the equivalent resistance then,

I = V/R

∴ V/R = V/R1 + V/R2 + V/R3

and

∴ 1/R = 1/R1 + 1/R2 + 1/R3

(b) Here series combination of R1 and R2 is joined to R3 in parallel arrangement. Hence the net resistance R between points A and B is given as

1/R = 1/R1 + R2 + 1/R3 = 1/4 + 4 + 1/8 = 1/8 + 1/8 = 1/4

⇒ r = 4Ω


Q9: (a) Establish a relationship to determine the equivalent resistance R of a combination of three resistors having resistances R1, R2 and R3 connected in series.
(b) Calculate the equivalent resistance R of a combination of three resistors of 2 Ω, 3 Ω and 6 Ω joined in parallel.  (2016)

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Ans: (a)
Previous Year Questions 2016

  • In series combination, the same current flows in all the resistances but the potential difference across each of the resistance is different.
  • According to Ohm’s law, we have
  • V1 = IR1, V2 = IR2, V3 = IR3
    • If the total potential difference between A and B is V, then
  • V = V1 + V2 + V3= IR1 + IR2 + IR3= I(R1 + R2 + R3)
    • Let the equivalent resistance be R, then
  • V = IRand hence IR = I(R1 + R2 + R3)⇒ R = R1 + R2 + R3.(b) Here R1 = 2Ω, R2 = 3Ω, and R3 = 6ΩThe equivalent resistance R for the parallel combination of resistors will be1/R = 1/R1 + 1/R2 + 1/R3= 1/2 + 1/3 + 1/6 = 1/1 ⇒ R = 1Ω
Also read: Practice Questions: Electricity

Previous Year Questions 2015

Q1: Identify the following symbols of commonly used components in a circuit diagram:   (2015)
Previous Year Questions 2015

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Ans: (a) A rheostat, (b) A closed plug key.


Q2: Define electric current.   (2015)

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Ans: Time rate of flow of charge through any cross-section of a conductor is called electric current.


Q3: State the SI unit of electric current and define it.   (2015)

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Ans: An ampere (A), which is defined as the rate of flow of 1 coulomb of charge per second.


Q4: In an electric circuit, state the relationship between the direction of conventional current and the direction of flow of electrons.   (2015)

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Ans: 

Key Points:

  • Conventional current: The flow of positive charge from the positive terminal to the negative terminal.
  • Electron flow: The movement of negatively charged electrons from the negative terminal to the positive terminal (opposite to conventional current).

Thus, the direction of conventional current is opposite to the direction of the flow of electrons.

In summary:

  • The conventional current flows from positive to negative (from high potential to low potential).
  • The flow of electrons occurs from negative to positive (from low potential to high potential).


Q5: What is meant by the statement “potential difference between points A and B in an electric field is 1 volt”?   (2015)

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Ans: Amount of work done to bring 1 C charge from point B to point A in the electric field is 1 joule.


Q6: Define kW h.   (2015)

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Ans: A kilowatt hour (kW h) is the commercial unit of electrical energy. It is the energy consumed when 1 kW (1000 W) power is used for 1 hour.


Q7: How many joules are equals to 1 kW h?   (2015)

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Ans: 3.6 x 106 J = 1 kW h.
1 kWh = 1000 × joule/sec × 60 × 60 sec
⇒ 1 kWh = 1000 × 60 × 60 joule/sec × sec
⇒ 1 kWh = 1000 × 60 × 60 joule
⇒ 1 kWh = 36, 00, 000 joule


Q8: Define an electric circuit. Draw a labelled, schematic diagram of an electric circuit comprising of a cell, a resistor, an ammeter, a voltmeter and a closed switch. Distinguish between an open and a closed circuit.   (2015)

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Ans: 

  • A continuous and closed path of an electric current is called an electric circuit.
  • A labelled, schematic diagram of an electric circuit showing a cell E, a resistor R, an ammeter A, a voltmeter V and a closed switch S is shown here.
  • An electric circuit is said to be an open circuit when the switch is in ‘off’ mode (or key is unplugged) and no current flows in the circuit.
  • The circuit is said to be a closed circuit when the switch is in ‘on’ mode (or key is plugged) and a current flows in the circuit.
    In summary:


Q9: (a) n electrons, each carrying a charge -e, are flowing across a unit cross section of a metallic wire in unit time from east to west. Write an expression for electric current and also give its direction of flow. Give reason for your answer.
(b) The charge possessed by an electron is 1.6 x 10-19 coulomb. Find the number of electrons that will flow per second to constitute a current of 1 ampere.   (2015)

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Ans: (a) Electric current in a circuit is defined as the time rate of flow of electric charge through any cross-section and its direction is opposite to that of flow of electrons. Hence in present case
electric current I = In summary:
As electrons are flowing from east to west, the direction of electric current is from west to east.
(b) Here current I = 1 A, time t = 1 s and charge on each electron e = 1.6 x 10-19 C.
Hence, number of electrons flowing n = In summary:


Q10: (a) List the factors on which the resistance of a cylindrical conductor depends and hence write an expression for its resistance.
(b) How will the resistivity of a conductor change when its length is tripled by stretching it?   (2015)

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Ans: (a) The resistance of a cylindrical conductor i.e., a wire (R) is (i) directly proportional to its length L, (ii) inversely proportional to its cross-section area A and (iii) depends on the nature of material of wire. Mathematically,
In summary:
Here, ρ is known as the resistivity of given material. It is defined as the resistance offered by a unit cube of given material when current flows perpendicular to the opposite faces.
(b) The resistivity of the conductor remains unchanged.


Q11: (a) V-I graphs for two wires A and B are shown in the Fig. If both the wires are made of same material and are of same length, which of the two is thicker? Give justification for your answer.
(b) A wire of length L and resistance R is stretched so that the length is doubled and area of cross-section halved.

How will (i) resistance change, and (ii) resistivity change?   (2015)

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Ans: (a) Resistance R of wire A is more than that of B (RA > RB) because slope of V-I graph for A is more.
The resistance of a wire is inversely proportional to its cross-section area.
Hence area of cross-section of wire B, of smaller resistance, must be more. Thus, wire B is thicker.
(b) (i) The new resistance of wire In summary:
Thus, resistance increases to four times its original value.
(ii) The resistivity remains unchanged because it does not depend on the dimensions of a conductor of given material.


Q12: Study the electric circuit of Fig. and find (i) the current flowing in the circuit, and (ii) the potential difference across 10 Ω resistor.   (2015)

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Ans: (i) Here V = 3 V, R1 = 10 Ω and R 2 = 20 Ω. Since R1 and R2 are connected in series, the effective resistance of the circuit
R = R1 + R2 = 10 + 20 – 30 Ω
∴ Current flowing in the circuit In summary:
(ii) The potential difference across R 1 = 10 Ω resistor is V1 = IR1 = 0.1 x 10 = 1.0 V.


Q13: Three resistors of 3 Ω each are connected to a battery of 3 V as shown in Fig Calculate the current drawn from the battery.   (2015)
In summary:

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Ans: In the electric circuit shown the series combination of R1 and R2 is joined in parallel to the resistance R3. Hence equivalent resistance R of the circuit is

1/R = 1/R1 + R2 + 1/R3 = 1/3 + 3 + 1/6 + 1/3 = 1/2

⇒ R = 2Ω

∴ Current drawn from the battery I = V/R = 3/2 = 1.5 A


Q14: A 5 Ω resistor is connected across a battery of 6 volts. Calculate:
(i) the current flowing through the resistor.
(ii) the energy that dissipates as heat in 10 s.   (2015)

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Ans: Here V = 6 V and R = 5 Ω
(i) The current flowing through the resistor I = In summary:
(ii) Energy dissipated as heat in time t = 10 s is
∴ H = I2Rt = (1.2)2 x 5 x 10 = 72 J


Q15: Calculate the amount of heat generated while transferring 90000 coulombs of charge between the two terminals of a battery of 40 V in one hour. Also determine the power expended in the process.   (2015)

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Ans: Here charge transferred Q = 90000 C, potential difference b etw een the terminals o f battery V = 40 V and time t = 1 h = 3600 s.
Current = In summary:
Amount of heat generated H = Vlt = 40 x 25 x 3600 = 3600000 J = 3.6 x 106 J
and power expended In summary:


Q16: How many 40 W; 220 V lamps can be safely connected to a 220 V, 5 A line?
Justify your answer.   (2015)

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Ans: The current drawn by a 40 W, 220 V electric lamp In summary:
As the electric line is o f rating 220 V, 5 A, hence we can connect n lamps in parallel where
In summary:
Thus, we can safely connect 27 lamps of 40 W, 220 V rating to a 220 V, 5 A line.


Q17: What is meant by electric current ? Name and define its SI unit. In a conductor electrons are flowing from B to A. What is the direction of conventional current ? Give justification for your answer.
A steady current of 1 ampere flows through a conductor. Calculate the number of electrons that flow through any section of the conductor in 1 second.
(Charge on electron — 1.6 x 10-19 coulomb)   (2015)

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Ans: 

  • Electric current is defined as the rate of flow of electric charge through a cross- section of a conductor. 
  • If Q charge passes through a section of a conductor in time t, then-current I = Q/t.
  • SI unit of electric current is an ampere (A). Current is said to be one ampere, if rate of flow of charge through a cross-section of conductor be 1 coulomb per second.
  • Direction of conventional current is taken as the direction of flow of positive charge or opposite to the direction of flow of negative charge. If negatively charged in a conductor flow from B to A then the direction of conventional current will be from A to B.
  • Here current I = 1 A, time t = 1 s and charge on electron e = 1.6 x 10-19 C
  • Let n electrons flow through a section of conductor so that charge passing through the section is Q= ne.
    ∴ In summary:


Q18: What is heating effect of electric current ? Find an expression for amount of heat produced. Name some appliances based on heating effect of current.   (2015)

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Ans: When a current flows through a conducting wire (resistance wire), heat is developed, and the temperature of the wire rises. It is known as the heating effect of electric current.
If V is the potential difference maintained across the ends of a wire then, by definition, the amount of work done for flow of 1 C charge through the wire is V.
∴ Work done for flow of Q charge
W = V/Q = V/It  [∵ Q = It]
where I is the current flowing in time t.
As V = IR, hence
W= V/It = (IR)It = I2Rt
This work done (i.e., electrical energy dissipated) is converted into heat. Hence, the amount of heat produced, Q = I2Rt J
This is known as Joule’s law of heating.
Incandescent lamps, electric irons, electric stoves, toasters, geysers, electric room heaters, etc., are appliances based on the heating effect of electric current.


Q19:  What is the minimum resistance which can be made using five resistors, each of 1 / 5Ω?
(a) 1 / 5Ω
(b) 1 / 25Ω
(c) 1 / 10Ω
(d) 25Ω (CBSE 2015, 13, 12)

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Ans: (b)
To achieve the minimum resistance, all resistors should be connected in parallel because the equivalent resistance of resistors in parallel is always less than any individual resistor.
Given:
Each resistor has a resistance of 1/5Ω.
There are five resistors.
When resistors are connected in parallel, the equivalent resistance Req is given by:

In summary:Therefore, the minimum resistance that can be made using five resistors, each of 1/5Ω, is 1/25Ω.

Previous Year Questions 2014

Q1: Should the resistance of an ammeter be low or high? Give reason. (CBSE 2014)

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Ans: 

The resistance of an ammeter should be low.

Reason:

An ammeter is used to measure the current flowing through a circuit. To ensure accurate measurement of current, the ammeter should have very low resistance. This is because if the ammeter had high resistance, it would impede the flow of current, and as a result, the current in the circuit would decrease, leading to an incorrect reading.

By having low resistance, the ammeter minimizes the potential drop across it, ensuring that it does not alter the current in the circuit significantly and provides an accurate measurement.

In summary, low resistance in an ammeter ensures that it doesn’t affect the current flow in the circuit, allowing for precise measurement.

10. Previous Year Questions: The Human Eye & Colorful World

Previous Year Questions 2025

Q1: An old person is suffering from an eye defect caused by weakening of ciliary muscles and diminishing flexibility of the eye lens. If the defect of vision is ‘a’ which can be corrected by lens ‘b’, then ‘a’ and ‘b’ respectively are : (1 Mark)

(a) hypermetropia and convex lens
(b) presbyopia and bifocal lens
(c) myopia and concave lens
(d) myopia and bifocal lens

Ans: (b)

Hide Answer  

With ageing, the ciliary muscles weaken and the flexibility of the eye lens decreases, reducing the power of accommodation. This causes presbyopia, which is corrected using bifocal lenses containing both concave and convex parts for distant and near vision respectively.


Q2: A person has to keep reading material much beyond 25 cm (say at 50 cm) from the eye for comfortable reading. Name the defect of vision he is suffering from. List two causes responsible for arising of this defect. Draw a labelled diagram showing correction of this defect using eye-glasses. Are these glasses convergent or divergent of light?  (3 Marks)

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Ans:
Defect
: Hypermetropia
Causes:

  • The focal length of the eye lens becomes too long, or
  • The eyeball becomes too small.

Correction:

  • Hypermetropia is corrected by using a convex lens (converging lens) of appropriate power.
  • The convex lens provides the additional focusing power required to form the image on the retina.

Diagram
Q2: A person has to keep reading material much beyond 25 cm (say at 50 cm) from the eye for comfortable reading. Name the defect of vision he is suffering from. List two causes responsible for arising of this defect. Draw a labelled diagram showing correction of this defect using eye-glasses. Are these glasses convergent or divergent of light?  (3 Marks)Glasses Type: Convergent (convex lens)


Q3: The students in a class took a thick sheet of cardboard and made a small hole in its centre. Sunlight was allowed to fall on this small hole and they obtained a narrow beam of white light. A glass prism was taken and this white light was allowed to fall on one of its faces. The prism was turned slowly until the light that comes out of the opposite face of the prism appeared on the nearby screen. They studied this beautiful band of light and concluded that it is a spectrum of white light.

(i) Give any one more instance in which this type of spectrum is observed. (1 Mark)
(ii) What happens to white light in the above case? (1 Mark)
(iii) (A) List two conditions necessary to observe a rainbow. (2 Marks)

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Ans:

(i) One more instance:
A similar spectrum is observed in a rainbow formed in the sky after a rain shower.

(ii) What happens to white light:
In the prism experiment, white light splits into its seven component colours — Violet, Indigo, Blue, Green, Yellow, Orange and Red (VIBGYOR). This process is called dispersion of light.

(iii) Conditions necessary to observe a rainbow:

  1. The Sun should be shining in one part of the sky (behind the observer).
  2. Tiny water droplets must be present in the atmosphere in the direction opposite to the Sun, to refract, disperse and internally reflect sunlight to form the rainbow.

OR
(iii) (B) Draw a ray diagram to show the formation of a rainbow. Mark on it, points (a), (b) and (c) as given below: (2 Marks)
(a) Where dispersion of light occurs.
(b) Where light gets reflected internally.
(c) Where final refraction occurs.

Hide Answer  (iii) (B) Diagram Description: A ray diagram for rainbow formation shows a sunlight ray entering a spherical water droplet. Label:

  • (a) Dispersion: At the point where the ray enters the droplet, white light splits into colors due to refraction and varying wavelengths.
  • (b) Internal Reflection: Inside the droplet, the dispersed light reflects off the inner surface (total internal reflection).
  • (c) Final Refraction: The light exits the droplet, refracting again, forming the spectrum seen as a rainbow.Q2: A person has to keep reading material much beyond 25 cm (say at 50 cm) from the eye for comfortable reading. Name the defect of vision he is suffering from. List two causes responsible for arising of this defect. Draw a labelled diagram showing correction of this defect using eye-glasses. Are these glasses convergent or divergent of light?  (3 Marks)


Q4: The Question consist of two statements – Assertion (A) and Reason (R). Answer the question selecting the appropriate option (a), (b), (c) and (d) as given below :

Assertion (A): White light is dispersed by a glass prism into seven colours. 
Reason (R): The red light bends the least while the violet the most when a beam of white light passes through a glass prism.  (1 Mark)

(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Hide Answer  

Ans: (a)
A glass prism disperses white light into seven colours because different colours (wavelengths) refract by different amounts in the prism — red bends least and violet bends most — causing the colours to separate.


Q5: The possible way to restore clear vision of those people whose eyeball has elongated is the use of suitable:  (1 Mark)

(a) bifocal lens
(b) concave lens
(c) converging lens
(d) convex lens

Hide Answer  

Ans: (b)
An elongated eyeball causes myopia (near-sightedness), where the image of distant objects forms in front of the retina.
Using a concave (diverging) lens shifts the image back onto the retina, restoring clear distant vision.


Q6: The part of human eye which controls the amount of light entering into it:  (1 Mark)
(a) Iris
(b) Cornea
(c) Ciliary muscles
(d) Pupil

Hide Answer  

Ans: (d)
The pupil regulates and controls the amount of light entering the eye. 


Q7: (A) Give reasons: 
(i) The sky appears dark to passengers flying at very high altitude. 
(ii) ‘Danger’ signal lights are red in colour.   (2 Marks)

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Ans (A):
(i) The sky appears dark to passengers flying at very high altitude:
At high altitudes, the atmosphere is very thin, and scattering of light is not prominent. Hence, there is no scattering of sunlight, and the sky appears dark instead of blue.
(ii) ‘Danger’ signal lights are red in colour:
Red light has the longest wavelength and is least scattered by fog, smoke, or dust. Therefore, it can be seen clearly from a distance, making it ideal for danger or warning signals.

OR

(b) What is a rainbow ? “We see a rainbow in the sky only after the rainfall.” Why ?  (2 Marks)

Hide Answer  

Ans: 
A rainbow is a natural spectrum of sunlight that appears in the sky after a rain shower. It is caused by the dispersion of sunlight by tiny water droplets present in the atmosphere.
We see a rainbow only after rainfall because water droplets act like small prisms — they refract, disperse, internally reflect, and then refract sunlight again when it emerges, forming the seven colours (VIBGYOR) visible to the observer in the direction opposite to the Sun.


Q8: A person uses lenses of +2.0 D power in his spectacles for the correction of his vision. 
(a) Name the defect of vision the person is suffering from. 
(b) List two causes of this defect. 
(c) Determine the focal length of the lenses used in the spectacles.   (3 Marks)

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Ans:
(a) Defect: Hypermetropia (far-sightedness).
(b) Two causes:

  1. The focal length of the eye lens is too long.
  2. The eyeball has become too small.

(c) Focal length of the lens:
Power  P = +2.0D.
Focal lengthQ8: A person uses lenses of +2.0 D power in his spectacles for the correction of his vision. (a) Name the defect of vision the person is suffering from. (b) List two causes of this defect. (c) Determine the focal length of the lenses used in the spectacles.   (3 Marks) (in metres).
Compute digit by digit: f = 1/2.0 = 0.5 metre = 50 cm.
So the lenses used have focal length 0.5 m (50 cm).


Q9: Most of the refraction for the light rays entering the eye occurs at:  (1 Mark)
(a) Iris
(b) Pupil
(c) Crystalline lens
(d) Outer surface of Cornea

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Ans: (d)
Most of the refraction of light entering the eye takes place at the outer surface of the cornea, while the crystalline lens provides only finer adjustments of focal length for clear vision.


Q10: The curvature of eye lens of human eye:  (1 Mark)
(a) is fixed.
(b) can be increased.
(c) can be decreased.
(d) increases or decreases as the case may be.

Hide Answer  

Ans: (d)
The curvature of the eye lens is controlled by the ciliary muscles.
It increases to view nearby objects (lens becomes thicker) and decreases to view distant objects (lens becomes thinner).


Q11: A person uses lenses of power -0.5 D in his spectacles for the correction of his vision. 
(a) Name the defect of vision the person is suffering from. 
(b) List two causes of this defect. 
(c) Determine the focal length of the lenses used in the spectacles.  (3 Marks)

Hide Answer  

Ans:
(a) Defect: Myopia (near-sightedness).(b)Two causes:

  1. Excessive curvature of the eye lens.
  2. Elongation (increased length) of the eyeball.

(c) Focal length of the lens:
Power 
P=−0.5 D.
Focal length Q11: A person uses lenses of power -0.5 D in his spectacles for the correction of his vision. (a) Name the defect of vision the person is suffering from. (b) List two causes of this defect. (c) Determine the focal length of the lenses used in the spectacles.  (3 Marks)Compute step by step: f = 1/- 0.5 = -2.0 meters = -200 cm.
So the spectacle lens has focal length −2.0 m (the negative sign shows it is a diverging lens).

Previous Year Questions 2024

Q1: The lens system of human eye forms an image on a light sensitive screen, which is called as:   (2024)
(a) 
Cornea
(b) 
Ciliary muscles
(c) 
Optic nerves
(d) 
Retina

Hide Answer  

Ans: (d)
The lens system of the human eye forms an image on a light-sensitive screen called the retina. The retina captures the light and sends the visual information to the brain through the optic nerves, allowing us to see.


Q2: Study the diagram given below and answer the questions that follow:   (2024)
(i) Name the defect of vision represented in the diagram. Give reason for your answer.
(ii) List two causes of this defect.
(iii) With the help of a diagram show how this defect of vision is corrected.

Hide Answer  

Ans: (i) Hypermetropia or Far-sightedness.
Reason – Image is formed behind the retina. Near point for the person is farther away from the normal near point (25 cm)  
(ii)

  • Focal length of the eye lens is too long. 
  • The eyeball has become too small.

(iii)
Previous Year Questions 2024N = Near point of a hypermetropic eye
N’= Near point of a normal eye


Q3: Consider the following statements in the context of human eye:   (2024)
(A) The diameter of the eye ball is about 2.3 cm. 
(B) Iris is a dark muscular diaphragm that controls the size of the pupil. 
(C) Most of the refraction for the light rays entering the eye occurs at the crystalline lens. 
(D) While focusing on the objects at different distances the distance between the crystalline lens and the retina is adjusted by ciliary muscles.
The correct statements are — 
(a)
 (A) and (B)
(b) (A), (B) and (C)
(c) (B), (C) and (D)
(d) (A), (C) and (D)

Hide Answer  

Ans: (a)
Adult eyeballs measure around 2.3 cm in diameter. The iris is the dark muscular diaphragm, whereas the ciliary muscle adjusts the focal length to focus images on the retina. The eye’s initial surface, where light transitions from air to the cornea, has the most substantial shift in its index of refraction. The cornea accounts for around 80% of refraction, whereas the inner crystalline lens accounts for 20%.


Q4: Assertion – Reason based questions:   (2024)
These questions consist of two statements — Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below :
Assertion (A): Myopic eye cannot see distant objects distinctly.
Reason (R): For the correction of myopia converging lenses of appropriate power are prescribed by eye-surgeons.
(a) 
Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

Hide Answer  

Ans: (c)

The assertion (A) states that a myopic eye, or near sighted eye, cannot see distant objects clearly, which is true. However, the reason (R) claims that converging lenses are used to correct myopia, which is false; instead, diverging lenses are prescribed for myopia to help the eye focus on distant objects. Thus, the answer is (c) because (A) is true, but (R) is false.


Q5: When do we say that a particular person is suffering from hypermetropia? List two causes of this defect. Name the type of lens used to correct this defect.   (CBSE 2024)

Hide Answer  

Ans: When he cannot see nearby objects distinctly but can see far object clearly.
2 causes: 
(i) Focal length of the eye lens is too long.
(ii) Eyeball becomes too small.
lens used to correct this defect is Convex or Converging lens


Q6: (a) Define the term power of accommodation of human eye. Write the name of the part of eye which plays a major role in the process of accommodation and explain what happens when human eye focuses (i) nearby objects and (ii) distant objects.
OR
(b) Draw a ray diagram to show the formation of a rainbow in the sky. On this diagram mark A — where dispersion of light occurs, B — where internal reflection of light occurs and C — where refraction of light occurs. List two necessary conditions to observe a rainbow.    (2024)

Hide Answer  Ans: (a) 

  • The ability of the eye lens to adjust its focal length, so as to clearly focus rays coming from distant as well as near objects on the retina, is called the power of accommodation of the eye.
  • Ciliary muscles play a major role in maintaining power of accommodation.

(i)  While focusing on nearby objects ciliary muscles contract, eye lens becomes thick and its focal length decreases.  
(ii) While focusing on distant objects ciliary muscles relax, eye lens becomes thin and its focal length increases.
OR
(b)
Previous Year Questions 2024

Two conditions:  
(i) The Presence of tiny water droplets in the atmosphere.
(ii) Position of Sun at the back(behind) of the observer. 


Q7: Name and explain the phenomenon of light due to which the path of a beam of light becomes visible when it enters a smoke filled room through a small hole. Also state the dependence of colour of the light we receive on the size of the particle of the medium through which the beam of light passes.    (CBSE 2024)

Hide Answer  

Ans: The phenomenon responsible is the Tyndall Effect, where light scatters due to tiny particles in the medium, making the beam visible in a smoke-filled room.
Dependence on Particle Size: 

  • Small Particles – Scatter shorter wavelengths (blue/violet) more, making the sky appear blue. 
  • Medium Particles – Scatter all wavelengths almost equally, making the light appear white. 
  • Large Particles – Scatter all wavelengths equally, giving a white or greyish appearance, like clouds. 


Q8: When a beam of white light passes through a region having very fine dust particles, the colour of light mainly scattered in that region is:    (2024)
(a)
 Red
(b) Orange
(c) Blue
(d) Yellow  

Hide Answer  

Ans: (c)
When a beam of white light passes through an area with fine dust particles, the color of light that gets scattered the most is blue. This happens because shorter wavelengths of light, like blue, scatter more easily than longer wavelengths, like red or orange. Therefore, blue light is not visible in such conditions.


Q9: Define the term power of accommodation of human eye. What happens to the image distance in the eye when we increase the distance of an object from the eye ? Name and explain the role of the part of human eye responsible for it in this case.      (2024)

Hide Answer  

Ans:

  • Power of Accommodation – The ability of the eye lens to adjust its focal length to focus objects at different distances on the retina.
  • Image Distance – Remains unchanged as the image always forms on the retina.
  • Role of Ciliary Muscles:
    • For Distant Objects – Ciliary muscles relax, lens becomes thin, focal length increases.
    • For Nearby Objects – Ciliary muscles contract, lens becomes thick, focal length decreases.


Q10: (a) Study the diagram given below and answer the questions that follow:      (2024)


(i) Name the defect of vision depicted in this diagram stating the part of the eye responsible for this condition. (ii) List two causes of this defect. (iii) Name the type of lens used to correct this defect and state its role in this case.
OR
(b) What is dispersion of white light? State its cause. Draw a diagram to show dispersion of a beam of white light by a glass prism. 

Hide Answer  

Ans:

(a) Defect of Vision

(i) Name & Responsible Part:

  • Hypermetropia (Farsightedness) – Caused by ciliary muscles/eye lens.

(ii) Causes:

  1. Focal length of the eye lens is too long.
  2. Eyeball is too small, forming the image behind the retina.

(iii) Correction:

  • Convex lens (Converging lens) decreases the focal length, helping focus the image on the retina.

(b) Dispersion of White Light

  • Definition: Splitting of white light into seven colours when passing through a prism.
  • Cause: Different colours bend at different angles due to varying wavelengths.
(b) Dispersion of White Light


Q11: For Q. Nos.11 and 12, two statements are given – One labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:    (2024)
Assertion (A): The rainbow is a natural spectrum of sunlight in the sky.
Reason (R): Rainbow is formed in the sky when the sun is overhead and water droplets are also present in air.
(a) 
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) 
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) 
Assertion (A) is true, but Reason (R) is false.
(d) 
Assertion (A) is false, but Reason (R) is true.

Hide Answer  

Ans: (c)

  • Assertion (A): True. A rainbow is a natural spectrum formed by the dispersion, refraction, and internal reflection of sunlight by water droplets.
  • Reason (R): False. Rainbows form when the sun is behind the observer at a low angle (typically ~42° above the horizon for the primary rainbow), not overhead, and water droplets are present in the atmosphere.


Q12: Assertion (A) and Reason (R), answer these questions selecting the appropriate option given below:    (CBSE 2024)
Assertion (A): Red light signals are used to stop the vehicles on the road.
Reason (R): Red coloured light is scattered the most so as to be visible from a large distance.
(a)
 Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

Hide Answer  

Ans: (c)
The assertion (A) is true because red light signals are indeed used to indicate that vehicles should stop. However, the reason (R) is false; red light is actually scattered the least among the visible colors due to longer wavelength , which is why it can be seen from a distance without much scattering. Therefore, the correct answer is (c), as (A) is true, but (R) is false.

Previous Year Questions 2023

Q1: Observe the following diagram and answer the questions following it :

Previous Year Questions 2023

(i) Identify the defect of vision shown.
(ii) List its two causes.
(iii) Name the type of lens used for the correction of this defect.   (2023)

Hide Answer  

Ans:  

  • Student is suffering from myopia.
  • The two possible reasons due to which the defect of vision arises are:
    • excessive curvature of the eye-lens
    • elongation of the eye-ball
  • A student with myopia has the far point nearer than infinity, thus, the image of a distant object is formed in front of the retina.
Previous Year Questions 2023

Correction of myopia: This defect can be corrected by using a concave lens of suitable power as it brings the image back on to the retina, thus the defect is corrected.

Previous Year Questions 2023


Q2: Observe the following diagram showing an image formation in an eye:

Previous Year Questions 2023

(a) Identify the defect of vision shown in the figure.
(b) List its two causes and suggest a suitable corrective lens to overcome this defect          (2023)

Hide Answer  

Ans: (a) Hypermetropia
(b) Hypermetropia is caused due to following reasons:
(i) Shortening of the eyeball
(ii) Focal length of crystalline lens is too long
It is corrected by using a convex lens which converges and shifts the image to the retina from behind.


Q3: Define the term dispersion of white light. State the colour which bends (i) the most, and (ii) the least while passing through a glass prism. Draw a diagram to show the dispersion of white light.   (2023)

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Ans: 
(i) The phenomenon of the splitting up of the white light into its constituent’s colours is called dispersion of light. Dispersion of light is caused due to different constituents and colours of light after different refractive indices to the material of the prism.
(ii) The formation of a rainbow is caused by the dispersion of the white sunlight into its constituent colours.
(iii) Based on the dispersion of white light into its constituent’s colours, we can conclude that
(a) The white light consists of seven colours.
(b) The violet light suffers maximum deviations and the red light suffers minimum deviation.


Q4: What is a rainbow? Draw a labelled diagram to show its formation.   (2023)

Hide Answer  

Ans: After a rain-shower, the sunlight gets dispersed by tiny droplets, present in the atmosphere. The water droplets act like small glass prisms. They refract and disperse the incident sunlight, then reflect it internally, and finally refract it again when it comes out of the raindrop. Due to dispersion or light and internal reflection, different colours reach the observer’s eye, which is called a rainbow.
Previous Year Questions 2023


Q5: The colour of the clear sky from the Earth appears blue but from space, it appears black. Why?   (2023)

Hide Answer  

Ans: As our earth has an atmosphere and space has no atmosphere, the sunlight scatters by the particles in the atmosphere. The amount of scattering of light is inversely proportional to wavelength thus violet, and blue colours scatter more than red and hence the colour of the sky appears blue from the earth.


Q6: A narrow beam XY of white light is passing through a glass prism ABC as shown in the diagram:

Previous Year Questions 2023

Trace it on your answer sheet and show the path of the emergent beam as observed on the screen PQ. Name the phenomenon observed and state its cause. (CBSE 2023)

Hide Answer  

Ans: 

Previous Year Questions 2023

The phenomenon of the splitting up of the white light into its constituent colours is called dispersion of light. Dispersion of light is caused when different constituent colours of light offer different refractive indices to the material of the prism.


Q7: Give reasons for the following. 
(A) Danger signals installed at airports and at the top of tall buildings are of red colour. 
(B) The sky appears dark to the passengers flying at very high altitudes.
(C) The path of a beam of light passing through a colloidal solution is visible. (CBSE 2023)

Hide Answer  

Ans: (A)  Red danger signals are used because red light has the longest wavelength in the visible spectrum and is scattered the least by atmospheric particles, fog, or smoke. This allows red light to be visible from greater distances, making it effective for warning signals.
(B) At higher altitude either the atmospheric medium is very rare or there are no particles present/no atmosphere, thus the scattering of light taking place is not enough at such heights or no scattering of sunlight takes place. Hence, the sky appears dark to the passengers flying at very high altitude. 
(C) Tyndall effect deals with the phenomenon of scattering of light by colloidal particles. When a fine beam of sunlight enters a room, the particles present in the room become visible due to scattering of light by these particles. When sunlight passes through a canopy of a dense forest, tiny water droplets in the mist scatter light.


Q8: It is observed that the power of an eye to see nearby objects as well as far off objects diminishes with age. 
(A) Give reason for the above statement. 
(B) Name the defect that is likely to arise in the eyes in such a condition. 
(C) Draw a labelled ray diagram to show the type of corrective lens used for restoring the vision of such an eye. (CBSE 2023)

Hide Answer  

Ans: (A) We know that the ability of the accommodation factor to alter the focal length declines with ageing. Without corrective spectacles, aged people have difficulty seeing surrounding objects clearly. 
(B) The person is suffering from presbyopia.
(C)Previous Year Questions 2023

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Previous Year Questions 2022

Q1: In the diagram given below. X and Y are the end colours of the spectrum of white light. The colour of ‘Y’ represents the    (2022)

(a) Colour of sky as seen from earth during the day
(b) Colour of the sky as seen from the moon
(c) Colour used to paint the danger signals
(d) Colour of sun at the time of noon.  

Hide Answer  

Ans: (c)
Sol: Red colour is used to paint the danger signals. As live know, visible spectrum has violet indigo, blue, green, yellow. orange and red colour and these colours are arranged with red on the top and violet at the bottom (near base of the prism) on the basis of wavelength and frequency, and among these colours red colour has largest wavelength (≈ 750 nm). Thus, here Y will be red.

Previous Year Questions 2021

Q1: Assertion (A): Sky appears blue in the day time. Reason
(R) : White light is composed of seven colours.   (2021)
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true, and R is not correct explanation of A
(c) (A) is true, but R is false.
(d) (A) is false but R is true.  

Hide Answer  

Ans: (b)
Sol: The blue colour of clear sky is due to  scattering of sunlight


Q2: Which of the following statements is NOT true for the scattering of light? 
(a) Colour of the scattered light depends on the size of particles of the atmosphere. 
(b) Red light is least scattered in the atmosphere. 
(c) Scattering of light takes place as various colours of white light travel with different speed in air. 
(d) The fine particles in the atmospheric air scatter the blue light more strongly than red. So the scattered blue light enters our eyes. (CBSE Term-1 2021)

Hide Answer  

Ans: (c)
(a) This statement is true. The color of scattered light depends on the size of the particles in the atmosphere. Smaller particles scatter shorter wavelengths (like blue) more than longer wavelengths (like red).
(b) This statement is true. Red light is least scattered in the atmosphere because it has a longer wavelength.
(c) This statement is NOT true. All colors of white light travel at the same speed in air. Scattering occurs due to the interaction of light with particles, not because of a difference in the speed of various colors in air.
(d) This statement is true. Fine particles in the atmosphere scatter blue light more strongly than red light, which is why the sky appears blue to our eyes.
Therefore, the correct answer is (c) Scattering of light takes place as various colours of white light travel with different speed in air.

Also read: The Human Eye

Previous Year Questions 2020

Q1: Why is the Tyndall effect shown by colloidal particles? State four instances of observing the Tyndall effect. (2020)

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Ans: The phenomenon of scattering of light by the colloidal particles gives rise to the Tyndall effect. When a beam of light strikes colloidal particles, the path of the beam becomes visible. This is known as the Tyndall effect. This phenomenon can be observed when
(i) Sunlight passes through a canopy of dense forest when tiny water droplets in the mist scatter light.
(ii) torch light is switched on in a foggy environment, light rays are visible after being scattered by the fog particles in the surrounding air.
(iii) a fine beam of sunlight enters a smoke-filled room through a small hole.
(iv) Shining a flashlight beam into a glass of dilated milk produces a Tyndall effect.

Previous Year Questions 2019

Q1: Define the term power of accommodation. Write the modification in the curvature of the eye which enables us to see the nearby objects. What are the limits of the accommodation power of a healthy normal human eye?  (CBSE 2019)

Hide Answer  

Ans:  

  • Accommodation power is the property of the eye lens to adjust its focal length to focus objects situated at different distances from the eye on the retina.
  • When the ciliary muscles are relaxed, the eye lens becomes thin and its focal length is maximum and equal to the diameter of the eyeball. In this condition, one can see distant objects.
  • At the time of looking at nearby objects, the ciliary muscles of the eye contract and the eye lens become thicker. Consequently, the focal length of the eye lens decreases and nearby objects are focussed at the retina.
  • There are definite limits to accommodation power for a healthy normal eye. The farthest distance, up to which an eye can see objects clearly, is called the far point of the eye and its value is infinity.
  • The minimum distance, up to which an eye can see distinctly, is known as the near point of the eye and its value is 25 cm for a normal eye.


Q2: When do we consider a person to be myopic? List two causes of this defect.
Explain using a ray diagram how can this defect of an eye be corrected.  (CBSE 2019)

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Ans: 

  •  A person is said to have a myopic vision if he can see objects situated near the eye clearly but cannot see distant objects. 
  •  If a person can see clearly up to a distance x from the eye then it means that the far point of the eye has shifted from infinity to a point O situated at distance V from the eye. 
  •  Light rays coming from a distant object (u = ∞) are focussed in front of the retina of the eye. 
  • Two possible causes of myopia are :
    •  Either the power of the eye lens has become more than its normal value due to excessive curvature of the cornea (focal length of eye lens has decreased) or 
    •  Elongation of the eyeball due to some genetic defect. 
  •  To rectify this defect a concave lens of focal length f is used, which may form the virtual image of the distant object at the far point of the defective eye (i.e., u = – ∞ and v = – x) so that now the defective eye may form the image at the retina. 
  •  Obviously, by using the lens formula, we have 

Previous Year Questions 2019⇒ f = -x
The ray diagram of the defective eye and its rectification are shown here.
Normal Eye 

Normal Eye Myopic Eye

Myopic EyeCorrection by concave lens 

Correction by concave lens 


Q3: Name two defects of vision. Mention their cause and the type of lenses used to correct them. (CBSE 2019)

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Ans: Two main defects of vision, their cause and correction are as follows :
(1) Myopia – In myopia or near-sightedness (or short-sightedness), a person can see nearby objects clearly but cannot see distant objects distinctly. For a myopic eye, the far point is not at infinity but has shifted nearer to the eye.
Myopia arises due to either (i) excessive curvature of the cornea, or (ii) elongation of the eyeball.
Myopia can be corrected by using a concave lens whose focal length has the same numerical value as the distance of the far point of the defective eye.Previous Year Questions 2019
(2) Hypermetropia -In hypermetropia or long-sightedness, a person can see distant objects distinctly but cannot see nearby objects so clearly. For a longsighted eye, the near point is not at 25 cm but has shifted away from the eye.
Hypermetropia arises either because
(i) the focal length of the eye lens is too large, or
(ii) contraction of the eyeball.
Hypermetropia can be corrected by using a suitable convex lens, which forms a virtual image of the object situated at 25 cm at the near point of the defective eye so that the eye lens can focus it on the retina.
Previous Year Questions 2019


Q4: What is a rainbow? Draw a labelled diagram to show the formation of a rainbow. (CBSE 2019)
Or
Describe the formation of the rainbow in the sky with the help of a diagram. 

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Ans: A rainbow is a natural spectrum appearing in the sky after a rain shower. A Rainbow is caused by the dispersion of sunlight by tiny water droplets hanging in the atmosphere after a rain shower. The water droplets act like small prisms. As shown in the figure, the water droplets refract and disperse the incident sunlight. These rays are then reflected internally and finally refracted again and come out of raindrops. Due to the dispersion and internal reflection of light different colours reach the observer’s eye and a rainbow is seen.
Previous Year Questions 2019An important point to be noted here is that a rainbow is always formed in a direction opposite to that of the Sun.


Q5: What is atmospheric refraction? Briefly explain. Why does the apparent position of a star appear different from its true position? (CBSE 2019)

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Ans: The atmospheric air layer just near the earth’s surface is comparatively denser and the upper layers of the atmosphere are successively rarer and more rarer. Hence, a light ray passing through atmospheric air undergoes refraction. Since the physical conditions of air are not stationary, the apparent position of the distant object, as seen through the air, fluctuates. It is known as an effect of atmospheric refraction.
Light coming from a distant star entering into the earth’s atmosphere gradually bends towards the normal on account of atmospheric refraction. Hence, the star appears slightly higher than its actual position when viewed near the horizon.


Q6: The stars appear higher from the horizon than they are. Explain why it is so. (CBSE 2019)

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Ans: As we go up and up in earth’s atmosphere, it goes on becoming rarer and more rarer. As a result, the atmospheric layer near the earth’s surface has a maximum refractive index and the refractive index gradually decreases with an increase in height.
When a light ray from a star enters into earth’s atmosphere, it travels from a rarer to denser medium and hence continues to bend towards the normal. As a result, an observer on Earth considers the apparent position of a star to be at a higher altitude.
Previous Year Questions 2019


Q7: What happens to the image distance in the normal human eye when we decrease the distance of an object, say 10 m to 1 m? Justify your answer.  (Delhi 2019)

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Ans: There is no change in the image distance in the eye. The eye lens can adjust its focal length called accommodation. When object distance decreases, ciliary muscles contract and lens becomes thick and its focal length decreases. It facilitates near vision.


Q8: Due to the gradual weakening of ciliary muscles and diminishing flexibility of the eye lens a certain defect of vision arises. Write the name of this defect. Name the type of lens required by such persons to improve their vision. Explain the structure and function of such a lens. (Delhi 2019)

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Ans: 

  • The defect of vision is Presbyopia.
  • A bifocal lens is required by such persons to improve their vision.

Structure and function of Bifocal lens 

  • To view far-off objects, the upper part of the bifocal lens is a Concave or Diverging lens. 
  • To facilitate or view nearby objects, the Lower part of the bifocal lens is Convex or Converging lens.


Q9: (A) State the role of ciliary muscles present in our eye.
(B) Identify the defect of vision in each of the given cases and suggest its corrective measure. 
(i) The eye lens has become milky and cloudy. 
(ii) The eye lens has excessive curvature. 
(iii) The eye lens has large focal length (longer than normal). 
(iv) Ciliary muscles have weakened.

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Ans: (A) Ciliary muscles relax and contract to adjust/modify the focal length of eye lens. 
(B) Eye defects and corrective measures:
Previous Year Questions 2019


Q10: (A) With the help of diagram explain Isaac Newton’s experiment that led to the idea that the sunlight is made up of seven colours. 
(B) What is atmospheric refraction? List two natural phenomena based on atmospheric refraction.

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Ans: (A) Issac Newton was the first to use a glass prism to obtain the spectrum of white light. He tried to split various colours of the spectrum of white light by using another similar prism, but he could not get any more colours. Thus he proved that sunlight is made of seven colours.

Previous Year Questions 2019

(B) Atmospheric Refraction: It is the refraction of light by the earth’s atmospheric layers having varying refractive indices. 
Two natural phenomena: 
(1) Twinkling of stars, 
(2) Advanced sunrise and delayed sunset.

Previous Year Questions 2018

Q1: (a) Write the function of each of the following parts of the human eye :
(i) Cornea
(ii) Iris
(iii) Crystalline lens
(iv) Ciliary muscles.
(b) Why does the sun appear reddish early in the morning? Will this phenomenon be observed by an astronaut on the Moon? Give a reason to justify your answer. (CBSE 2018)

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Ans: (a) The function of the given parts is stated below:
(i) The cornea is the outer bulged-out thin transparent layer of the eye and provides most of the refraction for the light entering into the eye.
(ii) The iris controls the size of the pupil of the eye.
(iii) The crystalline lens provides the finer adjustment of the focal length required to focus objects situated at different distances in front of the eye on the retina.
(iv) The ciliary muscles help in controlling the curvature of the crystalline lens and thus can change the power of the crystalline lens.

(b) In the early morning, the Sun is situated near the horizon. Light from the Sun passes through thicker layers of air and covers a larger distance in the Earth’s atmosphere before reaching our eyes. While passing through atmosphere blue light is mostly scattered away and the Sun appear reddish as shown in Fig.
Previous Year Questions 2018

The phenomenon is not observed by an observer on the Moon because the Moon has no atmosphere of its own and hence no scattering of light is possible.


Q2: (a) What is meant by the term power of accommodation? Name the component of the eye that is responsible for the power of accommodation.
(b) A student sitting at the back bench in a class has difficulty in reading. What could be his defect of vision? Draw a ray diagram to illustrate the image formation of the blackboard when he is seated at the (i) back seat and (ii) front seat. State two possible causes of this defect. Explain the method of correcting this defect with the help of a ray diagram. (CBSE 2018)

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Ans: (a) Power of accommodation: The ability of an eye lens to adjust its focal length to form a sharp image of the object at varying distances on the retina is called the power of accommodation. The ciliary muscles of the eye are responsible for changing the focal length of the eye lens.
(b) 

  • The student is suffering from myopia shortsightedness or nearsightedness.
  • When a student is seated in the back seat.
Previous Year Questions 2018

In this case, the student is suffering from myopia and has a short focal length of the eye lens.
(ii) When a student is seated in the front seat.
Previous Year Questions 2018

Causes:
(i) Excessive curvature of the eye lens
(ii) Elongation of eyeball.
The defect is corrected by using a concave lens of suitable power placed in front of the eye as shown below. It diverges the rays and forms a virtual image of a distant object at the far point of the myopic eye. These diverged rays enter the eye and form the image on the retina. Thus, the concave lens shifts the image back onto the retina instead of in front of it and the defect is corrected.
Previous Year Questions 2018


Q3: (a) What is presbyopia? State its cause. How is it corrected?
(b) Why does the sun appear reddish early in the morning? Explain with the help of a labelled diagram. (CBSE 2018C)

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Ans:  (a) (i) Presbyopia 

  • Presbyopia is a condition that occurs as a part of normal ageing.
    Due to loss of power of accommodation of the eye, with age, objects at a normal near working distance will appear blurry. The near point gradually recedes away. This defect of the eye is called Presbyopia. 
  • Sometimes, a person may suffer from both myopia and hypermetropia. 

(ii) Presbyopia is caused due to

  • weakening of ciliary muscles, and 
  • eye lens becomes less flexible and elastic, i.e. reducing the ability of the eye lens to change its curvature with the help of ciliary muscles. 

(iii) A bifocal lens will be required to see nearby as well as distant objects. For myopic defects, the upper part of the bifocal lens consists of a concave lens used for distant vision and to correct hypermetropia, the lower part of the bifocal lens consists of a convex lens. It facilitates near vision.
(b) At the time of sunrise/sunset, the sun is near the horizon, so the sun’s rays have to travel through a larger atmospheric distance. The fine particles of the atmosphere scatter away the blue component and other shorter wavelengths of the sunlight. As λb < λr, only red colour having a longer wavelength and is least scattered, reaches our eyes. Hence, the sun appears red at sunrise or sunset.
Previous Year Questions 2018

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Previous Year Questions 2017

Q1: What eye defect is hypermetropia? What are its two possible causes? Describe with a ray diagram how this defect of vision can be corrected by using an appropriate lens. (CBSE 2017)

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Ans: Hypermetropia or long-sightedness is that defect of vision in which the defective eye can see distant objects distinctly but is unable to see distinctly an object placed near his eye. For a nearby object, the image is formed behind the retina.
Two possible causes of this defect of vision are :
(i) The power of the eye lens is less (or the focal length of the eye lens is more) due to less curvature of the cornea.
(ii) The size of the eyeball is shortened.
The hypermetropia defect can be corrected by using a converging (convex) lens of appropriate power.
Ray diagrams showing the defect and its correction are

Previous Year Questions 2017
Previous Year Questions 2017
Previous Year Questions 2017
Previous Year Questions 2017


Q2: What is  dispersion of white light? Draw a neat diagram to show the dispersion of white light by a glass prism. What is the cause of dispersion? (CBSE 2017)

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Ans: When a beam of white light passes through a glass prism it splits up into its constituent seven colours. The splitting of white light into its constituent colours when light passes through a dispersive medium is called “dispersion of light”. The seven colours, usually expressed as ‘VTBGYOR’ constitute the spectrum of white light.
The ray diagram showing dispersion is given here
Previous Year Questions 2017Cause of Dispersion: When a beam of white light enters a glass prism (or any other dispersive medium), the light ray bends towards the normal one entering into the glass. However, different colours of light bend through different angles concerning the incident ray The red light bends the least while the violet light bends the most. So, rays of different colours emerge along different paths and, thus, become distinct. Hence, dispersion is caused and a spectrum is formed.


Q3:  State the cause of dispersion of white light by a glass prism. How did Newton, using two identical glass prisms, show that white light is made of seven colours? Draw a ray diagram to show the path of a narrow beam of white light, through a combination of two identical prisms arranged together in an inverted position concerning each other, when it is allowed to fall obliquely on one of the faces of the first prism of the combination.  (AI 2017)

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Ans: Cause of dispersion: From Snell’s law of refraction, the angle of refraction of light in a prism depends on the refractive index of the prism material. Moreover, the refractive index of the material varies inversely with the speed of light and also varies inversely with the wavelength of light. Hence, different colours of white light bend through different angles concerning the incident light, as they pass through the glass prism.
Newton Experiment: Consider a prism A. When a beam of white light falls obliquely on one of the faces of this prism, it splits up into seven constituent colours. The violet colour deviates the most and the red colour deviates the least.
If another identical prism B is placed in an inverted position concerning the first prism A, the constituent coloured rays that emerge out of prism A will be made to merge to come out as a beam of white light, as shown below.
Previous Year Questions 2017


Q4: Name the type of defect of vision a person is suffering from, if he uses convex lenses in his spectacles for the correction of his vision. If the power of the lenses is +0.5 D, find the focal length of the lenses. (AI2017C)

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Ans: The defect of vision is hypermetropia.
Focal length of lenses,
Previous Year Questions 2017


Q5: A student traces the path of a ray of light through a glass prism for different angles of incidence. He analyses each diagram and draws the following conclusion: 
(I) On entering prism, the light ray bends towards its base. 
(II)
 Light ray suffers  refraction at the point of incidence and point of emergence while passing through the prism. 
(III) Emergent ray bends at certain angle to the direction of the incident ray. 
(IV)While emerging from the prism, the light ray bends towards the vertex of the prism.

Out of the given inferences, the correct ones are: 
(a) (I), (II) and (III) 
(b) (I), (III) and (IV ) 
(c) (II), (III) and (IV ) 
(d) (I) and (IV ) (CBSE 2017)

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Ans: (a)
(I) On entering the prism, the light ray bends towards its base: This is correct. When light enters a denser medium (the prism) at an angle, it bends towards the base of the prism due to refraction.
(II) The light ray suffers refraction at the point of incidence and point of emergence while passing through the prism: This is correct. The ray undergoes refraction at both the surfaces of the prism — first when it enters the prism and again when it exits.
(III) The emergent ray bends at a certain angle to the direction of the incident ray: This is correct. The emergent ray is deviated from the original path of the incident ray, resulting in a measurable angle of deviation.
(IV) While emerging from the prism, the light ray bends towards the vertex of the prism: This is incorrect. When the ray emerges from the prism, it actually bends away from the normal (towards the base of the prism) as it moves from a denser to a rarer medium.
Thus, the correct inferences are (I), (II), and (III), making the correct answer (a) (I), (II) and (III).


Q6: What is scattering of light? Why is the colour of the clear sky blue? Explain. (CBSE 2017)

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Ans: Scattering of light is the phenomenon of change in the direction of light on striking an obstacle like an atom, a molecule, dust particle, water droplet, etc. The blue colour of the sky is due to the scattering of sunlight by a large number of molecules such as fine dust particles, gases, water vapour, etc. present in Earth’s atmosphere. Due to the very small size of the scatterer as compared to the wavelength of light, light of smaller wavelength, such as blue, is scattered the most as compared to light of longer wavelength.


Q7: Due to gradual weakening of ciliary muscles and diminishing flexibility of the eye lens a certain defect of vision arises. Write the name of this defect. Name the type of lens required by such persons to improve the vision. Explain the structure and function of such a lens. (CBSE 2017)

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Ans: The defect of vision that arises due to gradual weakening of ciliary muscles and diminishing flexibility of the eye lens is presbyopia. Most people find it difficult to see nearby objects comfortably and clearly without using corrective lenses. This is corrected by using convex lens of appropriate power. However, sometimes people may suffer from both myopia and hypermetropia. This can be corrected by using a bifocal lens. A bifocal lens consists of two parts – the upper part is concave lens and lower part is convex lens. The upper part is for viewing distant objects and the lower part facilitates near vision.


Q8: Varun instead of copying from the black board used to copy regularly from the notebook of his friend, Sudhir with whom he sat on the same desk. Sudhir told the teacher about it. The teacher asked Varun to get his eyes checked by a doctor and explained to the whole class the reason why Varun copied from Sudhir’s notebook. 
(A) What in your view, is wrong with Varun’s eyes and how can it be corrected? 
(B) If the doctor prescribes Varun to use lenses of power – 0.5 D, what is the type of the lenses? 
(C) Write the values displayed by Sudhir and his teacher.

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Ans: (A) Varun is suffering from the defect of vision called myopia. Myopia can be corrected by using concave lens of appropriate power.
(B) Power of lens = – 0.5 D. This means it is a concave lens. 
(C) Values displayed by Sudhir: Concerned, helpful, compassionate 
Values displayed by teacher: Concerned, Knowledgeable.


Q9: Curvature of eye lens of human eye can be modified by ciliary muscles to some extent so that its focal length is changed as per requirement. How will the curvature and focal length of eye lens change when: 
(A) a distant object is to be seen, and, 
(B) a nearby object is to be seen clearly? Write the reason why a normal eye is not able to see clearly, the objects placed closer than 25 cm, without any strain on the eye.

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Ans: (A) When we see a distant object, the muscles relax and the lens become thin and its focal length increases. 
(B) When we see a nearby object, the ciliary muscles contract, this increases the curvature of the eye lens and it becomes thicker. Consequently the focal length of the eye lens decreases. A normal eye is unable to clearly see the objects placed closer than 25 cm because the ciliary muscles of eyes are unable to contract beyond a certain limit.

Previous Year Questions 2016

Q1: Describe an activity to show that the colours of white light split by a glass prism can be recombined to get white light by another identical glass prism. Also, draw a ray diagram to show the recombination of the spectrum of white light. (AI 2016)

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Ans: Recombination of Colours: The colours of white light split by a glass prism can be recombined to get white light by another identical glass prism. Newton; demonstrated this phenomenon of recombination of the coloured rays of a spectrum to get back white light.
(а) A triangular prism ABC is placed on its base BC.
(b) A similar prism A ‘B ‘C ‘ is placed alongside its refracting surface in the opposite direction, i.e. in an inverted position concerning the first prism as shown in the figure.
Previous Year Questions 2016(c) A beam of white light entering the prism ABC undergoes refraction and is dispersed into its constituent seven colours.
(d) These constituent seven coloured rays are incident on the second inverted prism A’B’C’ and get further refracted.
(e) The second prism recombines them into a beam of white light that emerges from the other side of the second prism and falls on the screen.
(f) This is because the refraction or bending produced by the second inverted prism is equal and opposite to the refraction or bending produced by the first prism. This causes the seven colours to recombine.
(g) A white patch of light is formed on the screen placed beyond the second prism. This proves the phenomenon of recombination of the spectrum of white light.


Q2: What is the function of the retina in the human eye? (CBSE 2016)

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Ans. The retina behaves like a light-sensitive screen, on which real and inverted image of any object situated in front of the eye is formed.


Q3: What is the function of the crystalline lens of the eye? (CBSE 2016)

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Ans: It provides the fine adjustment of the focal length of the eye lens system to focus images of objects situated at different distances on the retina.


Q4: Write the function of the iris in the human eye. (CBSE 2016)

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Ans: Iris controls the size of the pupil


Q5: State the role of ciliary muscles in the accommodation of the eye. (CBSE 2016)

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Ans: Adjust/modify the shape (curvature) of the eye lens so that its focal length can be increased or decreased.


Q6: What are the values of (i) near point and (ii) far point of vision of a normal adult person? (CBSE 2016) 

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Ans: (i ) 25 cm, (ii ) infinity.


Q7: Name two possible causes of myopia. (CBSE 2016)

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Ans: (i) Excessive curvature of the eye lens.
(ii) Elongation of the eyeball.


Q8: An old person is unable to see nearby objects as well as distant objects. What defect of vision he is suffering from? (CBSE 2016)

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Ans: The old person is suffering from presbyopia.


Q9: Priya prefers to sit in the front row as she finds it difficult to read the blackboard from the last desk in her classroom. State the defect of vision she is suffering from. (CBSE 2016)

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Ans: Priya is suffering from myopia (near-sightedness).


Q10: Name the component of white light that deviates (i) the least and (ii ) the most while passing through a glass prism. (CBSE 2016)

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Ans: (e) Red light is deviated the least and  Violet light is deviated the most.


Q11: What will be the colour of the sky when it is observed from a place in the absence of any atmosphere? (CBSE 2016)

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Ans: Black (dark).


Q12: Why is the red colour selected for danger signal lights? (CBSE 2016)

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Ans: Red light is least scattered by fog or smoke and can be easily seen from a distance.


Q13: (a) What is the function of the iris and pupil of the eye? 
(b) How does the focal length of the eye lens change as per the distance of the object in front of the eye? (CBSE 2016)

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Ans: (a) The iris controls the size of the pupil. It adjusts in size, and therefore, helps in regulating the amount of light entering the eye through a variable aperture ‘pupil’. When the fight is very bright, the pupil becomes very small. However, in dim fight, it opens up completely through the relaxation of the iris.
(b) The crystalline eye lens consists of a fibrous, jelly-like material. Its curvature can be modified to some extent by the ciliary muscles. The change in the curvature of the eye lens can change its focal length. When the muscles are relaxed, the lens is thin and its focal length is more (about 2.5 cm). When the ciliary muscles contract the eye lens becomes thicker. Consequently, the focal length of the eye lens decreases.


Q14: What is a prism? Draw a neat diagram to show the refraction of a light ray through a triangular glass prism. Define the angle of deviation. (CBSE 2016)

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Ans: An optical prism has two triangular bases and three rectangular lateral surfaces, which are inclined to each other. The angle between its two lateral faces is called the angle of the prism.
The labelled diagram has been shown in Fig in which
∠PEN = ∠i = angle of incidence, ∠N’EF = ∠r = angle of refraction and ∠MFR = ∠e = angle of emergence
Previous Year Questions 2016Whenever refraction of light takes place through a prism, the emergent ray bends towards the base of the prism. The angle between the directions of the incident ray and the emergent ray is called the angle of deviation D.


Q15: Explain why the planets do not twinkle but the stars twinkle. (CBSE 2016)

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Ans: Stars are very far away from the earth and behave as almost a point object. The atmosphere is made of several layers and their refractive indices keep on changing continuously. So the light rays coming from the star keep on changing their paths continuously. As a consequence, the number of rays (or the light energy) entering the pupil of the eye changes with time and the stars appear twinkling.
A planet is comparatively nearer to the Earth and subtends a larger angle at the eye. So, it may be considered a collection of a large number of point-sized objects. Due to atmospheric refraction, the quantity of light coming from any one point-sized object changes with time but the total light entering the observer’s eye due to all these point objects remains almost the same. As a result, the planet does not twinkle.


Q16: Explain with the help of a diagram, how we are able to observe the sunrise about two minutes before the sun gets above the horizon. Hence, explain why does apparent duration of a day from sunrise to sunset is 4 minutes more than its actual duration. (CBSE  2016)

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Ans: The air becomes rarer as its height above the earth increases. Its refractive index decreases. A ray of light from the Sun when it enters the atmosphere at the horizon gets refracted from a rarer to a denser medium. The rays, therefore gradually bend towards the normal and the Sun appears to be raised.
As a result, the Sun is visible to an observer nearly two minutes before actual sunrise at the horizon. Similarly, even after actual sunset, the Sun is seen by us for about 2 minutes. Thus, in effect, the Sun is seen for 4 minutes more. It means that the apparent duration of the day (from sunrise to sunset) has increased by 4 minutes more than its actual duration.
Previous Year Questions 2016


Q17: (a) What is the dispersion of white light? State its cause.
(b) “Rainbow is an example of dispersion of sunlight” Justify this statement by explaining, with the help of a labelled diagram, the formation of a rainbow in the sky. List two essential conditions for observing a rainbow. (Foreign 2016)

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Ans: (a) Dispersion: The splitting up of white light into its component colours is called dispersion.
Cause of dispersion: From Snell’s law of refraction, the angle of refraction of light in a prism depends on the refractive index of the prism material. Moreover, the refractive index of the material varies with the speed of light. The different constituent colours of white light have different speeds in the transparent material of the prism. Hence for each colour/wavelength, the refractive index of prism material is different. Therefore, each colour bends (refracted) through a different angle concerning incident rays, as they pass through the prism. The red colour has maximum speed in the glass prism, so it is least deviated, while the violet colour has minimum speed so its deviation is maximum. Thus, the ray of each colour emerges along different paths and becomes distinct.
(b) Rainbow: It is an optical natural spectrum, produced by nature in the sky, in the form of a multicoloured arc. The rainbow is formed due to the dispersion of sunlight by water droplets suspended in the atmosphere after rainfall. These water droplets act like small prisms. The Sunlight enters the water droplets. At the point of incidence, it refracts and disperses then gets reflected internally and finally gets refracted again at the point of emergence as it comes out of the rain-drop.
Previous Year Questions 2016

Therefore, due to refraction, dispersion and internal reflection of the sunlight, different colours reach the observer’s eye along different paths and become distinct. It creates a rainbow in the sky.
Hence “Rainbow is an example of dispersion of sunlight.”
Necessary conditions for the formation of a rainbow.
(i) The presence of water droplets in the atmosphere, and
(ii) The sun must be at the back of the observer, i.e. the observer must stand with his back towards the sun.


Q18: What is atmospheric refraction? Use this phenomenon to explain the following natural events.
(a) Twinkling of stars
(b) Advanced sunrise and delayed sunset.
Draw diagrams to illustrate your answers. (AI 2016)

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Ans: Atmospheric Refraction: The refraction of light caused by the earth’s atmosphere due to gradual change in the refractive indices of its different layers by the varying conditions of it, is called atmospheric refraction.
(a) Twinkling of stars 
The hot layers (low densities) of air at a high altitude, behave as an optically rarer medium for the light rays, whereas the cold dense layers (high densities) of air near the earth’s surface, behave as an optically denser medium for the light rays. So, when the light rays (starlight) pass through the various layers of the atmosphere, they will deviate and bend toward the normal. As a result, the apparent position of the star is slightly different from its actual position. Thus, the stars appear slightly higher (above) than their actual positions in the sky.
Previous Year Questions 2016The fluctuation in the positions of the stars occurs continuously due to the changing amount of light entering the eye. The stars sometimes appear brighter and at some other times, they appear fainter. This causes the twinkling of stars.
(b) Advanced sunrise and delayed sunset: The sun is visible 2 minutes before sunrise and 2 minutes after sunset because of atmospheric refraction. This can be explained below.Previous Year Questions 2016

The figure shows the actual position of the sun S at the time of sunrise or sunset, just below the horizon while the apparent position S’ above the horizon appears to us.
When the sun is slightly below the horizon, the light rays move through the different layers of varying refractive indices of air and get bent towards the normal. These rays appear to come from S’ which is the apparent position of the sun. That is why, the sun is visible to us when it has  been actually below the horizon or before the actual crossing of the horizon by the sun at the time of sunrise or sunset. So, due to the atmospheric refraction, the phenomenon of advanced advance sunrise and delayed sunset is observed.


Q19: In an experiment to trace the path of a ray of light through a triangular glass prism, a student would observe that the emergent ray:
(a) is parallel to the incident ray. 
(b) is along the same direction of incident ray. 
(c) gets deviated and bends towards the thinner part of the prism. 
(d) gets deviated and bends towards the thicker part (base) of the prism. (CBSE 2016)

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Ans: (d)
In a triangular glass prism, when a ray of light passes through, it first bends towards the normal upon entering the prism (due to refraction into a denser medium). As it exits the prism into a less dense medium (air), the emergent ray bends away from the normal. This causes the emergent ray to deviate from its original path, and it generally bends towards the thicker part (base) of the prism.
Thus, the correct answer is (d) gets deviated and bends towards the thicker part (base) of the prism.

Also read: The Human Eye

Previous Year Questions 2015

Q1: What is a pupil? (CBSE 2005,2013,2015)

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Ans: The pupil is a small circular transparent aperture whose size is controlled by the iris.


Q2:  Name the condition resulting due to the eye lens becoming cloudy. (CBSE 2012,2013,2015)

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Ans: Cataract


Q3: What is the dispersion of light? (CBSE 2012,2015)

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Ans: Dispersion of light is the splitting of light into its component colours on passing through a dispersive medium e.g., a prism.


Q4:  List the factors on which the scattering of light depends. (CBSE 2012,2015)

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Ans: The scattering of light depends on the size of the scattering particle and the wavelength of light.

Previous Year Questions 2014

Q1: What is myopia? List two causes for the development of this defect. How can this defect be corrected using a lens? Draw ray diagrams to show the image formation in case of (i) defective eye and (ii) corrected eye. OR

A student is unable to see the words written on the blackboard placed at a distance of approximately 4 m from him. Name the defect of vision the boy is suffering from. Explain the method of correcting this defect. Draw a ray diagram for the:
(i) defect of vision and also
(ii) for its correction.  
OR 

What is myopia? State the two causes of myopia.
With the help of a labelled ray diagram show (a) the eye defect and (b) the correction of myopia.   (DoE)   (Foreign 2014)

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Ans:

Myopia (Nearsightedness):
Myopia is a defect of vision in which a person can see nearby objects clearly but cannot see distant objects distinctly. The image of a distant object is formed in front of the retina instead of on it.

Causes:

  1. Excessive curvature of the eye lens.
  2. Elongation of the eyeball.

Correction:

  • Concave lens (Diverging lens) is used to correct myopia.
  • It diverges the incoming light rays so that the eye lens forms the image on the retina instead of in front of it.

Ray Diagrams:
(i) Defective Eye – Image of a distant object forms in front of the retina.
(ii) Corrected Eye – Concave lens diverges light rays, shifting the image onto the retina.

Previous Year Questions 2014
Previous Year Questions 2014

09. Previous Year Questions: Light – Reflection & Refraction

Previous Year Questions 2025

Q1: Mirror ‘X’ is used to concentrate sunlight in a solar furnace, and Mirror ‘Y’ is fitted on the side of the vehicle to see the traffic behind the driver. Which of the following statements are true for the two mirrors?  (1 Mark)
(i) The image formed by mirror ‘X’ is real, diminished, and at its focus.
(ii) The image formed by mirror ‘Y’ is virtual, diminished, and erect.
(iii) The image formed by mirror ‘X’ is virtual, diminished, and erect.
(iv) The image formed by mirror ‘Y’ is real, diminished, and at its focus.
(A)
 (i) and (ii)
(B) (ii) and (iii)
(C) (iii) and (iv)
(D) (i) and (iv)

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Ans: (A) (i) and (ii)

  • Brief explanation: Mirror X is a concave mirror used to concentrate sunlight — for parallel rays from the Sun it forms a real, highly diminished image at the principal focus. Mirror Y is a convex (rear-view) mirror — it always gives a virtual, diminished, erect image.


Q2: An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm. Use the lens formula to find the position of the image formed in this case.  (2 Marks)

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Ans: 

Use lens formula

Previous Year Questions 2025

For a concave lens take u =−60 cm (object on left) and f = −30 cm. Substitute:
Previous Year Questions 2025

so v = -20 cm

The image is formed at 20 cm on the same side as the object (i.e. v = -20 cm.

Nature: the image is virtual, erect and diminished.


Q3: Draw ray diagrams to show the nature, position, and relative size of the image formed by a convex mirror when the object is placed (i) at infinity and (ii) between infinity and pole P of the mirror.  (3 Marks)

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Ans:

(i) Object at infinity:

For a convex mirror, when the object is at infinity, incident rays are parallel to the principal axis. After reflection, these rays diverge and appear to come from the focus behind the mirror.

Previous Year Questions 2025

  • Position: At F (behind the mirror)
  • Nature: Virtual and erect
  • Size: Highly diminished (point-like)

(ii) Object between infinity and pole P:

  • For any object in front of a convex mirror, the image is always formed between the pole and focus.
Previous Year Questions 2025
  • Position: Between P and F (behind the mirror)
  • Nature: Virtual and erect
  • Size: Diminished


Q4: (a) (i) The power of a lens ‘X’ is -2.5 D. Name the lens and determine its focal length in cm. For which eye defect of vision will an optician prescribe this type of lens as a corrective lens?
(ii) “The value of magnification ‘m’ for a lens is -2.” Using new Cartesian Sign Convention and considering that an object is placed at a distance of 20 cm from the optical centre of this lens, state:
(1) the nature of the image formed;
(2) size of the image compared to the size of the object;
(3) position of the image, and
(4) sign of the height of the image.
(iii) The numerical values of the focal lengths of two lenses A and B are 10 cm and 20 cm respectively. Which one of the two will show a higher degree of convergence/divergence? Give reason to justify your answer.  (5 Marks)

OR

(b)
(i) Draw a ray diagram to show the refraction of a ray of light through a rectangular glass slab when it falls obliquely from air into glass.

(ii) State Snell’s law of refraction of light. 
(iii) Differentiate between the virtual images formed by a convex lens and a concave lens on the basis of : (I) object distance, and  (II) magnification.  (5 Marks)

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Ans: 
(a) (i):
Power P = -2.5 D. From the chapter P = 1/f (with f in metres).
f = 1/P = 1/-2.5 = -0.4 m = -40 cm.
Sign: negative focal length ⇒ concave lens.
Eye defect corrected: Myopia (short-sightedness) — a concave lens diverges incoming rays so the image of distant objects forms at the myopic eye’s far point, enabling clear distant vision.
(a) (ii):
Previous Year Questions 2025Previous Year Questions 2025Previous Year Questions 2025

1. Nature of image: Real and inverted (since v is positive and m is negative).

2. Size of image: Image is twice the size of the object (since |m| = 2).

3. Position of image: 40 cm on the other side of the lens.Previous Year Questions 2025

(a) (iii):
Focal lengths: f= 10 cm = 0.10 m, f= 20 cm = 0.20 m.
Power P = 1/f
 so
Previous Year Questions 2025

The smaller the focal length, the greater the optical power (since P = 1/f).
xLens with shorter focal length bends rays through larger angles (greater convergence/divergence) and power is the reciprocal of focal length.
So lens A (10 cm) shows the higher degree of convergence/divergence because it has the larger power (10 D vs 5 D).

OR

(b)(i) 

(b) (ii) Snell’s law of refraction
(i) The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.

(ii)Previous Year Questions 2025 for a given pair of media and light colour. This constant is the refractive index of the second medium with respect to the first (Snell’s law).

(b)(iii) Difference between virtual images of a convex lens and a concave lens
Previous Year Questions 2025


Q5: An object is placed at a distance of 30 cm in front of a concave mirror of focal length 20 cm. Use the mirror formula to determine the position of the image formed in this case.  (2 Marks)

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Ans: The image is formed at 60 cm in front of the mirror
Given:

  • Object distance, u = -30 cm (negative due to the new Cartesian sign convention).
  • Focal length of concave mirror, f = -20 cm (negative for concave mirror).
  • Mirror formula: Previous Year Questions 2025

Rearranging for image distance v:
1/v = 1/f – 1/u
Substitute the values:
1/v = 1/(-20) – 1/(-30) = -1/20 + 1/30 = -3/60 + 2/60 = -1/60
v = -60 cm

The negative sign indicates the image is formed in front of the mirror (real image). Thus, the image is at 60 cm in front of the mirror.


Q6: An object is placed at a distance of 10 cm in front of a concave mirror of focal length 15 cm. Use the mirror formula to determine the position of the image formed by this mirror.  (2 Marks)

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Ans: Use the mirror formula
Previous Year Questions 2025For a concave mirror take u = -10 cm (object to the left) and f = -15 cm. Substitute:
Previous Year Questions 2025so, v = +30 cm.
The image is formed 30 cm behind the mirror. (i.e v = +30 cm.)
Nature: since the object lies between P and F, the image is virtual, erect and enlarged.


Q7: If the absolute refractive indices of two media X and Y are 6/5 and 4/3 respectively, then the refractive index of Y with respect to X will be:  (1 Mark)
(a) 10/9
(b) 9/10
(c) 9/8
(d) 8/9

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Ans: (a) 10/9
Refractive index of Y w.r.t X
Previous Year Questions 2025 


Q8: An object is placed at a distance of 30 cm from the pole of a concave mirror. If its real and inverted image is formed at 60 cm in front of the mirror, the focal length of the mirror is:  (1 Mark)
(a) -15 cm
(b) -20 cm
(c) +20 cm
(d) +15 cm

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Ans: (b) -20 cm

Using Previous Year Questions 2025Take u = -30 cm (object to the left) and v = -60 cm (real image in front of mirror, same side as object).
Previous Year Questions 2025

So f = -20 cm. 


Q9: A convex lens forms an 8.0 cm long image of a 2.0 cm long object which is kept at a distance of 6.0 cm from the optical centre of the lens. If the object and the image are on the same side of the lens, find (i) the nature of the image, (ii) the position of the image, and (iii) the focal length of the lens.  (3 Marks)

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Ans:

Given: h = 2.0 cm, h’ = 8.0 cm, u = -6.0 cm (object distance taken negative by the New Cartesian sign convention).

Step 1: Magnification: Previous Year Questions 2025Step 2: 
– Object is placed on the same side as the image.

– For a convex lens, object distance is usually negative if it is to the left (real object). Here, image is on the same side as object. Since both object and image are on the same side, the image is virtual.
Step 3: Using formula for magnification: 

v = −mu

Substitute:

v = −4 × (−6.0) = 24.0 cm

So, the image distance is +24 cm, meaning image is on the opposite side of the lens from the object. This conflicts with the problem statement that both are on the same side.

Since the question says both are on the same side of the lens, and magnification is positive (upright), then for convex lens object is at -6 cm, image should be on the same side (virtual image) with positive magnification.

If we take object distance u=+6 cm (because of sign conventions sometimes vary for virtual object), thenPrevious Year Questions 2025and
Previous Year Questions 2025Image distance =−24 cm indicates image is on the same side as object (virtual image).
Step 4: Previous Year Questions 2025
Substitute v=−24 cm and u=6 cm:
Previous Year Questions 2025


Q10: Absolute refractive index of water and glass is 4/3 and 3/2, respectively. If the speed of light in glass is 2 × 10^8 m/s, the speed of light in water is:  (1 Mark)
(a) 9/4 m/s
(b) 7/3 m/s
(c) 16/9 m/s
(d) 9/8 m/s

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Ans: (a)
Previous Year Questions 2025

Previous Year Questions 2025


Q11: A convex mirror used for rear view on an automobile has a focal length of 1.5 m. If a 3 m high bus is located at 6.0 m from the mirror, use the mirror formula to determine the position and size of the image of the bus as seen in the mirror.  (3 Marks)

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Ans: Given: convex mirror, f = +1.5m; object distance u = −6.0m; object height h= 3m.

Mirror formula

v>0 ⇒ image is behind the mirror (virtual).

Magnification

Result: The image is 1.2 m behind the mirror, virtual, erect, and diminished, with height 0.6 m.

Q12: To get an image of magnification -1 on a screen using a lens of focal length 20 cm, the object distance must be:  (1 Mark)
(a) Less than 20 cm
(b) 30 cm
(c) 40 cm
(d) 80 cm

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Ans: (c) 40 cm

Image on a screen ⇒ real image (thus a convex lens).
For a convex lens, magnification m=vu=−1 ⇒ v = −u.
Using Lens Formula:Previous Year Questions 2025

Solve for u:Previous Year Questions 2025 (Sign ‘−’ means the object is placed on the incoming-light side; distance = 40 cm.)

⇒ object is 40 cm from the lens.


Q13: An optical device ‘X’ is placed obliquely in the path of a narrow parallel beam of light. If the emergent beam gets displaced laterally, the device ‘X’ is:  (1 Mark)
(a) Plane mirror
(b) Convex lens
(c) Glass slab
(d) Glass prism

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Ans: (c) Glass slab

When a parallel beam of light passes obliquely through a rectangular glass slab, the emergent ray is parallel to the incident ray but laterally displaced — a characteristic property of a glass slab.


Q14: If we want to obtain a virtual and magnified image of an object by using a concave mirror of focal length 18 cm, where should the object be placed? Use the mirror formula to determine the object distance for an image of magnification +2 produced by this mirror to justify your answer.  (3 Marks)

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Ans: Use the mirror formula and magnification relation.
Given f = –18 cm (concave), required m = +2.
For mirrors Previous Year Questions 2025Previous Year Questions 2025Mirror formula:
Previous Year Questions 2025
Substitute v = -2u
Previous Year Questions 2025
Hence f = 2u so u = f/2  = -18/2 = -9 cm.
the object must be placed 9 cm in front of the mirror (i.e u = -9cm) which is between the pole and the principal focus.
The image will be virtual, erect and magnified (since m = +2); its position is v = -2u = +18 cm (i.e. 18 cm behind the mirror).


Q15: In order to obtain large images of the teeth of patients, the dentist holds the concave mirror in such a manner that the teeth are positioned:  (1 Mark)
(a) at the focus of the mirror
(b) between the pole and the focus of the mirror
(c) between the focus and the centre of curvature of the mirror
(d) at the centre of curvature of the mirror

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Ans: (b) between pole and focus of the mirrorWhen an object is placed between the pole and the focus of a concave mirror, the image formed is virtual, erect, and enlarged, which helps the dentist see a magnified view of the teeth.


Q16: The values of absolute refractive indices of kerosene and water are 1.44 and 1.33, respectively. Compare the two media on the basis of their:  (2 Marks)
(a) optical density
(b) mass density
(c) relative speed of propagation of light.
What do you infer on the basis of the above comparisons?

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Ans:
(a) Optical density: The optical density of a medium is directly related to its absolute refractive index. Since kerosene has a higher refractive index (1.44) compared to water (1.33), kerosene has a higher optical density.
(b) Mass density: Optical (refractive) index does not determine mass density. In fact, water is more mass-dense than kerosene (kerosene floats on water).
→ No direct relation between optical density and mass density.
(c) Relative speed of light: Speed ∝ 1 / refractive index, so light travels slower in kerosene than in water.
Numerical ratio:
Previous Year Questions 2025
Hence light in kerosene is about 0.924 times as fast as in water.


Q17: (A)  (a) Observe the following diagram and compare (i) speed of light and (ii) optical density of the three media A, B, and C. Also give justification for your answer of any one of the two cases in terms of refractive indices of A, B, and C.
(b) Redraw the path of a ray of light through the three media, if the ray of light starting from medium A falls on the medium B:
(i) Obliquely and the optical density of medium B is made more than that of A and C.
(ii) The ray falls normally from medium A to medium B. (5 Marks)
OR
(B) Analyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follow without doing any calculations:

(a) Determine the focal length of the lens. Give reason for your answer.
(b) Find magnification of the image formed in Observation No. 3.
(c) The numerical value of magnifications in cases of observation 1 and 2 is same. List two differences in the images formed in these two cases.  (5 Marks)

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Ans: 
(A) 
(a) 
In this problem, we are comparing the speed of light and optical densities of three media: A, B, and C. The speed of light in a medium is inversely proportional to its optical density. This means that the medium with the highest optical density will have the lowest speed of light.

The refractive index (n) of a medium is defined as the ratio of the speed of light in vacuum (c) to the speed of light in the medium (v):
Previous Year Questions 2025

Thus, a higher refractive index indicates a lower speed of light in that medium.

Let’s denote the refractive indices of the three media as follows:

– Refractive index of medium A: Previous Year Questions 2025

– Refractive index of medium B: Previous Year Questions 2025

– Refractive index of medium C: Previous Year Questions 2025

From the problem, if medium B has a higher optical density than A and C, it implies:
Previous Year Questions 2025This means that the speed of light in medium B is less than that in A and C:
Previous Year Questions 2025

Thus, we can conclude:
Speed of light: Medium B has the lowest speed of light, followed by A, and then C.
Optical densities: Medium B has the highest optical density, followed by A, and then C.
1. Speed of Light Comparison
:

  • Medium A: vA(higher speed)
  • Medium B: vB(lowest speed)
  • Medium C: vC (medium speed)

2. Optical Densities Comparison:

  • Medium A: lowest optical density
  • Medium B: highest optical density
  • Medium C: medium optical density

– Speed of Light: Previous Year Questions 2025

– Optical Densities: Previous Year Questions 2025

(b) 
(i) 
When a ray of light travels from medium A to medium B obliquely and medium B has a higher optical density than A, the ray will bend towards the normal upon entering medium B due to the change in speed.

Redraw the path: The ray will enter medium B at an angle less than the angle of incidence in medium A. Let’s denote:

  • Angle of incidence in medium A: iA 
  • Angle of refraction in medium B: rB

Using Snell’s law:Previous Year Questions 2025

⇒ The ray bends towards the normal when entering medium B obliquely.
(b) 
(ii)
 When the ray of light falls normally from medium A to medium B, it will not bend as the angle of incidence is 0 degrees. The light will continue in a straight line.

Redraw the path: The ray will enter medium B at a right angle (90 degrees) and continue straight without bending.

⇒ The ray continues straight into medium B without bending.

(B) 

(a) Focal length: From observation 4, u = -40 cm, v = + 40 cm.
Lens formula: Previous Year Questions 2025
= 1/40 – 1/(-40) = 1/40 + 1/40 = 2/40 = 1/20.f = 20 cm (positive, convex lens).
Reason: The data is consistent with a convex lens of focal length 20 cm, as verified across observations.

(b) Magnification for Observation 3: u = -30 cm, v = +60 cm.
m = v/u = 60/(-30) = -2.
The magnification is -2 (inverted, twice the size).

(c) Differences:

  • Observation 1: u = -15 cm, v = -60 cm. m = v/u = -60/(-15) = 4 (positive, virtual, erect image, same side).
  • Observation 2: u = -25 cm, v = +100 cm. m = v/u = 100/(-25) = -4 (negative, real, inverted image, opposite side).

Differences:

  1. Observation 1: Virtual, erect image; Observation 2: Real, inverted image.
  2. Observation 1: Image on same side as object; Observation 2: Image on opposite side.

Previous Year Questions 2024

Q1: At what distance from a convex lens should an object be placed to get an image of the same size as that of the object on a screen?    (2024)
(a) 
Beyond twice the focal length of the lens.
(b) At the principal focus of the lens.
(c) At twice the focal length of the lens.
(d) Between the optical centre of the lens and its principal focus.

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Ans: (c)

To get an image of the same size as the object using a convex lens, the object should be placed at twice the focal length of the lens. This distance is called “twice the focal length” because at this position, the light rays coming from the object will converge to form an image that is equal in size.


Q2: An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position of the image formed by the mirror.    (2024)

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Ans:

Given:

  • Object distance, u=−10 (negative, object in front of mirror, Cartesian sign convention).
  • Focal length, f=+15 (positive for convex mirror).

Using the mirror formula:
1f=1v+1u
Substitute:
115=1v+1−10

1v=115+110=2+330=530=16
v=+6

The image is formed 6 cm behind the convex mirror (positive v indicates a virtual image).


Q3: Source-based/case-based questions with 2 to 3 short subparts. Internal choice is provided in one of these sub-parts:    (2024)
Study the data given below showing the focal length of three concave mirrors A, B and C and the respective distances of objects placed in front of the mirrors :

(i) In which one of the above cases the mirror will form a diminished image of the object ? Justify your answer.
(ii) List two properties of the image formed in case 2. 
(iii) (A) What is the nature and size of the image formed by mirror C?
Draw ray diagram to justify your answer. 
OR
(iii) (B) An object is placed at a distance of 18 cm from the pole of a concave mirror of focal length 12 cm. Find the position of the image formed in this case.

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Ans: 

i.) Cases:

  • Mirror A: f=20,u=45 cm 
  • Mirror B: f=15,u=30 cm
  • Mirror C: f=30,u=20 cm 

For a concave mirror, a diminished image is formed when the object is beyond the centre of curvature (u>2f) or at the centre of curvature (u=2f).

  • Mirror A: Centre of curvature 2f=40cm. Since u=45 cm>40 cm, the image is real, inverted, and diminished.
  • Mirror B: Centre of curvature 2f=30 cm. Since u=30=2f, the image is real, inverted, and same-sized (considered diminished as it is not magnified).
  • Mirror C: u=20<f=30 cm. The object is between the pole and focus, producing a virtual, erect, and magnified image.

Thus, Mirrors A and B form diminished images.

ii.) For Mirror B: f=15,u=30 cm.

  • The object is at the centre of curvature (u=2f=30 cm).
  • Properties:
    1. Real and inverted: The image forms on the same side as the object and is inverted.
    2. Same size: The image is the same size as the object, located at the centre of curvature.
Previous Year Questions 2024

OR
(iii) (B) Here ƒ = –12 cm, u = –18 cm, n = ?

Mirror formula:

Previous Year Questions 2024v = -36 cm


Q4: (a) State two laws of refraction of light.
OR
(b) Define the term absolute refractive index of a medium. A ray of light enters from vacuum to glass of absolute refractive index 1.5. Find the speed of light in glass. The speed of light in vacuum is 3 x 108 m/s.    (2024)

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Ans: (a) Laws of Refraction of light :
(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
(ii) The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media.
OR
(b) Absolute refractive index of a medium is the ratio of speed of light in air/vacuum to the speed of light in the given medium.
Given:

c = 3 × 108 m/s; nm = 1.5; vm = ?

Absolute refractive index of a medium (nm):

nm = speed of light in vacuum (c)speed of light in medium (vm)

nm = cvm

vm = cnm = 2 × 108 m/s


Q5: The Phenomena of light involved in the formation of a rainbow in the sky are    (2024)
(a) 
Refraction, dispersion and reflection
(b) Refraction, dispersion and total internal reflection
(c) Dispersion, scattering and reflection
(d) Dispersion, refraction and internal reflection

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Ans: (b)

Refraction:

Light bends when entering a different medium, like when sunlight enters a raindrop. 

Dispersion:

Different colors of light bend at slightly different angles within the raindrop, causing them to separate. 

Total internal reflection:

Light is trapped inside the raindrop and reflected back out if the angle of incidence is greater than the critical angle. 


Q6: Absolute refractive index of glass and water is 3/2 and 4/3 respectively. If the speed of light in glass is 2 x 108 m/s, the speed of light in water is: (2024)  

(a) 2.25 × 108 m/s

(b) 52 × 108 m/s

(c) 73 × 108 m/s

(d) 169 × 108 m/s

Hide Answer  

Ans: (a)Previous Year Questions 2024


Q7: (a) The variation of image distance (v) with object distance (u) for a convex lens is given in the following observation table. Analyse it and answer the questions that follow:    (2024)

(i) Without calculation, find the focal length of the convex lens. Justify your answer.
(ii) Which observation is not correct ? Why? Draw ray diagram to find the position of the image formed for this position of the object.
(iii) Find the approximate value of magnification for u = – 30 cm.
OR
(b) (i) Define principal axis of a lens. Draw a ray diagram to show what happens when a ray of light parallel to the principal axis of a concave lens passes through it.
(ii) The focal length of a concave lens is 20 cm. At what distance from the lens should a 5 cm tall object be placed so that its image is formed at a distance of 15 cm from the lens? Also calculate the size of the image formed.

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Ans: (a) (i) S. No. 3, 2f is 50 cm. ∴ 2f = 50 cm, or  f = 25 cm.
Justification: Object distance(u) and image distance (v) are same so it implies that object is placed at 2F.
(ii) S. No. 6, is incorrect.
Reason: For u = −15 cm, sign of v must be – ve ( as the image is formed on the same side of the lens as the object)
Previous Year Questions 2024
(iii) Magnification: m = v/u
= +150-30 = – 5 .
OR
(b) (i) Principal axis: It is an imaginary line passing through the two centres of curvatures of a lens.  
Previous Year Questions 2024
(ii) f = − 20 cm; h = 5 cm; v = −15 cm

1u = 1v – 1f

or

1u = 1-15 – 1-20

1u = -115 + 120

1u = -460 + 360

1u = -160

u = -60 cm

or u = − 60 cm object is at a distance of 60 cm from the lens
Size of the image(magnification):m = h’h = vu
h’ = vu × h = -15-60 × 5 = 1.25 cm


Q8: How will the image formed by a convex lens be affected, if the upper half of the lens is wrapped with black paper?    (2024)
(a) The size of the image formed will be one-half of the size of the image due to the complete lens.
(b) The image of the upper half of the object will not be formed.
(c) The brightness of the image will reduce.
(d) The lower half of the inverted image will not be formed.

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Ans: (c)
If the upper half of a convex lens is wrapped in black paper, it blocks some light from passing through. As a result, the brightness of the image formed will reduce, but the size and shape of the image will remain the same since the lower half of the lens can still focus the light.


Q9: Case-based/data-based questions with 3 short  sub-parts. Internal choice is provided in one of these sub-parts.      (2024)
A highly polished surface such as a mirror reflects most of the light falling on it. In our daily life we use two types of mirrors  plane and spherical. The reflecting surface of a spherical mirrors may be curved inwards or outwards. In concave mirrors, reflection takes place from the inner surface, while in convex mirrors reflection takes place from the outer surface.
(a) Define the principal axis of a concave mirror.
(b) A ray of light is incident on a concave mirror, parallel to its principal axis. If this ray after reflection from the mirror passes through the principal axis from a point at a distance of 10 cm from the pole of the mirror, find the radius of curvature of the mirror.  
(c) (i) An object is placed at a distance of 10 cm from the pole of a convex mirror of focal length 15 cm. Find the position of the image.
OR 

(c) (ii) A mirror forms a virtual, erect and diminished image of an object. Identify the type of this mirror. Draw a ray diagram to show the image formation in this case.

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Ans: (a) The Principal axis of a concave mirror is an imaginary line that runs through the centre of the mirror, perpendicular to its surface. It is the line along which light rays parallel to it converge after reflection.
(b) Relation between the focal length (f) and the radius of curvature (R):R=2f

Given F = 10 cm, therefore R = 20 cm.

Radius of curvature ,R= 20 cm
(c) (i) u = -10 cm, f = +15 cm

1f = 1v + 1u

1v = 1f – 1u = 115 – 1-10

1v = 115 + 110

1v = 16

∴ v = +6 cm

The image is formed at a distance of +6 cm behind the mirror.OR
(c) (ii) The mirror that forms a virtual, erect, and diminished image is a convex mirror
Previous Year Questions 2024

Q10: (a) (i) Draw a ray diagram to show the path of the refracted ray in each of the following cases:   (CBSE 2024)
A ray of light incident on a concave lens 
(1) parallel to its principal axis, and 
(2) is directed towards its principal focus. 
(ii) A 4 cm tall object is placed perpendicular to the principal axis of convex lens of focal length 24 cm. The distance of object from the lens is 16 cm. Find the position and size of image formed. 
OR
(b) (i) Draw a ray diagram to show the path of the reflected ray in each of the following cases: A ray of light incident on a convex mirror 
(1) parallel to its principal axis, and 
(2) is directed towards its principal focus
(ii) A 1.5 cm tall candle flame is placed perpendicular to the principal axis of a concave mirror of focal length 12 cm. If the distance of the flame from the pole of the mirror is 18 cm, use mirror formula to determine the position and size of the image formed.  (CBSE 2024)

Hide Answer  Ans: (a) (i)
(1)  Parallel to its principal axis
Previous Year Questions 2024(2) is directed towards its principal focusPrevious Year Questions 2024(ii) Given u = –16 cm, f = + 24 cm, h = 4 cm
Lens formula:Previous Year Questions 2024

Magnification:Previous Year Questions 2024

Image height:Previous Year Questions 2024

Final Answer:

  • Position of image: −48−48cm (on same side as object → virtual and erect)
  • Size of image: +12+12cm (3 times taller than the object)


Q11: The color of light for which the refractive index of glass is minimum, is:   (CBSE 2024)
(a)
 Red
(b) Yellow
(c) Green
(d) Violet

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Ans: (a)

The refractive index of glass is lowest for red light, meaning that red light travels through glass the fastest compared to other colors. As a result, red light bends the least when it enters the glass, which is why it has the minimum refractive index.

Previous Year Questions 2023

Q1: The magnification produced when an object is placed at a distance of 20 cm from a spherical mirror is +1/2. Where should the object be placed to reduce the magnification to +1/3?    (2023)Ans: Initial Setup:

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  • Object distance: u=−20cm (negative, as the object is in front of the mirror).
  • Magnification: m=+1/2 (positive, suggesting a virtual, erect image, typical for a convex mirror).
  • Magnification formula:Previous Year Questions 2023The image is at  v=+10cm, virtual and behind the mirror.
    Find Focal Length:
    • Use the mirror formula
      Previous Year Questions 2023The positive focal length indicates a convex mirror, consistent with the positive magnification.
      • New Magnification Requirement:
        • New magnification: m′=+1/3
        • Magnification formula:Previous Year Questions 2023Mirror Formula: Previous Year Questions 2023Previous Year Questions 2023The new object distance is 40 cm from the mirror.
      • Verification:
        Calculate v′
        Previous Year Questions 2023Magnification:
        Previous Year Questions 2023The magnification is +1/3, as required.
        The object should be placed at a distance of 40 cm from the spherical mirror to reduce the magnification to +1/3+1/3.


Q2: Define the following terms in the context of a diverging mirror:    (2023)
(i) Principal focus
(ii) Focal length
Draw a labelled ray diagram to illustrate your answer.

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Ans:  For Diverging mirror (Convex Mirror)
(i) Principal focus: It is the point on the principal axis where rays incident parallel to the principal axis appear to diverge after reflection.
(ii) Focal length: The distance between the pole of the mirror and the principal focus is called focal length.
Previous Year Questions 2023


Q3: An object of height 10 cm is placed 25 cm away from the optical centre of a converging lens of focal length 15 cm. Calculate the image distance and height of the image formed.    (2023)

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Ans: Previous Year Questions 2023

  • Image distance: v=+37.5v=+37.5cm
  • Image height: h′=−15h′=−15cm


Q4: Define power of a lens. The focal length of a lens is -10 cm. Write the nature of the lens and find its power. If an object is placed at a distance of 20 cm from the optical center of this lens,  what will be the sign of magnification and nature of image in this case?    (2023)

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Ans: Power The ability of a lens to converge or diverge the rays of light is called power of lens. It is the reciprocal of focal length of lens.
P = 1f (cm) = 100f (cm)
Its Sl unit is dioptre.
If f = – 10 cm
P = 100f = -10010 = -10D
So, lens is concave in nature . u = – 20 cm, f = – 10 cm .
As the object placed beyond F image is virtual and thus magnification is +ve.


Q5: The ability of medium to refract light is expressed in terms of its optical density. Optical density has a definite connotation. It is not the same as mass density. On comparing two media, the one with the large refractive index is optically denser medium than the other. The other medium with a lower refractive index is optically rarer. Also the speed of light through a given medium is inversely proportional to its optical density.    (2023)
i. Determine the speed of light in diamond if the refractive index of diamond with respect to vacuum is 2.42. Speed of light in vacuum is 3 × 108 m/s.
ii. Refractive indices of glass, water and carbon disulphide are 1.5, 1.33 and 1.62 respectively. If a ray of light is incident in these media at the same angle (say θ), then write the increasing order of the angle of refraction in these media.
iii. (A) The speed of light in glass is 2 × 108 m/s and is water is 2.25 × 108 m/s.
(a) Which one of the two optically denser and why?
(b) A ray of light is incident normally at the water glass interface when it enters a thick glass container filled with water. What will happen to the path of the ray after entering the glass? Give reason.
OR
(B) The absolute refractive indices of glass and water are 4/3 and 3/2, respectively. If the speed of light in glass is 2 × 108 m/s, calculate the speed of light in (i) vacuum (ii) water.

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Ans: (i) Refractive index of diamond,

n = Speed of light in vacuumSpeed of light in diamond

n = cv

2.42 = 3 × 108Speed of light in diamond

Speed of light in diamond = 3 × 1082.42

= 1.25 × 108 m/s

(ii) rwater < rglass < rcarbon disulphide 
(iii) (A) (a) Since the speed of light is greater in water than in glass, glass is optically denser than water. This demonstrates that glass presents a greater barrier to light transmission than water.
(b) When a ray of light is incident normally at the water-glass interface, it passes straight through without deviation. This happens because the angle of incidence is 0∘0∘, and according to Snell’s law, there is no refraction at normal incidence. Only the speed of light decreases in the glass due to its higher refractive index.
OR
(B) Refractive index of glass, η= 4/3
Previous Year Questions 2023
Speed of light in vacuum Previous Year Questions 2023
Refractive index of water Previous Year Questions 2023
Previous Year Questions 2023
Speed of light in water = 1.73 x 108 m/s
Because the information provided is wrong, ideally the speed of light in vacuum is 3 × 108 m/s and the speed of light in water is 2.25 × 108 m/s.
The correct solution is

Refractive index of glass, ηg = 32

Refractive index of water, ηw = 43

Refractive index of glass, ηg = Speed of light in vacuumSpeed of light in glass

32 = Speed of light in vacuum2 × 108

Speed of light in vacuum = 3 × 2 × 1082 = 3 × 108 m/s

Refractive index of water, ηw = Speed of light in vacuumSpeed of light in water

43 = 3 × 108Speed of light in water

Speed of light in water = 3 × 3 × 1084

Speed of light in water = 2.25 x 108 m/s


Q6: Many optical instruments consists of a number of lenses. They are combined to increase the magnification and sharpness of the image. The net power (P) of the lenses places in contact is given by the algebraic sum of the powers of the individual lenses P1, P2, P3….as
P = P1 + P2 + P3
This is also termed as the simple additive property of the power of lens, widely used to design lens systems of cameras, microscopes and telescopes. These lens systems can have a combination of convex lenses and also concave lenses.    (2023)
(a) What is the nature (convergent/divergent) of the combination of a convex lens of power +4 D and a concave lens of power -2 D?
(b) Calculate the focal length of a lens of power -2.5 D.
(c) Draw a ray diagram to show the nature and position of an image formed by a convex lens of power +0.1 D, when an object is placed at a distance of 20 cm from its optical centre.

OR
(c) How is a virtual image formed by a convex lens different from that formed by a concave lens? Under what conditions do a convex and a concave lens form virtual image? 

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Ans: (a) Given: P1 = 4 D, P2 = -2 D
P = P1 + P2 = 4D – 2D
So the lens is convergent in nature.
(b) P = -2.5 D

P = 100f (cm) ⇒ −2.5 = 100f (cm) ⇒ f = −40 cm

(c) P = 0.1 D, u = −20 cm

P = 100f (cm) ⇒ 0.1 = 100f ⇒ f = 1000 cm

Previous Year Questions 2023
OR
(c) Virtual images formed by convex and concave lenses differ in several ways:

  • Convex Lens: Forms a virtual image when the object is within the focal length. The image is erectmagnified, and on the same side as the object.
  • Concave Lens: Always forms a virtual image. The image is erectdiminished, and on the same side as the object.


Q7: Hold a concave mirror in your hand and direct its reflecting surface towards the sun. Direct the light reflected by the mirror on to a white card-board held close to the mirror. Move the card-board back and forth gradually until you find a bright, sharp spot of light on the board. This spot of light is the image of the sun on the sheet of paper; which is also termed as “Principal Focus” of the concave mirror. (CBSE 2023)

Previous Year Questions 2023

(A) List two applications of concave mirror. 
(B) If the distance between the mirror and the principal focus is 15 cm, find the radius of curvature of the mirror. 
(C) Draw a ray diagram to show the type of image formed when an object is placed between pole and focus of a concave mirror. 
(D) An object 10 cm in size is placed at 100 cm in front of a concave mirror. If its image is formed at the same point where the object is located, find: 
(i) focal length of the mirror, and  
(ii) magnification of the image formed with sign as per Cartesian sign convention.

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Ans: (A) Applications of concave mirrors: 
(1) Concave mirror is used as a shaving mirror when the face is placed close to it so that it is within its focus and we get an erect and magnified image of the face.
(2) Doctors use concave mirror as a headmirror to concentrate parallel rays of light on its focus which enables them to examine body parts such as eye, throat, etc.
(B) Given, f = 15 cm
We know for a mirror,
R = 2f
R = 2 × 15 cm
R = 30 cm
(C)

Previous Year Questions 2023

(D) (i) We can use the mirror formula to find the focal length of the mirror:

1f = 1v + 1u

where, f is the focal length of the mirror, v is the image distance and u is the object distance. Since the image is formed at the same point as the object, v = u = –100 cm (Distances to the left of the mirror are negative). 
Substituting the values, we get:

1f = 1-100 + 1-100

1f = -2100

f = -50 cm

So the focal length of the mirror is –50 cm. (Negative sign indicates that it is a concave mirror). 
(ii) The magnification of the image is: 

m = – vu

where, m is the magnification of the image.
Substituting the values, we get:

m = – -100-100

m = -1

So the magnification of the image is 1. (Negative sign indicates that the image is real and inverted).


Q8: (A) Complete the following ray diagram to show the formation of image:

(B) Mention the nature, position and size of the image formed in this case. 
(C) State the sign of the image distance in this case using the Cartesian sign convention. (CBSE 2023)

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Ans: (A)
Previous Year Questions 2023(B) The image of object obtained in the convex mirror is erect and diminished. This is because a convex mirror always forms a virtual, erect and diminished image of an object. 
(C) The image distance is positive. This is because, the image is formed is behind the mirror

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Previous Year Questions 2022

Q1: An optical device forms an erect image of an object placed in front of it. If the size of the image is one half that of the object, the optical device is a    (2022)
(a) concave mirror
(b) convex mirror
(c) plane mirror
(d) convex lens.        

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Ans: (b)
Sol: The image formed by a convex mirror is always erect and of smaller in size than object.


Q2: The relation R = 2f is valid    (2022)
(a) for concave mirrors but not for convex mirrors
(b) for convex mirrors but not for concave mirrors
(c) neither for concave mirrors nor for convex mirrors
(d) for both concave and convex mirrors.  

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Ans: (d)
Sol: It is valid for both concave mirrors and convex mirrors.


Q3: In which of the following is a concave mirror used?    (2022)
(a) A solar cooker
(b) A rear view mirror in vehicles
(c) A safety mirror in shopping malls
(d) In viewing full size image of distant tall buildings                

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Ans: (a)
Sol: Concave mirrors are the mirrors best suited ir solar cookers because concave mirrors are convergent mirror and they are reflect sunlight towards a single focal point.


Q4: For the diagram shown, according to the new Cartesian sign convention the magnification of the image formed will have the following specifications :    (2022)

(a) Sign – Positive, Value – Less than 1
(b) Sign – Positive, Value – More than 1
(c) Sign – Negative, Value – Less than 1
(d) Sign – Negative, Value – More than 1              

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Ans: (b)
Sol: Magnification : Sign-positive, value-more the 1 because the object is placed between the focus and the pole. So, magnified image will be formed on other side of mirror. Hence, magnification of image formed will have i positive sign and value more than one.


Q5: The radius of curvature of a converging mirror is 30 cm. At what distance from the mirror should an object be placed so as to obtain a virtual image?    (2022)
(a) Infinity
(b) 30 cm
(c) Between 15 cm and 30 cm
(d) Between 0 cm and 15 cm  

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Ans: (d)
Radius of curvature of a converging mirror,  R = 30 cm
Focal Length, f = 30/2 cm = 15 cm
Thus, virtual image can be obtained from the mirror if an | object is placed between pole and focus, i.e., between 0 i cm and 15 cm.


Q6: Which of the following statements is not true in reference to the diagram shown above?    (2022)
(a) Image formed is real.
(b) Image formed is enlarged.
(c) Image is formed at a distance equal to double the focal length.
(d) Image formed is inverted. 

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Ans: (b)
Sol: Image formed is enlarged is not true. When  object is placed at C, image formed is real, inverted and of same size as object.

Previous Year Questions 2022


Q7:  An object of height 4 cm is kept at a distance of 30 cm from the pole of a diverging mirror. If the focal length of the mirror is 10 cm, the height of the image formed is    (2022)
(a) +3.0 cm
(b) +2.5 cm
(c) +1.0 cm
(d) +0.75 cm     

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Ans: (c)
Sol: Given, height of object (h) = +4 cm
Object distance (u) = -30 cm (object placed left side of the mirror)
Focal length, f = +10 cm
Mirror Formula,
1/f = 1/v + 1/u or v = uf/u – f
v = −30 × 10−30 × 10 or v = 304 = 7.5 cm

Now, Magnification (m) = −vu = h′h ; m = – −152 x 130 x = h′4

or h’ = 1 cm
Hence, height of the image formed is 1 cm.


Q8: If a lens and a spherical mirror both have a focal length of -15 cm, then it may be concluded that    (2022)
(a) both are concave
(b) the lens is concave and the mirror is convex
(c) the lens is convex and the mirror is concave
(d) both are convex.  

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Ans: (a)
Sol: As the focal length of a concave mirror and a concave lens is taken as negative, both are concave in nature.


Q9: A student determines the focal length of a device’ A’ by focusing the image of a far off object on a screen placed on the opposite side of the object. The device ‘A’ is    (2022)
(a) concave lens
(b) concave mirror
(c) convex lens
(d) convex mirror. 

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Ans: (c)
Sol: If the rays are travelling from far off distance and focused on opposite side of the lens, this is only possible in convex lens.


Q10: When light is incident on a glass slab, the incident ray, refracted ray and the emergent ray are in three media A, B and C. If n1, n2 and n3 are the refractive indices of A, B and C respectively and the emergent ray is parallel to the incident ray, which of the following is true ?     (2022)
(a) n< n2 < n3
(b) n> n2 > n3
(c) n1 < n2 = n3
(d) n= n< n2     

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Ans: (d)
Sol: Here, medium A and C is same and 8 is glass. n1 = n3 < n2


Q11: The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. According to new cartesian sign convention, if the image is three times the size of the flame, then the lens is    (2022)
(a) concave and magnification is +3
(b) concave and magnification is -3
(c) convex and magnification is -3
(d) convex and magnification is +3.

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Ans: (c)
Sol: The image of the candle flame is three times larger than the flame itself, which means the magnification is -3 according to the new Cartesian sign convention, where a negative sign indicates that the image is inverted. Since this type of magnification occurs with a convex lens, the correct answer is that the lens is convex and the magnification is -3.


Q12: The power of a combination of two lenses in contact is +1.0 D. If the focal length of one of the lenses of the combination is +20.0 cm, the focal length of the other lens would be    (2022)
(a) -120.0 cm
(b) +80.0 cm
(c) -25.0 cm
(d) -20.0 cm 

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Ans: (c)
Sol: Given, combined power = +1 D
Focal length of one lens, f1 = 20 cm
Combined focal length, fcombined = 1 m = 100 cm

1f = 1f1 + 1f2 ; 1100 = 120 + 1f2

Solving this equation, we get f2 = -25 cm Focal length of other lens = -25 cm


Q13: Study the diagram given below and identify the type of the lens XX’ and the position of the point on the principal axis OO’ where the image of the object AB appears to be formed    (2022)

Previous Year Questions 2022

(a) Concave; between O’ and Y
(b) Concave : between O and Y
(c) Convex; between O’ and Y
(d) Convex; between O and Y  

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Ans: (b)
Sol: As the ray after passing XX ‘is diverging therefore, XX’ is concave lens and image is formed between O and Y.


Q14: An object of height 3.0 cm is placed vertically on the principal axis of a convex lens. When the object i distance is -37.5 cm, an image of height -2.0 cm j is formed at a distance of 25.0 cm from the lens. I Next, the same object is placed vertically at 25.0 cm | from the lens. In this situation the image distance v and height h of the image is (according to the new j Cartesian sign convention)    (2022)
(a) v = +37.5 cm; h = +4.5 cm
(b) v = -37.5 cm; h = +4.5 cm  
(c) v = +37.5 cm; h = – 4.5 cm
(d) v = -37.5 cm; h = -4.5 cm  

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Ans: (c)
Sol: Given, object height = 3.0 cm
Let object distance is u and image distance is v.
Case-1: u = -37.5 cm and v = 25 cm
h = -2 cm (real and inverted)
From the lens formula, 1f = 1v – 1u = 125 – 1-37.5
f = +15 cm … (i)

Case2: u = -25 cm, f = +15 cm (from (i))
By lens formula:

1f = 1v – 1u

115 = 1v – 125

v = +37.5 cm

Now, h1 = 3 cm and h = ?

Magnification:

m = vu = hh1

∴ +37.5-25 = h+3

∴ h = -4.5 cm


Q15: An object is placed in front of a concave lens. For all positions of the object the image formed is always    (2022)
(a) Real, diminished and inverted
(b) Virtual, diminished and erect
(c) Real, enlarged and erect
(d) Virtual, erect and enlarged.

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Ans: (b)
Sol: The image formed by of a concave lens is always virtual, diminished and erect.


Q16: A ray of light starting from air passes through a medium A of refractive index 1.50, enters medium B of refractive index 1.33 and finally enters medium C of refractive index 2.42. If this ray emerges out in air from C, then for which of the following pairs of media the bending of light is least?    (2022)
(a) air-A
(b) A-B
(c) B-C
(d) C-air

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Ans: (b)
Sol: As refractive index of a medium increases, more bending of light takes place. For A – B, ratio of refractive indices of A and B is least, so least bending of light takes place for this pair of media.

Previous Year Questions 2022


Q17: A ray of light is incident as shown. If A, B and C are three different transparent media, then which among the following options is true for the given diagram?    (2022)

Previous Year Questions 2022

(a) ∠1 > ∠4
(b) ∠1< ∠2 
(c) ∠3 = ∠2
(d) ∠3 > ∠4  

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Ans: (c)
Sol: In A, B and C transparent media, ∠1 > ∠2 as light bends towards the normal and ∠3 < ∠4 as light bends away from the normal. ∠3 = ∠2 , because of alternate angles between two normals.


Q18: In the diagram shown above n1, n2 and n3 are refractive indices of the media 1, 2 and 3 respectively. Which one of the following is true in this case.    (2022)

Previous Year Questions 2022

(a) n1 = n2
(b) n1 > n2
(c) n2 > n3
(d) n3 > n 

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Ans: (d)
Sol: 

Previous Year Questions 2022

In medium 2, the light ray bends towards the normal, so n2 > n1.
Similarly, in medium 3, the light ray bends more towards the normal which indicate that refractive index of medium 3 is greater than medium 2. So, n3 > n1


Q19: The refractive index of medium A is 1.5 and that of medium B is 1.33. If the speed of light in air is 3 x 108 m/s, what is the speed of light in medium A and B respectively? (2022)
(a) 2 x 108 m/s and 1.33 x 1 08 m/s
(b) 1.33 x 108 m/s and 2 x 108 m/s
(c) 2.25 x 108 m/s and 2 x 108 m/s
(d) 2 x 108 m/s and 2.25 x 108 m/s  

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Ans: (d)
Sol: Given, refractive index of medium A, μA = 1.5
Refractive index of medium B, μB = 1.33
Speed of light in air, c = 3 x 108 m/s
μ = Speed of light in vacuum (c)Speed of light in medium (v)


For medium A:

μA = cvA ⇒ 1.5 = 3 × 108vA m/s

vA = 2 × 108 m/s

For medium B:

μB = cvB ⇒ 1.33 = 3 × 108vB m/s

VB = 2.25 x 108 m/s
Hence, speed of light in medium A is 2 x 108 m/s and in medium B is 2.25 x 108 m/s.


Q20: A student wants to obtain magnified image of an object AB as on a Screen. Which one of the following arrangements shows the correct position of AB for him/her to be successful?    (2022)
(a) 
(b) 
(c) 
(d) 

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Ans: (c)
Sol: To get real & magnified image, object should be kept either between 2F and F or at F. If it is kept at F, image will form at infinity which cannot be taken on screen, so object should be kept between 2F and F.


Q21: The following diagram shows the use of an optical device to perform an experiment of light. As per the arrangement shown, the optical device is likely to be a    (2022)

Previous Year Questions 2022

(a) concave mirror
(b) concave lens
(c) convex mirror
(d) convex lens   

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Ans: (b)
Sol: As per the arrangement, an optical device to perform an experiment of light is likely to be a concave lens.


Q22: If a lens can converge the sun rays at a point 20 cm j away from its optical centre, the power of this lens is    (2022)
(a) +2D
(b) -2D
(c) +5D
(d) -5D

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Ans: (c)
Sol: Converging point, f = 20 cm
Power P = ?
As we know that,

Power of lens = 1focal length (in m) = 1f (in m)

P = 10020 ⇒ P = +5 D

Hence, power of convex lens is +5 D.


Q23: A converging lens forms a three-times magnified image of an object, which can be taken on a screen. If the focal length of the lens is 30 cm, then the distance of the object from the lens is    (2022)
(a) -55 cm
(b) -50 cm
(c) -45 cm
(d) -40 cm

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Ans: (d)
Sol: –40 cm
Reason — Given, convex lens as converging lens
f = 30 cm
From formula,
m = v/u
-3 = v/u [-ve sign as real and inverted image is formed.]
v = -3u
Using formula,
1/f = 1/v – 1/u
We get,

130 = 1-3u – 1u

130 = -1 – 33u = -43u

3u = -4 × 30

u = -1203 = -40 cm

Hence, distance of the object from the lens = -40 cm.

Previous Year Questions 2021

Q1: The refractive index of glass is 1.50. What is the meaning of this statement?    (2021)

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Ans: It gives us the idea about the speed of light in the air and in the glass. It means that speed of light is 1.5 time more in air than the speed of light in the glass. 


Q2: A ray of light starting from air passes through medium A of refractive index 1.50, enters medium B of refractive index 1.33 and finally enters medium C of refractive index 2.42. If this ray emerges out in air from C, then for which of the following pairs of media the bending of light is least? 
(a) air-A 
(b) A-B 
(c) B-C 
(d) C-air (CBSE Term-1 2021)

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Ans: (b)
The bending of light depends on the difference in refractive indices between two media. The smaller the difference in refractive indices, the less the light will bend when passing from one medium to another.
For each pair of media:

  • Air-A: The difference in refractive indices is |1.00 – 1.50| = 0.50.
  • A-B: The difference in refractive indices is |1.50 – 1.33| = 0.17.
  • B-C: The difference in refractive indices is |1.33 – 2.42| = 1.09.
  • C-Air: The difference in refractive indices is |2.42 – 1.00| = 1.42.

The smallest difference is between A and B, which has a refractive index difference of 0.17. Therefore, the bending of light will be least when it passes between A and B.
Thus, the correct answer is (b) A-B.


Q3: A converging lens forms three times magnified image of an  object,  which  can be taken on a screen. If the focal length of the lens is 30 cm, then the distance of the object from the lens is: 
(a)
 –55 cm    
(b)
 –50 cm 
(c)
 –45 cm    
(d)
 –40 cm  (CBSE Term-1 2021)

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Ans: (d)
Given:
The lens forms a real, three times magnified image (m = -3 because the image is real and inverted).
Focal length f = 30cm.

For a lens, the magnification mm is given by: m = vu

where 
v is the image distance and 
u is the object distance. Since m = −3:

vu = -3 ⇒ v = -3u

Using the lens formula:

1f = 1v – 1u

Substitute f = 30cm and v = −3u:

130 = 1-3u – 1u

Simplify this equation:

130 = -1 – 33u = -43u

Now, solve for u:

u = -4 × 303 = -40 cm

Therefore, the distance of the object from the lens is –40 cm.


Q4: An object of height 4 cm is kept at a distance of 30 cm from the pole of a diverging mirror. If the focal length of the mirror is 10 cm, the height of the image formed is: 
(a)
 +3.0 cm    
(b)
 +2.5 cm 
(c)
 +1.0 cm    
(d)
 +0.75 cm  (CBSE Term-1 2021)

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Ans: (c)
Given:
Height of the object, h = 4cm
Object distance from the mirror, u = −30cm (distance is negative for mirrors according to the sign convention)
Focal length of the diverging (concave) mirror, f = −10cm

We need to find the height of the image h′.

1f = 1v + 1u

Substitute the given values into the formula:

1-10 = 1v + 1-30

Simplify the equation:

1v = -330 + 130 = -230 = -115

Therefore:

v = -15 cm

Calculating the magnification (m):

m = hiho = vu

Substitute the known values:

m = -15-30 = 12

Also, To find the height the value of hi

Using the magnification formula:

m = hiho

Substitute the values of m and ho:

12 = hi4

Solve for hi:

hi = 12 × 4 = 2 cm

So, the height of the image formed is 2 cm.

Previous Year Questions 2020

Q1: Define pole of a spherical mirror.     (2020)

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Ans: The pole of a spherical mirror define the geometrical center of the spherical surface of the mirror. It is the center of reflecting surface of spherical mirror and lies on the surface of spherical mirror.


Q2: The refractive index of a medium ‘x’ with respect to a medium ‘y’ is 2/3 and the refractive index of medium ‘y’ with respect to medium ‘z’ is 4/3. Find the refractive index of medium ‘z’ with respect to medium ‘x’. If the speed of light in medium ‘x’ is 3 x 108 m s-1, calculate the speed of light in medium ‘y.     (2020)

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Ans: Given, refractive index of medium x with respect to y, vμx = 23

Refractive index of medium y with respect to z, zμy = 43

∴ Refractive index of medium x with respect to z, zμx = vμx x zμy = 23 × 43 = 89

∴ Refractive index of medium z with respect to x, xμz = 1zμz = 98

Now, speed of light in x = 3 × 108 m/s

Speed of light in y, vy = ?

x = Speed of light in xSpeed of light in y
⇒ vy = 23 × 3 × 108 = 2 × 108 m/s


Q3: Study the ray diagram given below and answer the questions that follow:     (2020)
(a) Is the type of lens used converging or diverging? 
(b) List three characteristics of the image formed. 
(c) In which position of the object will the magnification be – 1?

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Ans:  (a) We have used a converging lens.
(b) The characteristics of the image formed:
(i) It is real.
(ii) It is inverted
(iii) It is enlarged.
(c) We get the magnification of object, m = – 1 at the  position 2F1.


Q4: Rishi went to a palmist to show his palm. The palmist used a special lens for this purpose.     (2020)
(i) State the nature of the lens and reason for its use.
(ii) Where should the palmist place/hold the lens so as to have a real and magnified image of an object?
(iii) If the focal length of this lens is 10 cm, the lens is held at a distance of 5 cm from the palm, use lens formula to find the position and size of the image. (2020)

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Ans: (i) The lens used here is a convex lens and it is used as a magnifying glass because at close range, i.e., when the object is placed between optical centre and principal focus it forms an enlarged, virtual and erect image of the object.
(ii) When this lens is placed such that the object is between the centre of curvature and the principal focus, the palmist obtain a real and magnified image.
(iii) Given, focal length, f = 10 cm and u = -5 cm 
According to lens formula,

1f = 1v − 1u or 1f = 1u + 1v

= 110 + 1−5 = −5 + 10−50

∴ v = −505 = −10 cm

Thus, the image will be formed at 10 cm on the same side of the palm and the size of the image will be enlarged.


Q5: Draw ray diagram in each of the following cases to show what happens after reflection to the incident ray when: 
(A) it is parallel to the principal axis and falling on a convex mirror. 
(B) it is falling on a concave mirror while passing through its principal focus.
(C) it is coming oblique to the principal axis and falling on the pole of a convex mirror. (CBSE 2020)

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Ans: (A) An incident ray parallel to the principal axis, after reflection appear to diverge from the principal focus, in the case of a convex mirror

Previous Year Questions 2020

A straight line XP is principal axis. AB is incident ray which is parallel to principal axis XP of a convex mirror MM’. The ray of light gets reflected at point B on the mirror and goes in the direction BD and it appears to be coming from the focus F of convex mirror. According to the laws of reflection, ∠i = ∠r. Therefore, the incident and reflected rays make equal angles with the normal. 
(B) An incident ray passing through the principal focus of a concave mirror, after reflection, will emerge parallel to the principal axis

Previous Year Questions 2020

(C) An incident ray coming obliquely to the principal axis towards a point (Pole of the mirror) on the convex mirror is reflected obliquely.

Previous Year Questions 2020


Q6: (A) A person suffering from myopia (nearsightedness) was advised to wear corrective lens of power  2.5 D. A  spherical lens of same focal length was taken in the laboratory. At  what  distance  should a student place an object from this lens so that it forms an image at a distance of 10 cm from the lens? 
(B)
     Draw a  ray diagram to show the position and nature of the image formed in the above case. (CBSE 2020)

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Ans: (A)

P = −2.5 D

f = ?

P = 1f

f = 1P = 1−2.5 D

f = −0.4 m = −40 cm

f = −40 cm

v = −10 cm

u = ?

1f = 1v − 1u

1u = 1v − 1f

1u = 1−10 − 1−40

= −110 + 140 = −4 + 140

= −340

u = −403 cm
= -13.33 cm

The object should be placed – 13.33 cm from the lens for the formation of an image at a distance of 10 cm from the lens.
(B)

Previous Year Questions 2020


Q7: Draw a ray diagram in each of the following  cases to show the formation of image, when the object is placed: 
(A) between optical centre and principal focus of a convex lens. 
(B) anywhere in front of a concave lens. 
(C) at 2F of a convex lens.
State the signs and values of magnifications  in the above mentioned cases (A) and (B). (CBSE 2020)

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Ans: (A) When an object is placed in front of the lens between optical centre and principal focus of a convex lens, the image is formed beyond 2F1 (on the same side of the object).

Previous Year Questions 2020

AB is the object and A’B’ is the image. The image formed is enlarged, virtual and erect. So the value of magnification will be greater than 1 and its sign will be positive. 
(B) When an object a placed anywhere infront of a concave lens. When we place an object between infinity and optical centre (O) of the concave lens, the image will be formed between focus (F1) and optical centre (O) on the same side of the lens.

Previous Year Questions 2020

AB is the object and A’B’ is the image. The image formed is diminished, virtual and erect. So the sign of magnification is positive (+) and the magnification will be less than 1.
(C) When the object is placed at 2F1 of convex  lens, the image is formed at 2F2 on the other side of the lens.

Previous Year Questions 2020The image formed is real, inverted and of the same size.

Also read: NCERT Textbook: Light – Reflection & Refraction

Previous Year Questions 2019

Q1: How far should an object be placed from a convex lens of focal length 20 cm to obtain its real image at a distance of 30 cm from the lens? Determine the height of the image if the object is 4 cm tall.      (2019)

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Ans: Focal length of convex lens (f) = 20 cm
Real image formed at a distance, (v) = 30 cm
Height of image (h1) = 4 cm
Let the object distance be u.

Lens formula,

1v − 1u = 1f ⇒ 130 − 1u = 120

u = −60 cm

From,

m = vu = 30−60 = −12

m = h’h = −12

h’ = −12 × 4

h’ = −2 cm

Hence, height of image of object is  -2 cm.


Q2: State laws of reflection of light. List four characteristics of the image formed by a plane mirror.    (2019) 

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Ans: The laws of reflection of light state that:

  • The incident ray, the reflected ray and the normal to the surface at the point of incidence all lie in the same plane.
  • The angle of incidence (i) is always equal to the angle of reflection (∠r) i.e. ∠i = ∠r
    Previous Year Questions 2019

Characteristics of the image formed by a plane mirror

  • Image formed in a plane mirror is virtual and erect.
  • It is the same size as the object.
  • The image formed is laterally inverted.
  • The image formed is at same distance behind the mirror as object is in front of mirror.


Q3: A student, holding a mirror in his hand, directed the reflecting surface of the mirror towards the sun. He then directed the reflected light on to a sheet of paper held close to the mirror.     (2019) 
(a) What should he do to burn the paper? 
(b) Which type of mirror does he use? 
(c) Will he be able to determine the approximate value of focal length of this mirror from this activity ? Give reason and draw ray diagram to justify your answer in this case. 

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Ans: 
(a) The student should adjust the distance between the mirror and the paper so that solar rays are sharply focussed on the paper.
(b) The mirror is a concave mirror.
(c) The student can find the approximate focal length by measuring the distance between the paper and the mirror.

Previous Year Questions 2019

As shown in Fig. 10.29, parallel rays from the sun are focussed on the paper at point A’ in focal plane of mirror such that PB’ = f.


Q4: A real image, 2/3 rd of the size of an object, is formed by a convex lens when the object is at a distance of 12 cm from it. Find the focal length of the lens.     (2019) 

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Ans: Here distance of the object from the lens u = – 12 cm and magnification of real image m = -2/3
As per relation m = v/u, we have v = mu =(-2/3) x (-12) = + 8 cm

So as per lens formula 1v – 1u = 1f

We have 1f = 1(+8) – 1(-12) = 18 + 112 = 524

Hence, f = 245 = 4.8 cm


Q5: Draw a ray diagram to show refraction through a rectangular glass slab. How is the emergent ray related to incident ray ? What is its lateral displacement ?    (2019) 

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Ans: A ray diagram showing refraction through a rectangular glass slab has been shown in adjoining Fig.

Previous Year Questions 2019

The emergent ray GH is exactly parallel to the incident ray EFNM. It means that ∠r2 = ∠i1.
However, the emergent ray is laterally (side ways) displaced as compared to the original path of light ray. In ray diagram, the lateral displacement is GN. Its value increases on increasing the width of glass slab.


Q6: An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm.    (2019) 
(i) Use the lens formula to find the distance of the image from the lens.
(ii) List four characteristics of the image (nature, position, size, erect/inverted) formed by the lens in this case.
(iii) Draw a ray diagram to justify your answer of the part (ii).

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Ans: We have, (i) Object distance, u= —60 cm Focal length of the concave lens, f= —30 cm Using lens formula,

So as per lens formula 1v – 1u = 1f

We have 1v = 1(-60) = 1(-30)

Next, 1v = -130 – 160

Then, 1v = -360

Thus, v = -20 cm
The image will be formed at a distance of 20 cm in front of the lens.
(ii) Nature of the image is virtual. The position of the image is between F1 and optical center 0. Size of the image is diminished. The image is Erect.
(iii)
Previous Year Questions 2019


Q7: (a) List four characteristics of the image formed by a convex lens when an object is placed between its optical centre and principal focus.
(b) Size of the image of an object by a concave lens of focal length 20 cm is observed to be reduced to 1/3 rd of its size. Find the distance of the object from the lens.    (2019) 

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Ans: (a) When an object is placed between the optical centre and principal focus of a convex lens, the image formed is virtual, erect and enlarged. Moreover, the image is formed on the same side of lens behind the object.
(b) Here magnification of given concave lens m = 1/3 and focal length of lens f = – 20 cm
As per relation m = v/u for a lens, we get

+13 = vu ⇒ vu = + u3

Therefore, as per sign convention followed, both u and v are -ve.

Using lens formula 1v – 1u = 1f, we have

1( -u/3 ) – 1( -u/20 ) ⇒ -3u + 1u = -120 ⇒ -2u = -120

Therefore, u = 40 cm

So the object is placed at a distance of 40 cm from the lens.

Previous Year Questions 2018

Q1: State the laws of refraction of light. Explain the term ‘absolute refractive index of a medium’ and write an expression to relate it with the speed of light in vacuum.    (2018) 

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Ans: Laws of refraction: 
a. The incident ray, refracted ray and normal to the point of incidence all lie in the same plane.
b. The ratio of sin of incident angle to sin of angle of refraction for a given pair of medium is constant.
Sin iSin r = Constant
Absolute refractive index of a medium is the ratio of speed of light in air or vacuum to speed of light in the medium.
Absolute refractive index
= Speed of light in air/vacuumSpeed of light in Medium


Q2: What is meant by power of a lens? Write its SI unit. A student uses a lens of focal length 40 cm and another of -20 cm. Write the nature and power of each lens.    (2018) 

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Ans: Power of lens is the ability of a lens to converge or diverge light rays passing through it. It is the reciprocal of the focal length. S.I. Unit: The Unit of power of a lens is Dioptre (D)

P = 1f(D)

Power of first lens: Focal length = +40 cm. Focal length is positive, hence it is a convex lens.

P1 = 100f(cm) = 10040 cm = +2.5 D

Power of second lens:

Focal length = -20 cm. Its focal length is negative, hence it is a concave lens.

f2 = -20100 m = -15 m

P2 = 1f2 = -5D


Q3: An object of height 4.0 cm is placed at a distance of 30 cm from the optical centre ‘O’ of a convex lens of focal length 20 cm. Draw a ray diagram to find the position and size of the image formed. Mark optical centre ‘O’ and principal focus ‘F’ on the diagram. Also find the approximate ratio of size of the image to the size of the object.    (2018) 

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Ans: Ray diagram: Position of O and F
Previous Year Questions 2018

1f = 1v – 1u ⇒ 120 = 1v – 1(-30)

1v = 120 – 130 ⇒ 1v = 360 – 260

Thus, v = 60 cm

m = hiho = vu ⇒ hi4 = 60(-30)hi = – 8 cm
Ratio = hi/ho is approximately 2:1

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Previous Year Questions 2017

Q1: If the image formed by a mirror for all positions of the object placed in front of it is always erect and diminished, what type of mirror is it ? Draw a ray diagram to justify your answer. Where and why do we generally use this type of mirror?    (2017) 

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Ans: Only a convex mirror always form an erect and diminished image behind the mirror between its pole and focus point for all positions of the object placed in front of the mirror.
Previous Year Questions 2017
A ray diagram showing the image formation of an object AB is shown here . A convex mirror is used as a rear view mirror in automobiles because it gives erect and diminished images of vehicles coming from behind. As a result, it helps the driver in having a much wider field of view.


Q2: An object 4 cm in height, is placed at 15 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror should a screen be placed to obtain a sharp image of the object. Calculate the height of the image.    (2017) 

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Ans: Here distance of object u = – 15 cm, height of object h = +4 cm and the focal length of concave mirror f = – 10 cm.

As per mirror formula 1v + 1u = 1f

We have 1v = 1(-10) = 1(-15) = 130 ⇒ u = -30 cm.

Thus, a screen is placed in front of the mirror at a distance of 30 cm from it.

Hence, magnification m = hiho = -vu

Thus, h = -vu × h = (-30)(-15) × 4 = -8 cm.

Thus, the image is an inverted image of height 8 cm.


Q3: An object is placed at a distance of 15 cm from a concave lens of focal length 30 cm. List four characteristics (nature, position, etc.) of the image formed by the lens.    (2017) 

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Ans: Here, object distance, u = -15 cm
Using lens formula,

1f = 1v – 1u

1-30 = 1v – 1-15

1v = 1-30 – 115

Thus, v = -10 cm
Magnification, m = u/v = -10/-15 = +23
Four characteristics of the image formed by the concave lens are :
(i) Image formed is virtual
(ii) Image is erect
(iii) Image is formed on the same side of the lens as the object
(iv) Image is smaller than the object


Q4: Give any two applications of a concave and convex mirror.    (2017) 

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Ans: (i) Concave mirrors:

  • Concave mirrors are used while applying make–up or shaving, as they provide a magnified image.
  • They are used in torches, search lights, and head lights as they direct the light to a long distance.

(ii) Convex mirrors:

  • Convex mirrors are used in vehicles as rearview mirrors because they give an upright image and provide a wider field of view as they are curved outwards.
  • They are found in the hallways of various buildings including hospitals, hotels, schools, and stores. They are usually mounted on a wall or ceiling where hallways make sharp turns.


Q5: Define power of a lens.     (2017) 

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Ans: Power is the ability of lens to converge or diverge the ray of light is called power of lens. It is equal to the reciprocal of focal length, i.e. P = 1/f

Q6: The magnification of an image formed by a lens is -1. If the distance of the image from the optical centre of the lens is 25 cm, where is the object placed? Find the nature and focal length of the lens. If the object is displaced 15 cm towards the optical centre of the lens, where would the image be formed? Draw a ray diagram to justify your answer.     (2017) 

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Ans: Magnification, m=−1
Image distance, v=25 

Calculate Object Position:
For a real image, m=−1
Magnification formula:

Previous Year Questions 2017So, the object is placed at u=−25.

v=+25cm indicates a real image (right of lens), uu is negative (object left of lens).Calculate Focal Length
Using the lens formula:
Previous Year Questions 2017The positive focal length indicates the lens is convex with a focal length of 12.5 cm.
New Image Position After Displacement
If the object is shifted 1515cm towards the optical centre, the new object distance is:u′=−25+15=−10.
Previous Year Questions 2017The negative sign indicates the image is formed on the same side as the object.Negative v′ means a virtual image on the object side.Object position: u=−25 cm (25 cm in front of lens).Focal length: f=+12.5 cm, convex lens.New image position: v′=−50 cm (50 cm on same side as object).

Q7: If the image formed by a lens for all positions of an object placed in front of it is always erect and diminished, what is the nature of this lens? Draw a ray diagram to justify your answer. If the numerical value of the power of this lens is 10 D, what is its focal length in the Cartesian system?    (2017) 

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Ans: It is a concave or diverging lens.
Previous Year Questions 2017
f = 1/P
P = – 10 D,
f = 1/-10D = -0.1 m
Or f = -10cm.


Q8: Define the term magnification as referred to spherical mirrors. If a concave mirror forms a real image 40 cm from the mirror, when the object is placed at a distance of 20 cm from its pole, find the focal length of the mirror. [Delhi 2017 C]

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Ans: Magnification of spherical mirror (m): It is equal to the ratio of size (height) o f the image to the size (height) of the object. Thus,
m = Size of Image (h2)/Size of Object (h1)
Given: For concave mirror u = – 20 cm,v = – 40 cm,
Using mirror equation,
1/f = 1/v + 1/u
or
1/f = 1/-40 + 1/-20
= -(1/40 + 1/20)
= -3/40

⇒ f = – 40/3


Q9: State Snell’s law of refraction of light. Express it mathematically. Write the relationship between absolute efractive index of a medium and speed of light in vacuum.    [AI 2017 C]

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Ans: Snell’s law: The ratio of sine of angle of incidence (i.e. sin i) to the sine of angle of refraction (i.e. sin r) is always constant for the light of given colour and for the given pair of media.
Mathematically, sin i/ sin r = constant = n21
The constant n21 is called refractive index of the second medium with respect to the first medium.
Absolute refractive index of the medium is given by nm = Speed of light in vacuum (c)/ Speed of light in medium  (v) = c/v


Q10: (a) What is the minimum number of rays required for locating the image formed by a concave mirror for an object? Draw a ray diagram to show the formation of a virtual image by a concave mirror.
(b) The linear magnification produced by a spherical mirror is +3. Analyze this value and state the (i) type of mirror and (ii) position of the object with respect to the pole of the mirror. Draw ray diagram to show the formation of image in this cased
(c) An object is placed at a distance of 30 cm in front of a convex mirror of focal length 15 cm. Write four characteristics of the image formed by the mirror.     (2017) 

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Ans: (a) Two rays are required.
Previous Year Questions 2017
(b) The linear magnification produced by a spherical mirror is +3. It shows that the size of image is three times the size of object, image is virtual and erect and formed behind the mirror.
Hence
(i) the mirror is a concave mirror, and (ii) the object is placed between the pole and the focus of a concave mirror.Previous Year Questions 2017(c) The four characteristics of the image formed by the convex mirror are virtual, erect, diminished and laterally inverted.


Q11: (a) To construct a ray diagram we use two rays which are so chosen that it is easy to know their directions after reflection from the mirror. List two such rays and state the path of these rays after reflection in case of concave mirrors. Use these two rays and draw ray diagram to locate the image of an object placed between pole and focus of a concave mirror.
(b) A concave mirror produces three times magnified image on a screen. If the object is placed 20 cm in front of the mirror, how far is the screen from the object?     (2017) 

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Ans: (a) Rays which are chosen to construct a ray diagram for reflection are:  A ray parallel to the principal axis and  A ray passing through the centre of curvature of a concave mirror.
Path of these light rays after reflection:
(i) It will pass through the principal focus of a concave mirror
(ii) It gets reflected back along the same path. When an object is placed between the pole and the principal focus of a concave mirror, a virtual, erect and enlarged image is formed behind the concave mirror as shown in the figure.
Previous Year Questions 2017
(b) m=−3, 
u=−20 cm , 
v=? 

Image Position
Previous Year Questions 2017

Distance Between Screen and Object

Answer:
Distance = 40 cm.


Q12: Analyse the following observation table showing variation of image-distance (v) with object-distance (u) in case of a convex lens and answer the questions that follow without doing any calculations:    (2017) 
(a) What is the focal length of the convex lens? Give reason to justify your answer.
(b) Write the serial number of the observation which is not correct. On what basis have you arrived at this conclusion? 
(c) Select an appropriate scale and draw a ray diagram for the observation at S.No. 2. Also find the approximate value of magnification.

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Ans: (a) From the observation 3, the radius of curvature of the lens is 40 cm as distance of object and the distance of the image is same.
We know, focal length, f = R/2 = 40/2 = 20 cm
(b) Observation 6 is not correct, because for this observation the object distance is between focus and pole and for such cases, the image formed is always virtual. But in this case real image is formed as the image distance is positive.
Previous Year Questions 2017
From the figure, object distance u = – 60 cm and image distance v = 30 cm.
We know, magnification = v/u = +30/-60 = -0.5


Q13: (a) If the image formed by a mirror for all positions of the object placed in front of it is always diminished, erect and virtual, state the type of the mirror and also draw a ray diagram to justify your answer. Write one use such mirrors are put to and why?
(b) Define the radius of curvature of spherical mirrors. Find the nature and focal length of a spherical mirror whose radius of curvature is +24 cm.    (2017) 

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Ans: (a) Convex (diverging) mirror
Previous Year Questions 2017

Reason: (i) It always produces a virtual and erect image.
(ii) The size of image formed is smaller than the object.

Therefore, it enables the driver to see a wide field view of the traffic behind the vehicle in a small mirror.
(b) Radius of Curvature: The separation between the pole and the centre of curvature or the radius of the hollow sphere, of which the mirror is a part, is called radius of curvature (R), i.e., PC = R.
Since focal length of the mirror is +24 cm. It indicates that nature of the given spherical mirror is convex/diverging mirror.
As R = 2f = 24 cm
Therefore, f = +12 cm


Q14: (a) Draw labelled ray diagrams for each of the following cases to show the position, nature and size of the image formed by a convex lens when the object is placed.     (2017) 
(i) between its optical center (O) and principal focus (F)
(ii) between F and 2 F
(b) How will the nature and size of the image formed in the above two cases, (i) and (ii) change, if the convex lens is replaced by a concave lens of same focal length?

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Ans: (a) A convex lens of focal length ‘f’ can form 
(i) a magnified and erect image only when the object is placed between its focus ‘F’ and optical centre ‘O’ of the lens.
Previous Year Questions 2017
(ii) a magnified and inverted image when an object is placed in the following positions: Between F1 and 2F1
Previous Year Questions 2017
(b) Whatever be the position o f object as given in case (i) and (ii),the image formed by the concave lens is always virtual, erect and diminished.


Q15: State the laws that are followed when light is reflected by spherical mirrors. Draw a ray diagram to show the formation of image of an object placed in front of a convex mirror. List two characteristics of the image formed. Briefly explain one use of convex mirrors.     (2017) 

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Ans: Laws of reflection
(i) The incident ray, reflected ray and normal to the reflecting surface at the incident point all lie in the same plane.
(ii) The angle of reflection is equal to the angle of incidence.
Previous Year Questions 2017

Characterstics of the Image:(1) The convex mirror always form virtual and erect images.
(2) The size of the image is always lesser than the object.
Use of convex mirror-
This mirror is used in vechiles mirror to see the nearer object to vechicle.


Q16: A student carries out the experiment of tracing the path of a ray of light through a rectangular glass slab for two different values of angle of incidence ∠i = 30º and ∠i = 45°. In the two cases the student is likely to observe the set of values of angle of refraction and angle of emergence as: 
(a) ∠r =30º, ∠e = 20º and ∠r = 45º, ∠e = 28º 
(b) ∠r =30º, ∠e = 30º and ∠r = 45º, ∠e = 45º 
(c) ∠r =20º, ∠e = 30º and ∠r = 28º, ∠e = 45º 
(d) ∠r =20º, ∠e = 20º and ∠r = 28º, ∠e = 28º (CBSE 2017)

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Ans: (c)
In an experiment involving a rectangular glass slab, when light passes through it, the angle of incidence (∠i) and angle of emergence (∠e) are equal. However, the angle of refraction (∠r) inside the glass slab is different due to the change in the medium’s refractive index.

For two different angles of incidence (∠i = 30° and ∠i = 45°):

  • When ∠i = 30°, the angle of refraction ∠r will be smaller due to the denser medium, resulting in ∠r ≈ 20°. As the ray emerges from the glass, the angle of emergence ∠e will be equal to the angle of incidence, so ∠e = 30°.
  • When ∠i = 45°, the angle of refraction ∠r will be approximately 28°. Again, the angle of emergence ∠e will match the angle of incidence, so ∠e = 45°.

Therefore, the correct set of values is (c) ∠r = 20º, ∠e = 30º and ∠r = 28º, ∠e = 45º.

Previous Year Questions 2016

Q1: What is a prism?    (2016) 

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Ans: A prism is an optical device with two triangular bases along with three rectangular lateral surfaces commonly inclined at an angle of 60°. 


Q2: Define the term reflection.    (2016) 

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Ans: The bouncing back of a ray of light in the same medium after striking on a surface of an object.


Q3: The nature, size and position of image of an object produced by a lens or mirror are as shown below. Identify the lens/ mirror (X) used in each case and draw the corresponding complete ray diagram, (size of the object about half of the image).    (2016) 

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Ans: a. Convex lens when object is in between F and C (2F).
Previous Year Questions 2016
b. Concave mirror when object is in between P and F its enlarged, erected and virtual image is formed.
Previous Year Questions 2016


Q4: (a) Calculate the distance at which an object should be placed in front of a convex lens of focal length 10 cm to obtain a virtual image of double its size.
(b) In the above given case, find the magnification, if image formed is real. Express it in terms of relation between v and u    (2016) 

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Ans: Given f = + 10 cm, u = ?
For virtual image
m = + 2
As
m = v/u or v/u = 2
v = 2u …(1)
1/v = 1/u = 1/f …(2)
Substituting (1) in (2)

12u – 1u = 110

12u = 110

u = -5 cm

For real image, f = 10 cm, m = -2

vu = -2, v = -2u

1v – 1u = 1f

1(-2u) – 1u = 110

-3/2u = 1/10 or u = -15 cm


Q5: One half of a convex lens is covered with a black paper.    (2016) 
a. Show the formation of image of an object placed at 2Fp of such covered lens with the help of ray diagram. Mention the position and nature of image. 
b. Draw the ray diagram for same object at same position in front of the same lens, but now uncovered. Will there be any difference in the image obtained in the two cases? Give reason for your answer.

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Ans: a. If the lower half of the lens is covered even then it will form a complete real, inverted image of same size at C2(2F2) with reduced intensity of image.
Previous Year Questions 2016
b. There will be no change in the nature and position of the object except in later case the image will be brighter.


Q6: State the relation between object distance, image distance and focal length of a concave or convex mirror. A concave mirror produces two times magnified real image of an object at 10 cm from it. Find the position of the image.    (2016) 

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Ans: For concave or convex mirrors the relation between u, v and f is given by
mirror formula,

1v – 1u = 1f

m = – 2
u = – 10 cm
m = – v/u = – 2 or v = 2u = -20 cm
v = – 20 cm


Q7: (A) One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations. 
(B) Name  the lens which can  be used as a magnifying glass. (CBSE 2016)

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Ans: (A) Yes, even when one half of the convex lens is covered with a black paper, the complete image of the object will be formed. When the upper half of the lens is covered: In this situation, rays of light coming from the object will be refracted by the lower half of the lens as shown in fig (a). These rays meet at the other side of the lens to form the image of the given object. When the lower half of the lens is covered: In this situation, rays of light coming from the object will be refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the figure (b). We will get a sharp image but the brightness of the image will be less now. 
Ray diagram:

Previous Year Questions 2016
Previous Year Questions 2016

(B) Convex lens can be used as a magnifying glass.


Q8: Suppose you have three concave mirrors A, B and C of focal lengths 10 cm, 15 cm and 20 cm. For each concave mirror you perform the experiment of image formation for three values of object distance of 10 cm, 20 cm and 30 cm. Giving reason answer the following: 
(A) For the three object distances, identify the mirror/mirrors which will form an image of magnification –1. 
(B) Out of the three mirrors identify the mirror which would be preferred to be used for shaving purposes/makeup. 
(C) For the mirror B draw ray diagram  for image formation for object distances 10 cm and 20 cm.   (CBSE 2016)

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Ans: (A) A real, inverted and same size image as that of object, formed by the concave mirror, will form an image of magnification –1. It is possible only when the object is placed at C (R = 2f). Hence, for the object distances of 20 cm and 30 cm, concave mirrors ‘A’ and ‘B’ will form the real, inverted and same size images as that of the object. Therefore, the mirrors ‘A’ and ‘B’ will form an image of magnification –1.
(B) Concave mirror of focal length 20 cm will be preferred to be used for shaving purpose or makeup. This is because when we bring our face within its focal length it forms a virtual, erect and enlarged image of our face. 
(C) Ray diagram for image formation by mirror ‘B’ 
(i) For object distance 10 cm.

Previous Year Questions 2016

(ii) For object distance 20 cm.

Previous Year Questions 2016

Previous Year Questions 2015

Q1: What is the magnification of the images formed by plane mirrors and why?    (2015) 

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Ans: Magnification of images formed by plane mirrors is unity because for plane mirrors, the size of the image formed is equal to that of the object.


Q2: What is meant by power of a lens?    (2015) 

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Ans: Power is the degree of convergence or divergence of light rays achieved by a lens.
It is defined as the reciprocal of its focal length. i.e, P = 1/f


Q3: Name the mirror that is used by a dentist in examining teeth.    (2015) 

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Ans: Dentist uses a concave mirror to see large images of the teeth of patients.


Q4: What is lateral displacement of a light ray passing through a glass slab?    (2015) 

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Ans: The shifting of the light ray sideways (though in the direction of original ray) on emergence from a rectangular glass slab is called “lateral displacement”.


Q5: Define power of a lens and write its SI unit.     (2015) 

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Ans: Reciprocal of focal length of a lens, expressed in meter, is called the power of that lens.
Its SI unit is 1 dioptre (1 D), where 1 D = 1 m-1.


Q6: Name the lens which can be used as a magnifying glass.    (2015) 

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Ans: A convex lens can be used as a magnifying glass so as to form magnified image of a tiny object placed near it.


Q7: Which type of lens has a negative power?    (2015) 

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Ans: A concave (diverging) lens has a negative power.


Q8: What is the difference between virtual image of an object formed by a convex lens and that formed by a concave lens?    (2015) 

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Ans: ‘Virtual image formed by a convex lens is always magnified but that formed by a concave lens is diminished one.


Q9: During its passage from one medium to another, where does a light ray change its path?    (2015) 

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Ans: During its passage from one medium to another a light ray changes its path at the boundary face separating the two media.


Q10: The power of a lens is + 5 D. Find its focal length in metres.     (2015) 

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Ans: Focal length  f = 1/P = +1/5 m = 0.2m


Q11: What are the units of power of a lens?    (2015) 

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Ans: If the focal length is measured in metre then the unit of power of a lens is dioptre.


Q12: If the image formed by a mirror for all positions of the object placed in front of it is always erect and diminished, what type of mirror is it? Draw a ray diagram to justify your answer. Where and why do we generally use this type of mirror?    (2015) 

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Ans: Only in convex mirror, for all positions of the object placed in front of it is always virtual, erected and diminished. Hence this mirror is convex mirror.
Previous Year Questions 2015
Previous Year Questions 2015
Convex mirrors are used in automobiles as a rear view mirror because of wider field of view and formation of erect image.


Q13: Name the type of mirror used in the following:    (2015) 
a. Solar furnace
b. Side/rear – view mirror of a vehicle.
Draw a labelled ray diagram to show the formation of image in each of the above two cases.
Which of these mirrors could also form a magnified and virtual image of an object? Illustrate with the help of a ray diagram.

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Ans: a. Concave mirror
Previous Year Questions 2015
b. Convex mirror
Previous Year Questions 2015
Concave mirror form magnified virtual image of an object.
Previous Year Questions 2015

Also read: NCERT Textbook: Light – Reflection & Refraction

Previous Year Questions 2014

Q1: Write down four important characteristics of image formed by a plane mirror.    (2014) 

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Ans: The characteristics of an image formed by a plane mirror are:

  • The image is virtual.
  • The image is erect.
  • The image is laterally inverted.
  • The size of the image is equal to that of the object.


Q2: Describe a spherical mirror.    (2014) 

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Ans: Spherical mirror is a part of a sphere. If reflection takes place from inside, it is said to be concave mirror, and if the reflection takes place from outside surface it is a convex mirror.
Previous Year Questions 2014


Q3: Define the following terms in relation to concave spherical mirror:    (2014) 
(a) Pole 
(b) Centre of curvature 
(c) Radius of curvature 
(d) Principal axis 
(e) Principal focus 
(f) Aperture 
(g)  Focal length

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Ans: (a) The mid point of mirror is known as pole.
(b) The centre of curvature of a spherical mirror is the centre of that sphere of which mirror is a part,
(c) The distance between pole and centre of curvature is called radius of curvature of the mirror.
(d) The straight line joining the pole and centre of curvature is called principal axis.
(e) The point on the principal axis through which parallel rays to the principal axis passes or appear to pass after reflection.
(f) The diameter of the mirror or size of the mirror is called aperture.
(g) The distance between focus and pole of a mirror is the focal length of the mirror.


Q4: With the help of ray diagram show that angle of incidence is equal to the angle of reflection when a ray is incident on the concave/convex mirror.    (2014) 

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Ans: 


Q5: A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of image and magnification. Describe what happens to the image as the needle is moved farther from the mirror.    (2014) 

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Ans: Given: u = – 12 cm
f = + 15 cm
h1 = 4.5 cm
v = ?, m = ?

Previous Year Questions 2014

  • Image distance: v ≈+6.67cm
  • Magnification: m≈+0.556
  • As needle moves farther, image remains virtual, erect, approaches 15 cm, and shrinks.

Previous Year Questions 2013

Q1: The path of a ray of light coming from air passing through a rectangular glass slab is traced by four students shown as A, B, C and D in the figures. Which one of these is correct?
(a) 

(b) 

(c) 

(d)    (CBSE 2013, 11)

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Ans: (b)
When light passes through a rectangular glass slab from air, it bends towards the normal upon entering the glass due to refraction. Inside the glass slab, the ray travels in a straight line parallel to the initial direction but displaced laterally. Upon exiting the glass slab into air, it bends away from the normal and emerges parallel to the original incident ray.

In the diagrams:

  • Student B shows the correct path of the light ray with lateral displacement and emergence parallel to the incident ray.
  • The other diagrams do not show the correct behavior of refraction through a rectangular glass slab.

Thus, the correct answer is (b) Student B.

08. Previous Year Questions: Heredity

Previous Year Questions 2025

Q1: If pea plants with round and green seeds (RRyy) are crossed with pea plants having wrinkled and yellow seeds (rrYY), the seeds developed by the plants of F₁ generation will be:  (1 Mark)
(A) 50% round and green
(B) 75% wrinkled and green
(C) 100% round and yellow
(D) 75% wrinkled and yellow

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Ans: (C) 100% round and yellow

According to Mendel’s experiment on the independent inheritance of two traits, round shape (R) and yellow colour (Y) are dominant traits, while wrinkled (r) and green (y) are recessive traits.Previous Year Questions 2025Hence, when RRyy (round, green) is crossed with rrYY (wrinkled, yellow), all F₁ offspring have the genotype RrYy, expressing the dominant traits — round and yellow seeds.


Q2: Question consist of two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below. (1 Mark)

Assertion (A): A human child bears all the basic features of human beings.
Reason (R): It looks exactly like its parents, showing very little variations.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.

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Ans: (C) A is true, but R is false.

A child bears all the basic features of a human being (so A is true), but it also says the child does not look exactly like its parents and human populations show a great deal of variation (so R is false).


Q3: Question consist of two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.  (1 Mark)
Assertion (A): A mango seed will germinate to form a mango tree.
Reason (R): Heredity determines the process by which traits and characteristics are reliably inherited from parents to offspring.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.

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Ans: (A) Both A and R are true, and R is the correct explanation of A.

Heredity ensures that characteristics and traits are reliably inherited from parents to offspring.
Hence, a mango seed germinates to form a mango tree because heredity passes on the basic body design and traits of the parent organism.


Q4: When a pure-tall pea plant is crossed with a pure-dwarf pea plant, the percentage of tall pea plants in F₁ and F₂ generation pea plants will be respectively:  (1 Mark)
(a) (100% ; 25%)
(b) (100% ; 50%)
(c) (100% ; 75%)
(d) (100% ; 100%)

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Ans: (c) (100% ; 75%)

According to Mendel’s monohybrid cross, when a pure tall (TT) pea plant is crossed with a pure dwarf (tt) plant, all F₁ plants are tall (Tt) — showing 100% tallness.
In the F₂ generation, when these F₁ plants self-pollinate, the ratio of tall to short plants is 3:1, i.e. 75% tall and 25% short.Previous Year Questions 2025


Q5: A tall pea plant with round seeds (TTRR) is crossed with a short pea plant with wrinkled seeds (ttrr). The F₁ generation will be:  (1 Mark)
(A) 25% tall with round seeds
(B) 50% tall with wrinkled seeds
(C) 75% tall with wrinkled seeds
(D) 100% tall with round seeds

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Ans: (D) 100% tall with round seeds

Previous Year Questions 2025


Q6: Mendel obtained F₂ generation by:  (1 Mark)
(A) Self-pollinating F₁ generation plants
(B) Cross-pollinating F₁ generation plants with plants having dominant trait
(C) Cross-pollinating F₁ generation plants with plants having recessive trait
(D) Cross-pollinating both the parents

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Ans: (A) Self-pollinating F₁ generation plants

Mendel obtained the F₂ generation by allowing the F₁ tall plants to self-pollinate. This produced both tall and short plants in the ratio of 3:1, showing that the F₁ plants carried both dominant and recessive traits.


Q7: (a) Explain how the proteins control the ‘characteristics’ in an organism with the help of an example of ‘short height’ trait in pea plant.
(b) Name the information source of making proteins in a cell.  (2 Marks)

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Ans: 

(a) Proteins control the characteristics of an organism because they determine how processes in the cell function.
For example, in pea plants, the height of the plant depends on a hormone that promotes growth.
If the enzyme involved in making this hormone works efficiently, the plant will produce more hormone and become tall.
If the gene for this enzyme has an alteration that makes the enzyme less efficient, less hormone will be produced, and the plant will be short.

(b) The information source for making proteins in a cell is DNA.


Q8: Pure-tall (TT) pea plants are crossed with pure-dwarf (tt) pea plants. The pea plants obtained in F₁ generation are then self-pollinated to produce F₂ generation.  (2 Marks)
(i) What do the plants of F₁ generation look like? Justify your answer.
(ii) What is the ratio of pure-tall plants to pure-dwarf plants in F₂ generation?

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Ans:

(i) All plants of the F₁ generation are tall.
This is because tallness (T) is a dominant trait, and shortness (t) is recessive.Previous Year Questions 2025When TT is crossed with tt, all offspring have the genotype Tt, which expresses the dominant tall trait.

(ii) In the F₂ generation, when F₁ plants (Tt) self-pollinate, the genotypic ratio obtained is:Previous Year Questions 20251 TT : 2 Tt : 1 tt
Therefore, the ratio of pure-tall (TT) to pure-dwarf (tt) plants is 1 : 1.


Q9: “Proteins control the expression of various characters.” Explain this statement by taking an example of “tallness” as a characteristic in plants.  (2 Marks)

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Ans: 

Proteins control the expression of characteristics because they determine how different biological processes function inside the organism.

For example, in pea plants, tallness depends on the presence of a plant hormone that promotes growth.
The production of this hormone is controlled by an enzyme, and the efficiency of that enzyme depends on the gene present.

  • If the gene produces an efficient enzyme, more hormone is made, and the plant becomes tall.
  • If the gene produces a less efficient enzyme, less hormone is made, and the plant becomes short.

Hence, proteins (enzymes) produced by genes control how a particular trait such as tallness is expressed.


Q10:(a) What are chromosomes?
(b) Explain in brief how stability of DNA content of a species is ensured in sexually reproducing organisms? 

 or 
Explain the mechanism of inheritance used by sexually reproducing organisms to ensure the stability of DNA of the species.  (3 Marks)

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Ans:

(a) Chromosomes are independent pieces of DNA that carry genes controlling the traits of an organism.
Each chromosome contains the genetic information inherited from the parents. In human beings, for example, chromosomes exist in pairs, with one set coming from the mother and the other from the father.

(b) In sexually reproducing organisms, each parent contributes one set of genes to the offspring through their germ cells.

  • Each cell of the body has two copies of each chromosome — one from the male parent and one from the female parent.
  • During the formation of germ cells (sperms and eggs), only one chromosome from each pair goes into a gamete.
  • When the male and female gametes fuse, the zygote formed again gets two sets of chromosomes — one from each parent.

This process ensures that the normal number of chromosomes and the stability of DNA in the species are maintained from generation to generation.


Q11: In one of Mendelian experiments, when F₁ generation pea plants with round yellow seeds were self-pollinated, pea seeds with the following combinations were obtained in F₂ generation:  (3 Marks)
Previous Year Questions 2025

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Ans:
Total seeds = 800 + 275 + 268 + 90 = 1433.
Observed distribution: 800 round yellow (55.83%)275 round green (19.19%)268 wrinkled yellow (18.70%)90 wrinkled green (6.28%).

The expected Mendelian dihybrid proportions are 9 : 3 : 3 : 1 (i.e. 9/16, 3/16, 3/16, 1/16). For 1433 seeds the expected numbers are approximately 806, 269, 269, 90 respectively. The observed counts (800, 275, 268, 90) are very close to these expected values, so the results conform to the 9:3:3:1 ratio.

Thus the data show that round (R) is dominant over wrinkled (r) and yellow (Y) is dominant over green (y), and that seed shape and seed colour are inherited independently, in agreement with Mendel’s law of independent assortment.


Q12: (a) How many chromosomes are present in human beings? Out of these how many are sex chromosomes?
(b) Explain how, in sexually reproducing organisms, the number of chromosomes in the progeny is maintained.  (3 Marks)

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Ans:

(a) In human beings, there are 46 chromosomes in total, arranged in 23 pairs. Out of these, one pair is the sex chromosomes — XX in females and XY in males — while the remaining 22 pairs are autosomes.

(b) In sexually reproducing organisms, the number of chromosomes in the progeny is maintained through the process of gamete formation and fertilisation. Each parent contributes one set of chromosomes through their germ cells (sperms and eggs), which contain only one chromosome from each pairPrevious Year Questions 2025During fertilisation, the male and female gametes fuse to form a zygote, restoring the diploid number of chromosomes. This ensures the stability of DNA content and the chromosome number in each new generation.


Q13: The lowest part of the ear called earlobe, is closely attached to the side of the head in some of us (Figure ‘X’), and not in others, called free earlobe (Figure ‘Y’). Attached and free earlobes are two variants found in human populations. The gene for free earlobe is dominant over attached earlobes.  (3 Marks)

Previous Year Questions 2025

(a) A man with attached earlobes marries a woman having free earlobes. 50% of their children have free earlobes and 50% have attached earlobes. Explain the inheritance of this trait and write the trait combinations of the progeny.
(b) Write the gene combinations of the father and the mother in the above case.

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Ans:

Let F = allele for free earlobe (dominant) and f = allele for attached earlobe (recessive).

(a) The observed 50% free : 50% attached children shows the mother must be heterozygous (Ff) and the father homozygous recessive (ff). Crossing Ff × ff gives gametes F or f from mother and f from father. Offspring genotypes: Ff (free) and ff (attached) in equal proportions — hence 50% Ff (free earlobe) and 50% ff (attached earlobe).

(Use of a small Punnett grid: mother → F, f; father → f
Offspring: Ff, ff → 1 Ff : 1 ff → 50% : 50%.)

(b) Father’s genotype = ff (attached earlobes).
Mother’s genotype = Ff (free earlobes, carrier of recessive allele).


Q14: A pure pea plant bearing terminal flowers was cross-pollinated with a pure plant having terminal flowers. In the F₁ generation, plants with axial flowers only were obtained. F₁ generation plants are self-pollinated and F₂ generation is obtained.  (3 Marks)
(a) Work out the pattern of inheritance in this case.
(b) What will be the ratio of plants obtained in F₂ generation?

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Ans: Let the gene for axial flowers be represented by A (dominant) and the gene for terminal flowers by a (recessive).

(a) When a pure axial (AA) pea plant is crossed with a pure terminal (aa) plant.Previous Year Questions 2025Previous Year Questions 2025
(b) When F₁ plants (Aa) are self-pollinated, the F₂ generation shows the genotypic ratio 1 AA : 2 Aa : 1 aa, giving a phenotypic ratio of 3 : 1 — that is, 3 plants with axial flowers and 1 plant with terminal flowers.Previous Year Questions 2025


Q15: A pure pea plant having round (R), yellow (Y) seeds is crossed with another pure pea plant having wrinkled (r), green (y) seeds. Subsequently F₁ progeny is self-pollinated to obtain F₂ progeny.  
(a) What do the seeds of F₁ generation look like?
(b) Give the possible combinations of traits in seeds of F₂ generation. Also give their ratio.
(c) State the reason of obtaining seeds of new combination of traits in F₂ generation. (3 Marks)
Ans: 

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(a) In the F₁ generation, all seeds are round and yellow because round (R) and yellow (Y) are dominant traits, while wrinkled (r) and green (y) are recessive. The F₁ plants therefore have the genotype RrYy.Previous Year Questions 2025(b) On self-pollinating the F₁ plants (RrYy × RrYy), the F₂ generation shows four possible trait combinations of seeds in the ratio 9 : 3 : 3 : 1 —Previous Year Questions 2025

(c) Seeds with new combinations of traits are obtained in the F₂ generation because genes separate and mix in new ways during gamete formation and fertilisation.This causes new combinations of characters to appear. 
This is called segregation and independent assortment of alleles during gamete formation and their random fertilisation.

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Q16: The gene combination of purple flowered pea plants is denoted as (WW) and that of white flowered pea plants as (ww), when these two plants are crossed F₁ generation is obtained.
(a) List two observations made by Mendel in F₁ generation plants.
(b) Give the 
(i) percentage of white flowered plants in F₂ generation and 
(ii) ratio of the gene combinations WW, Ww, and ww in F₂ generation.
(c) Write one difference between dominant and recessive trait.  (3 Marks)

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Ans:

(a) When Mendel crossed pure purple (WW) and pure white (ww) pea plants:Previous Year Questions 2025

  1. All plants of the F₁ generation had purple flowers — showing only one of the parental traits.
  2. There were no intermediate (light purple) flowers, indicating that the purple colour (W) is dominant over white (w).

(b)
(i) In the F₂ generation25% of the plants bear white flowers.
(ii) The ratio of gene combinations in F₂ generation is 1 WW : 2 Ww : 1 ww.Previous Year Questions 2025

(c)

  • Dominant trait: Expressed even when only one copy of the gene is present (e.g., purple flower – W).
  • Recessive trait: Expressed only when both copies of the gene are the same (e.g., white flower – ww).


Q17: Question is Case/data-based question with 2 or 3 subparts. Internal choice is provided in one of these sub parts.

In human beings, there are 23 pairs of chromosomes. Out of these 23 pairs of chromosomes (i.e. 46 chromosomes), 22 pairs of chromosomes are called autosomes, and one pair of chromosomes, i.e. two chromosomes, are called sex chromosomes. The sex chromosomes are of two types – ‘X’ chromosomes and ‘Y’ chromosomes. The sex of a child (i.e. progeny) is decided at the time of fertilisation. In other words, at the time of zygote formation, the sex chromosomes inherited from the parents of a child decide whether the newborn will be a boy or a girl.  ( 4 Marks)
(a) What are chromosomes?
(b) Why is the pair of sex chromosomes in human males called mismatched pair?
(c) (A) Show with the help of a flow chart that the statistical probability of getting a boy or a girl is 50:50.

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Ans:
(a) Chromosomes: Thread-like structures in the nucleus, composed of DNA and proteins, carrying genes that control hereditary traits.
(b) Mismatched Pair: Male sex chromosomes (XY) are mismatched because X and Y differ in size and gene content, unlike autosomes or female XX pairs.
(c) (A) Flow Chart:Previous Year Questions 2025

Previous Year Questions 2024

Q1: Consider the following statements: 
(i) The sex of a child is determined by what it inherits from the mother. 
(ii) The sex of a child is determined by what it inherits from the father. 
(iii) The probability of having a male child is more than that of a female child. 
(iv) The sex of a child is determined at the time of fertilisation when male and female gametes fuse to form a zygote.  (2 Marks) (2024)(2024)
The correct statements are: 
(a)
 (i) and (iii)
(b) (ii) and (iv)
(c) (iii) and (iv)
(d) (i), (iii) and (iv)

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Ans: (b)

The sex of a child is determined by the father, as the sperm can carry either an X or Y chromosome, while the mother’s egg always carries an X chromosome (statement ii).

The sex of a child is determined at the time of fertilization when the male and female gametes (sperm and egg) fuse to form a zygote (statement iv).


Q2: Source-based/case-based questions with 2 to 3 short subparts. Internal choice is provided in one of these sub-parts: Mendel worked out the rules of heredity by working on garden pea using a number of visible contrasting characters. He conducted several experiments by making a cross with one or two pairs of contrasting characters of pea plant. On the basis of his observations, he gave s     ome interpretations which helped to study the mechanism of inheritance.
(i) When Mendel crossed pea plants with pure tall and pure short characteristics to produce progeny, which two observations were made by him in F1 plants?
(ii) Write one difference between dominant and recessive traits.
(iii) (A) In a cross with two pairs of contrasting characters
Mendel observed 4 types of combinations in F2 generation. By which method did he obtain F2 generation? Write the ratio of the parental combinations obtained and what conclusions were drawn from this experiment.
OR
(iii) (B) Justify the statement: “It is possible that a trait is inherited but may not be expressed.” (4 to 5 Marks) (2024)

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Ans: (i)  

  • In F1 generation, all plants were tall / No short plants were observed
  • No medium height plants / No halfway characteristics were observed / Only dominant parental traits were seen and not the mixture of the two.

(ii)Previous Year Questions 2024(iii) (A) Self-pollination / Self-fertilisation / Selfing of F1 plants
Ratio – Round Yellow : Wrinkled Green = 9   :  1
Traits are inherited independently.
OR
(iii)
 (B) If pea plants with yellow seeds are crossed with plants of green seeds, it is found that in F1 generation all the plants have yellow seeds. When F1 plants are self-pollinated, it is found that in F2 generation, plants with yellow seeds and plants with green seeds are obtained. This shows that both the traits are inherited but only one trait is visible in F1 progeny while the other remains unexpressed. 


Q3: Assertion – Reason based questions : These questions consist of two statements — Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below:  (1 Mark) (2024)
Assertion (A): Human female has a perfect pair of sex chromosomes.
Reason (R): Sex chromosome contributed by the human male in the zygote decides the sex of a child.
(a)
 Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

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Ans: (b)
The answer is (b) because both the assertion and reason are true, but they explain different things. The human female has two X chromosomes (a perfect pair), and while it’s true that the male’s sperm decides the sex of the child, this fact is not directly explaining the assertion about the female’s sex chromosomes.


Q4: In an experiment to study independent inheritance of two separate traits: shape and colour of seeds, the ratio of the different combinations in F2 progeny would be (1 Mark) (CBSE 2024)
(a)
 1 : 3
(b) 1 : 2 : 1
(c) 9 : 3 : 3 : 1
(d) 9 : 1 : 1 : 3

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Ans: (c)

The 9:3:3:1 ratio occurs when two traits are inherited independently, as in Mendel’s dihybrid cross experiment. It represents the combinations of dominant and recessive traits for both seed shape and color in the F2 generation.


Q5: (a) List any two pairs of visible contrasting characters of garden pea plants used by Mendel for his experiments stating the dominant and recessive characters in each pair.

OR
(b) In human beings, the probability of getting a male or a female child is 50%. Explain with the help of a flow diagram only. (3 Marks) (2024)

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Ans: (a) 2 visible characters of garden pea plants are: 

  • Tallness (dominant) , Dwarfness (recessive) 
  • Yellow seeds (dominant) , Green seeds (recessive)

OR
(b)
Previous Year Questions 2024


Q6: The survival of a species is promoted through creation of variations. Illustrate with an example.  (1 Mark) (2024)

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Ans: Example: A population of bacteria living in temperate waters that can withstand heat due to the rise in temperature due to global warming will survive better in a heat wave than the non-variant bacteria having no capacity to tolerate heat wave. Thus, suitable variations promote survival.


Q7: List two differences between dominant traits and recessive traits. What percentage of pea plants in the F2 generation was with yellow seeds in Mendel’s cross between the pea plants having yellow (YY) and green coloured (yy) seeds?  (3 Marks) (2024)

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Ans:
 Previous Year Questions 202475% yellow seeds


Q8: Mendel crossed pure tall pea plants (TT) with pure short pea plants (tt) and obtained F1 progeny. When the plants of F1 progeny were self-pollinated, plants of Fprogeny were obtained. (3 Marks) (2024)
(a) What did the plants of F1 progeny look like? Give their gene combination.  
(b) Why could the gene for shortness not be expressed in plants of Fprogeny?  
(c) Write the ratio of the plants obtained in F2 progeny and state the conclusion that can be drawn from this experiment. 

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Ans: (a) All Plants Tall
Gene combination: Tt
(b) It is a recessive trait / it cannot be expressed in presence of dominant trait.
(c) Tall: Short 3 : 1
Conclusion: Tall trait is dominant and short trait is recessive. 


Q9: A cross made between two pea plants produces 50% tall and 50% short pea plants. The gene combination of the parental pea plants must be   (1 Mark) (2024)
(a) 
Tt and Tt    
(b) TT and Tt
(c) Tt and tt    
(d) TT and tt

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Ans: (c)

In the cross Tt (tall) and tt (short), 50% of the offspring inherit Tt (tall) and 50% inherit tt (short). This results in a 1:1 ratio of tall to short plants, explaining why option (c) is the correct answer.


Q10: A cross between two tall pea plants resulted in offspring having a few dwarf plants. The gene-combination of the parental plants must be (1 Mark) (2024)
(a) 
Tt and Tt
(b) Tt and tt
(c) TT and tt
(d) TT and Tt

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Ans: (a)

The parental plants must be Tt (heterozygous tall) and Tt (also heterozygous tall) because this combination can produce offspring with both tall (TT and Tt) and dwarf (tt) plants. In this case, the expected ratio would be 3 tall to 1 dwarf, which explains the presence of dwarf plants among the offspring.

Previous Year Questions 2023

Q1: A cross between pea plant with white flowers (w) and pea plant with violet flowers (VV) resulted in Fprogeny in which ratio of violet (VV) and white (w) flowers will be (1 Mark) (2023)
(a) 1 : 1
(b) 2 : 1
(c) 3 : 1
(d) 1 : 3  

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Ans: (c)

In this cross, the parent with violet flowers (VV) is homozygous dominant, while the white flower parent is homozygous recessive (ww). When they are crossed, all offspring in the F1 generation will have violet flowers (Vw). If F1 plants are crossed with each other, the F2 generation will have a 3:1 ratio of violet (Vw) to white (ww) flowers, making option (a) the correct answer.


Q2: Assertion (A): In humans, if gene (B) is responsible for black eyes and gene (b) is responsible for brown eyes, then the colour of eyes of the progeny having gene combination bb or Bb will be black only.
Reason (R): The black colour of the eyes is a dominant trait.   (1 Mark) (2023)
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true. 

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Ans: (d)
The assertion (A) is false because the progeny with the gene combinations Bb (black eyes) and bb (brown eyes) would not have black eyes; only BB would. However, the reason (R) is true because the black color is indeed a dominant trait. Therefore, the correct answer is (d), which states that (A) is false but (R) is true.


Q3: Consider the following two statements:
(i) The trait that expresses itself in the F1 generation.
(ii) The trait that keeps on passing from one generation to another.
The appropriate terms for the statements (i) and (ii) respectively are  (1 Mark) (2023)
(a) Recessive trait; Dominant trait
(b) Dominant trait; Recessive trait
(c) Dominant trait; Inherited trait
(d) Recessive trait; Inherited trait      

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Ans: (c)
In the given statements, the trait that expresses itself in the F1 generation is called a “dominant trait,” while the trait that can be passed down through generations, even if it doesn’t show up in every generation, is referred to as an “inherited trait.” Therefore, the correct answer is (c) “Dominant trait; Inherited trait.”


Q4: Assertion (A): Human population shows a great deal of variations in traits.
Reason (R): All variations in a species have equal; chances of surviving in the environment in which they live.  (1 Mark) (2023)
(a) Both Assertion (A) and Reason (R) are true and j Reason (R) is the correct explanation of the Assertion (A)
(b) Both Assertion (A) and Reason (R) are true, but; Reason (R) is not the correct explanation of the Assertion (A)
(c) Assertion (A) is true, but Reason (R) is False.
(d) Assertion (A) is false, but Reason (R) is true  

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Ans: (c)
Sol: Variations get accumulated or discarded as combined effect of environmental factors and reproduction process.


Q5: The most obvious outcome of the reproductive process is the generation of individuals of similar design, but in sexual reproduction, they may not be exactly alike. The resemblances as well as differences are marked. The rules of heredity determine the process by which traits and characteristics are reliably inherited. Many experiments have been done to study the rules of inheritance.
(i) Why an offspring of human being is not a true copy of his parents in sexual reproduction?
(ii) While performing experiments on inheritance in plants, what is the difference between F1 and Fgeneration?
(iii) Why do we say that variations are useful for the survival of a species over time? (4 to 5 Marks) (2023)

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Ans:  (i) In sexual reproduction offspring of human being is not a true copy of his parents because it inherits half of its genetic material from each parent. During the formation of gametes (sperm and egg cells), the genetic material undergoes recombination that leads to shuffling or mixing of genetic material from both parents, resulting in offspring with unique combination of genes.
(ii) F1 generation is the first filial generation of the offspring from the parents whereas F2 generation is the second filial generation of the offspring, generated through inbreeding of F1 generation. F1 generation can be distinctly different from the parental type whereas Fgeneration always exhibit some parental genotypes.
(iii) Variations allow genetic diversity that is important for the development of new traits and increasing adaptability towards changes in the environment. It makes the organisms better adapted for survival under changing conditions.


Q6: In some families, either rural or urban, females are tortured for giving birth to a female child. They do not seem to understand the scientific reason behind the birth of a boy or a girl. In fact the mother is not responsible for the sex of the child and it has been genetically proved that the sex of a newborn is determined by what the child inherits from the father.
(a) State the basis on which the sex of a newborn baby is determined in humans.
(b) Why is the pair of sex chromosomes called a mismatched pair in males?
(c) How is the original number of chromosomes present in the parents restored in the progeny?
OR
(c) Explain by giving two examples of the organisms in which the sex is not genetically determined. (4 to 5 Marks) (2023)

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Ans: (a) Sex of child is determined by what it inherits from the father. A child who inherits X chromosome from her father will be a girl and one who inherits a Y chromosome from father will be a boy.
(b) Human beings have 22 pairs of autosomes and one pair of sex chromosomes. The females possess two homomorphic sex chromosomes, named XX while males contain two heteromorphic sex chromosomes, i.e., XY. The Y chromosome is shorter than X chromosome. Therefore a pair of sex chromosomes in human beings is called mismatched pair in terms of type and size.
(c) Gametes contain half the number of chromosomes of parent. But, when the two gametes (male and female) fuse to form the zygote, the normal diploid condition is restored. Hence, the formation of gametes by meiosis and fusion of male and female gamete help to maintain the number of chromosomes in organism.

OR

(c) Two animals in which sex is not determined genetically are turtle and crocodile. The type of sex ir them is determined by environmental factors. In turtles the temperature of egg incubation has a significant effect on the sex of developing embryos. Males are predominant below 28°C, females above 33°C and equa number of the two sexes between 28-33°C. In crocodiles high temperature induces maleness and low temperature femaleness.


Q7: The statement that correctly describes the characteristic(s) of a gene is:  
(a) In individuals of a given species, a specific gene is located on a particular chromosome.
(b) A gene is not the information source for making proteins in the cell. 
(c) Each chromosome has only one gene located all along its length. 
(d) All the inherited traits in human beings are not controlled by genes.       (1 Mark) (CBSE 2023)

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Ans: (a)
(a) This statement is correct. In individuals of a species, each gene has a specific location (locus) on a particular chromosome.
(b) This statement is incorrect. A gene is indeed the information source for making proteins in the cell; it contains instructions for protein synthesis.
(c) This statement is incorrect. Each chromosome contains many genes, not just one, arranged along its length.
(d) This statement is incorrect. While some traits are influenced by environmental factors, genes control most inherited traits in human beings.
Therefore, the correct answer is (a) In individuals of a given species, a specific gene is located on a particular chromosome.


Q8: Assertion (A): Genes inherited from the parents decide the sex of a child.   (1 Mark) (CBSE 2023)
Reason (R): X chromosome in a male child is inherited from his father.
(a)
 Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true. 
(e) Both (A) and (R) are false.

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Ans: (e)
Assertion (A): “Genes inherited from the parents decide the sex of a child.” This is partly true, as the sex of a child is determined by the specific combination of sex chromosomes inherited from the parents. However, the statement lacks precision because it is specifically the presence of an X or Y chromosome from the father that determines the sex (not just “genes” in general). Thus, this statement can be considered false or misleading.
Reason (R): “X chromosome in a male child is inherited from his father.” This is incorrect. In humans, a male child inherits the X chromosome from the mother and the Y chromosome from the father. Therefore, this statement is false.
Since both Assertion (A) and Reason (R) are incorrect, the correct answer is (e) Both (A) and (R) are false.


Q9: In order to trace the inheritance of traits Mendel crossed pea plants having one contrasting character or a pair of contrasting characters. When he crossed pea plants having round and yellow seeds with pea plants having wrinkled and green seeds, he observed that no plants with wrinkled and green seeds were obtained in the F1 generation. When the F1 generation pea plants were cross-bred by self-pollination, the F2 generation had seeds with different combinations of shape and colour also. 
(A) Write any two pairs of contrasting characteristics of pea plant used by Mendel other than those mentioned above. 
(B) Differentiate between dominant and recessive traits. 
(C) State the ratio of the combinations observed in the seeds of Fgeneration (in the above case). What do you interpret from this result?
OR 
Given below is a cross between a pure violet flowered pea plant (V) and a pure white flowered pea plant (v). Diagrammatically explain what type of progeny is obtained in F1 generation and F2 generation: 
Pure violet flowered plant (VV) × Pure white-flowered plant (vv).  (4 to 5 Marks) (CBSE 2023)

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Ans: (A) Mendel selected seven pairs of contrasting characters of pea plants for his experiments. 
(1) Pea shape: Round or Wrinkled. 
(2) Pea colour: Green or Yellow. 
(3) Pod shape: Constricted or Inflated.
(B) Dominant traits: 
(1) 
When both alleles of a gene are dominant, or when one allele is dominant and the other is recessive, it manifests. 
(2) A capital letter indicates the dominant allele. 
(3) Humans who are right-handed are an example. 
Recessive traits: 
(1) 
Only when a gene’s two recessive alleles are present is it expressed. 
(2) The recessive allele is hidden from view if one of the two alleles is dominant. 
(3) A small letter designates the recessive allele. 
(4) Example: People who are left-handed. 
(C) (1) Pea plants with rounded green seeds (RRyy) and pea plants with wrinkled yellow seeds (rrYY) were crossed by Mendel. 
(2) Since pea plants with green round seeds and pea plants with yellow wrinkled seeds are crossed to get F1 plants, both of these characters will appear in the F1 generation. But since we already know that round seeds and yellow seeds are dominant traits, the F1 plants will have yellow round seeds. 
(3) The F2 progeny was discovered to have yellow round seeds, green round seeds, yellow wrinkled seeds, and green wrinkled seeds in the ratio 9:3:3:1. This F1 progeny was then self-pollinated.
An illustration of a dihybrid cross:

Previous Year Questions 2023OR

Previous Year Questions 2023

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Previous Year Questions 2022

Q1: Justify the statement “Sex of the children will be determined by what they inherit from their father”.         (2022)

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Ans: The statement “Sex of the children will be determined by what they inherit from their father” is based on the principle of genetic inheritance. The determination of an individual’s sex is influenced by the combination of sex chromosomes inherited from both parents.
In humans, females have two X chromosomes (XX) while males have one X and one Y chromosome (XY). The father contributes either an X or a Y chromosome to the offspring, while the mother always contributes an X chromosome. Therefore, it is the father’s genetic contribution that ultimately determines the sex of the child.
During fertilization, if the father’s sperm carries an X chromosome, the resulting combination of XX chromosomes will develop into a female child. On the other hand, if the father’s sperm carries a Y chromosome, the combination of XY chromosomes will lead to the development of a male child.
It is important to note that the sex of the child is determined by the father’s genetic contribution, but other factors such as environmental influences and chance also play a role in the development of an individual’s sex characteristics. Therefore, while the father’s genetic contribution is a significant factor, it is not the sole determinant of a child’s sex.


Q2: “Sex chromosomes in human males and females are X Y and X X respectively. The statistical probability of getting either a male or a female child is 50%. Justify this statement giving a reason.             (2022)

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Ans: Human female (XX) produces all gametes (ova) with X-chromosomes, while human male (XY) produces 50% gametes (sperms) with X-chromosome while 50% gametes with Y-chromosome.
If sperm having X chromosome fertilises the ovum with X chromosome then a female child will be produced, otherwise a male child will be produced.
Previous Year Questions 2022

Sex of the child (offspring) is determined by the type of sperm that fuses with ovum at the time of fertilisation. Therefore, there is 50% chance of a male child being born and a 50% chance of a female child being born.


Q3: What is variation? List two main reasons that may lead to variation in a population.           (2022)

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Ans: Variation is the degree of differences in the progeny (off springs) and between the progeny and parents. Two main reasons of variations are mutations and genetic recombination during sexual reproduction.


Q4: (a) Name the two types of gametes produced by men
(b) Does a male child inherit X chromosome from his father? Justify
(c) How many types of gametes are produced by a human female?     (2022)

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Ans: (a) The two types of gametes produced by men are sperm cells (spermatozoa) and X or Y sex chromosomes. Sperm cells carry either an X or a Y sex chromosome, determining the sex of the offspring upon fertilization with a female’s egg. 
(b) No, a male child does not inherit an X chromosome from his father. In humans, sex determination is based on the X and Y chromosomes. Females have two X chromosomes (XX), while males have one X and one Y chromosome (XY). When a child is conceived, they inherit one sex chromosome from each of their parents. The mother always passes on an X chromosome, while the father can pass on either an X or Y chromosome. If the father passes on an X chromosome, the child will be female (XX), and if the father passes on a Y chromosome, the child will be male (XY).
(c) A human female typically produces one type of gamete, which is the egg or ovum.


Q5: Sex of an individual is determined by different factors in various species. Some animals rely entirely on the environmental cues, while in some other animals the individuals scan change their sex during their life time indicating that sex of some species is not genetically determined. However, in human beings, the sex of an individual is largely determined genetically.
(a) In what way are the sex chromosomes ‘X’ and ‘Y different in size? Name the mismatched pair of sex chromosome in humans. 
(b) Write the number of pair/pairs of sex chromosomes present in human beings. In which one of the parent (male/female) perfect pair/pairs of sex chromosomes are present?
 (c) Citing two examples, justify the statement “Sex of an individual is not always determined genetically”.           (2022)

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Ans: (a) X chromosome is morphologically distinct from Y chromosome. Y chromosome is smaller than X chromosome. Hence, they are dissimilar or heteromorphic. Men have mismatched pair of sex chromosome in humans in which one is normal size X while other is a short one called Y.
(b) Human beings have 22 pair of autosomal chromosomes and one pair of sex chromosome. Women have a perfect pair of sex chromosomes both called X.
(c) (i) Sex of the individual is not always determined genetically. In some organisms, gender may be determined by environmental factors. For example snails, turtles and lizards sex is determined by the temperature at which fertilised egg are kept.


Q6: (i) In a cross between violet flowered plants and white flowered plants, state the characteristics of the plants obtained in the F1 progeny. 
(ii) If the plants of Fprogeny are self-pollinated, then what would be observed in the plants of F1 progeny? 
(iii) If 100 plants are produced in F1 progeny, then how many plants will show the recessive trait?         (2022)

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Ans: (i) In the F1 progeny, the plants obtained will have violet flowers.
(ii) If the plants of F1 progeny are self-pollinated, the plants obtained in the F2 progeny will show a phenotypic ratio of 3:1, with 75% of the plants having violet flowers and 25% having white flowers.
(iii) 
If 100 plants are produced in the F1 progeny, approximately 25 plants will show the recessive trait of white flowers.


Q7: A cross was made between green-stemmed tomato plants denoted by (GG) and purple-stemmed tomato plants denoted as (gg) to obtain F1 progeny. 
(a) What colour of the stem would you expect in their F1 progeny and why? 
(b) Give the percentage of purple-stemmed plants if F1 plants are allowed to self-pollinate to produce F2 progeny. 
(c) Write the ratio between GG and gg plants in the F2 progeny        (2022)

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Ans: (a) F1 progeny will have green stemmed tomato plants as green is dominant over purple stemmed tomato plants. 
(b) If F1 plants are self pollinated, then the percentage of purple stemmed plant in F2 progeny will be 25%.
Previous Year Questions 2022

(c) Ratio of GG and gg plant in F2 generation will be 1:1.


Q8: The mechanism by which the sex of an individual is determined is called sex-determination. In human beings, sex of a newborn is genetically determined, whereas in some others it is not. There are 46 (23 pairs) chromosomes in human beings. Out of these, 44 (22 pairs) control the body characters and 2 (one pair) are known as sex chromosomes. The sex chromosomes are of two types – X chromosome and Y chromosome. At the time of fertilisation, depending upon which type of male gamete fuses with the female gamete, the sex of the newborn child is decided.
(a) Why is a pair of sex chromosomes in human beings called a mismatched pair in terms of type and size?
(b) If the gametes always have half the number of chromosomes, then how is the original number of chromosomes restored in the organism?
(c) Name two animals whose sex is not genetically determined. Explain the process of their sex determination.

OR

(c) With the help of a flowchart only, show how sex is genetically determined in human beings.  (2022)

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Ans: (a) Human beings have 22 pairs of autosomes an one pair of sex chromosomes. The females possess tw homomorphic sex chromosomes, called XX while male have a mismatched pair in which one is a normal sized while the other is a short-one called Y.
(b) Gametes contain half the number of chromosome But, when the gametes fuse to form the zygote, the normal diploid condition is restored. Hence, the formation of gametes by meiosis helps to maintain the number of chromosomes in organism. (c) Two animals in which sex is not determined genetically are turtle and crocodile. The type of sex in them is determined by environmental factors. In turtles, the temperature of egg incubation has a significant effect on the sex of developing embryos. Males are predominant below 28°C, females above 33°C and equal number of the two sexes between 28-33°C. In crocodiles, high temperature induces maleness and low temperature induces femaleness.

OR

(c) Flow chart of sex determination in human beings is as follows

Previous Year Questions 2022


Q9: Mendel blended his knowledge of Science and Mathematics to keep the count of the individual exhibiting a particular trait in each generation. He observed a number of contrasting visible characters controlled in pea plants in a field. He conducted many experiments to arrive at the laws of inheritance.
(a) What do the F1 progeny of tall plants with round seeds and short plants with wrinkled seeds look like?
(b) Name the recessive traits in above case.
(c) Mention the type of the new combinations of plants obtained in F2 progeny along with the ir ratio, if F1 progeny w as allowed to self pollinate.

OR
(c) If 1600 plants were obtained in F2 progeny, write the number of plants having traits:
(i) Tall with round seeds
(ii) Short with wrinkled seeds
Write the conclusion of the above experiment.     (2022)

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Ans: (a) The F1 progeny will be tall plant with round seeds.
(b) Recessive traits are short plant and wrinkled seed.
(c) The new combinations will be tall plant with wrinkled seeds and short plant with round seeds. The ratio will be 9 : 3 : 3 : 1:: Tall plant with round seed : Tall plant with wrinkled seeds: Short plant with round seed : Short plant with wrinkled seed.

OR

(c) (i) The F2 ratio will be 9: 3: 3:1:: Tall plant with round seed : Tall plant with wrinkled seeds : Short plant with round seed : Short plant with wrinkled seed.
Tall plant with round seeds = 9/16 x 1600 = 900
(ii) Short plant with wrinkled seed = 1/16 x 1600 = 100

Previous Year Questions 2021

Q1: What is heredity? (2021 C, AI 2014)

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Ans: The inheritance of characters (or traits) from the parents to their off springs is called heredity.

Also read: Important Equations and Definitions: Heredity

Previous Year Questions 2020

Q1: How many pairs of chromosomes are present in human beings? (2020 C)

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Ans: In human beings, 23 pairs of chromosomes are present in each cell. Out of 23 pairs, 22 pairs of chromosomes carry genes which control somatic traits, these chromosomes are called autosomes. The 23rd pair is called sex chromosomes.


Q2: (a) Why did Mendel carry out an experiment to study inheritance of two traits in garden pea?
(b) What were his findings with respect to inheritance of traits in F1 and F2 generation?
(c) State the ratio obtained in the F2 generation in the above mentioned experiment. (2020)

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Ans: (a) Mendel carried out crosses with two traits to see the interaction and basis of inheritance between them. In a dihybrid cross given by Mendel, it was observed that when two pairs of characters were considered each trait expressed independent of the other.

(b) For Example, a cross between round yellow and wrinkled green parents.
Previous Year Questions 2020

In F1 generation, all plants are with round yellow seeds. But in F2 generation, we find all types of plants : Round yellow, Round green, Wrinkled yellow, Wrinkled green.
F2 generation ratio : Round-yellow = 9 : Round- green = 3 : Colour of stem in F1 progeny Wrinkled- yellow = 3 : Wrinkled-green = 1


Q3: A green stemmed rose plant denoted by GG and a brown stemmed rose plant denoted by gg are allowed to undergo a cross with each other.
(a) List your observations regarding :
(i) Colour of stem in their F1 progeny
(ii) Percentage of brown stemmed plants in Fprogeny if plants are self pollinated.
(iii) Ratio of GG and Gg in the F2 progeny.
(b) Based on the findings of this cross, what conclusion can be drawn? (2020)

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Ans: (a) (i) Colour of stem in F1 progeny:
Previous Year Questions 2020

The colour in the F1 progeny is green stemmed as green stem colour is dominant.

(ii) F1 progeny on self pollination:
Previous Year Questions 2020

F2 generation Green stemmed: Brown stemmed
14 or 25% of F2 progeny are brown stemmed rose plant.

(iii) Ratio of GG and Gg in F2 progeny:
Genotype of F2 progeny – GG : Gg
1 : 2

(b) This is a monohybrid cross. This shows that out of two contrasting traits only one dominant trait appears in F1 generation and the trait which does not express is recessive. On selfing the F1 plants, both the traits appear in next generation but in a definite proportion.

Previous Year Questions 2019

Q1: Define genetics. Why is a decrease in the number of surviving tigers a cause of concern from the point of view of genetics? Explain briefly. (AI 2019)

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Ans: Genetics is the branch of biology that deals with the study of heredity and variations. The term genetics’ was coined by William Bateson in 1906. When a population is small, the number and scope of variations is limited and hence diversity and traits are reduced. Small numbers of surviving tigers are a cause of worry from the point of genetics because of the following reasons:

  • Their loss would cause a loss of gene pool, i.e., many genes will be eliminated from a gene pool.
  • Tigers are surviving in limited numbers, so if some natural calamity kills these small population of tigers, they will suddenly become extinct as per genetic drift phenomenon,
  • A disease may wipe out the leftover population, if the entire population is susceptible to the disease. This can cause sudden extinction of the tiger species and loss of their genes forever, thus, adversely affecting the diversity of nature.


Q2:  Name the plant Mendel used for his experiment. What type of progeny was obtained by Mendel in the F1 and F2 generations when he crossed the tall and short plants? Write the ratio he obtained in F2 generation plants. (Delhi 2019)

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Ans: Mendel selected garden pea (Pisum sativum) for his series of hybridization experiments.
He first selected two pure line plants (tall plant having gene TT and short plant having gene tt) and then crossed such plants having contrasting characters. In the F1 generation, he observed that only one of the two contrasting character appeared, he called this character as dominant and the one which does not get expressed in F1 was called as recessive. He later self pollinated the F1 plants and observed that both the traits appear in next generation but in a definite proportion. This can be explained by the following cross :Previous Year Questions 2019So, the plants of F1 generation will be all tall plants and after self pollinating the ratio of tall and dwarf plants that Mendel obtained in F2 generation plants is 3 : 1.


Q3: List in tabular form the distinguishing features between acquired traits and inherited traits, with one example of each.
Or
List two differences between acquired traits and inherited traits by giving an example of each. (Delhi 2019)

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Ans:
Previous Year Questions 2019


Q4: Explain, with the help of an example each, how the following provide evidence in favour of evolution:    (Delhi 2017, AI 2019)
(а) Homologous organs
(b) Analogous organs
(c) Fossils

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Ans: (а) Homologous organs are those organs in different groups of organisms, which are similar in their basic structure/anatomy but are different in their functions. Such a similarity indicates that they are inherited from a common ancestor and the two species are closely related. For example, forelimbs of vertebrates like humans, wings of binds.
(b) Analogous organs are those organs/structures in different groups of organisms, that are similar in their function, but are dissimilar in their basic structural plan design and origin. Such organs do not indicate common ancestry of the species. For example, wings of birds and those of bats.
(c) Fossils are the preserved traces of organisms that lived in the past. Fossils indicate the time periods when the different groups of organisms lived in the earth. For example, Dinosaurs are reptilian: fossils, some of which show resemblance to birds in having feathers.


Q5: (a) What are homologous structures? Give an example.
(b) “The sex of a newborn child is a matter of chance and none of the parents may be considered responsible for it.” Justify this statement with the help of a flow chart showing sex-determination in human beings.    (Allahabad 2019)

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Ans: (a) Homologous organs are those organs in different groups of organism which are similar in their basic structure, but are modified to perform different functions.
e.g. forelimbs of mammals, and those of reptiles and amphibians.
(b) Sex of a child depends on what happens during fertilization:

Previous Year Questions 2019

(i) The female gamete, ova always contributes an X chromosome during fertilization.
(ii) The male gamete, sperm contributes either X or Y chromosome during fertilization. Whether sperm will contribute the X chromosome or Y chromosome is a matter of chance and the man does not have any control on it.
(iii) If a sperm carrying X chromosome fertilizes an egg which always carries a X chromosome, then the child born will be a girl. But if a sperm carrying Y chromosome fertilizes an egg which always carries X chromosome, then the child born will be a boy.
(iv) Thus, sex of a new born child is a matter of chance and none of the parents may be considered responsible for it.


Q6: In a pea plant, the trait of flowers bearing purple colour (PP) is dominant over white colour (pp). Explain the inheritance pattern of F1 and F2 generations with the help of a cross following the rules of inheritance of traits. State the visible characters of F1 and F2 progenies.(CBSE 2019)

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Ans: Let purple trait be represented by: PP 

Previous Year Questions 2019

Visible characters of F1 progeny all flowers are purple coloured and in F2 progeny 3 are purple coloured and 1 is white coloured flower.


Q7: With the help of Mendel’s experiments, show that: 
(A) traits may be dominant or recessive, and 
(B) traits are inherited independently.(CBSE 2019, 17)

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Ans: (A) (1) Mendel crossed pure tall pea plants with pure short pea plants. 
(2) All tall plants were produced in the F1 generation. 
(3) When F1 tall plants were selfpollinated, Mendel got both tall and short plants in the ratio of 3 (Tall) : 1 (Short). 
(4) This clearly indicates that tall character is dominant over short character, which although present would not be expressed in F1 generation.

Previous Year Questions 2019

(B) When pea plants with two different characteristics like plants with round and green seeds and the plants with wrinkled and yellow seeds; were bred with each other, the F1 generation had plants with round and yellow seeds (dominant character). 
On self-pollination of F1 generation plants, F2 generation obtained was a mixture of round yellow, round green, wrinkled yellow and wrinkled green in the ratio 9:3:3:1, thus, showing that the traits are inherited independently.

Previous Year Questions 2018

Q1: A Mendelian experiment consisted of breeding pea plants bearing violet flowers with pea plants bearing white flowers. W hat will be the result in F1 progeny? (2018)

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Ans: According to the Mendelian experiment, violet colour (W) is a dominant trait while white colour (vv) is a recessive trait. Hence, the colour of the flower in F1 progeny will be violet (Vv).


Q2: Define “artificial selection.” Comment on the purpose why farmers selected the following vegetables to cultivate:   (CBSE 2011, CBSE Sample Paper 2018-19)
(a) Cabbage
(b) Broccoli
(c) Cauliflower
(d) Kohlrabi through artificial selection

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Ans: Human beings have artificially selected certain variants that arose in nature by chance. This led to evolution of different species.
For example, wild cabbage was cultivated and its variants were selected due to different advantages, by artificial selection.
(а) Short distances between leaves – led to formation of modem day cabbage.
(b) Arrested flower development – Broccoli.
(c) Sterile flowers – Cauliflower
(d) Swollen parts – Kohlrabi
(e) Large leaves – Kale


Q3: Sometimes, accidently a dead body or its parts get buried under depositing sediments and are preserved. These are fossils. How can the estimation of the age of fossils be done?    (CBSE 2013, 2017-18, C, 2018-19)

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Ans: It can be done in two ways:
(а) The fossils found in upper layers are recent and the ones that are deeper are older.
(b) By radio-active carbon dating.


Q4: (a) What is variation? How is variation created in a population? How does the creation of variation in a species promote survival?
(b) Explain how, offspring and parents of organisms reproducing sexually have the same number of chromosomes.    (CBSE 2018 C)

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Ans:
(a) Variation refers to the differences in the characteristics among the individuals of a species.
1. Variation is created in a population by:

(i) Errors in DNA copying
(ii) Recombination during reproduction

2. In case of a drastic change in the environment of the niche of the population, at least some variants would have chances of survival.
(b)
1. There are special lineages of cells in specialized organs in multicellular organisms.
2. Such cells undergo a special type of cell division, called meiosis, and the germ cells (gametes) formed have only half the amount of chromosomes as the parent cell.
3. When two such germ cells (with half the number of chromosomes) fuse, a zygote/ new individual is formed with the reestablishment of the number of chromosomes as in parent organism.

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Previous Year Questions 2017

Q1: What are acquired traits? Why are these traits generally not inherited over generations? Explain.    (CBSE 2017-18 C)

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Ans: The traits that a person acquires during one’s life time and not by virtue of his/her genes, are known as acquired traits.
These traits are not present on our genes in reproductive cells. Hence they cannot be inherited.
Since these changes are in the non-reproductive cells and thus cannot be passed on to the germ-cells, hence are not inheritable by the future generations. An organism acquires them, for himself, in his life time, due to his environment/experiences.


Q2: What is speciation? List four factors responsible for speciation.    (CBSE 2017-18 C)

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Ans:  Speciation is the formation of new species from the pre-existing ones, due to accumulation of changes in such a way that two groups of individuals of one species can no longer interbreed. This results in formation of two new, independent species.
Factors responsible for speciation are:
(i) Natural selection – Nature selects due to survival advantage.
(ii) Genetic drift – Due to accumulation of changes over generations in sub-population.
(iii) Geographical isolation – In different geographical locations, the natural selection acts differently on different sub-population because of varying abiotic/biotic factors.
(iv) Mutation – Mutation is the change in the DNA. The large mutation in DNA can also result in speciation.


Q3: Briefly explain the role of natural selection and genetic drift in speciation by citing an example.    (CBSE 2017-18C)
Or
With the help of an example, explain how new species are produced.

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Ans: Let us study the example of beetles.  Let the original beetle population be red. If due to variation a green beetle was produced it would have survival advantage over red beetle. Thus green beetles would be naturally selected, and will grow in number.
If few of these green beetles move to some other area, adapt to changed environment and after few generations, would vary greatly from the original population of red beetles. These two populations may not be able to interbreed due to accumulation of variation.
Thus, due to Natural selection and Genetic Drift, a new species of beetles is formed.


Q4: What are homologous organs? Give one example. Can the wings of a butterfly and the wings of a bat be regarded as homologous? Give reasons in support of your answer.    (AI 2017C)

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Ans:  Homologous organs are those organs in different groups of organism which are similar in their basic structure, but are modified to perform different functions.
Previous Year Questions 2017

Example: Forelimbs of mammals, and those of Reptiles and Amphibians.
2. The wings of a butterfly and those of a bat cannot be considered as homologous, because they have a common function (flying), but their origin and basic structure are different.
3. They are analogous organs because they have a common function, though the basic structure is different.


Q5: “Evolution and classification of organisms are interlinked.” Give reasons to justify this statement.    (AI 2017)

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Ans:
1. We group organisms into groups based on the similarities in their characteristics.
2. Certain basic characteristics are shared by most or all the organisms, like the cell is the basic unit of life in all organisms.
3. The characteristics at the next level of classification would be shared by most organisms, but not by all organisms.
4. By taking the fundamental design differences, a hierarchy is developed that allows making of classification groups.
5. We can work out the evolutionary relationships of the species by identifying the hierarchies of characteristics between them.
6. The more characteristics two species will have in common, the more closely related they are.
7. The more closely related the two species, they would have had a recent common ancestor.
8. Thus, classification of species is a reflection of their evolutionary relationship.


Q6: If we cross-bred a tall (dominant) pea plant with a pure-bred dwarf (recessive) pea plant, we will get plants of F1 generation. If we now self-cross the pea plants of the F1 generation, we obtain pea plants of the F2 generation. (A12017C, 13,12)
(i) What do the plants of the F1 generation look like?
(ii) State the ratio of tall plants to dwarf plants in the F2 generation.
(iii) State the type of plants not found in the F1 generation but that appeared in the F2 generation. Write the reason for the same.

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Ans:
(i) The plants of F1 generation will be tall like the dominant parent.
(ii) Tall plants 3 : Dwarf plants 1, i.e., 3 : 1.
(iii) Dwarf plants are not found in F1 generation.
It is because, when two copies of a gene (alleles) exist together in the F1 plants, only the trait; tallness is expressed, i.e. it is dominant.
The other trait dwarfness remains hidden as it is a recessive trait.


Q7: How did Mendel explain that it is possible that a trait is inherited but not expressed in an organism?    (AI 2017)

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Ans:
1.  Mendel crossed a tall pea plant with a short pea plant.
2. All the plants produced in the Fgeneration were tall.
3. When the F1 tall plants were self-pollinated, the F2 generation consisted of both tall and short plants.
4. It explains that the dominant trait expresses itself in the F1 plants, where the recessive trait (shortness) is hidden.
5. The appearance of short plants in the F2 indicates that the trait shortness has been inherited by the F1 plants, but not expressed.


Q8:  Define variation in a species. How does it increase the survival chance of a species? Why do environmentalists get worried due to the small population of a species?    (CBSE 2013, 2017-18C)

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Ans: Any deviation from original trait in a species is variation.
Sometimes these variations may give added advantage of being of the nature so as to give better adaptability in changed conditions. Hence, it increases the chances of survival.
Smaller population of species would enforce inbreeding within the population which would result in fewer variations. This would also result in inbreeding depression and expression of recessive traits, thereby making it prone to diseases or extinction.
Hence, environmentalists are worried about small population size.


Q9: What is meant by the trait of a species ? Distinguish between acquired and inherited traits by giving an example of each.    (CBSE 2013,2016,2017)

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Ans: The typical characters present in all individuals of a species are said to be trait of that species.

Previous Year Questions 2017


Q10: What are the various evidences in favour of evolution?    (CBSE 2015,2017)

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Ans: Evidences in favour of evolution are:
(i) Homologous organs: Such organs which perform different functions but have similar structure and origin are called homologous organs. For example, forelimbs of bird, forelimb of man and frog perform different functions, but have similar basic structure. Presence of such organs indicate that all these vertebrates had common ancestors.
Previous Year Questions 2017Skeleton of forelimbs of (a) Human (b) Dog  (c) Bird (d) Whale showing homologous features.

Previous Year Questions 2017 (ii) Analogous organs: Such organs which perform similar functions but are structurally different are called analogous organs. For example, wings of a bird and wing of an insect. Presence of such organs show that these organisms have different origin.

(iii) Evidences from embryology: Early embryos of different vertebrates show striking similarities such as presence of tail. This indicates common origin and ancestry of different vertebrates.

Previous Year Questions 2017

Comparison of early Development stages (a) Fish (b) Salamander (c) Tortoise (d) Chicken (e) Human

Previous Year Questions 2017

(iv) Vestigial organs : These are the organs which appear functionless in one organism and functional in some others. For example, Vermiform appendix of the large intestine is nonfunctional in human beings but functional in herbivorous, ruminant animals. Presence of such organs also show common ancestry.
(v) Evidences from fossils : Archaeopteryx a fossil that resembles reptiles but has some bird like features. This shows that birds have been evolved from reptiles.
The dinosaur skull fossil shown was found only a few years ago in the Narmada valley.Previous Year Questions 2017


Q11: What are homologous organs? Give one example. Can the wings of a butterfly and the wings of a bat be regarded as homologous? Give reasons in support of your answer.    (AI 2017C)

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Ans: Homologous organs are those organs in different groups of organism which are similar in their basic structure, but are modified to perform different functions.
Previous Year Questions 2017

Example: Forelimbs of mammals, and those of Reptiles and Amphibians.
2. The wings of a butterfly and those of a bat cannot be considered as homologous, because they have a common function (flying), but their origin and basic structure are different.
3. They are analogous organs because they have a common function, though the basic structure is different.


Q12: “Evolution and classification of organisms are interlinked.” Give reasons to justify this statement.    ( AI 2017)

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Ans:
1. We group organisms into groups based on the similarities in their characteristics.
2. Certain basic characteristics are shared by most or all the organisms, like the cell is the basic unit of life in all organisms.
3. The characteristics at the next level of classification would be shared by most organisms, but not by all organisms.
4. By taking the fundamental design differences, a hierarchy is developed that allows making of classification groups.
5. We can work out the evolutionary relationships of the species by identifying the hierarchies of characteristics between them.
6. The more characteristics two species will have in common, the more closely related they are.
7. The more closely related the two species, they would have had a recent common ancestor.
8. Thus, classification of species is a reflection of their evolutionary relationship.


Q13: If we cross-bred a tall (dominant) pea plant with a pure-bred dwarf (recessive) pea plant, we will get plants of F1 generation. If we now self-cross the pea plants of the F1 generation, we obtain pea plants of the F2 generation. (A12017C, 13,12)
(i) What do the plants of the F1 generation look like?
(ii) State the ratio of tall plants to dwarf plants in the F2 generation.
(iii) State the type of plants not found in the F1 generation but that appeared in the F2 generation. Write the reason for the same.

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Ans:
(i) The plants of F1 generation will be tall like the dominant parent.
(ii) Tall plants 3 : Dwarf plants 1, i.e., 3 : 1.
(iii) Dwarf plants are not found in F1 generation.
It is because, when two copies of a gene (alleles) exist together in the F1 plants, only the trait; tallness is expressed, i.e. it is dominant.
The other trait dwarfness remains hidden as it is a recessive trait.


Q14: How did Mendel explain that it is possible that a trait is inherited but not expressed in an organism?    (AI 2017)

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Ans:
1.  Mendel crossed a tall pea plant with a short pea plant.
2. All the plants produced in the Fgeneration were tall.
3. When the F1 tall plants were self-pollinated, the F2 generation consisted of both tall and short plants.
4. It explains that the dominant trait expresses itself in the F1 plants, where the recessive trait (shortness) is hidden.
5. The appearance of short plants in the F2 indicates that the trait shortness has been inherited by the F1 plants, but not expressed.


Q15: (a) Can the wing of a butterfly and the wing of a bat be regarded as homologous? Why?
(b) What is speciation? State any two factors which could lead to speciation.
(c) Name the vegetables made from wild cabbage by artificial selection when farmers:
(i) opted for swollen stems
(ii) opted for sterile flowers.
(iii) opted for arrested flowers.
(iv) opted for large leaves.    (AI 2017C)

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Ans:
(a) No, the wing of a butterfly and the wing of a bat cannot be considered homologous organs because they have a common function of flying but their origin and basic structural designs are not common. So, they are analogous organs.
(b) Speciation refers to the phenomenon in which new species are formed form the existing species.
The factors leading to speciation are: (i) genetic drift and (ii) natural selection
(c) (i) Kohlrabi
(ii) Cauliflower
(iii) Broccoli
(iv) Kale


Q16: How do Mendel’s experiments show that the    (Delhi 2017)
(a) traits may be dominant or recessive,
(b) traits are inherited independently?

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Ans: Mendel’s Experiments on Inheritance of Traits. Mendel used a number of visible contrasting characters of garden pea like round/wrinkled seeds, tall/short plants, white/violet flowers, etc.

Previous Year Questions 2017Independent inheritance of two separate traits, shape and colour of seeds(i) Traits may be dominant or recessive:

  • Mendel used a number of visible contrasting pairs of characters in garden pea.
  • He made crosses between pea plants with different characters; there were no halfway or intermediate characters.
  • Only one of the parental traits appeared in the F1 generation; it is called dominant trait and the trait which remains hidden, is called recessive trait.
  • When the F1 plants were self-pollinated, the F2 progeny consists of plants with the dominant trait and recessive trait in the ratio of 3 : 1; it proves that traits may be dominant or recessive.

(ii) Traits are inherited independently:

  • When a cross is made between a tall plant with round seeds, (when inheritance of two traits is considered), with a short plant with wrinkled seeds, the F1 progeny plants were all tall with round plants.
  • When the F1 plants are self-pollinated, the F2 progeny consisted of some tall plants with round seeds and some short plants with wrinkled seeds; these two are the parental types of combinations of traits.
  • There were also some new combinations like tall plants with wrinkled seeds and short plants with round seeds.
  • Thus it is clear that the tall and short traits and round and wrinkled seed traits are inherited independently of each other.

Previous Year Questions 2016

Q1: Define speciation. Mention the factors due to which this can happen.    (CBSE 2016)

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Ans: Speciation – It is the process of formation of new species from an existing one. Factors that can lead to speciation are
(i) Natural selection.
(ii) Geographical isolation.
(iii) Migration – Genetic drift.


Q2: List two differences in tabular form between dominant traits and recessive traits. What percentage/proportion of the plants in the F2 generation/progeny were round, in Mendel’s cross between round and wrinkled pea plants?    (CBSE 2016)

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Ans:
Previous Year Questions 2016

75% of the plants in F2 generation were round in Mendel’s cross between round and wrinkled pea plants.


Q3: How does Mendel’s experiment show that traits are inherited independently?    (Al 2016)

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Ans: Mendel performed an experiment in which he took two different traits like tall and dwarf plant and round and wrinkled seeds. In second (F2) generation, some plants were tall with round seeds and some were dwarf with wrinkled seeds. There would also be dwarf plants having round seeds. Thus, the tall/short traits and round/wrinkled seed traits are independently inherited.


Q4: (a) If we cut the tail of a mouse, will the tail occur in the next generation of that mouse? Give reasons to support your answer.
(b) What are the features that Archaeopteryx had in common with the reptiles?    (CBSE 2016)

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Ans: (a) Even after cutting tail of a mouse its progeny continues to have tail. This is because ‘no tail’ is an acquired trait. The mouse continues to have information for presence of tail in its DNA and hence the progeny will have tail.
(b) Archaeopteryx has reptilian features as presence of tail, vertebra, teeth etc.


Q5: Name two homologous structures in vertebrates. Why are they named so? What is the significance of these structures in the study of evolution?    (CBSE 2016)

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Ans:
(i) Homologous structures in vertebrates are wings in birds and forelimbs of lizard.
(ii) They are so named as they have same structural design but different function.
(iii) Such structures give us idea about common ancestry.


Q6: Give two uses of fossils. How does the study of fossils provide evidence in favour of organic evolution?    (CBSE 2008, 2012, 2016-17C)

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Ans:  Two uses of fossils are:
(а) To help study evolution of plants and animals.
(b) To know past climatic conditions.
(c) To help calculate geological time etc. (Any two)
Evidence in favour of organic evolution:
(i) Fossils help to identify an evolutionary relationship between apparently different species.
(ii) The older fossils, present deeper, are simpler in body design, as compared to those present in upper layers which are more recent.
This clearly provides evidence in favour of organic evolution.


Q7: What is DNA?    (Delhi 2016)

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Ans: Deoxyribonucleic Acid (DNA) is a molecule which carry the hereditary characters or traits in a coded form from one generation to the next in all the organisms.


Q8: “A trait may be inherited, but may not be expressed.” Justify this statement with the help of a suitable example.    (CBSE 2016)

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Ans: A trait may be inherited, but if it is recessive, it will not be expressed unless it is homozygous. Example:
Previous Year Questions 2016Hence, genotype Tt with a recessive gene is expressed as phenotype of Tall. This shows that only dominant gene is expressed as the trait (T) while ‘t’ is not expressed.


Q9: What are chromosomes? Explain how, in sexually reproducing organisms, the number of chromosomes in the progeny is maintained.     (CBSE 2016)

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Ans: Chromosomes are long DNA strands, presents in nucleus, carrying genes which code for a trait. Hence, they are the hereditary material.
In an organism, each cell has two copies of a chromosome, one each from a male and female parent. In sexually reproducing organisms, where fusion of gametes (germ-cells) takes place, one chromosome from each pair is taken up in formation of a germ-cell (these may be either maternal or paternal in origin) by a special cell division called meiosis. When two germ cells fuse, they restore the original number of chromosomes in the progeny.


Q10: Define evolution. How does it occur? Explain how fossils provide evidences in support of evolution.    (CBSE 2016)

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Ans: The inbuilt tendency of variation, either due to errors in DNA copying or due to sexual reproduction, both result in some changes in the existing population of an organism. This continuous change ultimately leads to Evolution.
It occurs due to changes in DNA which keep accumulating over generations, ultimately giving rise to new species – Natural selection, genetic drift, mutation etc.
Fossils provide evidence for evolution as they are preserved traces of once living organisms. The fossils are formed when on death of an organism, the body doesn’t decompose, instead it gets trapped in the environment where it gets preserved. For example, an organism getting trapped in volcanic lava will not decompose. On cooling the lava will harden and retain the impression of the body parts of that organisms, as fossil.


Q11: What is speciation? List four factors that could lead to speciation. Which of these cannot be a major factor in the speciation of a self-pollinating plant species? Explain.     (Delhi 2016)

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Ans: Speciation: Speciation is the evolution of reproductive isolation among once-interbreeding population. Factors which can lead to speciation are:
(i) Genetic drift: Over generation, genetic drift may lead to the accumulation of different changes which lead to speciation.
(ii) Natural selection: Natural selection may work differently in different location which may give rise to speciation.
(iii) Severe DNA change: Variations during of DNA copying often leads to speciation.
(iv) A variation may occur which does not allow sexual act between two groups.
Out of these variation, severe DNA change is not a major factor in the speciation of a self- pollinating plant species, because:
(i) Variation is the differences in the characters among the individuals of species. In self- pollinating species, pollen grains fall on the stigma of the same flower or another flower of the same plant. Since self pollination is taking place in the same plant, so changes among the flowers of the same plant is negligible and hence variation in self-pollinating plant do not have any major effect in speciation of a self- pollinating plant.
(ii) Due to severe DNA change, individuals may vary from each other. In case of self-pollinating plants, pollination take place within the same plant and hence severe DNA change among individual plants do not have any major impact.


Q12: (a) What is meant by natural selection? Explain.
(b) Why are thorn of Bougainvillea plant and a tendril of Passiflora a plant considered homologous.    (CBSE 2016)

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Ans:
(a) Natural selection – It is selection of certain traits in nature in an individual in a population of a particular species.
This leads to survival advantage and hence variation leading ultimately to speciation.
(b) Thom of Bougainvillea and tendril of Passiflora both are modified stem, i.e., both have similar structure, but different function.
Thorn of Bougainvillea protects plants from being grazed while tendril of Passiflora helps the plants to climb up a support.
Hence, they are homologous organs.

Test: Genetics – 1Start Test

Previous Year Questions 2015

Q1:  “We cannot pass on to our progeny the experiences and qualifications earned during our lifetime.” Justify the statement by giving reasons and examples.    (Delhi 2015)

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Ans: Experiences of life and qualifications we earn do not make any change in the genes of the individual. Changes made in the gene are only passed on from one generation to the next. These qualities are acquired by an individual in his life, and are called acquired traits which cannot be passed on to future progeny. For example, if a person reads a book on birds, the knowledge he earns by reading the book does not make any change in his genes. Hence, this knowledge will not get automatically transmitted to his next generation.


Q2: A pea plant with a blue-coloured flower, denoted by BB, is cross-bred with a pea plant with a white flower, denoted by ww.
(a) What is the expected colour of the flowers in their F1 progeny?
(b) What will be the percentage of plants bearing white flowers in F2 generation, when the flowers of F1 plants are self-pollinated?
(c) State the expected ratio of the genotypes BB and Bw in the F2 progeny.    (AI 2015)

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Ans:
Previous Year Questions 2015(a) All flowers in F1 progeny will be blue in colour.
(b) When F1 progeny are self pollinated, 25% of the flowers in F2 progeny will be white.
(c) Expected ratio of the genotype BB and Bw will be 1:2.


Q3: “It is possible that a trait is inherited but may not be expressed.” Give a suitable example to justify this statement.    (Foreign 2015)

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Ans: The statement “It is possible that a trait is inherited but may not be expressed” can be explained with the help of Mendel’s experiment on pea plant with one visible contrasting character.
Mendel took pure breeding pea plant with one visible contrasting character viz, height of the plant (tall and short plant). The pure breed tall and short plant were crossed and it was found that all the plants in the F1 progeny were tall. Mendel then allowed the F1 progeny plants for self-pollination. It was found that all the F2 progeny plants are not tall, some are short. This indicates that both tallness and shortness traits were inherited separately in the Fprogeny but shortness trait was not expressed in the F1 progeny.


Q4: (i) We see eyes in Planaria, insects, octopuses, and vertebrates. Can eyes be grouped together in case of the above-mentioned animals to establish a common evolutionary origin? Why?
(ii) State one evidence to prove that birds have evolved from reptiles.    (Delhi 2015)

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Ans:
(i) Yes, eyes can be grouped together, which have evolved over generation from imperfect eyes in Planaria to perfect eyes in vertebrates.
(ii) Dinosaur is a type of reptile which has wings. Birds also have wings, so it can be opined that birds have evolved from reptiles.


Q5: Explain the following:    (AI 2015)
(a) Speciation
(b) Natural Selection

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Ans:
(a) Speciation: It is the evolution of reproductive isolation among once-interbreeding populations, i.e. the development of one or more species from an existing species.
(b) Natural Selection: It is the process, according to Darwin, which brings about the evolution of new species of animals and plants.

  • It was noted that the size of any population tends to remain constant despite the fact that more off-springs are produced than are needed to maintain.
  • Darwin found that variations existed between individuals of the population and concluded that disease, competition and other forces acting on the population eliminated those individuals which are less well-adapted to their environment.
  • The surviving population would pass the hereditary advantageous characteristics to their off-springs.


Q6:  Define the following:    (CBSE 2015)
(a)  Natural selection
(b) Reproduction isolation.

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Ans:
(a) Natural selection – Nature selects the best traits in a species, leading to survival of fittest and evolution of species. This phenomenon is known as natural selection.
(b) Reproduction isolation – It refers to the mechanism which checks the organisms of two different groups from interbreeding.


Q7: (a) Give evidence that birds have evolved from reptiles.    (CBSE 2015)
(b) Insects, octopuses, planarians, and vertebrates possess eyes. Can we group these animals together on the basis of the eyes that they possess? Justify your answer by giving a reason.

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Ans: 
(a) Fossils are important evolutionary evidence to show what kind of organisms existed earlier.
Archaeopteryx is a fossil dinosaur with wings. This proves that it has features of both reptiles as well as birds. Hence we can say that birds evolved from reptiles.
(b) These organisms cannot be grouped together as the structure of eye in each is very different. This means that they have separate evolutionary origin.

Previous Year Questions 2014

Q1: All the variations in a species do not have equal chances of survival. Why? (NCERT, Foreign 2014)

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Ans: All the variations do not have equal chances of survival in the environment in which they live. Depending on the nature of variations, different individuals would have different kinds of advantages. The organisms which are most adapted to the environment will survive.


Q2: Name the information source for making proteins in the cells. (Delhi 2014)

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Ans: Deoxyribonucleic acid (DNA) present in the chromosomes of cell nucleus is the information source for making proteins.


Q3: What is a gene?   (At 2014)

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Ans: A gene is a unit of DNA on a chromosome which governs the synthesis of particular protein that controls specific characteristics (or traits) of an organism.

07. Previous Year Questions: How do Organisms Reproduce?

Previous Year Questions 2025

Q1: The correct/true statement(s) for a bisexual flower is/are:  (1 Mark)
(i) They possess both stamen and pistil.
(ii) They possess either stamen or pistil.
(iii) They exhibit either self-pollination or cross-pollination.
(iv) They cannot produce fruits on their own.
(a)
 (i) only
(b) (iv) only
(c) (i) and (iii)
(d) (i) and (iv)

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Ans: (c)

bisexual flower contains both stamen (male part) and pistil (female part), allowing it to undergo self-pollination or cross-pollination. This is clearly mentioned in the chapter where examples like Hibiscus and mustard are given.

Q2: (a) (i) Write the functions of the following parts of human female reproductive system:  (5 Marks)
(I) Ovary, 
(II) Fallopian tube, 
(III) Uterus. 
(ii) State briefly two contraceptive methods used by human males. 
OR 
(b) (i) Differentiate between self-pollination and cross-pollination. 
(ii) Identify A, B, and C in the diagram given below and write one function of each.

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Ans (a):

(i) Functions of the parts of human female reproductive system:

(I) Ovary:

  • The ovary produces female germ-cells or eggs (ova).
  • It also secretes female hormone estrogen that regulate the reproductive cycle.

(II) Fallopian tube (Oviduct):

  • It carries the egg from the ovary to the uterus.
  • It is the site of fertilisation, where the male sperm meets the female egg.

(III) Uterus:

  • The uterus is the site where the embryo gets implanted after fertilisation.
  • It provides nourishment and protection to the developing embryo or foetus until birth.

(ii) Two contraceptive methods used by human males:1. Use of condoms:

  • Acts as a mechanical barrier preventing sperm from entering the female body.
  • Also helps reduce the risk of sexually transmitted diseases (STDs).

2. Surgical method (Vasectomy):

  • The vas deferens is blocked to prevent sperm from mixing with semen.
  • Thus, fertilisation cannot occur.

Ans (b):

(i) Difference between Self-Pollination and Cross-Pollination:

Q2: (a) (i) Write the functions of the following parts of human female reproductive system:  (5 Marks)(I) Ovary, (II) Fallopian tube, (III) Uterus. (ii) State briefly two contraceptive methods used by human males. OR (b) (i) Differentiate between self-pollination and cross-pollination. (ii) Identify A, B, and C in the diagram given below and write one function of each.

(ii) Parts and Functions (Flower Diagram):

A. Stigma: Receives pollen grains during pollination.
B. Pollen tube: Carries male gametes from the pollen grain to the ovule.
C. Female germ-cell / Egg cell: Fuses with the male gamete to form a zygote.


Q3: Bryophyllum produces new plant through:  (1 Mark)
(a) Apical buds formed on the tip of the plant
(b) Vegetative buds produced in the notches of the leaf
(c) Flowers produced in the notches of the branches
(d) Fruits formed on the branches of the plant

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Ans: (b)

In Bryophyllum, small buds develop in the notches along the leaf margins.
When these buds fall on the soil, they grow into new plants.
This is a form of vegetative propagation, an example of asexual reproduction in plants.Q2: (a) (i) Write the functions of the following parts of human female reproductive system:  (5 Marks)(I) Ovary, (II) Fallopian tube, (III) Uterus. (ii) State briefly two contraceptive methods used by human males. OR (b) (i) Differentiate between self-pollination and cross-pollination. (ii) Identify A, B, and C in the diagram given below and write one function of each.


Q4: The number of chromosomes in a cell division is halved. This kind of cell division is observed in:  (1 Mark)
(a) Only testis
(b) Only ovary
(c) Ovary and testis both
(d) All cells of the body

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Ans: (c)

The halving of chromosomes occurs during the formation of germ-cells (gametes) — sperm in males and eggs in females.
This special type of cell division takes place in the testes and ovaries.


Q5: Which one of the following statements is not correct for the plants raised by vegetative propagation?  (1 Mark)
(a) Can bear flowers and fruits earlier than those produced from seeds.
(b) Those plants that have lost the capacity to produce seeds can be grown.
(c) As compared to the parent plant, vegetatively propagated plants show more variations.
(d) All the plants produced in this way are genetically similar to the parent plant.

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Ans: (c)
Plants produced by vegetative propagation are genetically similar to the parent plant and show no variation.
Hence, statement (c) is incorrect.


Q6: Which of the following is the correct sequence of parts of female reproductive system of flowering plants in terms of their placement?  (1 Mark)
(a) Stigma, ovule, ovary, style
(b) Ovule, stigma, ovary, style
(c) Style, stigma, ovule, ovary
(d) Stigma, style, ovary, ovule

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Ans: (d)

In the female reproductive part (pistil) of a flower:

  • The stigma is the top sticky part that receives pollen.
  • The style is the elongated tube below the stigma.
  • The ovary is the swollen base.
  • Inside the ovary are the ovules, each containing an egg cell.Q6: Which of the following is the correct sequence of parts of female reproductive system of flowering plants in terms of their placement?  (1 Mark)(a) Stigma, ovule, ovary, style(b) Ovule, stigma, ovary, style(c) Style, stigma, ovule, ovary(d) Stigma, style, ovary, ovule


Q7: When a girl is born, the ovaries already contain thousands of immature eggs. On reaching puberty, some of these start maturing. One matured egg is released every month by one of the ovaries. The two oviducts unite into an elastic bag-like structure known as uterus.
(a) Write the site of fertilization in human female.  (1 Mark)
(b) How does the uterus prepare itself to receive and nurture the growing embryo? Explain.  (1 Mark)
(c) (i) What happens when the egg is not fertilized?  (2 Marks)
OR
(c) (ii) How does the developing embryo get nutrition from the mother’s blood? Explain.  (2 Marks)

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Ans:
(a) The site of fertilisation in the human female is the oviduct or fallopian tube.
It is here that the male sperm meets the female egg to form the zygote.
(b) The uterus prepares itself by thickening its lining and becoming richly supplied with blood to nourish the growing embryo after implantation.
(c) (i) When the egg is not fertilised:

  • The unfertilised egg lives for about one day.
  • The thickened lining of the uterus breaks down and comes out through the vagina as blood and mucus.
  • This process is called menstruation, which lasts for about two to eight days.Q7: When a girl is born, the ovaries already contain thousands of immature eggs. On reaching puberty, some of these start maturing. One matured egg is released every month by one of the ovaries. The two oviducts unite into an elastic bag-like structure known as uterus.(a) Write the site of fertilization in human female.  (1 Mark)(b) How does the uterus prepare itself to receive and nurture the growing embryo? Explain.  (1 Mark)(c) (i) What happens when the egg is not fertilized?  (2 Marks)OR(c) (ii) How does the developing embryo get nutrition from the mother`s blood? Explain.  (2 Marks)

OR
(c) (ii) Nutrition of the developing embryo:

  • The embryo receives nutrition through a special tissue called the placenta.
  • The placenta is embedded in the uterine wall and has villi on the embryo’s side and blood spaces on the mother’s side.
  • This arrangement provides a large surface area for glucose and oxygen to pass from the mother to the embryo and for wastes to be removed into the mother’s blood. 


Q8: Match Column-I with Column-II and select the correct option from the choices provided.  (1 Mark)

Q8: Match Column-I with Column-II and select the correct option from the choices provided.  (1 Mark)

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Ans:

The correct matching is a-(iii), b-(ii), c-(i), d-(iv).

  • Fertilisation occurs in the oviduct (fallopian tube).
  • The embryo implants in the uterus.
  • sperm enter through the vagina.
  • The placenta removes waste from the embryo’s blood into the mother’s blood.


Q9: Plants like rose and banana have lost the capacity to produce:  (1 Mark)
(a) Flowers
(b) Buds
(c) Seeds
(d) Fruits

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Ans: (c)
Plants such as rose and banana have lost the capacity to produce seeds.
They are grown by vegetative propagation, where new plants arise from stems, roots, or leaves instead of seeds.


Q10: In a bisexual flower the male gametes are present in the:  (1 Mark)
(a) Anther
(b) Ovary
(c) Stigma
(d) Filament

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Ans: (a)
In a bisexual flower, the stamen is the male reproductive part.
Its anther produces pollen grains, which contain the male gametes.


Q11: The modes of reproduction in Spirogyra and Planaria respectively are:  (1 Mark)
(a) Regeneration and budding
(b) Regeneration and fragmentation
(c) Fragmentation and regeneration
(d) Budding and regeneration

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Ans: (c)

  • Spirogyra reproduces by fragmentation, where the filament breaks into pieces, and each piece grows into a new individual.
  • Planaria reproduces by regeneration, where each cut piece of the body develops into a complete organism.

Q12: Select the option having correct matching of the organism given in Column I with the mode of reproduction in Column II:  (1 Mark)

(a) P-4, Q-2, R-1, S-3
(b) P-3, Q-4, R-5, S-2
(c) P-3, Q-4, R-1, S-2
(d) P-4, Q-3, R-2, S-1

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Ans: (c)Q12: Select the option having correct matching of the organism given in Column I with the mode of reproduction in Column II:  (1 Mark)

Q13: Give reasons:  (2 Marks)
(a) The male reproductive organ responsible for formation of germ cells is located outside the abdominal cavity. 
(b) The roles of the glands, present along the path of the vas-deferens, are very significant.

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Ans:

(a) The testes, which form the male germ-cells (sperm), are located outside the abdominal cavity in the scrotum because sperm formation requires a lower temperature than the normal body temperature.
The scrotum helps maintain this lower temperature, which is essential for healthy sperm production. 
(b) The glands present along the vas deferens, such as the prostate gland and seminal vesiclesadd their secretions to the sperm.
These secretions:

  • Provide nutrition to the sperm, and
  • Make their transport easier by forming a fluid medium. 

Q14: (a) Define fertilisation. 
(b) What happens to Zygote, Ovule, Ovary, and Stamens after fertilisation in a flowering plant?  (3 Marks)

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Ans:
(a) Fertilisation is the process of fusion of the male and female germ-cells (gametes) to form a zygote.
In flowering plants, the male germ-cell from the pollen grain fuses with the female gamete present in the ovule.
(b) After fertilisation in a flowering plant:

Q14: (a) Define fertilisation. (b) What happens to Zygote, Ovule, Ovary, and Stamens after fertilisation in a flowering plant?  (3 Marks)


Q15: Differentiate between a gamete and a zygote. State their significance in sexual reproduction.  (3 Marks)

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Ans:
Significance in sexual reproduction:

  • Gametes ensure genetic variation by combining DNA from two individuals.
  • The zygote is the first cell of the new organism; it divides repeatedly to form an embryo, which develops into a new individual.


Q16: Differentiate between self-pollination and cross-pollination. Which one of the two is better for the survival of species? Give reason to justify your answer.  (3 Marks)

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Ans:

Q16: Differentiate between self-pollination and cross-pollination. Which one of the two is better for the survival of species? Give reason to justify your answer.  (3 Marks)

Q17: (a) (i) Identify the parts ‘X’ and ‘Y’ in the figure given below:   (5 Marks)
(ii) Name the yellowish coloured structures produced by the part labelled as ‘Y’.
 (iii) Write the name of the process by which these are transferred to the part labelled as ‘X’.
(iv) Explain the process of seed formation in a flowering plant. 
OR 
(b) 
(i) Name the type of asexual mode of reproduction shown in the given figure.
(ii) Identify the unicellular organism in the diagram. 
(iii) List any two advantages of asexual reproduction over sexual reproduction. 
(iv) Name and explain any one mode of asexual reproduction observed in Hydra.  (5 Marks)

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Ans: (a)
(i) Parts: X = Stigma, Y = AntherQ17: (a) (i) Identify the parts `X` and `Y` in the figure given below:   (5 Marks)(ii) Name the yellowish coloured structures produced by the part labelled as `Y`. (iii) Write the name of the process by which these are transferred to the part labelled as `X`.(iv) Explain the process of seed formation in a flowering plant. OR (b) (i) Name the type of asexual mode of reproduction shown in the given figure.(ii) Identify the unicellular organism in the diagram. (iii) List any two advantages of asexual reproduction over sexual reproduction. (iv) Name and explain any one mode of asexual reproduction observed in Hydra.  (5 Marks)(ii) Yellowish Structures: Pollen grains
(iii) Process: Pollination
(iv) Seed Formation: After pollination, pollen grains on the stigma germinate, forming a pollen tube that grows through the style to the ovary. The male gamete fuses with the egg in the ovule (fertilization), forming a zygote. The ovule develops into a seed, with the zygote becoming the embryo, the ovule wall forming the seed coat, and the ovary developing into the fruit.

Ans (b):
(i) Type: Binary fission
(ii) Organism: Leishmania
(iii) Advantages:

  1. Faster reproduction, as it does not require a mate or complex processes.
  2. Produces genetically identical offspring, ensuring successful traits are passed on in stable environments.

(iv) Budding in Hydra: In budding, a small outgrowth (bud) forms on the Hydra’s body due to cell division. The bud grows, develops tentacles and a mouth, and eventually detaches as a new Hydra.

Q18: (A) (a) Define Puberty. List any two changes seen in boys at the time of puberty.
(b) Why are testes in human males located outside the abdominal cavity in scrotum?
(c) List any three techniques of contraception used by humans. Which one of these is not meant for males?   (5 Marks)
OR 
(a) Name the part performing the following functions in human female reproductive system: 
(i) production of eggs, 
ii) site of fertilization, 
(iii) site of implantation, 
(iv) entry of the sperm. 
(b) What changes are observed in the uterus: 
(i) subsequent to implantation of zygote and 
(ii) if an egg does not get fertilized?  (5 Marks)

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Ans: (A)
(a) Puberty and changes in boys:

Puberty is the period during adolescence when the reproductive tissues mature, and the body becomes capable of reproduction.

Two changes in boys at puberty:

  1. Growth of thick hair on the face (beard and moustache) and other body parts.
  2. Voice begins to crack and becomes deeper. 

(b) Location of testes:

The testes are located outside the abdominal cavity in the scrotum because sperm formation requires a lower temperature than the normal body temperature.
The scrotum maintains this lower temperature, ensuring proper sperm production. 

(c) Three techniques of contraception:

  1. Condoms – act as a mechanical barrier to prevent sperm from entering the female body.
  2. Oral pills – change the hormonal balance so that eggs are not released.
  3. Copper-T (loop) – placed in the uterus to prevent pregnancy by stopping implantation.

Not meant for males:
Oral pills and Copper-T are for females


OR 

(a) Parts and Functions:

Q18: (A) (a) Define Puberty. List any two changes seen in boys at the time of puberty.(b) Why are testes in human males located outside the abdominal cavity in scrotum?(c) List any three techniques of contraception used by humans. Which one of these is not meant for males?   (5 Marks)OR (a) Name the part performing the following functions in human female reproductive system: (i) production of eggs, ii) site of fertilization, (iii) site of implantation, (iv) entry of the sperm. (b) What changes are observed in the uterus: (i) subsequent to implantation of zygote and (ii) if an egg does not get fertilized?  (5 Marks)

(b) Uterus Changes:

Q18: (A) (a) Define Puberty. List any two changes seen in boys at the time of puberty.(b) Why are testes in human males located outside the abdominal cavity in scrotum?(c) List any three techniques of contraception used by humans. Which one of these is not meant for males?   (5 Marks)OR (a) Name the part performing the following functions in human female reproductive system: (i) production of eggs, ii) site of fertilization, (iii) site of implantation, (iv) entry of the sperm. (b) What changes are observed in the uterus: (i) subsequent to implantation of zygote and (ii) if an egg does not get fertilized?  (5 Marks)

Previous Year Questions 2024

Q1: (A) (i) Name three techniques/devices used by human females to avoid pregnancy. Mention the side effects caused by each.   (4 to 5 Marks) (2024)
(ii) What will happen if in a human female (a) fertilization takes place, (b) an egg is not fertilized?
OR
(B) (i) Draw a diagram showing spore formation in Rhizopus and label the (a) reproductive and (b) non-reproductive parts. Why does Rhizopus not multiply on a dry slice of bread?
(ii) Name and explain the process by which reproduction takes place in Hydra.

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Ans: (A)  (i)

  • Chemical Method/Oral pills Side effects: Change the hormonal balance of the body. 
  • Barrier method / Loop / Copper–T Side effects: Irritation in uterus. 
  • Surgical method / Fallopian tube in female is blocked; Side effects – may cause infections.

(ii) (a) Fertilized egg/zygote gets implanted in the lining of uterus and starts dividing.
(b) If the egg is not fertilized, the thick and spongy lining of the uterus breaks and comes out through the vagina as blood and mucous.
(B) (i)
Previous Year Questions 2024

(a) Reproductive part – Sporangia  
(b) Non-reproductive part – Hypha/Hyphae.
Dry slice of bread does not provide moisture and nutrients necessary for the germination and multiplication of Rhizopus.
(ii) Budding: Hydra uses regenerative cells for reproduction. A bud develops as an outgrowth due to repeated cell division at one specific site and develop into tiny individuals. On maturation, these buds detach from the parent and become new individuals.
Alternate answer: Regeneration: It is carried out by specialised cells. If hydra is cut or broken into many pieces, many of these pieces grow into separate individuals.


Q2: Identify the mode of asexual reproduction in the following organism:   (1 Mark) (2024)
(a) 
Fragmentation    
(b) Multiple fission
(c) Budding    
(d) Binary fission

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Ans: (c)
In the organism shown, you can observe a small new organism growing from the body of the larger one. This is typical of budding, where a new organism develops from an outgrowth or bud due to cell division at one particular site. This method is common in yeast and some invertebrates like hydras.


Q3: Explain the events that take place once a sperm reaches the oviduct till it becomes a foetus. Write the role of the placenta in pregnancy.  (3 Marks) (2024)

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Ans: 

  • In the oviduct, sperm encounters the egg and fertilisation takes place.
  • The fertilized egg (zygote) starts dividing and forms a ball of cells or embryo.  
  • Embryo is implanted in the lining of the uterus, where it continues to grow and develops organs to become a foetus.
  • Role of Placenta: 
    (i) Provides a large surface area for glucose and oxygen to pass from the mother to the embryo.
    (ii) Waste generated by the embryo will be removed by transferring them into the mother’s blood. 


Q4: Part(s) of a flower which attracts insects for pollination is (are)   (1 Mark) (2024)
(a)
 petals and Sepals
(b) anther and Stigma
(c) petals only
(d) sepals only

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Ans: (c)
Petals are often brightly colored and fragrant, which helps attract insects for pollination. They serve as visual cues that guide insects to the flower, encouraging them to visit and transfer pollen from one flower to another, aiding in plant reproduction.


Q5: (a) (i) What are spores? On which structures are they formed? How do they overcome unfavourable conditions? Name the organism which multiplies with the help of these structures. (4 to 5 Marks) (2024)
(ii) Give two reasons why some plants are grown by the method of vegetative propagation. List two methods used to grow plants vegetatively.
OR
(b) (i) Study the diagram given below and name the parts marked as A, B and C. What happens when B reaches C in the ovary ? Mention its significance.
(ii) Write the post fertilisation changes that occur in a flower.

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Ans: (a) (i)

  • Spores are reproductive structures that detach from the parent and give rise to a new individual.
  • Sporangium / Sporangia
  • Covered by thick walls to protect them from unfavourable conditions.
  • Rhizopus

(ii)

  • Plants which have lost the capacity to produce seeds.
  • Plants bear flowers and fruits earlier so as to reduce time.
  • To get genetically similar plants.

Methods: Layering and Grafting
OR
(b) (i)

  • A – Male Germ Cell/Male Gamete; B – Pollen tube; C – Female Germ Cell / Female Gamete.
  • B carries A (male germ cell) and this germ cell fuses with C (female germ cell) to form a zygote.
  • Significance: Zygote is capable of growing into a new plant.


Q6: The plants that can be raised by the method of vegetative propagation are:   (1 Mark) (2024)
(a)
 Sugarcane, roses, grapes  
(b) Sugarcane, mustard, potato  
(c) Banana, orange, mustard  
(d) Papaya, mustard, potato   

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Ans: (a)
Vegetative propagation is a method of asexual reproduction where new plants are grown from parts of existing plants, such as stems, roots, or leaves. Sugarcane, roses, and grapes can be easily propagated using this method, allowing for the rapid production of new plants with desirable traits.


Q7: A zygote is formed by the fusion of a male gamete and a female gamete. The number of chromosomes in the zygote of a human is:  (1 Mark) (2024) 
(a) 
23  
(b) 44  
(c) 46  
(d) 92 

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Ans: (c)
In humans, a zygote is formed when a male gamete (sperm) and a female gamete (egg) fuse together. Each gamete contains 23 chromosomes, so when they combine, the zygote has a total of 46 chromosomes, which is the normal number for human cells.


Q8: Name the type of nutrition exhibited by Amoeba. Explain how food is taken in and digested by this organism.    (2 Mark) (2024)

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Ans: 

  • Heterotrophic /Holozoic Nutrition   
  • Amoeba takes in food using temporary finger-like projections/pseudopodia of the cell which fuse over the food particle forming a food vacuole. Inside the food vacuole complex substances are broken down into simpler substances. 


Q9: (a) Explain with the help of a labelled diagram, the process of reproduction in Hydra by budding. Name the cells used for reproduction in this process. (4 to 5 Marks) (2024)
OR
(b) List two roles of each of the following in human reproductive system :
(i) Seminal vesicles and prostate gland
(ii) Oviduct  
(iii) Testis 

Hide Answer  

Ans: (a) In hydra, a bud develops as an outgrowth due to repeated cell division at one specific site. These buds develop into tiny individuals and when fully mature, detach from the parent body and become new independent individuals.
Regenerative cells.
Previous Year Questions 2024

OR
(b) (i) Seminal vesicles and prostate glands: 

  • Secrete a fluid for nourishment of sperm. 
  • Secrete a fluid which makes the transport of the sperm easier

(ii)  Oviduct: 

  • Egg is carried from ovary to the womb or uterus.
  • Site of Fertilization

(iii) Testis: 

  • Produces sperm 
  • Secretion of hormone – testosterone


Q10: The incorrect statement about placenta is:   (1 Mark) (2024)
(a)
 It is a disc embedded in the uterine wall.
(b) It contains villi on the embryo’s side of the tissue.
(c) It has a very small surface area for glucose and oxygen to pass from mother to the embryo.
(d) The embryo gets nutrition from the mother’s blood through it.

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Ans: (c)
This statement is incorrect because the placenta actually has a large surface area, which allows for efficient transfer of nutrients like glucose and oxygen from the mother’s blood to the embryo. The placenta is designed to maximize the exchange of these essential substances to support the developing fetus.


Q11: Some unicellular organisms such as Plasmodium and Leishmania differ in the manner in which they reproduce. Name and explain the reproductive process taking place in them.   (1 Mark) (2024)

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Ans: (i) Plasmodium: Multiple fission- A single cell divides into many daughter cells simultaneously.
(ii) Leishmania: Binary fission- Splitting of one cell into two daughter cells in definite orientation.


Q12: Case-based/data-based questions with 3 short  sub-parts. Internal choice is provided in one of these sub-parts.     (4 to 5 Marks)(2024)
Pollination is an important process in sexual reproduction of plants. It is an essential process that facilitates fertilisation in plants. Pollinating agents can be wind, water, insects and birds. Several changes take place in the flower after the fertilization has taken place.
(a) Write the main difference between self-pollination and cross-pollination.
(b) Name the part of the flower which attracts insects for pollination. What happens to this part after fertilisation?
(c) (i) Define fertilisation. What is the fate of ovules and the ovary in a flower after fertilisation?
OR
(c) (ii) In a germinating seed, which parts are known as future shoot and future root? Mention the function of cotyledon.

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Ans: (a)
Previous Year Questions 2024(b) Petals, they dry and fall off.  
(c) (i) Fusion of male and female gametes to form a zygote
Ovule – Seed,
Ovary – fruit        
OR
(c) (ii) Future shoot – Plumule,
Future root – Radicle
Cotyledon – Stores food.


Q13: List two roles of each of the following in human reproductive system:  (3 Marks) (2024)
(i) Seminal vesicles and prostate gland    
(ii) Oviduct  
(iii) Testis 

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Ans: (i) Seminal vesicles and prostate glands: 

  • Secrete a fluid for nourishment of sperm. 
  • Secrete a fluid which makes the transport of the sperm easier 

(ii) Oviduct: 

  • Egg is carried from ovary to the womb or uterus. 
  • Site of Fertilization

(iii) Testis: 

  • Produces sperm 
  • Secretion of hormone – testosterone  


Q14: Chromosomes: 
(i) carry hereditary information from parents to the next generation. 
(ii) are thread-like structures located inside the nucleus of an animal cell. 
(iii) always exist in pairs in human reproductive cells. 
(iv) are involved in the process of cell division.  (2 Marks) (2024)
The correct statements are : 
(a) 
(i) and (ii)
(b) (iii) and (iv)
(c) (i), (ii) and (iv)
(d) (i) and (iv)

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Ans: (c)
(i) carry hereditary information from parents to the next generation: True. Chromosomes carry genetic information from parents to offspring, which determines inherited traits.

(ii) are thread-like structures located inside the nucleus of an animal cell: True. Chromosomes are thread-like structures made of DNA and proteins, located in the nucleus of animal cells, especially visible during cell division.

(iii) always exist in pairs in human reproductive cells: False. Human reproductive cells (sperm and egg) contain only one set of chromosomes (haploid), not pairs. The chromosome pairs are restored during fertilization when sperm and egg combine.

(iv) are involved in the process of cell division: True. Chromosomes ensure that genetic material is accurately copied and distributed to daughter cells during both mitosis and meiosis.

Thus, the correct combination is (i), (ii), and (iv).


Q15: Assertion (A) and Reason (R), answer these questions selecting the appropriate option given below: (1 Mark) (2024)
Assertion (A): Offsprings produced by asexual reproduction are genetically similar to the parents.
Reason (R): Asexual reproduction involves a single parent.
(a) 
Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

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Ans: (a)
Offspring produced by asexual reproduction are genetically identical to their parent because asexual reproduction involves only one parent, resulting in no mixing of genetic material. This is why the assertion about the genetic similarity of offspring and the reason about single-parent involvement are both true and directly related.


Q16: Offsprings formed as a result of sexual reproduction produce more variations because: 
(a) genetic material is contributed by many parents. 
(b) sexual reproduction is a lengthy process. 
(c) genetic material is contributed by two individuals of same species to produce a new generation. 
(d) DNA copying is not accompanied by the creation of cellular apparatus. (1 Mark) (CBSE 2024)

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Ans: (c)
In sexual reproduction, offspring inherit genetic material from two parents, each contributing half of the genetic information. This combination of genetic material introduces genetic variation, as the offspring have a mix of traits from both parents. This variation is essential for the evolution and adaptation of species.
The other options are incorrect:
(a) Genetic material is not contributed by many parents but by two parents.
(b) The length of the process does not directly cause variations.
(d) DNA copying in sexual reproduction is accompanied by cellular processes that support the formation of the zygote.
Therefore, the correct answer is (c) genetic material is contributed by two individuals of the same species to produce a new generation.


Q17: (A) Name any two sexually transmitted diseases. 
(B) Prenatal sex determination is prohibited by law. Why? 
(C) Name any three methods of contraception stating one side-effect of each. (4 to 5 Marks) (CBSE 2024)

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Ans: (A) Gonorrhoea, HIV-AIDS and Human Papillomavirus (HPV). (Any two) 
(B) Prenatal sex determination has been prohibited by law because these are sometimes misused by people who do not want a particular child, as in case of illegal sex-selective abortion of female foetuses. For a healthy society, the female-male sex ratio must be maintained. Due to reckless female foeticides, child sex ratio is declining at an alarming rate. 
(C) (i) Pills: These act by changing the hormonal balance of the body so that eggs are not released and fertilisation cannot occur. 
Side-effects: Causes severe hormonal imbalance. 
(ii) Copper-T or loop: Placed in the uterus to prevent pregnancy. 
Side effect: Causes irritation of the uterus. 
(iii) Surgical methods: Vas deferens in male or fallopian tube in female are blocked to stop the sperm and egg transfer in males and females respectively. 
Side effects: Surgery can cause infections and other problems if not performed properly.

Previous Year Questions 2023

Q1: (i) What happens when:
(1) Leaves of Bryophyllum fall on the soil?
(2) Planaria is cut into many pieces.
(3) Sporangia of Rhizopus on maturation liberate spores?
Mention the modes of reproduction in each of the above three cases.
(ii) Write the changes that occur in a flower once the fertilization has taken place.   (5 Marks) (2023)

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Ans: (i) (1) When the Bryophyllum leaf falls on the wet soil, the buds present in the notches along the leaf margin develop into new plants. This is one of the example of vegetative propagation by leaves.
(2) When Planaria accidentally cut into many pieces then its each piece grows into a complete organism. This is one of the example of regeneration.
(3) The sporangia of Rhizopus contain cells or spores that can eventually develop into new Rhizopus individuals when it bursts on maturation. Rhizopus reproduce asexually by the formation of the spores (sporulation).
(ii) After fertilization, the fertilized egg (or zygote) divides several times to form an embryo within the ovule. The ovule develops a tough coat around it and is gradually converted into a seed. The ovary of flower develops and becomes a fruit (with seeds inside it). The other parts of flower like sepals, petals, stamens, stigma and style dry up and fall off. Only the ovary is left behind. So, at the place on plant where we had a flower originally, we now have a fruit (which is the ovary of the flower containing seeds). A fruit protects its seeds.


Q2: Select the INCORRECT match (between the plant and its vegetative part) from the following: 
(a) Bryophyllum, leaf 
(b) Potato, stem 
(c) Money-plant, stem 
(d) Rose, root  (1 Mark) (CBSE 2023)

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Ans: (d)
(a) Bryophyllum, leaf: Correct match. Bryophyllum reproduces vegetatively through its leaves, which can develop buds that grow into new plants.
(b) Potato, stem: Correct match. Potato reproduces vegetatively through its modified stem, called a tuber.
(c) Money-plant, stem: Correct match. The money plant reproduces through its stem cuttings.
(d) Rose, root: Incorrect match. Rose plants are typically propagated vegetatively through stem cuttings, not roots.
Therefore, the correct answer is (d) Rose, root because it is an incorrect match.


Q3: (A) Name the parts of a bisexual flower that are not directly involved in reproduction. 
(B) Differentiate between self-pollination and cross-pollination. List any two significance of pollination. 
(C) What is the fate of ovules and ovary after fertilisation in a flower?   (4 to 5 Marks) (CBSE 2023)

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Ans: (A) The sepals and petals of a flower are non-essential parts. They specialise in protecting the reproductive organs and draw pollinators like insects and birds. The gynoecium and androecium are the flower’s essential parts that are directly involved in reproduction.
(B)

Previous Year Questions 2023

Significance: 
(i) Plant reproduction and genetic diversity. 
(ii) Food production. 
(C) The ovules develop into the seed and the ovary develops into the fruit after the fertilisation in a flower.

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Previous Year Questions 2022

Q1:  (a) Which of the following flowers will have higher possibility of self-pollination?
Mustard, Papaya, Watermelon, Hibiscus 

(b) List the two reproductive parts of a bisexual flower.   (2022)

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Ans: (a) Mustard and Hibiscus will have higher possibility of self pollination, since these are bisexual flowers i.e., produced on the same plant
(b) The two reproductive parts of a bisexual flower are stamens (male reproductive part) and carpels (female reproductive part).


Q2:  In flowering plants, the pollen grains are transferred to stigma by pollination but the female germ cells are present in the ovary. Explain with the help of a labelled diagram (only concerned parts), how the male germ cells reaches the ovary.   (2022)

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Ans: The pollen particle receives water and nutrients after touching a suitable stigma. It creates a tube known as a pollen tube. Pollen tube develops in the style and travels to the ovary. Two male gametes, or sperm cells, and a tube nucleus may be found at its tip. The advancing pollen tube penetrates the inside of the embryo sac of an ovule, often through the micropyle. The tube ruptures at this point, releasing its two male gametes. To create a zygote, one male gamete unites with one egg.

Previous Year Questions 2022

The action is referred to as syngamy. Endosperm is created when the second male gamete combines with binucleate central cells. Triple fusion is the name given to the process since it includes the fusing of three haploid nuclei. Accordingly, triple fusion is the union of a male gamete with two polar nuclei inside the angiosperm’s embryo sac.


Q3: Give reasons: 
(i) Placenta is extremely essential for fetal development. 
(ii) Uterine lining becomes thick and spongy after fertilisation.    (2022)

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Ans: (i) Placenta is a special tissue by means of which embryo gets nutrition from the mother’s blood. The villi present in placenta provides larger surface area for glucose and oxygen to pass from the mother’s blood to the embryo. Also, the waste generated by the developing embryo is removed by transferring them into the mother’s blood through placenta. Hence, placenta is extremely essential for fetal development.
(ii) The uterine lining becomes thick after fertilisation and is richly supplied with blood to nourish the growing embryo.


Q4: What Is puberty? Mention any two changes that are common to both boys and girls in early teenage years (2022)

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Ans: Puberty is the age of human males and females at which the reproductive organs become functional, gonads start producing gametes and sex hormones, and the boys and the girls become sexually mature. Changes that are common to both boys and girls in early teenage years are as follows :

  • Hair grows in the pubic area and armpits
  • Increase in height and acquisition of muscle mass.


Q5: Name the part/organ of the human female reproductive system     (2022)
(a) where contraceptive devices such as loop or copper-T are placed to prevent pregnancy.
(b) which is blocked to prevent the transfer of eggs.
(c) where formation of green cells as ova takes place.
(d) from where the embryo gets nutrition from the mother’s blood.                  

Hide Answer  

Ans: (a) The part/organ of the human female reproductive system where contraceptive devices such as loop or copper-T are placed to prevent pregnancy is the uterus or womb. These devices are inserted into the uterus to inhibit fertilization and implantation of a fertilized egg.
(b) The part/organ of the human female reproductive system that is blocked to prevent the transfer of eggs is the fallopian tubes. These tubes connect the ovaries to the uterus, and they are blocked or sealed off in certain contraceptive methods or sterilization procedures to prevent the eggs from reaching the uterus for fertilization.
(c) The part/organ of the human female reproductive system where the formation of green cells as ova takes place is the ovaries. Ovaries are responsible for producing and releasing mature eggs, or ova, during the menstrual cycle.
(d) The part/organ of the human female reproductive system from where the embryo gets nutrition from the mother’s blood is the placenta. The placenta is a temporary organ that develops during pregnancy and attaches to the uterine wall. It facilitates the exchange of nutrients, oxygen, and waste products between the mother’s blood and the developing fetus.

Previous Year Questions 2021

Q1: Define fragmentation. (2021 C)

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Ans: Fragmentation is the mode of reproduction in which parent body breaks into two or more fragments and each fragment develops into a new individual. E.g., Spirogyra.


Q2: Mention the changes that occur in the following after fertilisation in a flower:
(a) Petals
(b) Zygote
(c) Ovary
(d) Ovule (Term-11,2021 C)

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Ans: (a) Petals: After fertilization, the petals of the flower may start to wither and fall off. This is because their primary function of attracting pollinators has been fulfilled, and they are no longer needed for reproduction.
(b) Zygote: After fertilization, the zygote is formed. The zygote undergoes mitotic divisions and develops into an embryo. It is the beginning of the new plant’s life cycle.
(c) Ovary: After fertilization, the ovary develops into a fruit. The ovary wall thickens and matures, protecting the developing seeds inside. The fruit helps in the dispersal of seeds and ensures the survival of the next generation.
(d) Ovule: After fertilization, the ovule develops into a seed. The ovule contains the fertilized egg, which develops into an embryo. The ovule undergoes changes in structure and becomes the seed coat, which protects the embryo and provides nutrients for its growth and development.


Q3: Name the reproductive parts of an angiosperm. Where are these parts located? Explain the structure of its male reproductive part. (Term II, 2021)

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Ans: 

  • The reproductive parts of an angiosperm are the flower, which is located at the tip of the stem. The male reproductive part of an angiosperm is called the stamen. The stamen consists of two main parts: the filament and the anther.
  • The filament is a slender stalk that supports the anther. The anther is a sac-like structure located at the top of the filament. Inside the anther, there are pollen sacs that contain pollen grains. These pollen grains are the male gametes, which are responsible for fertilizing the female reproductive cells. The anther releases these pollen grains when they are mature and ready for pollination.
  • Overall, the structure of the male reproductive part of an angiosperm, the stamen, is specialized for the production and release of pollen grains for fertilization.


Q4: (a) Name the reproductive and non-reproductive parts of bread mould (Rhizopus).
(b) List any two advantages of vegetative propagation. (Term II, 2021)

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Ans: (a) Reproductive part of bread mould or Rhizopus— Sporangia, a bob like structure which contain spores. Non-reproductive part – Hyphae, which are thread like structures developed on bread.
(b) Advantages of vegetative propagation:
(i) It produces a large number of plants in shortest time.
(ii) All plants produced are genetically similar to parent, preserves purity, resistance and good qualities.


Q5: (a) Name the process shown below and define it:
(b) Name the type of cells present in the organisms which exhibit this process. (Term II, 2021)

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Ans: (a) The process shown is regeneration in Planaria. Regeneration is the ability to give rise to a new individual from any broken or injured body part.
(b) Regeneration in Planaria is carried out by specialised cells known as neoblasts (stem cells).


Q6: What is vegetative propagation? List with brief explanation three advantages of practising this process for growing some types of plants. Select two plants from the following which are grown by this process: Banana, Wheat, Mustard, Jasmine, Gram (2021)

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Ans: Vegetative propagation is an asexual method of reproduction in plants. In this method, new plants are obtained from the parts of old plants (like stems, roots and leaves), without the help of any reproductive organs.
Advantages of vegetative propagation are as follows: 

  • Vegetative propagation is usually used for the propagation of those plants which produce either very few seeds or do not produce viable seeds.
  • Seedless plants can be obtained by artificial vegetative propagation.
  • Grafting is a propagation method which is very useful for fruit trees and flowering bushes. It enables to combine the most desirable characteristics of two plants. Banana and jasmine are generally grown through vegetative propagation method.

Previous Year Questions 2020

Q1: What are chromosomes? Explain how in sexually reproducing organisms the number of chromosomes in the progeny is maintained.    (2020)

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Ans: ‘Chromosomes’ are long thread-like structures which contain hereditary information of the individual and are thereby the carriers of genes. Chromosomes are located in the nucleus of a cell.
The parents are diploid (2n) as each of them has two sets of chromosomes. They form haploid (In) male and female gametes through the process of meiosis. The haploid gametes have one set of chromosomes. These two gametes fuse during fertilisation and the offspring become diploid (2n) which is same as parents chromosome number.


Q2: (a) What provides nutrition to human sperm? State the genetic constitution of a sperm.
(b) Mention the chromosome pair present in a zygote which determines the sex of (i) a female child and (ii) a male child. (2020)

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Ans: (a) The secretions of seminal vesicles and prostate gland provides nutrition to the human sperm and also make their further transport easier. The genetic constitution of a 50% sperm have X chromosome and 50% have Y chromosome.
(b) (i) X X – Female child
(ii) X Y – Male child


Q3: (a) Name the mode of reproduction of the following organisms and state the important feature of each mode: 
(i) Planaria 
(ii) Hydra 
(iii) Rhizopus 
(b) We can develop new plants from the leaves of Bryophyllum. Comment. 
(c) List two advantages of vegetative propagation over other modes of reproduction.  (2020)

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Ans: (a) (i) Planaria – Regeneration 

  • Regeneration of organism from its cut body parts occurs by the process of growth and development. 
  • Regeneration is an asexual mode of reproduction common in lower plants and animals.

(ii) Hydra – Budding 

  • In budding, a small part of the body of the parent organism grows out as a bud which on detaching from a new organism.

(iii) Rhizopus – Spores Spores are usually produced in sporangia. Spore formation is a common method of an asexual reproduction in bacteria and most of the fungi. 

(b) The leaves of a Bryophyllum have special type of buds in their margins. These buds may get detached from the leaves fall to ground and then grow to produce new Bryophyllum plants. The buds can also drop to the ground together with the leaf and then grow to produce new plants.
(c) Advantages of vegetative propagation are: 

  • The new plants produced by artificial vegetative propagation are exactly like the parent plants.
  • Many plants can be grown from one plant  via vegetative propagation.


Q4: Draw a neat diagram showing fertilisation in a flower and label
(a) pollen tube 
(b) Male germ cell and 
(c) Female germ cell on it. 
Explain the process of fertilisation in a flower. 
What happens to the (i) ovary and (ii) ovule after fertilisation? (2020)

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Ans: Diagram showing fertilisation in a flower is as follows:
Previous Year Questions 2020Fertilisation, in plants, occurs when the male gamete present in pollen grain fuses with the female gamete (or egg) present in ovule. When a pollen grain falls on the stigma of the carpel, it bursts open and grows a pollen tube downwards through the style towards the female gamete in the ovary. Male gametes move down the pollen tube. The pollen tube enters the ovule in the ovary. The tip of pollen tube bursts and male gametes comes out of pollen tube. In ovary, the male gamete of pollen combines with the female gamete or egg present in ovule to form a fertilised egg. After fertilisation,(i) ovule develops into seed
(ii) ovary develops into fruit.


Q5: (a) Identify the modes of asexual reproduction in each of the following organisms:
(i) Hydra
(ii) Planaria
(iii) Amoeba
(iv) Spirogyra
(v) Rhizopus
(b) List three advantages of vegetative propagation.
(c) Why cannot fertilisation take place in flowers if pollination does not occur? (2020)

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Ans: (a) The modes of asexual reproduction in each of the following organisms are:
(i) Hydra – Budding
(ii) Planaria – Fragmentation and Regeneration
(iii) Amoeba – Binary Fission
(iv) Spirogyra – Fragmentation and Asexual spore formation
(v) Rhizopus – Spore formation and Fragmentation
(b) Three advantages of vegetative propagation are:

  • Genetic uniformity: Vegetative propagation produces offspring that are genetically identical to the parent plant. This ensures that desirable traits, such as high yield or disease resistance, are preserved in the offspring.
  • Faster propagation: Vegetative propagation allows for rapid multiplication of plants compared to sexual reproduction, which requires the development of seeds and germination.
  • Preservation of unique characteristics: Some plants have unique and desirable characteristics that cannot be reliably passed on through seeds. Vegetative propagation allows for the preservation and multiplication of these characteristics.

(c) Fertilisation cannot take place in flowers if pollination does not occur because pollination is the transfer of pollen grains from the anther to the stigma of a flower. The pollen grains contain the male gametes (sperm cells), while the stigma contains the female gametes (egg cells). Fertilisation can only occur when the sperm cells reach the egg cells. Without pollination, there is no transfer of pollen to the stigma, and thus, the sperm and egg cells cannot meet for fertilisation to take place.


Q6: (a) In the female reproductive system of human beings, state the functions of:
(i) Ovary 

(ii) Oviduct.
(b) Mention the changes which the uterus undergoes, when
(i) it has to receive a zygote.
(ii) no fertilisation takes place.
(c) State the functions of placenta.   (2020)

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Ans: (a) (i) The ovaries in female are primary sex organs (or female gonads) which perform the dual function – production of female gametes (eggs or ova) and secretion of female sex hormones (estrogen and progesterone).
(ii) Oviducts or fallopian tube are paired tubes originating near the ovaries of their respective sides and extend upto uterus. The terminal part of fallopian tube is funnel-shaped with finger-like projections called fimbriae lying near ovary. Fimbriae pick up the ovum released from ovary and push it into fallopian tube. Fertilisation also takes place in the oviduct.
(b) (i) As the ovary releases one egg every month, the uterus also prepares itself, every month to receive fertilised egg by making its lining thick and spongy to nourish the zygote if fertilisation takes place.
(ii) When the female gamete/egg is not fertilised, this lining is not needed any longer. So, the lining slowly breaks and comes out through vagina as blood and mucus through menstrual cycle that takes place every month.
(c) Placenta performs the following functions: 
(i) All nutritive elements from maternal blood pass into the fetus through it.
(ii) Placenta helps in respiration i.e., supply of oxygen and removal of CO2 from fetus to maternal blood.
(iii) Fetal excretory products diffuse out into maternal blood through placenta and are excreted by mother.
(iv) Placenta also secretes various hormones during pregnancy.


Q7: (a) What is puberty?
(b) Describe in brief the functions of the following parts in the human male reproductive system.
(i) Testes
(ii) Seminal vesicle
(iii) Vas deferens
(iv) Urethra
(c) Why are testes located outside the abdominal cavity?
(d) State how sperm move towards the female germ cell.   (2020)

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Ans: (a) The age at which the sex hormones begin to be produce and the boy and girl becomes sexually mature, i.e., able to reproduce is called puberty.
(b) (i) Testes: The two testes in male are the sites where male gametes, i.e., sperm are formed. Testes also produce the male sex hormone called testosterone.
(ii) Seminal vesicles: Seminal vesicles are one pair of sac-like structures near the base of bladder. Seminal fluid is a watery alkaline fluid that contains nutrients (fructose) which serve as a source of energy for the sperm. Each seminal vesicle releases its contents into the ejaculatory duct during ejaculation.
(iii) Vas deferens: This is a straight tube, about 40 cm long, which carries the sperm to the seminal vesicles, where mucus and a watery alkaline fluid containing fructose, mix with the sperm. (iv) Urethra : It is a long tube that arises from urinary bladder. Urethra carries urine from the bladder as well as sperm from the vas deferens, through the penis.
(c) Testes are located outside the abdominal cavity because sperm formation requires a lower temperature than normal body temperature. The temperature of the testes in the scrotum is about 2-2.5°C lower than normal body temperature. This temperature is ideal for sperm formation and development.
(d) The sperm present in the testes of man are introduced into the vagina of the woman through penis during copulation. Millions of sperm are released into the vagina at one time. The sperm are highly active and mobile. They travel from here upward through the uterus at the top of fallopian tube within five minutes.


Q8: Based on the given diagram answer the questions given below:
(a) Label the parts A, B, C and D.(b) Name the hormone secreted by testis and mention its role.(c) State the functions of B and C in the process of reproduction. (2020)

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Ans: (a) A-Ureter
B – Seminal vesicle
C – Urethra
D – Vas deferens
(b) Testes produce male sex hormone testosterone Hormone testosterone brings about the development of secondary sexual characters during puberty in boys like growth of facial hair, deepening of voice, build up of muscle mass and also regulates formation of sperm.
(c) Seminal vesicles (B) release its contents into the ejaculatory duct during ejaculation. Urethra (C) carries sperm from the vas deferens through the penis.


Q9: The growing size of the human population is a cause of concern for all people. The rate of birth and death in a given population will determine its size. Reproduction is the process by which organisms increase their population. The process of sexual maturation for reproduction is gradual and takes place while general body growth is still going on. Some degree of sexual maturation does not necessarily mean that the mind or body is ready for sexual acts or for having and bringing up children. Various contraceptive devices are being used by human beings to control the size of population.
(i) List two common signs of sexual maturation in boys and girls.
(ii) What is the result of reckless female Foeticide?
(iii) Which contraceptive method changes the hormonal balance of the body?
(iv) Write two factors that determine the size of a population. (2020)

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Ans: (i) Two common signs of sexual maturation in boys and girls are:
(a) Growth of pubic hair and extra hair in the armpits.
(b) Development of oily skin and pimples.
(ii) Female Foeticide is reducing the number of girls drastically in our country, due to which male-female sex ratio is also declining.
(iii) Chemical contraceptive method changes the hormonal balance of the body.
(iv) The rate of birth and death in a given population will determine the size of a population.


Q10: (a) List three different categories of contraceptive methods.
(b) Why has Government of India prohibited prenatal sex determination by law? State its benefits in the long run.
(c) Unsafe sexual act can lead to various infections. Name two bacterial and two viral infections caused due to unsafe sex. (2020)

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Ans: (a) Three different categories of contraceptive methods are:
(i) Barrier methods, i.e., use of condoms, etc.
(ii) Oral contraceptive methods, i.e., use of oral pills etc.
(iii) Surgical methods, i.e., vasectomy and tubectomy.
(b) Prenatal sex determination was banned in India in 1994. This was done to prevent sex selective abortion. It is being used to kill the normal female fetus. This killing of the unborn girl child is called female feticide which is reducing the number of girls drastically in some societies of our country. Due to reckless female feticide, male-female sex ratio is declining at an alarming rate. Its benefit in the long run is that the female-male ratio could be maintained for a healthy society.
(c) Bacterial diseases due to unsafe sex are gonorrhoea and syphilis. Viral diseases due to unsafe sex are AIDS and genital herpes.

Also watch: Sexual Reproduction In Flowering Plants

Previous Year Questions 2019

Q1: After examining a prepared slide under the high power of a compound microscope, a student concludes that the given slide shows the various stages of binary fission in a unicellular organism. Write two observations on the basis of which such a conclusion may be drawn. (2019)

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Ans: The two observations that was taken by the student that concludes that the given slide shows the various stages of binary fission in a unicellular organism are as follows:

  • Presence of unicellular and uninucleate organism showing irregular outline.
  • Division of nucleus of parent cell into two equal halves.


Q2: Define multiple fission. Give its one example. (2019, Foreign 2014)

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Ans: Multiple fission is an asexual mode of reproduction in which the parent organism splits to form many new organisms at the same time. Multiple fission occurs in Plasmodium.


Q3: Draw labelled diagram to show the following parts in an embryo of a pea seed: Cotyledon, Plumule, Radicle (2019)

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Ans: Embryo of pea seed is shown as follows:
Previous Year Questions 2019


Q4: What are testes? List two functions performed by testes in human beings. (2019 C)

Hide Answer  

Ans: Testes are the primary reproductive organs in males. Testes produce male gametes (sperm) and male sex hormones (testosterone)


Q5: What are sexually transmitted diseases? List two examples each of diseases caused due to (i) bacterial infection and (ii) viral infection. Which device or devices may be used to prevent the spread of such diseases? (2019)

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Ans: The diseases that are spread by sexual contact with an infected person are called sexually transmitted disease (STDs).
(i) Bacterial infection causes gonorrhoea, syphilis.
(ii) Viral infection causes AIDS, genital herpes. STDs can be prevented by using male and female condoms.


Q6: Define pollination. Explain the different types of pollination. List two agents of pollination. How does suitable pollination lead to fertilisation? (Delhi 2019)

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Ans: The process of transfer of pollen grains from anther of a flower to the stigma of the same flower or another flower of the same species is known as pollination. Pollination may be of two major types- (i) self pollination and (ii) cross pollination.

  • Self pollination is the transfer of pollen grains from the anther to the stigma of the same flower, or to the stigma of another flower of the same plant. This pollination generally takes place in bisexual flowers because they have both male and female gametes in them.
  • Cross pollination is the transfer of pollen grains from the anther of a flower of one plant to the stigma of a flower of another plant of the same species. This occurs in unisexual as well as bisexual flowers. Two agents of pollination are wind and water. Pollination results in the deposition of related pollen grains over the receptive stigma of the carpel. Pollen grains after landing on stigma, absorb water, swell and then germinate to produce pollen tubes.
    Many pollen tubes grow into the stigma, but only one passes through the style and then moves towards the ovary. Two non- motile male gametes are formed inside the tube during its growth through the style. After reaching the ovary, pollen tube enters the ovule through the micropyle. The tip of the tube finally pierces the micropylar end of the embryo sac. After penetration, the tip of pollen tube ruptures releasing two male gametes into the embryo sac. The mature embryo sac consists of an egg apparatus (one haploid egg and two synergids), two polar nuclei and three antipodal cells. During the act of fertilisation, one male gamete fuses with the egg to form the diploid zygote.


Q7: (a) Draw a diagram of human female reproductive system and label the parts: 
(i) which produce an egg
(ii) where fertilisation takes place
(b) List two bacterial diseases which are transmitted sexually. 
(c) What are contraceptive devices? Give two reasons for adopting contraceptive devices in humans. (Al 2019)

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Ans:  (a) The sectional view of human female reproductive system is as follows:
Previous Year Questions 2019(b) Gonorrhoea and syphilis are two bacterial diseases which are transmitted sexually.
(c) Contraceptive devices are those devices which are used to prevent pregnancy. It includes diaphragm, condom and intrauterine devices. Contraceptive methods are adopted:
(i) to avoid unwanted birth.
(ii) to keep the population of a country under control.


Q8: (a) Identify the given diagram. Name the parts 1 to 5
(b) What is contraception? List three advantages of adopting contraceptive measures. (NCERT Exemplar, Delhi 2019)

Hide Answer  

Ans: (a) The given diagram is the sectional view of human female reproductive system. The labelled parts are:
1. Funnel of fallopian tube or oviduct
2. Ovary
3. Uterus or womb
4. Cervix
5. Vagina
(b) Contraception is the avoidance of pregnancy. Three advantages of adopting contraceptive methods are:
(i) They prevent frequent or unwanted pregnancies.
(ii) They prevent the transfer of sexually transmitted infections.
(iii) They help to regulate the population growth.

Previous Year Questions 2018

Q1: Write one difference between asexual and sexual mode of reproduction. Which species is likely to have comparatively better chances of survival – the one reproducing asexuaily or the one reproducing sexually? Give reason to justify your answer. (CBSE 2018)

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Ans: Differences between asexual and sexual mode of reproduction.
Previous Year Questions 2018The species reproducing sexually will have better chances of survival because genetic variation is created during sexual reproduction; in case of an adverse environmental change, atleast some variants will survive and continue the race.


Q2: (a) Write the functions of the following parts in human female reproductive system:
(i) Ovary
(ii) Oviduct
(iii) Uterus
(b) Describe the structure and function of placenta.    (CBSE 2018)

Hide Answer  

Ans: (a) (i) Ovary

  • It produces the female gametes or germ cells, called ova.
  • It secretes the female sex hormones such as oestrogen and progesterone.

(ii) Oviduct

  • It transports the ova from the ovary to uterus/womb.
  • Fertilisation occurs in the oviduct.

(iii) Uterus

  • Implantation of the embryo occurs in the lining of uterus and; the complete development of foetus occurs here.
  • the contractions of the muscles of uterus help in child birth.

(b) Structure of placenta:

  • Placenta is a disc-like structure embedded in the uterine wall.
  • It contains villi on the embryo’s side and on the mother’s side there are blood spaces, which surround the villi; this arrangement provides a large surface area for exchange of materials.

Functions of placenta: 

  • It transfers glucose and oxygen from the mother’s blood to the foetus.
  • It also removes the wastes (CO2 and nitrogenous wastes) generated by the foetus to the mother’s blood.
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Previous Year Questions 2017

Q1: Newly formed DNA copies may not be identical at times. Give one reason.  (NCERT,AI 2017)

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Ans: When a cell reproduces, DNA replication occurs which results in formation of two similar copies of DNA. The process of copying the DNA leads to some variations each time. As a result, the DNA copies produced are similar to each other but sometimes may not identical.


Q2: When a cell reproduces, what happens to its DNA? (AI 2017)

Hide Answer  

Ans: When a cell reproduces, DNA replication occurs  which forms two similar copies of DNA.


Q3: Name the method by which Spirogyra reproduces under favourable conditions. Is this method sexual or asexual? (Delhi 2017)

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Ans: The method by which Spirogyra reproduces under favorable conditions is fragmentation. This is an asexual mode of reproduction.


Q4: How does Plasmodium reproduce? Is this method sexual or asexual? (Delhi 2017)

Hide Answer  

Ans: Plasmodium reproduces through multiple fission method. In this method, the parent organism splits to form many new organisms at the same time. This is an asexual method of reproduction.


Q5: Reproduction is one of the most important characteristic of living beings. Give three reasons in support of the statement.  (Al 2017)

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Ans: Reproduction is one of the most important characteristics of living beings because:
(i) it is essential for existence and continuity of a species.
(ii) it helps to pass genetic information to next generation.
(iii) it brings variations in next generation which is the basis for evolution.


Q6: State the basic requirement for sexual reproduction. Write the importance of such reproductions in nature.    (Delhi 2017)

Hide Answer  

Ans: The basic requirement for sexual reproduction is involvement of both sexes, i.e., male and female, to produce an offspring. It takes place by the combination of gametes which come from two different parents.
The importance of sexual reproduction in nature are:
(i) Fusion of male and female gametes coming from two different and sexually distinct individuals, exhibit diversity of characters in offspring.
(ii) Meiosis during gametogenesis provides opportunities for new combination of genes, which leads to variation required for evolution and plays a prominent role in the origin of new species. Variations lead to the appearance of such characters, which fit to the changing environment, resulting in the survival of the species.


Q7: List any two steps involved in sexual reproduction and write its two advantages. (Delhi 2017)

Hide Answer  

Ans: The two main steps involved in sexual reproduction are:
(i) formation of male and female gametes.
(ii) fusion of a male gamete with a female gamete to form a new cell called zygote by the process of fertilisation.
The two important advantages of sexual reproduction are:
(i) It promotes diversity of characters in the offspring through genetic variations.
(ii) It plays an important role in continuous evolution of better organisms that may lead to the origin of new species.


Q8: State the changes that take place in the uterus when:
(a) Implantation of embryo has occurred.
(b) Female gamete/egg is not fertilised. (Delhi 2017)

Hide Answer  

Ans: (a) Implantation is the close attachment of the blastocyst (young multicellular embryo) to the uterine wall. It is followed by a number of developmental changes in the thickened wall of uterus. An intimate connection between the fetal membrane and the uterine wall called placenta is formed. This is a disc which is embedded in the uterine wall. The placenta serves as the nutritive, respiratory and excretory organ of the fetus.
(b) When the female gamete/egg is not fertilised, this lining is not needed any longer. So, the lining slowly breaks and comes out through vagina as blood and mucus. This cycle takes place every month and is known as menstrual cycle.


Q9: List three techniques that have been developed to prevent pregnancy. Which one of these techniques is not meant for males? How does the use of these techniques have a direct impact on the health and prosperity of a family? (NCERT Exemplar, Al 2017)

Hide Answer  

Ans: Methods developed to prevent pregnancy are:
(i) barrier method, i.e., use of condoms, diaphragm, etc.
(ii) oral contraceptive method, i.e., use of oral pills.
(iii) surgical method, i.e., vasectomy and tubectomy. Out of these methods, chemical method is not meant for males.
Use of these techniques help to keep control over number of children in a family, which directly affects prosperity of a family. One of the most common reason for deterioration of women’s health is frequent conception. Controlled childbirth will directly affect women health and this will indirectly affect the prosperity of family and nation.


Q10: What happens when
(a) accidently, Planaria gets cut into many pieces
(b) Bryophyllum leaf falls on the wet soil
(c) on maturation sporangia of Rhizopus bursts? (NCERT Exemplar, Delhi 2017)

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Ans: (a) When Planaria accidently gets cut into many pieces then its each piece grows into a complete  organism. This is known as regeneration.
(b) When the Bryophyllum leaf falls on the wet soil, the buds present in the notches along the leaf margin develop into new plants. This is known as vegetative propagation.
(c) The sporangia of Rhizopus contain cells or spores that can eventually develop into new Rhizopus individuals when it bursts on maturation.


Q11: Describe reproduction by spores in Rhizopus. (Al 2017)

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Ans: Fungus Rhizopus reproduces by spore formation. During the growth of Rhizopus, small rounded, bulb-like structures develop at the top of the erect hyphae. Such structures are called sporangia. Inside each sporangium, : nucleus divides several times. Each nucleus gets surrounded by a little amount of cytoplasm to become spore. Large number of spores are formed inside each sporangium. After sometime sporangium bursts and spores are released in the air. When these spores land on food or soil, under favourable conditions, they germinate into new individuals.


Q12: What is vegetative propagation? State two advantages and two disadvantages of this method. (Al 2017)

Hide Answer  

Ans: Vegetative propagation is a type of asexual reproduction in which the plant parts other than seeds are used as a propagule. 

Advantages of vegetative propagation :

  • Desirable character of the plant can be preserved through generation.
  • Seedless plants can be grown via this method.

Disadvantages of vegetative propagation:

  • Plants produced by this method posses less vigour and are more prone to diseases.
  • Plants produced by this method show no genetic variation.


Q13: (a) Name the organ that produces sperm as well as secretes a hormone in human males. Name the hormone it secretes and write its functions. 
(b) Name the parts of the human female reproductive system where fertilisation occurs. 
(c) Explain how the developing embryo gets nourishment inside the mother’s body.     (Delhi 2017)

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Ans: (a) The male organ is testis. It secretes the hormone testosterone  which regulates the formation of sperm. — It brings about changes in the appearance of boys at the time of puberty. 
(b) Fertilisation occurs in the oviduct. 
(c) The developing embryo gets nourishment from the mother’s blood with the help of a special tissue, called placenta. The placenta provides a large surface area for the passage of glucose and oxygen from the mother’s blood to the embryo.


Q14: (a) What is variation ? How is variation created in a population ? How does the creation of variation in a species promote survival ?
(b) Explain how, offspring and parents of organisms reproducing sexually have the same number of chromosomes. (CBSE 2017-18 C)

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Ans: (a) Variation is occurrence of differences between organisms.
Variations are created in a population in different ways :
(i) There may be minor changes/errors during DNA copying mechanism which happens before any cell division. These variations may go on accumulating from previous generations, ultimately leading to visible changes.
(ii) In sexually reproducing organisms, the traits of two individuals combine and give rise to new combination in the progeny. This also leads to variation.
(iii) Thus combining variations from two or more individuals would thus create new combinations of variations.
Organisms with suitable variations will have better chances of survival. Depending on the nature of variation, different individuals would have different kind of advantages.
(b) In sexually reproducing organisms each cell has two copies of each chromosome, one each from the male and female parent. During gamete formation, one chromosome from each pair goes to a gamete. Hence, the gametes have half the number of chromosomes, but one chromosome of each pair. When two gametes combine, they restore the normal number of chromosome in the progeny.


Q15: Sneha was taught by her teacher that ‘‘Variation is useful for the survival of species.’’ She passed on the same information to her friend, Abdul. Support the view of both Sneha and her teacher by giving a suitable justification for the same. (CBSE 2017)

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Ans: Variations that are favourable, increase the chances of survival of the species. If an organism can withstand a higher temperature, then the variation goes on accumulating in its future generations. Hence, these organisms can survive sudden rise in the temperature. This ensures the survival of the species. But other organisms (variants) without this variation may not survive due to sudden rise in temperature. So, variation is beneficial to the species, but not a necessity for an individual.

Previous Year Questions 2016

Q1: Name the part of Bryophyllum where the buds are produced for vegetative propagation. (Delhi 2016)

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Ans: Bryophyllum propagates vegetatively by the buds produced at the margins of leaves.


Q2: What happens when a mature Spirogyra filament attains considerable length? (Al 2016)

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Ans: When a mature Spirogyra filament attains considerable length it simply breaks into two or more fragments and each fragment, then grows into a new Spirogyra.


Q3: What are all organisms called which bear both the sex organs in the same individual? Give one example of such organism. (Al 2016)

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Ans: Organisms which bear both male and female sex organs in the same individual are called bisexual. For example, Hibiscus.


Q4: List two unisexual flowers. (Foreign 2016)

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Ans: Flowers of papaya and cucumber are unisexual.


Q5: Why is fertilisation not possible without pollination? (NCERT Exemplar, Foreign 2016)

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Ans: The process of pollination (in plants) ensures that male gametes bearing structure called pollen comes in contact with the female reproductive structure of the plant. Once the male and female gametes are in close vicinity, they fuse and fertilisation is accomplished. Hence, fertilisation cannot take place without pollination.


Q6: List two functions of ovary of human female reproductive system. (AI 2016)

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Ans: Two functions of ovary of human female are:
(i) production of female gametes, i.e., ova
(ii) secretion of female hormones, i.e., estrogen and progesterone.


Q7: Define reproduction. How does it helps in providing stability to the population of species? (NCERT Exemplar, Al 2016)

Hide Answer  

Ans: The production of new organisms by the existing organisms of the same species is known as reproduction. It is linked to the stability of population of a species. DNA replication during reproduction ensures transfer of specific characters or body design features that is essential for an individual of a population to live and use that particular niche. Some variations present in a few individuals of population caused due to reproduction which also helps in their survival at changing niches.


Q8: What is multiple fission? How does it occur in an organism? Explain briefly. Name one organism which exhibits this type of reproduction. (Delhi 2016)

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Ans: Multiple fission refers to the process of asexual reproduction in which many individuals are formed from a single parent. This method of reproduction occurs in unfavourable conditions. The unicellular organism develops a protective covering called cyst, over the cell. The nucleus of the cell divides repeatedly producing many nuclei. Later on, each nucleus is surrounded by small amount of cytoplasm and many daughter cells are produced within the cyst. When conditions are favourable, the cyst breaks and small offspring are liberated. This type of reproduction is seen in some protozoans, e.g., malarial parasite (Plasmodium).


Q9: Explain the term “regeneration” as used in relation to reproduction of organisms. Describe briefly how regeneration is carried out in multicellular organisms like Hydra. (AI 2016)

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Ans: The process of formation of entire organism from the body parts of a fully differentiated organism is called regeneration. It occurs by process of growth and development. Simple animal like Hydra shows regeneration. When a small piece of Hydra breaks off it grows into complete new Hydra. During regeneration, the cells of cut body part of the organism divide rapidly to make a mass of cells. The cells here move to their proper places within the mass where they have to form different types of tissues. In this way complete organism is regenerated.


Q10: In the context of reproduction of species state the main difference between fission and fragmentation. Also give one example of each. (Al 2016)

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Ans: The main differences between fission and fragmentation are as follows:Previous Year Questions 2016


Q11: What happens when 
(a) Planaria gets cut into two pieces 
(b) a mature Spirogyra filament attains considerable length 
(c) on maturation sporangia burst?  (Foreign 2016, Delhi 2016)

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Ans: (a) When Planaria is cut into two pieces then each piece grows into a complete organism. This is known as regeneration.
(b) When a mature Spirogyra filament attains a considerable length it breaks into small pieces called fragments. These fragments grow into new individuals and this mode of reproduction is called fragmentation.
(c) When a sporangia burst, large number of spores are released in the air. When these spores land on food or soil, under favourable conditions they germinate into new individuals.


Q12: List and explain briefly any three methods of contraception. (CBSE 2016-17 Cl)

Hide Answer  

Ans: Three methods of contraception :
(i) Mechanical Barrier – Condoms on penis or vagina is a covering worn so as to create a barrier such that sperm do not reach the egg.
(ii) Oral Contraceptive Pills – They change the hormonal balance o f the body so that eggs are not released and fertilisation does not take place.
(iii) Intra-Uterine Device – Devices like copper-T or loop are placed in the uterus which change the pH of uterus and hence sperm cannot survive, thereby preventing pregnancy.
(iv) Surgical Method – Blocking vas deferens in males and fallopian tube in females, would allow respective gametes to be released and hence act as contraceptive.


Q13: How do organisms, whether reproduced asexually or sexually maintain a constant chromosome number through several generations? Explain with the help of a suitable example. (CBSE 2016)

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Ans: When an organism reproduces sexually, it produces gametes through a special type of division called meiosis or reductional division, in which the original number of chromosome becomes half. In sexual reproduction, male gametes and female gametes are combined to form zygote, and the original number of chromosomes is restored.
Previous Year Questions 2016

In asexual reproduction, only mitotic divisions are involved and the chromosome number remains the same.
Previous Year Questions 2016

Example: Humans have 46 chromosomes or 23 pairs of chromosomes. When the gametes formed through meiosis, the chromosomes number becomes half. Thus, an organism maintains a constant chromosome number through several generations.

Previous Year Questions 2015

Q1: Name the life process of an organism that helps in the growth of its population. (AI 2015)

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Ans: Reproduction is a life process that helps in multiplication of an organism and growth of its population.


Q2: Name two simple organisms having the ability of regeneration. (Al 2015)

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Ans: Hydra and Planaria are two organisms that have the ability to regenerate.


Q3:  Name the causative agent of the disease “Kala-azar” and its mode of asexual reproduction. (Foreign2015)

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Ans: Causative agent of the disease Kala-azar is Leishmania. It reproduces asexually by binary fission.


Q4: Why is DNA copying an essential part of the process of reproduction? What are the advantages of sexual reproduction over asexual reproduction? (2015)

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Ans: DNA copying is an essential part of the process of reproduction as it results in passing of nearly same genetic information from parents to the off springs. DNA replication also ensures that same number of chromosomes are passed from parents to offspring. Advantages of sexual reproduction over asexual reproduction is that sexual reproduction provides variations which is a major factor for evolution that helps in survival of species in changing environment.


Q5: Draw a diagram of the longitudinal section of a flower exhibiting germination of pollen on stigma and label (i) ovary, (ii) male germ cell, (iii) female germ cell and (iv) ovule on it. (2015)

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Ans. The diagram of the longitudinal section of flower is as follows:
Previous Year Questions 2015


Q6: The diagram shown depicts the process of pollination. Choose the options that will show a maximum variation in the offspring. 

(a) A, B and C 
(b) B and D 
(c) B, C and D 
(d) A and C (CBSE 2015, 11)

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Ans: (b)
In the context of pollination, maximum variation in the offspring is achieved through cross-pollination, which involves the transfer of pollen between two different plants. Cross-pollination increases genetic diversity compared to self-pollination (where pollen from the same plant fertilizes its own flowers).
From the diagram:

(i) B and D indicate cross-pollination between different plants, which would result in greater genetic variation in the offspring.
(ii) A and C or other combinations involving the same plant would indicate self-pollination, which results in less variation.
Thus, (b) B and D is the correct option as it will show maximum variation in the offspring.

Also watch: Sexual Reproduction In Flowering Plants

Previous Year Questions 2014

Q1: No two individuals are absolutely alike in a population. Why?   (Delhi 2014)

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Ans: No two individuals are absolutely alike in a population because sexual reproduction promotes diversity of characters in the offspring by providing genetic variation.


Q2: Describe in brief the function of the various parts of the female reproductive part of a bisexual flower.   (2014)

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Ans: A carpel is made of three parts : stigma, style and ovary. The top part of carpel is called stigma. Stigma is for receiving the pollen grains during pollination. Stigma is sticky so that pollen can stick to it. The middle part of carpel is called style. Style is a tube which connects stigma to the ovary. The swollen part at the bottom of a carpel is called ovary. The ovary contains ovules. Ovules contain the female gametes or female sex cells (egg) of the plant. There are usually many ovules in the ovary.


Q3: Name the two reproductive parts of a bisexual flower which contain the germ cells. State the location and function of its female reproductive part.   (2014)

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Ans: The two reproductive parts of a bisexual flower which contain the germ cells are carpel (female reproductive part) and stamen (male reproductive part). Carpel is situated in the centre of the flower as a flask-shaped structure. A carpel is made up of three parts-stigma, style and ovary. The distal part of a carpel is called stigma. Stigma is responsible for receiving pollen during pollination. Style is an elongated tubular structure which connects stigma with ovary. The basal swollen part of carpel is ovary. Ovary bears several ovules. After fertilisation, ovules form seeds and ovary forms the fruit.

Previous Year Questions 2011

Q1: Name the method by which Spirogyra reproduces under favourable conditions. Is this method sexual or asexual? (CBSE 2011)

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Ans: The method by which Spirogyra reproduces under favourable conditions is fragmentation. It is an asexual mode of reproduction.

06. Previous Year Questions: Control & Coordination

Previous Year Questions 2025

Q1: The plant hormone whose concentration stimulates the cells to grow longer on the side of the shoot which is away from light is:  (1 Mark)
(a) 
Cytokinins
(b) Gibberellins
(c) Adrenaline
(d) Auxins

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Ans: (d) Auxins

Explanation: Auxins collect on the shaded side of the shoot. This increases cell elongation on that side, so the shoot bends towards light.

Q2: The plant hormone present in greater concentration in the areas of rapidly dividing cells is:  (1 Mark)
(a) Auxin
(b) Cytokinins
(c) Gibberellins
(d) Abscisic acid

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Ans: (b) Cytokinins

Explanation: Cytokinins are concentrated in areas of rapid cell division, like fruits and seeds. They promote cell division.

Q3: Question consist of two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.  (1 Mark)

Assertion (a): In our actions of writing or talking, our nervous system communicates with the muscles.
Reason (R): Cranial nerves and spinal nerves form the peripheral nervous system.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

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Ans: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation: Our actions of writing or talking involve the nervous system communicating with muscles (true). Cranial and spinal nerves do form the peripheral nervous system (true), but this does not explain communication with muscles.

Q4: The growth of the pollen tubes towards ovules is an example of: (1 Mark)
(a) Phototropism
(b) Hydrotropism
(c) Geotropism
(d) Chemotropism

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Ans: (d) Chemotropism

Explanation:  Growth of pollen tubes towards ovules happens due to chemical attraction from the ovule. Hence, it is chemotropism.

Q5: The plant hormones promoting rapid cell division in seeds and wilting of leaves respectively are: (1 Mark)
(a) Auxins and Abscisic acid
(b) Cytokinins and Abscisic acid
(c) Gibberellins and Auxins
(d) Abscisic acid and Gibberellins

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Ans: (b) Cytokinins and Abscisic acid

Explanation: Cytokinins promote rapid cell division (like in seeds). Abscisic acid promotes wilting of leaves and dormancy.


Q6: Electrical impulse travels in a neuron from:  (1 Mark)
(a) Nerve ending → Axon → Cell body → Dendrite
(b) Dendrite → Cell body → Axon → Nerve ending
(c) Cell body → Dendrite → Axon → Nerve ending
(d) Dendrite → Axon → Nerve ending → Cell body

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Ans: (b) Dendrite → Cell body → Axon → Nerve ending

Previous Year Questions 2025

Q7: Study the given diagram and write the type of movement exhibited by:  (2 Marks)
(a) Root, and
(b) Shoot, mentioning the stimulus in each case.

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Ans:
(a) Root: Geotropism (positive); Stimulus: Gravity.
(b) Shoot: Phototropism (positive); Stimulus: Light.

  • Root (Geotropism): Roots exhibit positive geotropism, growing downward in response to gravity. This is mediated by auxins, which accumulate on the lower side of the root, inhibiting elongation there and causing downward growth.
  • Shoot (Phototropism): Shoots show positive phototropism, growing toward light. Auxins accumulate on the shaded side, promoting cell elongation and causing the shoot to bend toward light.

Mechanism:

  • In roots, gravity causes auxin redistribution, directing growth downward to anchor the plant and access water.
  • In shoots, light triggers auxin movement to the shaded side, enhancing growth toward light for photosynthesis.

Significance: These tropic movements optimize plant survival by ensuring roots access resources and shoots maximize light capture.


Q8: A person is making a list to purchase few things from a nearby market. Explain how the fore-brain plays an important role in performing this activity.  (2 Marks)Ans: 

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Ans: The fore-brain is the main thinking part of the brain. When a person makes a list to purchase things, the fore-brain receives sensory information, interprets it, thinks about the needs, and helps in decision making. It enables reasoning, memory, and voluntary action like writing the list.

Q9: (a) How is brain protected in our body?
(b) A doctor finds in one of his patients that he is not maintaining a proper posture and balance of his body. State the region of brain and also the part of brain which is responsible for it. (2 Marks)

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Ans: (a) The brain is protected by the skull and cerebrospinal fluid (CSF).

Brain Protection:

  • Skull: The bony cranium provides a hard protective barrier against physical injury.
  • Cerebrospinal Fluid (CSF): Surrounds the brain, absorbing shocks and maintaining pressure stability.

(b) The patient is not maintaining posture and balance → controlled by the hind-brain. The part responsible is the cerebellum.

Q10: Plants have neither a nervous system nor muscles, even then they respond to stimuli. For example, the leaves of chhui-mui (touch-me-not) plant when touched begin to fold up and droop.
(a) How is the information communicated in “touch-me-not” plants?
(b) What enables the plant cells to bring out the observable response?
(c) Differentiate the movement mentioned above from the movement of tendrils in a pea plant.  (3 Marks)

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Ans: (a)

  • Plants like chhui-mui (Mimosa pudica) do not have a nervous system, but when touched, the stimulus is passed from cell to cell through electrical-chemical means.
  • This is how information about the stimulus is transmitted in the plant body.

(b)

  • The signal causes a change in water movement in the plant cells.
  • As water leaves the cells, they lose their turgidity (swollen state).
  • The sudden loss of turgidity makes the leaves fold up and droop, which is the visible response.

(c)Previous Year Questions 2025


Q11: A hormone ‘X’ is secreted in blood when a person is under scary situation.  (3 Marks)
(a) Identify the hormone ‘X’ and the gland that secretes it.
(b) Explain its role in dealing with scary or emergency situations.

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Ans: (a)

  • The hormone ‘X’ is Adrenaline.
  • It is secreted by the Adrenal glands (located above the kidneys).

(b) Role in Scary Situations

  • Adrenaline is called the emergency hormone because it prepares the body to face stress, fear, or danger.
  • It increases the supply of oxygen and glucose to the brain and muscles so that the body can respond quickly.
  • It causes:
    1. Faster heartbeat and breathing
    2. Increased blood pressure
    3. More blood flow to skeletal muscles
    4. Less blood supply to skin and digestive system

This prepares the body for “fight or flight” – i.e., to either face the situation or run away.

Q12: (a) Define hormone.
(b) “Hormones should be secreted in precise quantities. We have a feedback mechanism through which this is done.” With the help of an example justify the statement. (3 Marks)

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Ans: (a) Hormones are chemical messengers secreted by endocrine glands. They are poured directly into the blood and act on specific target organs to regulate various body functions.
(b) Feedback Mechanism (Example: Thyroxin): T

  • Hormones should be secreted in precise quantities.
  • If there is too much or too little hormone, it can disturb body functions.
  • To control this, the body has a feedback mechanism.

Example – Regulation of Thyroxin:

  • The thyroid gland secretes thyroxin, which regulates metabolism of carbohydrates, proteins, and fats.
  • The secretion of thyroxin is controlled by TSH (Thyroid Stimulating Hormone) from the pituitary gland.
  • If thyroxin level in blood becomes high, the pituitary stops releasing TSH → secretion of thyroxin decreases.
  • If thyroxin level becomes low, the pituitary releases more TSH → secretion of thyroxin increases.

Thus, by feedback mechanism, the timing and amount of hormone secretion are kept under control.

Q13: (a) What is meant by the reflex arc? Where are they formed in the human body?
(b) Why have reflex arcs evolved in animals? (3 Marks)

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Ans: (a) Reflex arc:

  • reflex arc is the pathway through which nerve impulses pass during a reflex action.
  • It consists of receptor → sensory neuron → spinal cord → motor neuron → effector (muscle/gland).
  • Reflex arcs are formed in the spinal cord of the human body.Previous Year Questions 2025

Reflex arcs have evolved in animals:

  • Reflex arcs have evolved because they enable quick, automatic, and involuntary responses to harmful stimuli.
  • This helps protect the body from injury without waiting for instructions from the brain.
  • Example: Immediately withdrawing hand from a hot object.

Q14: State two limitations of electrical impulses in multicellular organisms. Why is chemical communication better than electrical impulses as a means of communication between cells in multicellular organisms?  (3 Marks)

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Ans: In multicellular organisms, electrical impulses have certain limitations:

  1. The electrical impulse will reach only those cells that are connected by the nervous tissue, not every cell of the body.
  2. After an electrical impulse is generated in a cell and transmitted, the cell needs some time to reset before it can generate another impulse.

Because of these limitations, electrical impulses alone cannot perform all communication functions.

Hence, chemical communication through hormones is better because:

  • Hormones are secreted directly into the blood and can reach all cells of the body, not just nerve-connected ones.
  • The effect of hormones, though slower, is widespread and longer lasting compared to short-lived electrical impulses.

Q15: When a seed germinates, the root grows downwards and a small shoot grows upward. This shoot is known as:  (1 Mark)
(a) 
Radicle
(b) Stem
(c) Cotyledon
(d) Plumule

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Ans: (d) Plumule
Explanation:

When a seed germinates:

  • The root grows downward due to geotropism → called the radicle.
  • The small shoot grows upward due to phototropism → called the plumule.

Q16: (a) Analyse the given situations and interpret the possible reasons for each:  (5 Marks)
(i) Iodine deficiency in diet increases the possibility of a disease of swollen neck in a person.
(ii) Some people in the population may have very short heights (dwarfs).
(iii) Thick facial hairs develop in boys at the age of 10–12 years.
(b) Explain two reasons that necessitate the need for chemical communication in multicellular organisms.

OR
(b) (i) Differentiate between voluntary and involuntary action.
(ii) Define reflex action. With the help of a flow diagram, show the correct sequence of path of nerve impulse from place of its origin.

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Ans:
(a) Situations and Reasons:

(i) Iodine deficiency in diet causes goitre (swollen neck) because iodine is necessary for the production of thyroxin by the thyroid gland.

(ii) Some people have very short heights (dwarfs) because of deficiency of growth hormone, secreted by the pituitary gland.

(iii) Thick facial hairs develop in boys around age 10–12 due to secretion of testosterone hormone by the testes during puberty.
(b) Need for Chemical Communication:

  • Electrical impulses cannot reach every cell in the body — hormones (chemical messengers) can reach all cells through blood.
  • The effect of hormones is longer-lasting, whereas electrical impulses are short-lived.

OR (b) (i)
Previous Year Questions 2025(ii)A reflex action is a quick, automatic, and involuntary response to a stimulus. It is controlled by the spinal cord and helps protect the body from sudden harmful changes in the environment.
Example: Withdrawing the hand immediately on touching a hot object.
Previous Year Questions 2025

Stimulus (Receptor in skin) → Sensory neuron → Spinal cord (Interneuron/Relay neuron) → Motor neuron → Effector (muscle/gland)

Q17: A person, while climbing up a rocky hill, comes into a panic state and fear. His body starts reacting in a “fight-or-flight” condition to adjust to the dangerous and stressful situation.
(a) (i) Name the hormone secreted in the blood of the person in this situation.  (2 Marks)
OR
(a) (ii) Name the source gland of the hormone secreted in this condition. (2 Marks)
(b) State any two responses in the body of the person as a result of the secretion of this hormone. (1 Mark)
(c) How does the action of the chemical signal in terms of hormones differ from the electrical impulses via nerve cells? (1 Mark)

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Ans:

(a) (i) The hormone secreted in this condition is Adrenaline.
OR
(a) (ii) The source gland is the Adrenal glands (located above the kidneys).
(b) Two responses in the body due to adrenaline secretion:

  1. Increases heartbeat and blood pressure, supplying more oxygen to muscles.
  2. Increases breathing rateand diverts more blood to skeletal muscles, preparing the body for “fight-or-flight”.

(c) Difference between hormonal action and electrical impulses:

Previous Year Questions 2025

Q18: In life there are certain changes in the environment called ‘stimuli’ to which we respond appropriately. Touching a flame suddenly is a dangerous situation for us. One way is to think consciously about the possibility of burning and then moving the hand. But our body has been designed in such a way that we save ourself from such situations immediately.
(i) Name the action by which we protect ourself in the situation mentioned above and define it.  (1 Mark)
(ii) Write the role of (a) motor and (b) relay neuron. (1 Mark)
(iii) (a) What are the two types of nervous system in the human body? Name the components of each of them. (2 Mark)
OR
(iii) (b) Which part of the human brain is responsible for: (1 Mark)
(a) thinking
(b) picking up a pencil
(c) controlling blood pressure
(d) controlling hunger

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Ans: (i) Action: Reflex action
Definition: Reflex action is an automatic, immediate, and involuntary response to a stimulus without the involvement of the conscious part of the brain.

Example: Withdrawing the hand instantly when touching a hot flame.
(ii) Role of neurons:

  • (a) Motor neuron: It carries the message from the spinal cord to the effector organ (like muscles or glands) to produce a response.
  • (b) Relay neuron (interneuron): It connects the sensory neuron to the motor neuron inside the spinal cord and helps in transmitting the impulse.

(iii) (a) Types of Nervous System and Components:

Central Nervous System (CNS):

  • Components: Brain and spinal cord
  • Function: Processes information and controls responses

Peripheral Nervous System (PNS):

  • Components: Nerves arising from the brain (cranial nerves) and spinal cord (spinal nerves)
  • Function: Connects CNS to the rest of the body

OR
(iii) (b) Brain Parts:

(a) Thinking → Cerebrum

(b) Picking up a pencil → Cerebellum

(c) Controlling blood pressure → Medulla oblongata

(d) Controlling hunger → Separate part of fore-brain

Previous Year Questions 2024

Q1: In a nerve cell, the site where the electrical impulse is converted into a chemical signal is known as:  (2024)
(a)
 Axon
(b) Dendrites
(c) Neuromuscular junction
(d) Cell body

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Ans: (c)
This is the place where the electrical impulse traveling along a nerve cell is converted into a chemical signal to communicate with a muscle cell. At this junction, neurotransmitters are released, allowing the nerve cell to signal the muscle to contract.


Q2: (A) How is the movement of the leaves of a sensitive plant different from the downward movement of the roots?  (2024)
OR
(B) There is a hormone which regulates carbohydrate, protein and fat metabolism in our body. Name the hormone and the gland which secretes it. Why is it important for us to have iodised salt in our diet?

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Ans: (A) 
Previous Year Questions 2024OR
(B)

  • The hormone that regulates carbohydrate, protein, and fat metabolism is thyroxine.
  • It is secreted by the thyroid gland.
  • Iodine is crucial for the thyroid gland to produce thyroxine.
  • A lack of iodine can lead to a condition called goitre, which causes swelling in the neck.


Q3: Define reflex action. With the help of a flow chart show the path of a reflex action such as sneezing.  (2024)

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Ans: Reflex action is a quick and automatic response to a stimulus in the environment. An example of this is sneezing.

The process of a reflex action can be summarised as follows:

  • Stimulus: An irritant, such as dust, enters the nose.
  • Receptors: Nerve endings in the nasal passages detect the irritant.
  • Signal Transmission: A sensory neuron sends a signal to the spinal cord.
  • Reflex Arc: The signal is processed in the spinal cord, and a response is sent through a motor neuron.
  • Effector: The muscles in the diaphragm and chest contract, causing a sneeze.
Previous Year Questions 2024


Q4: Select out of the following a gland which does NOT occur as a pair in the human body :  (2024)
(a) 
Pituitary    
(b) Ovary
(c) Testis    
(d) Adrenal

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Ans: (a)

  • The pituitary gland is a single gland located at the base of the brain.
  • It is often called the “master gland” because it regulates many other hormone glands in the body.
  • In contrast, the ovaries, testes, and adrenal glands exist in pairs, one on each side of the body.


Q5: (a) (i) Distinguish between hormonal co-ordination in plants and animals. (2024)
(ii) Which part of the brain is responsible for —
(1) Intelligence
(2) riding a bicycle
(3) vomiting
(4) controlling hunger
(iii) How is the brain and spinal cord protected against mechanical injuries?
OR
(b) (i) What are tropic movements? Give an example of a plant hormone that (1) inhibits growth and (2) promotes cell division.
(ii) Explain the directional movement of a tendril in a pea plant in response to touch. Name the hormone responsible for this movement.

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Ans: (i)
Previous Year Questions 2024

(ii) (1) Cerebrum/forebrain,
(2) cerebellum i.e part of hindbrain
(3) medulla in hindbrain
(4) Separate part of forebrain.
(iii) Brain – Bony box/skull/cranium/fluid filled balloon in skull, Spinal cord – Backbone/Vertebral column.
OR
(b) (i) Plant growth movements in response to stimuli in a particular direction/directional movements due to light, gravity etc.
(1) Plant growth inhibitor: Abscisic Acid
(2) Promotes cell division: Cytokinins
(ii)  When the tendrils come in contact with any support, auxins move away from the point of contact of the support. More growth occurs on the side away from the support. As a result, unequal growth occurs on its two sides and thus tendrils coil/ circle around the support.
Auxins


Q6: Select from the following a plant hormone which promotes cell division.  (CBSE 2024)
(a) 
Gibberellins
(b) Auxins
(c) Abscisic Acid
(d) Cytokinins

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Ans: (d)

  • Cytokinins are plant hormones that promote cell division.
  • They stimulate the growth of shoots and leaves.
  • Cytokinins play a vital role in:
    • Fruit development
    • Delaying aging in plant tissues


Q7: A plant growth inhibitor hormone that causes wilting of leaves is  called:    (2024)
(a) 
Auxin  
(b) Cytokinin  
(c) Abscisic acid  
(d) Gibberellin  

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Ans: (c)

Abscisic acid is a plant hormone that inhibits growth and plays a crucial role in how plants respond to stress, such as drought. Its main effects include:

  • Causing the stomata to close, which reduces water loss.
  • Leading to the wilting of leaves when the plant is under stress.

This hormone is essential for helping plants manage their resources during challenging conditions.


Q8: Where are auxins synthesized? How do they promote phototropism?  (CBSE 2024)

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Ans: Auxins are synthesized primarily at the shoot tip and root tip of plants.

They promote phototropism through the following process:

  • When light shines on one side of the plant, auxins move towards the shady side.
  • This increased concentration of auxins causes the cells on the shady side to grow longer.
  • As a result, the plant bends towards the light.


Q9: (a) (i) Define a reflex arc. Why have reflex arcs evolved in animals? Trace the sequence of events which occur, when you suddenly touch a hot object. (2024)
(ii) Name the part of the nervous system which helps in communication between the central nervous system and other parts of the body. What are the two components of this system?
OR
(b) (i) Leaves of ‘chhui-mui’ plant begin to fold up and droop in response to a stimulus. Name the stimulus and write the cause for such a rapid movement. Is there any growth involved in the movement?
(ii) Define geotropism in plants. What is meant by positive and negative geotropism? Give one example of each type.

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Ans: (a) (i) 

  • Reflex arc: The pathway through which impulses travel during a reflex action.
  • Evolution: Reflex arcs have developed in animals because the brain’s thinking processes are not quick enough for immediate responses, which helps avoid injury.
  • Sequence of events when touching a hot object:
    • Stimulus detected by receptors in the skin.
    • Impulse travels through the sensory neuron to the spinal cord.
    • Relay neuron transmits the impulse to the motor neuron.
    • Motor neuron sends a message to the muscle, causing a quick withdrawal from the hot object.
      Previous Year Questions 2024

(ii) Peripheral Nervous System Components : Cranial Nerves; Spinal Nerves
OR
(b)  (i) 

  • Stimulus: Touch
  • The leaves fold due to changes in water content.
  • No growth is involved in this rapid movement.

(ii)  Geotropism: The growth of a plant part in response to gravity.

  • Positive geotropism: Movement towards gravity (e.g., roots growing downwards).
  • Negative geotropism: Movement away from gravity (e.g., shoots growing upwards).


Q10: The correct sequence of events when someone’s hand touches a hot object unconsciously:  (2024)
(a)
 Receptors in skin → Motor neuron → Relay neuron → Sensory neuron → Effector muscle in arm
(b) Receptors in skin → Relay neuron → Sensory neuron → Motor neuron → Effector muscle in arm
(c) Receptors in skin → Sensory neuron → Relay neuron → Motor neuron → Effector muscle in arm
(d) Receptors in skin → Sensory neuron → Effector muscle in arm → Motor neuron → Relay neuron

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Ans: (c)
Receptors in skin → Sensory neuron → Relay neuron → Motor neuron → Effector muscle in arm. 
When you touch something hot, the receptors in your skin detect the heat and send a signal through sensory neurons to the relay neurons in the spinal cord. Then, relay neurons send the signal to motor neurons, which activate the effector muscles in your arm to pull away quickly, helping you avoid injury.


Q11: Sense organ in which olfactory receptors are present is:  (2024)
(a) 
Nose  
(b) Skin  
(c) Tongue  
(d) Inner ear

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Ans: (a)
Olfactory receptors, which are responsible for the sense of smell, are located in the nasal cavity of the nose. When you inhale, these receptors detect different scents and send signals to the brain, allowing you to identify various smells.


Q12: Taking the example of any two animal hormones along with their gland of secretion, explain how these hormones help (i) in growth and development and (ii) regulate metabolism, in the body.    (2024)

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Ans: (i)  Growth and Development:

  • Growth Hormone: Secreted by the pituitary gland.
  • Stimulates growth in all organs, promoting overall development.

(ii)  Regulation of Metabolism:

  • Thyroxin: Produced by the thyroid gland.
  • Regulates the metabolism of carbohydrates, proteins, and fats, essential for healthy growth.
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Previous Year Questions 2023

Q1: The sensory nerve of a reflex arc carries information from the receptor cells to the  (2023)
(a) spinal cord
(b) brain
(c) muscles of the effector organ
(d) bones of the receptor organ               

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Ans: (a)
Sol: A Reflex arc is the pathway that nerve impulses follow during a reflex action. It involves:

  • Information travelling from receptor organs (like the skin) to the spinal cord.
  • Then, signals are sent from the spinal cord to effector organs (like muscles) to produce a response.


Q2: Name the part of the brain which is responsible for the following actions:   (2023)
(i) Maintaining posture and balance
(ii) Beating of heart
(iii) Thinking
(iv) Blood pressure    

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Ans: 

 (i) Cerebellum: Responsible for maintaining posture and balance.

(ii) Medulla oblongata: Controls the beating of the heart.

(iii) Cerebrum: Involved in thinking processes.

(iv) Medulla oblongata: Regulates blood pressure.


Q3: (a) How are the brain and spinal cord protected in human beings?  
(b) State one main function of
(i) Medulla and
(ii) Cerebellum. (2023)

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Ans: (a) Brain and spinal cord are protected by skeleton- brain by the cranium and spinal cord by vertebral column,
(b) (i) Medulla – It controls involuntary actions like heart beat, respiration, etc.
(ii) Cerebellum – It maintains posture and balance of body


Q4: (a) Where are auxins synthesised in a plant? 
(b) Which organ of the plant shows:
(i) Positive phototropism
(ii) Negative geotropism
(iii) Positive hydrotropism       (2023)

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Ans: Auxin is synthesised at tip of the stem apices or shoots.
(i) Shoots
(ii) Shoots
(iii) Roots (bend towards water)


Q5: Name a plant hormone responsible for bending of a shoot of a plant when it is exposed to unidirectional light. How does it promote phototropism?      (2023)

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Ans: 

  • Auxin is responsible for bending of a shoot of a plant when it is exposed to unidirectional light.
  • In the presence of sunlight, the auxin present in the stem starts accumulating in the region away from sunlight, due to which the concentration of auxin in the region of the stem increases, resulting in bending of a shoot of a plant when it is exposed to unidirectional light.
  • Thus, auxin promotes phototropism that is the regulation of physiology or development in response to day length.


Q6: Select from the following the correct statement about tropic movement in plants. 
(a) It is due to the stimulus of touch and temperature. 
(b) It does not depend upon the direction of stimulus received. 
(c) It is observed only in roots and not in stem. 
(d) It is a growth-related movement. (CBSE 2023)

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Ans: (d)

  • Tropic movements in plants are growth-related movements that occur in response to specific directional stimuli, such as light, gravity, water, or chemicals.
  • These movements are typically slow and involve growth in a particular direction, either towards (positive tropism) or away from (negative tropism) the stimulus.
  • Therefore, the correct answer is (d) It is a growth-related movement.


Q7: (A) Write the role of insulin in regulating blood sugar levels in the human body. Mention the disease caused due to it. 
(B) How is the timing and the amount of release of insulin in the blood regulated? (CBSE 2023)

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Ans: (A) 

  • Glucose enters the blood stream as soon as meals containing carbohydrates is digested by the body.
  • As a result, the body’s blood glucose levels rise.
  • In order to direct all of the body’s cells to take in the glucose, the pancreas releases chemical signals in the form of insulin.
  • The majority of this glucose is consumed by the cells as energy.
  • The liver and muscle cells take the extra glucose in the bloodstream and convert it to glycogen for later usage.
  • In the end, insulin secretion reduces blood sugar levels.
  • Diabetes is either caused due to the insufficient production of insulin by the pancreas or it may occur if the cells of the body are not effectively responding to the insulin produced.

(B) 

  • Hormones are chemicals that our body secretes in a precise quantity and at a certain time. 
  • The release of hormones and the timing of their release are controlled by particular regulatory centres in our bodies.
  • For instance:
    •  Regulation of insulin release – the pancreas secretes insulin, which regulates blood glucose levels. 
    • The higher blood glucose concentration affects how much insulin is produced.
    • When blood glucose levels are really high, the pancreas releases more insulin.

Previous Year Questions 2020

Q1: Define geotropism. Draw a labelled diagram of a plant showing the geotropic movement of its parts. (2020)

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Ans: Geotropism is the growth of plant parts in response to gravity. It can be classified into two types:

  • Positive geotropism: This occurs when roots grow downwards, following the direction of gravity.
  • Negative geotropism: This happens when shoots grow upwards, against the pull of gravity.
Previous Year Questions 2020


Q2: A squirrel is in a scary situation. Its body has to prepare for either fighting or running away. State the immediate changes that take place in its body so that the squirrel is able to either fight or run. (2020)

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Ans: When a squirrel finds itself in a scary situation, its body undergoes several immediate changes to prepare for either fighting or running away. These changes are triggered by the nervous system, which stimulates the adrenal glands to release more adrenaline into the bloodstream. The key changes include:

  • Increased heartbeat: This helps pump more blood to the muscles.
  • Faster breathing: This increases oxygen intake.
  • Enhanced blood flow: Blood is redirected to the muscles, improving their performance.
  • More glucose in the blood: The liver releases stored glucose, providing extra energy.

These responses collectively prepare the squirrel’s body for quick action, whether it needs to fight or flee.


Q3: Why is chemical communication better than electrical impulses as a means of communication between cells in a multicellular organisms? (2020)

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Ans: 

  • In animals, the message communicated in the form of nerve impulses, from receptors to central nervous system and from effectors is very quick.
  • But nerve impulses can reach only those animal cells which are connected by the nervous tissue.
  • These cells after generation and transmission of nerve impulses, take some time to reset their mechanism before a new impulse is generated and transmitted.
  • It means, cells cannot continuously generate and transmit electrical impulses.
  • This is the reason most multicellular organisms use another means of communication called chemical communication.
  • In chemical communication, information spreads out throughout the body by blood and its effects lasts longer.
  • Chemical communication is however slow, but it can reach all the cells of body regardless of nervous connections.


Q4: A cheetah, on seeing a prey moves towards him at a very high speed. What causes the movement of his muscles? How does the chemistry of cellular components of muscles change during this event? (2020)

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Ans: The cheetah’s movement towards its prey is initiated by its ability to sense the prey through photoreceptors. This information is transmitted to the central nervous system via neurons. The hormonal system also plays a crucial role in this process.

When the cheetah spots its prey, the adrenal glands release the hormone adrenaline into the bloodstream. This triggers several physiological responses:

  • Increased heart rate: This helps pump more blood throughout the body.
  • Enhanced breathing: This increases oxygen intake.
  • Improved blood flow: Blood is directed to the leg muscles, providing them with the necessary energy.
  • Release of glucose: The liver releases stored glucose into the bloodstream, supplying energy for rapid movement.

These combined actions of adrenaline generate a significant amount of energy, enabling the cheetah to run at high speeds.


Q5: What constitutes the central and peripheral nervous systems? How are the components of central nervous system protected? Which signals will get disrupted in case of a spinal cord injury? (2020 C)

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Ans: Central nervous system consists of brain and spinal cord. It contains centres for controlling various activities of the body. Peripheral nervous system consists of cranial nerves and spinal nerves. Components of the central nervous system are protected by:
(i) Brain is contained within the skull while the spinal cord is encircled by a series of vertebrae.
(ii) Meninges are a succession of tissue layers that reside within these bony structures.
(iii) Cerebrospinal fluid (CSF) is a clear, colourless, slightly alkaline fluid that surrounds the brain and spinal cord that protects from infections.
The following signals will get disrupted in case of a spinal cord injury:

  • Reflex action 
  • Impulses from various body parts will not be conducted to brain. 
  • Message from brain will not be conducted to various organs of the body.


Q6: (a) Name the hormone secreted by (i) Pituitary, and (ii) Thyroid stating one main function of each. Name the disorder a person is likely to suffer from due to the deficiency of the above mentioned hormones.
(b) How is the timing and amount of hormone released regulated ? Explain with an example. (2020 C)

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Ans: (a) (i)  

  • Pituitary secretes hormones GH, TSH, ACTH, FSH, LH, PRL, MSH, oxytocin and vasopressin. 
  • Growth hormone (GH) or somatotropic hormone controls the overall development of body, muscles, bones and tissues. 
  •  Disorder related to the thyroid gland: Goitre (enlarged thyroid gland) can occur due to inadequate dietary intake of iodine which results in low level of thyroid hormones (since iodine is essential to make thyroid hormones). 
  •  Low levels of thyroid hormones stimulate the pituitary gland to secrete more TSH, which causes thyroid gland enlargement. 
  • Hyposecretion of growth hormone (GH) during growth years results in dwarfism. 
  •  This could happen due to the hyperactivity of the pituitary gland. 

(ii)

  •  The thyroid gland secretes three hormones: thyroxine (T4), triiodothyronine (T3), and calcitonin. 
  • Thyroxine and triiodothyronine maintain the basal metabolic rate (BMR) of the body by regulating the rate of oxidation of carbohydrates, fats, and proteins and production of energy in our body. 
  •  They promote growth of body tissues and development of mental faculties. 

(b) 

  • Both the timing and amount of hormone released are regulated by feedback mechanism.
  • For example, the pancreas is involved in the secretion and release of the hormone insulin when the positive feedback is signalled by the increased blood glucose level.
  • The insulin sequesters the glucose in adipose tissue and liver cells as glucose and fats respectively.
  • When the blood glucose level decreases, the negative feedback will signal the pancreas to stop the secretion of insulin.


Q7: (a) Name one organ each where growth hormone is synthesized in man and plant.
(b) List the sequence of events that occur when a plant is exposed to unidirectional light, leading to bending of a growing shoot. Also name the hormone and the type of movement. (2020)

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Ans: (a) In humans, growth hormone is synthesized in the pituitary gland. In plants, growth hormone is synthesized in the apical meristem.
(b) When a plant is exposed to unidirectional light, the following sequence of events occur leading to bending of a growing shoot:
1. Photoreceptors in the shoot tip detect the direction and intensity of light.
2. Auxin, a plant hormone, is produced in higher concentrations on the shaded side of the shoot tip.
3. The higher concentration of auxin promotes cell elongation on the shaded side, causing the shoot to bend towards the light source.
4. The bending of the shoot towards the light is known as phototropism.
5. The bending helps the plant optimize its exposure to light, allowing for efficient photosynthesis.


Q8: Write in tabular form the location and function of the hormones secreted by each of the following glands present in the human body: (A) Pituitary gland 
(B) Thyroid gland 
(C) Pancreas (CBSE 2020)

Ans: (A) Pituitary Gland

  • Location: At the bottom of the hypothalamus at the base of the brain.
  • Function: Secretes growth hormone, which regulates the growth and development of the body.

(B) Thyroid Gland

  • Location: In the neck region.
  • Function: Secretes thyroxine hormone, which regulates the metabolism of carbohydrates, proteins, and fats in the body.

(C) Pancreas

  • Location: Just below the stomach.
  • Function: Secretes insulin hormone, which lowers the blood sugar level.
Also read: Diagram Based Questions: Control and Coordination

Previous Year Questions 2019

Q1: What is a nerve impulse? State the direction followed by a nerve impulse while travelling in the body of an organism. (2019)

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Ans: A nerve impulse is a gradual change in the nerve fibre’s membrane that occurs after it is stimulated. This impulse travels in the following direction:

  • From the dendrite to the cell body
  • Along the axon to its end

At the end of the axon, the impulse triggers the release of chemicals that cross the synapse, starting a new electrical impulse in the next neuron.


Q2: Draw a diagram of neuron and name and label the part
(a) where information is acquired,
(b) through which information travels as an electric impulse, and
(c) where the electric impulse must be converted into a chemical signal for onward transmission. (2019 C)

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Ans: Diagrammatic representation of a neuron:

(a) The part where information is acquired is the dendrite.

(b) Information travels as an electrical impulse through the axon.

(c) The electric impulse is converted into a chemical signal at the synapse.

Previous Year Questions 2019


Q3: Why does the flow of signals in a synapse from axonal end of one neuron to dendritic end of another neuron take place but not in the reverse direction? Explain. (NCERT.AI 2019)

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Ans: At the synapse (functional junction between neurons), axon terminal comes in close proximity to the dendron terminal of next neuron. Axon terminal is expanded to form pre-synaptic knob and the other dendrite terminal forms post-synaptic depression. In between the two, lies a narrow fluid filled space called synaptic cleft. As the nerve impulse reaches the pre-synaptic knob, the synaptic vesicles get stimulated to release neurotransmitter in the synaptic cleft. The neurotransmitter molecules diffuse across the gap to come in contact with post-synaptic membrane. In this way, nerve impulse passes across the minute gap to stimulate dendron of another neuron. The synapse acts as a one-way valve to conduct impulse in one direction only. This is so because chemical substance called neurotransmitter is secreted only on one side of the gap, i.s., on axons side. It carries impulse across the synapse and passes it to the dendron of the other neuron. In this way, impulses travel across the neurons only in one direction, i.e., from axon of one neuron to dendron of other neuron through a synapse.


Q4: List in tabular form three distinguishing features between cerebrum and cerebellum. (2019)

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Ans: Distinguishing features between cerebrum and cerebellum:Previous Year Questions 2019


Q5: How do auxins promote the growth of a tendril around a support? (2019)

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Ans: 

  • When tendrils touch a support, the part in contact grows more slowly than the part away from it. This is due to the action of the auxin hormone.
  • The side of the tendril touching the support has less auxin.
  • The free side has more auxin, which promotes growth.
  • This difference in growth causes the tendril to coil around the support.


Q6: State the function of each of the following plant hormones:
(a) Gibberellins
(b) Auxins
(c) Abscisic acid   (2019 C)

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Ans: (a) In genetically dwarf plants, the length of internode is very much reduced but the number of nodes remains the same. Gibberellins overcome the phenotypic expression of dwarfism in certain plants. Besides general increase in stem length, gibberellins specifically induce internodal growth in some genetically dwarf varieties of plants like pea and maize. Gibberellins, however, have little or no effect when they are applied to the normal plant.
(b) Auxin is the plant hormone which promote cell enlargement and cell differentiation in plants. It also plays a role in apical dominance (i.e., the phenomenon in which presence of apical bud does not allow the nearby lateral buds to grow). When the apical bud is removed, the lateral buds sprout. This produces dense bushy growth. The phenomenon is widely used in tea plucking. Apical bud inhibits the growth of lateral buds by release auxin.
(c) Abscisic acid or ABA is a plant hormone or phytohormone which acts as growth inhibitor. It promotes dormancy in seeds and buds, abscission (falling of leaves) and senescence in leaves and thus inhibits growth of plant.


Q7: What are plant hormones? Name the plant hormones responsible for the following:
(i) Growth of stem
(ii) Promotion of cell division
(iii) Inhibition of growth
(iv) Elongation of cells   (NCERT, Delhi 2019)

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Ans: Plant hormones or phytohormones are chemical substances produced naturally in plants and capable of translocation and regulating one or more physiological processes when present in low concentration. These are also known as plant growth substances or plant growth regulators.
The plant hormones responsible for different functions are as follows:
(i) Growth of stem: Gibberellins (Gibberellic acid) promote growth in stems.
(ii) Promotion of cell division: Cytokinins promote cell division in plants.
(iii) Inhibition of growth: ABA (Abscisic acid) promotes dormancy in seeds as well as in buds and thus inhibits growth.
(iv) Elongation of cells: Auxin and cytokinin both cause cell elongation.

Previous Year Questions 2018

Q1: (a) What are sensory and motor neurons? Write their function.    (CBSE 2018C)
(b) Different parts of brain are associated with specific functions. Name the part of human brain which perform the following functions:
(i) Sensation of feeling full
(ii) Vomiting

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Ans: (a) Two types of neurons are:
(i) Sensory neurons – They transmit information from the receptors to the central nervous system.
(ii) Motor neurons – They transmit information from the brain to the effector organs.
(b) 

(i) Forebrain (a centre for hunger)
(ii) Hindbrain (medulla)

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Previous Year Questions 2017

Q1: (a) Name the part of human brain which controls:  (2017)
(i) voluntary actions (ii) involuntary actions.
(b) State the significance of peripheral nervous system. Name the components of this nervous system and distinguish between their origins.

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Ans: (a) (i) Voluntary actions are controlled by the cerebrum (the forebrain).
(ii) Involuntary actions are controlled by mid-brain (the medulla oblongata).
(b) Peripheral nervous system facilitates the communication between central nervous system and other body parts.
Components of peripheral nervous system are:
(i) Cranial Nerves: Originating from brain.
(ii) Spinal Nerves: Originating from spinal cord.


Q2: (a) Name the gland that secrete:
(i) insulin, (ii) thyroxin.
(b) Explain with an example how the timing and amount of hormone secreted are regulated in a human body.  (2017)

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Ans: (a) Gland that secretes:
(i) Insulin – Pancreas
(ii) Thyroxine – Thyroid gland
(b) The timing and amount of hormone secreted are regulated by the “Feedback mechanisms”. For example:
(i) High glucose level in blood induces the pancreatic cells to produce more insulin which converts glucose to glycogen.
(ii) Low glucose level in the blood does not induce the pancreatic cells to produce insulin so that less conversion of glucose to glycogen occurs.


Q3: (a) State the role performed b y plant hormones. Name a plant hormone which is essential for cell division.
(b) Name and explain the role of plant hormone involved in phototropism.  (2017)

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Ans: (a) Role of plant hormones:
Chemical control in the plants is performed through plant hormones. They help to coordinate growth, development and responses to the environment.
Plant hormones auxin and cytokinin help in cell division.
(b) In phototropism, the hormone auxin plays a crucial role. It is produced at the shoot tip and helps cells grow longer. When a plant senses light from one side, auxin moves to the shaded side, causing those cells to elongate. This results in the plant bending towards the light.


Q4: What is geotropism? Draw a labelled diagram of a potted plant showing positive geotropism and negative geotropism  (2017)

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Ans: Geotropism refers to the directional growth of plants in response to gravity.

  • Positive geotropism: This is when roots grow downwards towards the Earth’s centre.
  • Negative geotropism: This is when shoots grow upwards away from the ground.
Previous Year Questions 2017


Q5: How does the feedback mechanism regulate hormone secretion? Explain with the help of an example.  (2017)

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Ans: The feedback mechanism is a regulatory system that controls hormone secretion based on their levels in the body. This process can either promote or inhibit the production of hormones.

An example of this is the regulation of thyroxine production:

  • If blood levels of thyroxine are high, the hypothalamus detects this and reduces the production of thyroid-stimulating hormone (TSH).
  • With less TSH, the thyroid gland produces less thyroxine, leading to a decrease in its blood levels.
  • Conversely, if thyroxine levels are low, the hypothalamus increases TSH production.
  • This TSH travels through the bloodstream to the thyroid, prompting it to secrete more thyroxine.


Q6: State the source of secretion and function of the following hormones:  (2017)
(i) Thyroxin 
(ii) Insulin 
(iii) Growth hormone.

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Ans: Previous Year Questions 2017


Q7: What are trophic levels? Make a food chain of four trophic levels.  (2017)

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Ans: Trophic levels refer to each step in a food chain where energy is transferred. 
They illustrate how energy moves through an ecosystem
Example: (i) Plants → Grasshopper Frog → Snake
(ii) Plants Deer → Hyena Tiger


Q8: Different parts of brain are associated with specific functions. Name the part of human brain which performs the following functions:    (CBSE 2017-18 C)
(a) Sensation of feeling full
(b) Vomiting
(c) Picking up a pencil
(d) Riding a bicycle.

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Ans: (a) Hypothalamus – Forebrain
(b) Medulla – Hind brain
(c) Cerebellum – Hind brain
(d) Cerebellum – Hind brain


Q9: (a) What are phytohormones? List four types of phytohormones. Where are these hormones synthesised?
(b) What happens when a growing plant detects light? Explain in brief. (Board Term I, 2017)

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Ans: (a) Phytohormones are chemical substances that regulate various physiological processes in plants. They are produced naturally by plants and control growth, development, and responses to environmental stimuli. 
Four types of phytohormones are:

  • Auxins: These hormones regulate cell elongation, apical dominance, and phototropism. They are mainly synthesized in the apical meristem, young leaves, and developing seeds.
  • Gibberellins: These hormones promote stem elongation, seed germination, and fruit development. They are primarily synthesized in the apical meristem, young leaves, and developing seeds.
  • Cytokinins: These hormones promote cell division and differentiation, delay senescence, and regulate apical dominance. They are synthesized in actively growing tissues such as roots, embryos, and fruits.
  • Abscisic Acid: This hormone controls seed dormancy, stomatal closure, and stress responses. It is synthesized in mature leaves, fruits, and seeds.

(b) When a growing plant detects light, it triggers a series of responses known as photomorphogenesis. The plant’s photoreceptors, specifically phytochrome and cryptochrome, detect the light signal and initiate various physiological changes. These changes include:

  • Phototropism: The plant bends towards the light source, which is controlled by auxin distribution in the stem.
  • Chloroplast movement: Chloroplasts within the plant cells move to optimize light absorption for photosynthesis.
  • Stomatal opening: Light stimulates the opening of stomata, allowing for gas exchange and transpiration.
  • Flowering: Light is crucial for triggering the flowering process in many plants.

Overall, the detection of light by plants is essential for their growth, development, and ability to respond to their environment. 


Q10: Draw a diagram of the cross-sectional view of human brain as given below on your answer sheet and label:    (CBSE 2017-18 C)
(а) The part that helps in performing voluntary actions. 
(b) The part that controls salivation and vomiting. 
(c) The largest part of forebrain. 
(d) A fluid that protects the brain. 
(e) Meninges.

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Ans: (a) Cerebrum 
(b) Medulla
(c) Cerebrum
(d) Cerebrospinal fluid
(e) Meninges or three membranes covering the brain.Previous Year Questions 2017

Previous Year Questions 2016

Q1: Name the two components of central nervous system. How are they protected? Name the component which is considered as highest coordinating centre of the body. Describe its three regions.   (CBSE 2016)

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Ans: 

  • Brain and spinal cord are the two components of central nervous system. 
  •  Human brain is protected by a bony skull, also called as cranium. Internal to the skull it is covered from outside by three membranes called meninges. In between the meninges there is a fluid called as cerebrospinal fluid. Beneath the membranes, the brain is subdivided into three major parts namely fore-brain, midbrain, and hind brain. 
  • Brain is considered as the highest coordinating centre of the body. 
  • Its three regions are given below:
    • Fore-brain is the main thinking part of the brain. It has different regions or areas such as:
      • Sensory area – It receives sensory impulses from various receptors.
      • Auditory area – It is concerned with hearing.
      • Olfactory area – It is concerned with smell.
      • Gustatory area – It is concerned with taste.
      • Optic area – It is concerned with sight.
      • Association area – This area interprets various sensory informations and makes decision how to respond.
      • Motor area – Decisions made by association area are passed on to this area which control the movement of voluntary muscles.
    • Midbrain comprises of four lobes. So it is also known as copra quadrigemina. Actions of involuntary muscles are controlled by the mid-brain and hind-brain. 
    • Hind-brain comprises of cerebellum, pons varolii, and medulla oblongata. Cerebellum is the second largest part of brain that coordinates muscular activity of the body as well as maintains body equilibrium or posture. Pons connects cerebellum and medulla oblongata and functions as a relay centre among different parts of brain. It possesses pneumotaxic area of respiratory centre. Medulla oblongata lies below the cerebellum and continues behind the spinal cord. Involuntary actions like blood pressure, salivation, and vomiting are controlled by the medulla in the hind-brain. 


Q2: With the help of suitable examples explain the terms phototropism, geotropism and chemotropism.    (CBSE 2016)

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Ans: The stimulus oriented movement of plant or plant part is called tropism. The direction of movement is related to the direction of stimulus. These movements are always growth oriented movements. They are of different types namely phototropism (growth movement in response to light), hydrotropism (water), geotropism (force of gravity), thigmotropism (contact) and chemotropism (chemicals).

  • Phototropism is a directional growth movement which occurs in response to unidirectional exposure to light. The region of photoreception is shoot apex where auxin is produced. Auxin moves from the illuminated region to the shaded region. This causes more growth on the dark side of the stem causing it to bend towards the source of light. Movement of shoot towards light is called phototropism.
  • Roots are positively geotropic and negatively phototropic. Shoots are positively phototropic and negatively geotropic.
  • Growth of pollen tube inside the style, ovary and ovule in response to the chemicals produced by them is an example of chemotropism.


Q3: (a) Define hormone. Write four characteristics of hormones in humans.
(b) Name the disorder caused by the following situations:
(i) Under secretion of growth hormone.
(ii) Over secretion of growth hormone.
(iii) Under secretion of insulin.
(iv) Deficiency of iodine.     (CBSE 2016)

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Ans: (a) Hormones are chemical informational molecules that are required in minute quantities and are directly poured in the blood stream by the glands. They act on a specific organ called target organ. So their site of production varies from site of action.
Their characteristics are:

  • They are poured into blood stream.
  • Are generally proteinaceous or steroid.
  • Their secretion is regulated by feedback mechanism.
  • Their site of action and secretion is different.

(b) 
(i) Dwarfism
(ii) Gigantism
(iii) Diabetes
(iv) Goitre


Q4: Name the hormone which is released into the blood when its sugar level rises. Name the organ which produces this hormone and its effect on blood sugar level. Also, mention the digestive enzymes secreted by this organ with one function of each.     (CBSE 2016)

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Ans: When blood sugar level rises in blood, a hormone named ‘insulin” is produced by β-cells of islet of Langerhans in the organ Pancreas. Pancreas also produces pancreatic juice which contains pancreatic enzymes such as trypsin, pancreatic amylase and pancreatic lipase.
Insulin promotes glucose absorption by individual cells and absorption and formation of glycogen in liver and muscles. This reduces glucose level in blood. As soon as the blood sugar comes to normal the pancreatic cells stops secretion of insulin.
Pancreas secretes slightly alkaline pancreatic juice which contains three major following enzymes:

  • Trypsin which digests proteins.
  • Lipase which digests fats.
  • Amylase which digests carbohydrates.


Q5: What are phytohormones? List four types of phytohormones. Where are these synthesised?     (CBSE 2016)

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Ans: Phytohormones are chemical substances that help plants coordinate their growth and respond to environmental stimuli. They can either promote or inhibit growth.
Five major types of plant growth hormones are auxins, gibberellins, cytokinins, abscisic acid and ethylene.
These plant growth regulators are also known as phytohormones.
These are synthesised at shoot apices, leaves, developing embryos, root tips etc.


Q6: Name the plant growth hormone which is synthesized at shoot tip. Explain with the help of a diagram why does a plant bend towards light during growth.     (CBSE 2016)

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Ans: (i) Auxin is a phytohormone which is synthesized at shoot tip.
(ii) When a plant is exposed to light coming from one side of the plant then auxin located at the shoot tip diffuses towards the shaded side of the shoot. High concentration of  auxin in the shaded region stimulates the cells  to grow longer in comparison to the region  exposed to light. So the shoot tends to bend towards the light.

Previous Year Questions 2016


Q7: Name the hormone which regulates carbohydrate, protein and fat metabolism in our body. Which gland secretes this hormone? Why is it important for us to have iodised salt in our diet?     (CBSE 2016)

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Ans: 

  • Thyroxine regulates the metabolism of carbohydrates, proteins, and fats in our body.
  • It is secreted by the thyroid gland.
  • A deficiency of iodine in our diet can lead to a condition called goitre, where the thyroid gland enlarges as it tries to absorb more iodine.
  • Iodine is essential for the production of thyroxine.
  • To ensure we get enough iodine, it is added to salt, making it iodised salt.


Q8: Name the two components of central nervous systems in humans.     (CBSE 2016)

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Ans: The two components of the Central Nervous System in humans are the brain and spinal cord.


Q9: In the absence of muscle cells, how do plant cells show movements?     (CBSE 2016)

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Ans: 

  • Plants do not have a nervous system and muscles. 
  • Even then they exhibit well coordinated and controlled movements.
  • Plants possess chemical coordination.
  • They respond to stimuli by secreting chemical substances called plant growth regulators.
  • They either stimulate or retard growth.
  • Five major types of plant growth hormones are:
    • Auxins
    • Gibberellins
    • Cytokinins
    • Abscisic acid
    • Ethylene
  • Certain changes in the orientation of plant parts in relation to other parts caused by intrinsic or external stimuli are non-directional, growth independent movements.
  • They are referred to as nastic movements.
  •  Thus, plants show movements due to growth or change in turgor pressure. 


Q10: (a) Write two points of differences between enzymes and hormones.     (CBSE 2016)
(b) Name one endocrinal gland in our body which performs dual function. Write the functions.

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Ans: (a) Differences between enzymes and hormones:Previous Year Questions 2016

(b) Pancreas performs dual function. It produces enzymes like Trypsin, Lipase etc., which aid in digestion. It also produces insulin hormone which regulates blood-sugar level.


Q11: Name the system which facilitates the communication between the central nervous system and the other body parts of the body.    (CBSE 2016)

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Ans: Peripheral nervous system.

Also read: Diagram Based Questions: Control and Coordination

Previous Year Questions 2015

Q1: What is synapse? In a neuron cell how is an electrical impulse created and what is the role of synapse in this context?   (CBSE 2015)

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Ans: 

  • Synapse is the junction between two nerve cells, specifically between the axon ending of one neuron and the dendrite of another. 
  •  The process of transmitting a nerve impulse involves several steps: 
  •  When information reaches the end of a neuron’s dendrite, it triggers a chemical reaction. 
  •  This reaction generates an electrical impulse that travels from the dendrite to the cell body and along the axon. 
  •  At the axon’s end, the electrical impulse causes the release of chemicals. 
  •  These chemicals cross the synapse and initiate a similar electrical impulse in the next neuron’s dendrite. 
  •  In summary, the synapse plays a crucial role in transmitting impulses from one neuron to another, facilitating communication throughout the body. 


Q2: “As the blood sugar level in our body falls insulin secretion is reduced.” Justify this statement in the reference of feedback mechanism that regulates the timing and amount of hormone released.   (CBSE 2015)

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Ans: 

  • Insulin is a hormone that plays a crucial role in regulating blood sugar levels.
  • It helps convert excess sugar in the blood into glycogen for storage.

When blood sugar levels are high:

  • The pancreas detects the increase.
  • It responds by producing more insulin.

Conversely, when blood sugar levels fall:

  • Insulin secretion is reduced.

This is part of a feedback mechanism where blood sugar levels directly influence insulin production.

This regulation ensures that our body maintains balanced blood sugar levels, preventing harmful effects from both high and low sugar levels.


Q3: State the role played by ovaries in a human female.   (CBSE 2015)

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Ans: Roles of ovary:

  • Ovaries produce female gametes, known as ova.
  • They also produce the female sex hormone, oestrogen, which is vital for:
  • The development of secondary sex organs.
  • The emergence of secondary sexual characteristics, such as a high-pitched voice and the growth of mammary glands.


Q4: What is meant by reflex-action? With the help of a labelled diagram trace the sequence of events which occur when we touch a hot object.   (CBSE 2015)

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Ans: Reflex action is defined as an unconscious, automatic and involuntary response of effector, i.e. muscle and gland to a stimulus which is monitored through the spinal cord.

Previous Year Questions 2015Sequence of events when we touch a hot object are:

  • Receptor organ skin receives the stimulus and activates a sensory nerve impulse.
  • Sensory neuron carries the message in the form of sensory impulse to the spinal cord.
  • The spinal cord acts as a modulator. The neurons of the spinal cord transmit the sensory nerve impulses to the motor neuron.
  • Motor nerve conducts these impulses to the effector organ hand which responds by pulling back the hand away from the hot object.


Q5: Write the names of the hormones secreted by the pituitary gland and adrenal gland. State their functions in the body.   (CBSE 2015)

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Ans: 

05. Previous Year Questions: Life Processes

Previous Year Questions 2025

Q1: Secretion of less saliva in mouth will affect the conversion of:  (1 Mark)
(a) 
Proteins into amino acids
(b)
 Fats into fatty acids and glycerol
(c) 
Starch into simple sugars
(d) Sugars into alcohol

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Ans: (c) Starch into simple sugars

Saliva in the mouth contains the enzyme salivary amylase, which helps in the breakdown of starch into simple sugars. If less saliva is secreted, this conversion will be affected.

Q2: The breakdown of glucose has taken the following pathway:  (1 Mark)

The sites ‘a’ and ‘b’ respectively are:
(a) 
Mitochondria and Oxygen deficient muscle cells
(b) Cytoplasm and Oxygen rich muscle cells
(c) Cytoplasm and Yeast cells
(d) Cytoplasm and Oxygen deficient muscle cells

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Ans: (d) Cytoplasm and Oxygen deficient muscle cells

The breakdown of glucose into pyruvate occurs in the cytoplasm. Under oxygen-deficient conditions in muscle cells, pyruvate is converted into lactic acid with the release of a small amount of energy.Previous Year Questions 2025

Q3: The opening and closing of stomata is regulated by:  (1 Mark)
(a) 
CO₂ concentration in stomata
(b) Temperature in guard cells
(c) O₂ concentration in stomata
(d) Amount of water in guard cells

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Ans: (d) Amount of water in guard cells

The guard cells control the opening and closing of stomata. When they are full of water (turgid), the stomata open; when they lose water (flaccid), the stomata close.Previous Year Questions 2025

Q4: One-cell thick blood vessels are known as:  (1 Mark)
(a)
 Alveoli
(b) Capillaries
(c) Arteries
(d) Veins

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Ans: (b) Capillaries

Capillaries are the one-cell thick blood vessels that allow exchange of materials like oxygen, carbon dioxide, nutrients, and wastes between blood and body tissues.

Q5: The essential element used in the synthesis of proteins is:  (1 Mark)
(a)
 Nitrogen
(b) Phosphorus
(c) Iron
(d) Magnesium

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Ans: (a) Nitrogen

Nitrogen is an essential element required by plants for the synthesis of proteins and other important compounds like nucleic acids.

Q6: Select TRUE statements about lymph from the following:  (1 Mark)
A. Lymph vessels carry lymph through the body and finally open into larger arteries.
B. Lymph contains some amount of plasma, proteins, and blood cells.
C. Lymph contains some amount of plasma, proteins, and red blood cells.
D. Lymph vessels carry lymph through the body and finally open into larger veins.
The true statements are:
(a) 
A and B
(b) B and D
(c) A and C
(d) C and D

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Ans: (b) B and D

Lymph contains plasma, proteins, and white blood cells (not red blood cells).
Lymph vessels carry lymph and open into larger veins, helping in the return of tissue fluid to the blood.

Q7: Select the correct option from the following statements about the functioning of the human heart:  (1 Mark)
(a) 
Right atrium receives deoxygenated blood from different parts of the body and sends it to pulmonary veins.
(b) Left atrium sends oxygenated blood to right ventricle which pumps it to different parts of the body.
(c) Right atrium receives deoxygenated blood from the body and sends it to the right ventricle.
(d) Left atrium receives oxygenated blood from the pulmonary arteries and sends it to the left ventricle.

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Ans: (c) Right atrium receives deoxygenated blood from the body and sends it to the right ventricle.

The right atrium collects deoxygenated blood from all parts of the body through veins and passes it to the right ventricle, which then pumps it to the lungs for oxygenation.

Q8: Parasitic mode of nutrition is observed in:  (1 Mark)
(a) 
Bryophyllum
(b) Hibiscus
(c) Cuscuta
(d) Helianthus

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Ans: (c) Cuscuta

Cuscuta (also known as Amarbel) shows a parasitic mode of nutrition, as it obtains food from the host plant on which it climbs by absorbing nutrients through specialized structures called haustoria.

Q9: Question consist of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option (a), (b), (c) and (d) as given below:  (1 Mark)
Assertion (a): In large animals, oxygen can reach different parts of the animal’s body easily.
Reason (R): Respiratory pigments take up oxygen from the air and carry it to body tissues.
(a) 
Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

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Ans: (d) A is false, but R is true.In large animals, oxygen cannot reach all body parts easily by diffusion alone. Hence, respiratory pigments like haemoglobin are needed to carry oxygen to tissues — making the reason true but the assertion false.

Q10: Question consist of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option (a), (b), (c) and (d) as given below:  (1 Mark)
Assertion (a): Xylem tissue moves water and minerals obtained from the soil by the roots.
Reason (R): Xylem tissue is found only in the roots of a plant.
(a) 
Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

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Ans: (c) A is true, but R is false.

The xylem indeed transports water and minerals from the roots to other parts of the plant, but it is not found only in the roots — it is present throughout the plant, including the stems and leaves.

Q11: Which one of the following is not an excretory product in plants?  (1 Mark)
(a)
 CO₂
(b) Starch
(c) Resins and gums
(d) Dead cells

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Ans: (b) Starch

Starch is a stored form of carbohydrate in plants, not an excretory product.
Plants excrete substances like CO₂resinsgums, and dead cells as waste materials.

Q12: The basic filtration unit of the excretory system in human beings is:  (1 Mark)
(a) 
Nephron
(b) Urethra
(c) Neuron
(d) Urinary bladder

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Ans: (a) Nephron

The nephron is the basic structural and functional unit of the human kidney.
It is responsible for the filtration of blood and the formation of urine.

Q13: In human alimentary canal, the digestive juice secreted by the gastric glands are:  (1 Mark)
(a) 
Bile, Trypsin, Pepsin
(b) Hydrochloric acid, Pepsin, Mucus
(c) Lipase, Bile, Mucus
(d) Salivary amylase, Pepsin, Bile

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Ans: (b) Hydrochloric acid, Pepsin, Mucus
Gastric Glands Secretions:

  • Hydrochloric acid (HCl): Creates an acidic environment (pH 1.5–3.5) to activate pepsin and kill pathogens.
  • Pepsin: Digests proteins into peptides (activated from pepsinogen by HCl).
  • Mucus: Protects the stomach lining from acid and pepsin damage.

Other Options:

  • Bile: Secreted by the liver, stored in the gallbladder, acts in the small intestine.
  • Trypsin, Lipase: Secreted by the pancreas, not gastric glands.
  • Salivary amylase: Secreted in saliva, not stomach.

Q14: Listed below are the steps of nutrition in Amoeba. Select the correct sequence of these steps:  (1 Mark)
(i) Diffusion of simple nutrients into cytoplasm
(ii) Food vacuole formation
(iii) Formation of finger-like temporary extensions of cell surface
(iv) Complex substances broken to simpler ones
(v) Undigested material thrown out of the cell surface
(a)
 (iv), (i), (ii), (iii), (v)
(b) (iii), (iv), (i), (v)
(c) (iii), (ii), (iv), (i), (v)
(d) (iii), (iv), (i), (ii), (v)

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Ans: (c) (iii), (ii), (iv), (i), (v)

In Amoeba, nutrition occurs in this sequence:

(iii) Formation of pseudopodia (finger-like extensions) to engulf food,
(ii) Formation of a food vacuole,
(iv) Digestion of complex food into simpler substances,
(i) Absorption of nutrients into the cytoplasm,
(v) Egestion of undigested material from the cell.
Correct sequence: (iii) → (ii) → (iv) → (i) → (v)

Q15: In aerobic respiration, the steps are: breakdown of glucose to pyruvate and its further conversion to carbon dioxide. Both processes respectively occur in:  (1 Mark)
(a) 
Vacuole and Cytoplasm
(b) Chloroplast and Mitochondria
(c) Mitochondria and Cytoplasm
(d) Cytoplasm and Mitochondria

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Ans: (d) Cytoplasm and Mitochondria

In aerobic respirationglucose is first broken down into pyruvate in the cytoplasm, and then pyruvate is further broken down into carbon dioxide and water in the mitochondria, releasing energy.Previous Year Questions 2025

Q16:  (a) Besides minimising the loss of blood, why is it essential to plug any leak in a blood vessel? Name the component of blood which helps in this process and state how this component performs this function.  (2 Marks)
OR
(b) (i) The transport system in plants is relatively slower than in animals. Give reasons.
(ii) State the role of phloem in the transport of materials in plants.  (2 Marks)

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Ans: (a) It is essential to plug any leak in a blood vessel to prevent blood loss and maintain blood pressure so that oxygen and nutrients can continue to reach all parts of the body.

The component of blood that helps in this process is platelets.
Platelets release certain enzymes that help in the clotting of blood, forming a clot or scab at the site of injury, thereby sealing the wound and preventing further bleeding.

OR
(b) (i) The transport system in plants is slower than in animals because:

  • Plants do not have a pumping organ like the heart.
  • The movement of materials in plants depends mainly on physical processes such as diffusionosmosis, and transpiration pull.

(ii) The phloem transports the prepared food (sucrose) from the leaves (source) to other parts of the plant (sink) like roots, fruits, and storage organs — a process known as translocation.

Q17: State the main function of arteries. Why do they have thick and elastic walls?  (2 Marks)

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Ans: The main function of arteries is to carry oxygenated blood away from the heart to different parts of the body.

They have thick and elastic walls because:

  • The blood pumped by the heart flows through them at high pressure, and
  • The elasticity helps them withstand and maintain this high pressure without bursting.

Q18: “Plants use a variety of techniques to get rid of waste material.” Justify this statement giving any four ways.  (2 Marks)

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Ans: Plants get rid of their waste materials in several ways, such as:

  1. Gaseous wastes like CO₂ and O₂ are removed through stomata in leaves and lenticels in stems.
  2. Excess water is lost by transpiration through stomata.
  3. Resins and gums are stored in old xylem or other dead tissues and later excreted.
  4. Wastes may also be stored in leaves, bark, or fruits, which fall off from the plant.

Thus, plants use different methods to manage and remove their wastes efficiently.

Q19: Based on the characteristics of the processes given in the brackets in each case, differentiate between the following:
(a) Products of breakdown of pyruvate in aerobic and anaerobic respiration in human beings (product(s) of the processes)
(b) Respiration and photosynthesis in plants (gas released)
(c) Respiration in terrestrial animals and fishes (organs involved)  (3 Marks)

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Ans: 

(a) In aerobic respiration, the breakdown of pyruvate produces carbon dioxide and water, while in anaerobic respiration in human beings, it produces lactic acid.

(b) In respirationcarbon dioxide is released, whereas in photosynthesisoxygen is released.

(c) In terrestrial animalslungs are the organs of respiration, while in fishesgills are used for respiration.

Q20: (a) Enlist any two nitrogenous waste products removed from the blood of human kidney.
(b) Name the capillary cluster formed by the branch of renal artery in the Bowman’s capsule.
(c) Depict in the form of a flow chart the path of the urine formed in each kidney until it is finally passed out through the urethra.  (3 Marks)

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Ans: 

(a) Two nitrogenous waste products removed from the blood by human kidneys are urea and uric acid.

(b) The capillary cluster formed by a branch of the renal artery in the Bowman’s capsule is called the glomerulus.

(c) Flow chart showing the path of urine:

Previous Year Questions 2025

Q21: Why is blood circulation in vertebrates known as “double circulation”? Trace its path in the form of a flow chart.  (3 Marks)

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Ans:  Blood circulation in vertebrates is called double circulation because the blood passes through the heart twice during one complete cycle — once for oxygenation and once for supplying oxygenated blood to the body.
This ensures efficient separation of oxygenated and deoxygenated blood.

Flow chart of Double circulation:Previous Year Questions 2025

Q22: What is the first step of cellular respiration? In which part of the cell does it occur? Write the equation for the process of breakdown of glucose in a human cell:  (3 Marks)
(i) in the presence of oxygen
(ii) due to lack of oxygen

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Ans:

The first step of cellular respiration is the breakdown of glucose into pyruvate, and it occurs in the cytoplasm of the cell.Previous Year Questions 2025

(i) In the presence of oxygen (Aerobic respiration):

Pyruvate is completely broken down into carbon dioxide and water in the mitochondria, releasing energy.

(ii) Due to lack of oxygen (Anaerobic respiration):

In the absence of oxygen, pyruvate is converted into lactic acid in the muscle cells, releasing less energy.

Q23: Name the blood vessel that brings (i) oxygenated blood (ii) deoxygenated blood, to the human heart. Also name that chamber of the heart which receives deoxygenated blood and state how deoxygenated blood from this chamber is sent to lungs for oxygenation.  (3 Marks)

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Ans:

(i) The blood vessel that brings oxygenated blood to the heart is the pulmonary vein.
(ii) The blood vessel that brings deoxygenated blood to the heart is the vena cava (superior and inferior vena cava).

The right atrium of the heart receives deoxygenated blood from all parts of the body through the vena cava.
From the right atrium, the blood passes into the right ventricle, which then pumps the deoxygenated blood to the lungs for oxygenation through the pulmonary artery.

Q24: (a) (i) “The length of the small intestine in various animals depends on the food they eat.” Justify the statement.
(ii) Discuss the role of the pancreas and bile juice in the digestion of food in human beings.
(iii) How is the small intestine designed to absorb digested food? (5 Marks)
OR
(b) (i) State the role of rings of cartilage present in the throat.
(ii) Discuss the role of the ribs and diaphragm when air is taken in during the breathing cycle.
(iii) Why do we get muscle cramps during heavy exercise? Explain.  (5 Marks)

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Ans:
(a) (i) The length of the small intestine varies in different animals according to their diet.

  • In herbivores, such as cows and rabbits, the small intestine is longer because plant food (cellulose) is hard to digest, and it needs more time for complete digestion and absorption.
  • In carnivores, like tigers and lions, it is shorter because animal flesh is easier to digest.

(ii) The role of the pancreas and bile juice in the digestion of food in human beings is as follows:

  • The pancreas secretes pancreatic juice which contains enzymes such as trypsin (for digesting proteins) and lipase (for breaking down fats).
  • The bile juice, produced by the liver and stored in the gall bladder, helps in the emulsification of fats—breaking them into smaller droplets for easier enzyme action.

(iii) The small intestine is highly adapted for absorption of digested food because:

  • Its inner walls have finger-like projections called villi, which increase the surface area for absorption.Previous Year Questions 2025
  • Each villus contains blood capillaries, allowing digested nutrients to quickly enter the bloodstream and be transported to different parts of the body.

OR

(i) The rings of cartilage present in the trachea prevent it from collapsing when there is no air inside, keeping the airway open for smooth breathing.Previous Year Questions 2025(ii) During inhalation, the ribs move upward and outward, and the diaphragm contracts and moves downward, increasing the space in the chest cavity. This creates low pressure, allowing air to enter the lungs.
(iii) During heavy exercise, there is a lack of oxygen in the muscles. Hence, anaerobic respiration takes place, producing lactic acid. The accumulation of lactic acid causes muscle cramps and fatigue.

Q25: The following question are Source-based/Case-based questions. Read the case carefully and answer the question that follow.

The maintenance functions of all living organisms must go on even when they are not doing anything particular. Even when we are just sitting in a class or even asleep, this maintenance job has to go on. These maintenance processes require energy to prevent damage and break-down of cells and tissues, which is obtained by the individual organism from the food prepared by the autotrophs, called producers.

(a) Name and define the process by which green plants prepare food.  (1 Mark)
(b) Write chemical equation involved in the above process. (1 Mark)
(c) (i) State in proper sequence the events that occur in synthesis of food by desert plants. (2 Marks)
OR
(c) (ii) Explain giving reasons what happens to the rate at which the green plants will prepare food: (2 Marks)
(I) during cloudy weather, and
(II) when stomata get blocked due to dust.

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Ans: 

(a) The process by which green plants prepare food is called photosynthesis.
It is the process by which green plants use sunlight, carbon dioxide, and water to produce food (glucose) and release oxygen.

(b) Chemical equation of photosynthesis:

Previous Year Questions 2025

(c) (i) In desert plants, photosynthesis occurs as follows:

  1. Stomata remain closed during the day to prevent water loss.
  2. Carbon dioxide is absorbed at night and converted into an intermediate compound.
  3. During the daytime, this compound is broken down to release CO₂ internally.
  4. This CO₂ is then used for photosynthesis in the presence of sunlight.

OR

(c) (ii)
(I) During cloudy weather, sunlight is less available, so the rate of photosynthesis decreases.
(II) When stomata get blocked due to dustCO₂ cannot enter the leaves, and hence the rate of photosynthesis decreases or may stop completely depending on blockage.

Previous Year Questions 2024

Q1: A stomata closes when:    (1 Mark)   (2024)
(i) It needs carbon dioxide for photosynthesis.
(ii) It does not need carbon dioxide for photosynthesis.
(iii) water flows out of the guard cells.
(iv) water flows into the guard cells.
The correct reason(s) in this process is/are:
(a) 
(i) only
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (ii) and (iv)

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Ans: (c)

Stomata are small openings on the surface of leaves that allow gases to enter and exit the plant. They close when the plant doesn’t need carbon dioxide for photosynthesis (which is when it’s not producing food) and when water flows out of the guard cells (which helps the plant conserve water). Therefore, the correct answers are (ii) and (iii).


Q2: State one role of each of the following in the human digestive system:   (2 Marks)   (2024)
(i) Hydrochloric acid
(ii) Villi
(iii) Anal Sphincter
(iv) Lipase

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Ans: 
(i) Hydrochloric acid: Creates an acidic environment that helps enzymes work and kills harmful microorganisms.

(ii) Villi: Increases the surface area of the intestine, enhancing the absorption of nutrients.
(iii) Anal Sphincter:
 Controls the release of waste from the body through the anus.
(iv) Lipase:
 Breaks down fats into smaller molecules, aiding in their digestion.


Q3: (i) Why is respiratory pigment needed in multicellular organisms with large body sizes?     (3 Marks)  (2024)
(ii) Give reasons for the following:
(a) Rings of cartilage are present in the throat.
(b) Lungs always contain a residual volume of air.
(c) The diaphragm flattens and ribs are lifted up when we breathe in.
(d) Walls of alveoli contain an extensive network of blood vessels.

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Ans: (i) In large animals, diffusion alone cannot deliver enough oxygen throughout the body. Instead, respiratory pigments are essential because they:

  • Absorb oxygen from the air in the lungs.
  • Transport oxygen to tissues that need it.
  • Release oxygen where it is required.

In humans, the main respiratory pigment is haemoglobin, found in red blood cells, which has a strong affinity for oxygen. Carbon dioxide is mostly carried in a dissolved form in the blood.

(ii) Reasons:
(a) Rings of cartilage in the throat prevent the air passage from collapsing.
(b) Lungs retain a residual volume of air to allow time for oxygen absorption and carbon dioxide release.
(c) When we breathe in, the diaphragm flattens and the ribs lift, making the chest cavity larger, which draws air into the lungs.
(d) The walls of alveoli contain many blood vessels to facilitate gas exchange.


Q4: Which of the following statement(s) is (are) true about human heart?  (1 Mark)   (CBSE 2024)
(A) Right atrium receives oxygenated blood from lungs through pulmonary artery.
(B) Left atrium transfers oxygenated blood to left ventricle which sends it to various parts of the body.
(C) Right atrium receives deoxygenated blood through vena cava from upper and lower body.
(D) Left atrium transfers oxygenated blood to aorta which sends it to different parts of the body.
(a)
 (A)    
(b) (A) and (D)
(c) (B) and (C)    
(d) (B) and (D)

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Ans: (c)

  • (A) is incorrect because the right atrium receives deoxygenated blood from the body through the vena cava, not oxygenated blood from the lungs. The pulmonary artery carries deoxygenated blood from the right ventricle to the lungs, not to the right atrium.
  • (B) is correct because the left atrium transfers oxygenated blood to the left ventricle, which then pumps it to various parts of the body through the aorta.
  • (C) is correct because the right atrium receives deoxygenated blood through the vena cava from the upper and lower parts of the body.
  • (D) is incorrect because it is the left ventricle, not the left atrium, that transfers oxygenated blood to the aorta, which then sends it to different parts of the body.

Therefore, the correct answer is (c) (B) and (C).


Q5: Photosynthesis takes place in the leaves and the food prepared by it reaches other parts of the plants. Name the process involved and explain it.    (1 Mark) (2024)

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Ans: 

  • The process is translocation.
  • It involves the movement of food, mainly sugars like sucrose, from the leaves (where photosynthesis occurs) to other parts of the plant through the phloem.
  • This ensures all parts of the plant receive nutrients for growth, storage, and energy.


Q6: Which of the following statement (s) is (are) true about human heart?  (1 Mark)  (2024)
(A) Right atrium receives oxygenated blood from lungs through pulmonary artery. 
(B) Left atrium transfers oxygenated blood to left ventricle which sends it to various parts of the body. 
(C) Right atrium receives deoxygenated blood from different parts of the body through vena cava. 
(D) Left atrium transfers oxygenated blood to aorta which sends it to different parts of the body. 
(a) 
(B) only
(b) (A) and (D)
(c) (B) and (C)
(d) (B) and (D)

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Ans: (c)
The correct statements about the human heart are (B) and (C). The left atrium does transfer oxygenated blood to the left ventricle, which then pumps it to the body. The right atrium receives deoxygenated blood from the body through the vena cava, while (A) is incorrect because the right atrium actually receives deoxygenated blood, not oxygenated, and (D) is misleading as the left atrium transfers blood to the left ventricle, not directly to the aorta.


Q7: Case/Source based questions.  (4 & 5 Marks) (CBSE 2024)
Human digestive system is a tube running from mouth to anus. Its main function is to breakdown complex molecules present in the food which cannot be absorbed as such into smaller molecules. These molecules are absorbed across the walls of the tube and the absorbed food reaches each and every cell of the body where it is utilised for obtaining energy. 
(a) Name the glands present in the buccal cavity and write the components of food on which the secretion of these glands act upon. 
(b) Two organs have a sphincter muscle at their exit. Name them. 
(c) What will happen if: 
(i) mucus is not secreted by the gastric glands. 
(ii) Villi are absent in the small intestine. 
OR
(c) “Bile juice does not contain any enzyme, yet it has important roles in digestion.” Justify the statement.

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Ans: (a) Salivary glands; Starch / Carbohydrate
(b) Stomach, Anus
(c) The consequences of specific conditions are:

(i) If mucus is not secreted by the gastric glands, the inner lining of the stomach will be unprotected from acid, leading to potential damage.

(ii) If villi are absent in the small intestine, the absorption of digested food will be significantly reduced.

OR
(c) Emulsification of fats.
Acidic medium has to be made alkaline for the pancreatic enzymes to act.  


Q8: In human beings, when the process of digestion is completed, the (i) proteins, (ii) carbohydrates, and (iii) fats are respectively finally converted into:  (1 Mark)  (CBSE 2024)
(a)
 (i) Amino acids, (ii) glucose and (iii) fatty acids
(b) (i) Amino acids, (ii) glucose, (iii) fatty acids and glycerol
(c) (i) Glucose, (ii) fatty acids and glycerol, (iii) amino acids
(d) (i) Sugars, (ii) amino acids, (iii) fatty acids and glycerol

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Ans: (b)

During digestion in humans:

  • Proteins are broken down into amino acids.
  • Carbohydrates are converted into glucose.
  • Fats are transformed into fatty acids and glycerol.

The correct answer is (b) as it accurately describes the final products of these macronutrients after digestion.


Q9: We need to water the soil in plants on a regular basis. But it ultimately reaches the leaves of the plant. Explain how this takes place.  (1 Mark)  (CBSE 2024)

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Ans: Water reaches the leaves through the xylem.

  1. Roots absorb water from the soil by osmosis.
  2. Water moves upward through the xylem due to transpiration pull, capillary action, and cohesion of water molecules.

This ensures water supply to the leaves for photosynthesis and other processes.


Q10: The process in which transport of soluble products of photosynthesis takes place in plants is known as:  (1 Mark) (CBSE 2024)
(a)
 Transpiration    
(b) Evaporation
(c) Conduction    
(d) Translocation

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Ans: (d)

The process of transporting soluble products of photosynthesis in plants is called translocation. This involves:

  • Moving substances like sugars from the leaves, where they are produced, to other parts of the plant.
  • Utilising the phloem, a type of vascular tissue, for this transport.
  • Delivering essential nutrients, including amino acids, to storage organs such as roots, fruits, and seeds.
  • Facilitating movement in both upward and downward directions through sieve tubes and companion cells.

Translocation is vital for distributing energy throughout the plant, ensuring that all parts receive the necessary nutrients for growth and development.


Q11: (a) Sometimes while running, the athletes suffer from muscle cramps. Why? How is the respiration in this case different from aerobic respiration?
OR
(b) Write the other name given to lymph. State its two functions.   (3 Marks) (CBSE 2024)

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Ans: (a)

i. The formation of lactic acid in muscles leads to cramps.

ii. Aerobic respiration occurs with oxygen, while cramps result from anaerobic respiration, which happens without oxygen.

iii. End products of aerobic respiration are carbon dioxide, water, and energy, whereas anaerobic respiration produces lactic acid and energy.

OR
(b) Tissue fluid / Extracellular fluid Functions:
i. Carries digested and absorbed fats from the intestine.  
ii. Drains excess fluid from extracellular space back into the blood.
iii. Fight against infections.


Q12: (a) Design an experiment to demonstrate that carbon dioxide is essential for photosynthesis. Write the observation and conclusion of the experiment.  (3 Marks) (CBSE 2024)
OR
(b) (i)
In the experimental set-up shown above in diagram (I) atmospheric air is being passed into lime water with a syringe while in diagram (II) air is being exhaled into lime water. The time taken for the lime water to turn milky in both the test tubes is different. Give reason.
(ii) Draw the diagram of an open stomatal pore and label  (I) Guard cells, and (II) Chloroplast on it. Mention two functions performed by stomata.

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Ans: (a) 

  • Take two healthy potted plants, A and B of nearly the same  size.
  • Keep them in darkness for three days. (Destarch the plant)
  • Place a watch glass containing potassium hydroxide by the side of potted plant A but not in potted plant B.
  • Cover both the plants with separate bell jars and seal the bottom of the jars with Vaseline.
  • Keep both the plants in sunlight for two hours.
  • Pluck one leaf each from both the plants and test for the presence of starch with iodine solution.
  • Observation: The leaf of the potted plant A with KOH did not turn blue-black, indicating that starch was not present. The leaf of the potted plant B turns blue.
  • Conclusion: KOH absorbs CO2 so photosynthesis did not occur in potted plant A.

OR
(b) (i) In set up (I) lime water turns milky in more time as compared to set up (II) because the air we exhaled contains high percentage of CO2 as compared to atmospheric air.  
(ii)
Previous Year Questions 2024
Two labellings: (I) Guard Cells
(II) Chloroplast Two functions performed by stomata:
Gaseous exchange
Transpiration


Q13: In human respiratory system, when a person breathes in, the position of ribs and diaphragm will be:   (1 Mark) (CBSE 2024)
(a)
 lifted ribs and curve/dome shaped diaphragm.
(b) lifted ribs and flattened diaphragm.
(c) relaxed ribs and flattened diaphragm.
(d) relaxed ribs and curve/dome shaped diaphragm.

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Ans: (b)
The correct answer is (b) lifted ribs and flattened diaphragm. When a person breathes in, the diaphragm contracts and flattens, while the ribs are lifted upward and outward. This movement creates more space in the chest cavity, allowing air to fill the lungs.


Q14: For Q. Nos., two statements are given – One labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:  (1 Mark) (CBSE 2024)
Assertion (A): The rate of breathing in aquatic organisms is much faster than in terrestrial organisms.
Reason (R): The amount of oxygen dissolved in water is very high as compared to the amount of oxygen in air.
(a) 
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) 
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) 
Assertion (A) is true, but Reason (R) is false.
(d) 
Assertion (A) is false, but Reason (R) is true.

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Ans: (c)
The correct answer is (c) Assertion (A) is true, but Reason (R) is false. Aquatic organisms do breathe faster than terrestrial organisms because they need to extract enough oxygen from water, which has less oxygen available than air. Therefore, while the assertion is correct, the reason given is incorrect because it states that water has a high amount of dissolved oxygen, which is not true compared to air.

Previous Year Questions 2023

Q1: Opening and closing of stomata is due to (1 Mark) (2023)
(a) High pressure of gases inside the cells
(b) Movement of water in and out of the guard cells
(c) Stimulus of light in the guard cells
(d) Diffusion of CO2 in and out of the guard cells.

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Ans: (b)
The opening and closing of stomata are indeed regulated by the movement of water in and out of guard cells. Here’s a brief explanation of how it works:

Mechanism of Stomatal Movement:

Guard Cells and Stomata:

  • Stomata are small openings on the surfaces of leaves and stems that allow for gas exchange (CO2, O2, and water vapor) between the plant and the atmosphere.
  • Each stoma is flanked by a pair of guard cells, which control the opening and closing of the stoma.

Opening of Stomata:

  • When guard cells take up water by osmosis, they become turgid (swollen).
  • The turgidity of guard cells is primarily driven by the active transport of potassium ions (K⁺) into the guard cells. This lowers the water potential inside the guard cells, causing water to enter by osmosis.
  • As the guard cells swell, they bow outward, causing the stomatal pore to open.

Closing of Stomata:

  • When the guard cells lose water, they become flaccid (less swollen).
  • The loss of turgor pressure in the guard cells is usually due to the active transport of potassium ions (K⁺) out of the guard cells. This raises the water potential inside the guard cells, causing water to exit by osmosis.
  • As the guard cells lose turgor and shrink, the stomatal pore closes.


Q2: Assertion (A): The inner walls of the small intestine have finger like projections called villi which are rich in blood. (1 Mark) (2023)
Reason (R): These villi have a large surface area to help the small intestine in completing the digestion of food.
(a) 
Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.

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Ans: (c)
Assertion (A):

  • This statement is true. The small intestine is lined with villi, which are small, finger-like projections rich in blood vessels. These vessels help absorb nutrients from digested food into the bloodstream.

Reason (R):

  • This statement is false. While it is true that the villi increase the surface area of the small intestine, their primary role is to aid in the absorption of nutrients, not in the digestion of food. Digestion is primarily completed by enzymes within the small intestine.

Conclusion: (A) is true, but (R) is false.


Q3: Water in the root enters due to  (1 Mark) (2023)
(a) The function of the root to absorb water
(b) Difference in the concentration of ions between the root and the soil
(c) Excess water present in the soil
(d) Diffusion of water in the roots

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Ans: (b)

  • Roots absorb water from the soil through a process called osmosis.
  • This process occurs because of a difference in ion concentration between the roots and the soil.
  • As a result, water moves into the roots to balance this concentration difference.


Q4: As compared to terrestrial organisms, the rate of breathing in aquatic organism is   (1 Mark) (2023)
(a) Faster because they need more oxygen for their survival
(b) Faster because the amount of dissolved oxygen in water is fairly low
(c) Slower because the amount of dissolved oxygen in water is fairly low
(d) Slower because the capacity of water of dissolving atmospheric air is limited

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Ans: (b)
 In aquatic organisms, the rate of breathing is higher.
These organisms utilize oxygen dissolved in water.
The amount of dissolved oxygen in water is relatively low compared to the oxygen available in the air.
Therefore, aquatic organisms need to breathe faster to obtain sufficient oxygen for their survival.


Q5: Observe the following diagram and identify the process and its significance from the following options:   (1 Mark) (2023)
Previous Year Questions 2023(a) Evaporation : maintains water contents in leaf cells.
(b) Transpiration : creates a suction force which pulls water inside the plant.
(c) Excretion : helps in excreting out waste water from the plant.
(d) Translocation : helps in transporting materials from one cell to another.

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Ans: (b)

 Transpiration is a vital process in plants that involves the loss of water vapour from the aerial parts, primarily leaves. This process plays a crucial role in several ways:

  • Water Movement: It creates a suction force that pulls water from the roots through the xylem.
  • Nutrient Transport: Transpiration aids in the upward movement of water and dissolved minerals from the roots to the leaves.
  • Temperature Regulation: It helps maintain optimal temperatures within the plant.

During the day, when the stomata are open, transpiration becomes the main driving force for water movement. At night, root pressure assists in this transport. Overall, transpiration is essential for maintaining the plant’s hydration and nutrient supply.


Q6:  The process in which loss of water in the form of vapours from the aerial parts of plants takes place is X, which helps in Y. Here, X and Y respectively are  (1 Mark) (2023)
(a) Transpiration and photosynthesis
(b) Transpiration and temperature regulation
(c)  Translocation and movement of soluble products of photosynthesis in phloem
(d) Translocation and absorption of water and minerals from soil by roots.

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Ans: (b)

  • Process (X): Transpiration 
    Transpiration is the process where water is lost in the form of vapours from the aerial parts of plants, primarily through the stomata in the leaves.
  • Significance (Y): Temperature Regulation 
    Transpiration helps in temperature regulation by cooling the plant. As water evaporates from the leaf surfaces, it dissipates heat, thus maintaining optimal temperature conditions for various physiological processes, including photosynthesis.


Q7: Assertion (A) : The walls of atria are thicker than those of the ventricles.  (1 Mark) (2023)
Reason (R) : Ventricles have to pump blood into various organs at high pressure.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A)
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A)
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

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Ans: (d)
Assertion (A): The walls of atria are thicker than those of the ventricles.

  • This statement is false. In reality, the walls of the ventricles are thicker than those of the atria.
  • Ventricles have thicker muscular walls because they pump blood out of the heart to the lungs and the rest of the body, which requires more force (pressure).

Reason (R): Ventricles have to pump blood into various organs at high pressure.

  • This statement is true. Ventricles indeed pump blood into the arteries and throughout the body at high pressure to ensure blood reaches all organs and tissues.

Therefore, while Reason (R) correctly explains the physiological need for thicker ventricular walls due to high-pressure pumping, Assertion (A) is incorrect because the ventricular walls are thicker than atrial walls, not the other way around.


Q8: Two green plants are kept separately in oxygen-free containers, one in the dark and the other in sunlight. It was observed that plants kept in the dark could not survive longer. Give a reason for this observation.  (2 Marks) (2023)

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Ans: The plant kept in the dark could not survive longer because:

  • It cannot perform photosynthesis, which is essential for producing food.
  • Photosynthesis requires light to convert carbon dioxide and water into glucose and oxygen.
  • Without light, the plant lacks the energy needed for this process, leading to its inability to produce the oxygen necessary for respiration.

As a result, the plant in the dark cannot sustain itself and eventually dies.


Q9: List the events that take place during the process of photosynthesis in the proper sequence.  (2 Marks) (2023)

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Ans: The three events that occur during the process of photosynthesis are;
(i) Absorption of light energy by chlorophyll.
(ii) Conversion of light energy to chemical energy (in the form of ATP and NADPH) and splitting of water molecules into hydrogen and oxygen.
(iii) Reduction of carbon dioxide to carbohydrates (carbon assimilation). 


Q10: (i) How does Paramecium obtain its food?  (3 Marks) (2023)
(ii) List the role of each of the following in our digestive system: 
(a) Hydrochloric acid
(b) Trypsin
(c) Muscular walls of the stomach
(d) Salivary amylase

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Ans: (i) In Paramecium, a unicellular protozoan, the hair-like outgrowth cilia are present on the entire surface and help collect the food. They sweep the food inside the body through the oral groove.
(ii) (a) Role of hydrochloric acid: It helps to maintain the acidic pH in the stomach required for activation of the enzyme pepsin that digests proteins in the stomach.
(b) Role of trypsin in our digestion system: Trypsin breaks down protein into smaller peptides in the duodenum of the small intestine. It helps in digesting dietary protein by breaking the chain of amino acids. It also activates some proenzymes present in pancreatic juice.
(c) Role of muscular walls in the stomach in our digestive system: The muscular walls in the stomach contract periodically and thereby help in the churning and mixing of the food with the digestive enzymes and HCI. It helps in chemical digestion.
(d) Role of salivary amylase in our digestive system: Salivary amylase found in saliva breaks down the starch and converts it into the simplest sugar. 


Q11: (a) With the help of an activity, explain the action of saliva on the food we eat.   (3 Marks) (2023)
(b) Why is bile juice important in the process of digestion? 

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Ans: (a)  The action of saliva on food can be demonstrated through a simple activity:

  • Take two test tubes, A and B.
  • In test tube A, add one teaspoon of boiled rice.
  • In test tube B, add one teaspoon of boiled rice that has been chewed for 3 to 5 minutes.
  • Add 3.4 mL of water to both test tubes.
  • Add a few drops of iodine solution to each test tube.

Results:

  • In test tube A, the rice changes colour due to the absence of enzymes.
  • In test tube B, there is no colour change because the amylase enzyme in saliva breaks down the starch into simple sugars.

(b)  Bile juice plays a crucial role in digestion:

  • It breaks down fats into fatty acids, making them easier to absorb.
  • Bile also provides an alkaline fluid that neutralises the acidic food from the stomach.


Q12: What is the other name of ’tissue fluid’? Write its two functions.   (2 Marks) (2023)

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Ans: Tissue fluid is also called lymph. The functions of tissue fluid are:
(i) It carries digested and absorbed fat from the intestine.
(ii) It drains excess fluid from extracellular space back into the blood. 


Q13: What will happen if:   (2 Marks) (2023)
(a) Xylem tissue in a plant is removed. 
(b) We are injured and start bleeding?

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Ans: (a) If the xylem is removed, transport of water and minerals from the soil would not occur, which leads to the wilting of leaves and, ultimately, the death of the plant.
(b) In case of any injury, when bleeding occurs, platelets circulate around the body and form a mesh-like network or clot at the site of injury. 


Q14: (i) What is double circulation?  (3 Marks) (2023)
(ii) Why is the separation of the right side and the left side of the heart useful? How does it help birds and mammals?  

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Ans: (i)  Double circulation is a system where blood passes through the heart twice during one complete cycle. It consists of two main pathways:

  • Systemic circulation: Carries oxygen-rich blood from the heart to the body.
  • Pulmonary circulation: Transports deoxygenated blood from the heart to the lungs for oxygenation.


Previous Year Questions 2023(ii) The separation of the heart into right and left sides is beneficial because:

  • It prevents mixing of oxygenated and deoxygenated blood.
  • This separation allows for a more efficient supply of oxygen to the body.
  • Birds and mammals, being warm-blooded, have high energy needs and require a constant oxygen supply.
  • Mixing blood could lead to inefficient oxygen delivery, which is critical for their complex metabolism.

In summary, the distinct sides of the heart ensure that oxygen-rich blood is effectively circulated to all body parts, supporting the energy demands of these animals.


Q15: Write one specific function of each of the following organs in relation to excretion in human beings:  (2 Marks) (2023)
(i) Renal Artery
(ii) Urethra
(ii) Glomerulus
(iv) Tubular part of nephron

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Ans: (i) Renal artery: Carries blood from the aorta to the kidneys for filtration.
(ii) Urethra: Transports urine from the bladder to the outside of the body.

(iii) Glomerulus:  Filters blood to initiate the formation of urine.
(iv) Tubular part of nephron: Enables selective reabsorption of essential substances like glucose, amino acids, salts, and water into the blood.


Q16: Explain in brief two ways by which leaves of a plant help in excretion.  (2 Marks) (2023)

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Ans: Leaves of a plant assist in excretion in the following ways:

  • Stomata: Most carbon dioxide is released through tiny openings called stomata in the leaves.
  • Transpiration: Plants lose excess water through a process known as transpiration, which is the evaporation of water from the leaf surface.
  • Waste Storage: Many waste products are stored in leaves, which are eventually shed, allowing the plant to excrete these materials.


Q17: An organism which breaks down the food material outside the body and then absorbs it is: 
(a) a plant parasite, Cuscuta 
(b) an animal parasite, Tapeworm 
(c) a bacteria, Rhizobium 
(d) a fungi, Rhizopus    (1 Mark) (CBSE 2023)

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Ans: (d)
(i) Rhizopus is a fungus that performs extracellular digestion. It secretes digestive enzymes outside its body to break down food material into simpler substances, which are then absorbed.
(ii) Cuscuta is a plant parasite that absorbs nutrients directly from the host plant.
(iii) Tapeworm is an animal parasite that lives inside the host’s body and absorbs pre-digested nutrients.
(iv) Rhizobium is a nitrogen-fixing bacterium associated with legume roots; it does not break down food outside the body for absorption.
Therefore, (d) a fungi, Rhizopus is correct, as it digests food material outside its body and then absorbs it.


Q18: Consider the following statements about small intestine and select the one which is NOT correct: 
(a) The length of the small intestine in animals differ as it depends on the type of food they eat. 
(b) The small intestine is the site of complete digestion of food. 
(c) The small intestine receive secretions from liver and pancreas. 
(d) The villi of the small intestine absorb water from the unabsorbed food before it gets removed from the body via the anus.    (1 Mark) (CBSE 2023)

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Ans: (d)
(a) This statement is correct. The length of the small intestine varies among animals depending on their diet. Herbivores generally have longer small intestines than carnivores to allow more time for digesting plant material.
(b) This statement is correct. The small intestine is the primary site for the complete digestion of food, where nutrients are broken down and absorbed.
(c) This statement is correct. The small intestine receives secretions from the liver (bile) and the pancreas (pancreatic enzymes) that aid in digestion.
(d) This statement is incorrect. The main function of the villi in the small intestine is to absorb nutrients, not water. Water absorption primarily occurs in the large intestine, not the small intestine.
Therefore, the correct answer is (d), as it is the statement that is NOT correct.


Q19: Name the type of blood (oxygenated/ deoxygenated) transported by each of the following mentioning the path i.e., from one organ (which place) to another (which place). 
(A) Vena cava 
(B) Pulmonary artery    (3 Marks) (CBSE 2023)

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Ans: (A) Vena cava:  The vena cava consists of two main veins:

  • Superior vena cava: Carries deoxygenated blood from the upper body and arms to the right atrium.
  • Inferior vena cava: Transports deoxygenated blood from the legs and abdominal area to the right atrium.


(B) Pulmonary artery: 

  • It carries deoxygenated blood from the right ventricle to the lungs.
  • In the lungs, the blood becomes oxygenated.


Q20: (A) State the role of ATP in cellular respiration. 
(B) What ensures sufficient exchange of gases in plants? 
(C) State the conditions on which the direction of diffusion of gases in plant depend upon. (3 Marks) (CBSE 2023)

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Ans: (A) ATP in Cellular Respiration:

  • ATP is the energy currency of cells.
  • Most ATP is produced in the mitochondrial matrix during cellular respiration.
  • Each glucose molecule can generate around 32 ATP molecules.

(B) Gas Exchange in Plants:

  • Plants exchange gases through stomata.
  • Large intercellular spaces ensure all cells contact air.
  • Carbon dioxide and oxygen are exchanged by diffusion.

(C) Conditions Affecting Gas Diffusion:

  • The direction of diffusion depends on environmental conditions.
  • It also relies on the needs of the plant.
  • At night, carbon dioxide is mainly released; during the day, oxygen is produced.
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Previous Year Questions 2022

Q1: In human alimentary canal, the specific enzyme/ juice secreted in locations (i), (ii) and (iii) are  (2022)

(a) (i) Amylase (ii) Pepsin (iii) Bile
(b) (i) Amylase (ii) Bile (iii) Trypsin
(c) (i) Lipase (ii) Amylase (iii) Pepsin
(d) (i) Trypsin (ii) Bile (iii) Amylase        
         

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Ans: (b)
Saliva is secreted in the mouth (i) and contains the enzyme salivary amylase, which helps digest carbohydrates. The liver (ii) secretes bile to emulsify fats, breaking them into smaller globules for easier digestion. The pancreas (iii) produces pancreatic juice, which contains enzymes like trypsin for protein digestion.


Q2: Read the following and answer the questions from (i) to (iv).  (2022)
Take a healthy potted plant with elongated leaves. Select a leaf and insert about one half of this leaf in a test tube containing KOH and make it air tight. Place the set-up in sun for two hours. Take out the leaf from the test tube and dip it in boiling water for a few minutes. Put this leaf in a beaker with alcohol and boil it in a water bath. Wash the leaf with water and then dip the leaf in iodine solution for a few minutes. The portion of the leaf dipped in KOH solution will not show any change when dipped in iodine solution.

Previous Year Questions 2022

(i) The function of KOH taken in the test tube is to absorb
(a) Released water vapours
(b) Released CO2
(c) Released O2
(d) Chlorophyll.  

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Ans: (b)
 In the given experiment, KOH (Potassium hydroxide) in the test tube absorbs carbon dioxide; thus, due to the absence of CO2, the leaf fails to produce starch which proves that carbon dioxide is necessary for photosynthesis.

(ii) On the basis of this activity, we may conclude that the factor for photosynthesis is
(a) Carbon dioxide
(b) Oxygen
(c) Chlorophyll
(d) Water vapour.     
             

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Ans: (a)
 The given experiment demonstrates the requirement of carbon dioxide for photosynthesis. When KOH absorbs the available CO2 from the portion of leaf dipped in it, formation of sugar and starch is inhibited and the portion of leaf dipped in KOH did not show any change when dipped in iodine solution. Thus, it is proved that CO2 is necessary for photosynthesis.

(iii) The event that does not occur in photosynthesis is 
(a) Absorption of light energy by chlorophyll
(b) Reduction of carbon dioxide to carbohydrates
(c) Oxidation of carbon to carbon dioxide
(d) Conversion of light energy to chemical energy.   

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Ans: (c)
Following are the three events that occur during the process of photosynthesis
(i) Absorption of light energy by chlorophyll.
(ii) Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen.
(iii) Reduction of carbon dioxide to carbohydrates.

(iv) Iodine solution gives blue-black colour with
(a) Starch
(b) Proteins
(c) Glucose
(d) Fats. 

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Ans: (a)
 The brown coloured iodine will turn blue-black when it reacts with starch.


Q3: The correct statements with reference to single celled organisms are  (2022)
(i) complex substances are not broken down into simpler substances
(ii) simple diffusion is sufficient to meet the requirement of exchange of gases
(iii) specialised tissues perform different functions in the organism
(iv) entire surface of the organism is in contact with the environment for taking in food.
(a) (i) and (iii)    
(b) (ii) and (iii)
(c) (ii) and (iv)    
(d) (i) and (iv).                  

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Ans: (c)
 Unicellular organisms are one-celled and perform all the life processes that are essential for maintaining the life of cell or organisms like nutrition, respiration, reproduction, excretion, etc. In single-celled organisms such as Amoeba, complex substances are broken into simpler substances.


Q4: Which one of the following conditions is true for the state of stomata of a green leaf shown in the given diagram?

Previous Year Questions 2022

(a) Large amount of water flows into the guard cells.
(b) Gaseous exchange is occurring in large amount.
(c) Large amount of water flows out from the guard cells.
(d) Large amount of sugar collects in the guard cells.  

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Ans: (c)
 The opening and closing of the stomatal pore depend on the turgidity of the guard cells. When guard cells lose water, they become flaccid, leading to a closed stomatal pore. Conversely, when they swell with water, the pore opens.


Q5: Assertion (A): Nitrogen is an essential element for plant growth and is taken up by plants in the form of inorganic nitrates or nitrites.    (2022)
Reason (R) : The soil is the nearest and richest source of raw materials like nitrogen, phosphorus and other minerals for the plants. 
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.                 

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Ans: (b)
The correct answer is (b) Both (A) and (R) are true but (R) is not the correct explanation of (A). Nitrogen is indeed essential for plant growth, and plants absorb it mainly as inorganic nitrates or nitrites. However, while the soil is a rich source of nutrients for plants, the reason provided does not directly explain why nitrogen is taken up in that specific form, so it’s not a correct explanation of the assertion.


Q6: A student was asked to write a stepwise procedure to demonstrate that carbon dioxide is necessary for photosynthesis. He wrote the following steps. The wrongly worded step is   (2022)

Previous Year Questions 2022

(a) Both potted plants are kept in dark room for at least three days
(b) Bottom of the bell jars is sealed to make them air tight
(c) Both potted plants are kept in sunlight after the starch test
(d) A leaf from both the plants is taken to test the presence of starch.                   

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Ans: (c)
Both potted plants should be kept in sunlight for about two hours before the starch test.


Q7: The length of small intestine in a deer is more as compared to the length of small intestine of a tiger. The reason for this is  (2022)
(a) Mode of intake of food
(b) Type of food consumed
(c) Presence or absence of villi in intestines
(d) Presence or absence of digestive enzymes.  

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Ans: (b)
 The length of the small intestine in a deer is greater than that of a tiger due to their differing diets.

  • Herbivores, like deer, consume grass, which is high in cellulose. This requires a longer small intestine for effective digestion.
  • Carnivores, such as tigers, eat meat, which is easier to digest, allowing for a shorter small intestine.

The structure of the small intestine is adapted to the dietary needs of each animal, reflecting their evolutionary differences.


Q8: The sequence of anaerobic respiration in our muscle cells during heavy exercise is  (2022)
(a) 
(b) 
(c) 

(d)                    

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Ans: (a)
 During heavy exercise, when oxygen is scarce in our muscle cells, the process of anaerobic respiration occurs. Here’s a simplified breakdown:

  • Glucose (a six-carbon molecule) is broken down into pyruvate (a three-carbon molecule).
  • This conversion leads to a build-up of lactic acid in muscles, which can cause cramps.
  • In the absence of oxygen, pyruvate is converted into lactic acid and energy.
Previous Year Questions 2022


Q9: The energy released during cellular respiration is used to synthesise  (2022)
(a) Ribosomes 
(b) RBC
(c) ATP 
(d) mitochondria.                  

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Ans: (c)
The energy released during cellular respiration is immediately used to synthesise a molecule called ATP which is energy currency of the cell and is used as fuel for cellular activities.


Q10: Which of the following statements are correct in reference to the role of A (shown in the given diagram) during a breathing cycle in human beings?  (2022)

Previous Year Questions 2022

(i) It helps to decrease the residual volume of air in lungs.
(ii) If flattens as we inhale.
(iii) It gets raised as we inhale.
(iv) It helps the chest cavity to become larger. 
(a) (ii) and (iv) 
(b) (iii) and (iv)
(c) (i) and (ii)
(d) (i), (ii) and (iv)                  

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Ans: (a)

During inhalation, we lift our ribs and flatten our diaphragm (A), which enlarges the chest cavity. This creates a vacuum that pulls air into the lungs, filling the expanded alveoli.When we exhale, our ribs and diaphragm return to their normal positions, causing the chest cavity to shrink. This pushes air out of the lungs.

  • The lungs always retain a residual volume of air, ensuring there is enough time for oxygen absorption and carbon dioxide release.
  • This residual volume remains constant under normal conditions.


Q11: Assertion (A): The rate of breathing in aquatic organisms is much slower than that seen in terrestrial organisms.  (2022)
Reason (R): The amount of oxygen dissolved in water is very low as compared to the amount of oxygen in air.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.  

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Ans: (d)

The rate of breathing in aquatic organisms is much slower than that seen in terrestrial organisms.The rate of breathing in aquatic organisms is actually much faster than in terrestrial organisms due to the following reasons:

  • The amount of dissolved oxygen in water is relatively low compared to the oxygen available in the air.
  • Aquatic animals must breathe rapidly to obtain enough oxygen from the water.
  • Fish, for example, take in water through their mouths and push it past their gills, where oxygen is absorbed into their blood.


Q12: The function of the lining of mucus in the nasal passage of human beings is to  (2022)
(a) Increase the temperature of inhaled air
(b) Move the air in and out
(c) Filter the air that we breathe in
(d) Absorb oxygen from the air.  

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Ans:  (c)
 The mucus lining in the nasal passage serves several important functions:

  • It helps to moisten the air we breathe.
  • It warm the inhaled air.
  • It traps dust particles and other impurities, acting as a filter.


Q13: In living organisms during respiration which of the following products are not formed if oxygen is not available?  (2022)
(a) Carbon dioxide + Water
(b) Carbon dioxide + Alcohol
(c) Lactic acid + Alcohol
(d) Carbon dioxide + Lactic Acid                  

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Ans: (a)
In aerobic respiration, glucose is completely broken down to CO2 and H2O with the production of  a large amount of energy (ATP).

Previous Year Questions 2022

Thus, CO2 and water are formed as a result of aerobic ; respiration and will not formed in the absence of oxygen (anaerobic respiration).


Q14: Respiratory structures of two different animals-a fish and a human being are shown.  (2022)
Observe (A) and (B) and select one characteristic that hold true for both of them.

Previous Year Questions 2022

(a) Both are placed internally in the body of animal.
(b) Both have thin and moist surface for gaseous exchange.
(c) Both are poorly supplied with blood vessels to conserve energy.
(d) In both the blood returns to the heart after being oxygenated.                 

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Ans: (b)

  • Both fish and humans have respiratory structures that facilitate gas exchange.
  • Both possess thin and moist surfaces for efficient gas exchange.
  • In fish, this occurs in the gills, while in humans, it takes place in the alveoli.
  • This structure allows for the diffusion of oxygen into the blood and the removal of carbon dioxide.


Q15: Observe the diagram of an activity given below. What does it help to conclude, when the person exhales into the test-tube?  (2022)

Previous Year Questions 2022

(a) Percentage of carbon dioxide is more in inhaled air.
(b) Fermentation occurs in the presence of oxygen.
(c) Percentage of carbon dioxide is more in the exhaled air.
(d) Fermentation occurs in the presence of carbon dioxide.              

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Ans: (c)

  • When a person exhales into a test tube containing lime water, the following occurs:
  • The lime water turns milky due to the formation of a white precipitate.
  • This change indicates that the air we exhale contains a higher level of carbon dioxide.
  • The reaction occurs because exhaled air has more carbon dioxide compared to inhaled air.


Q16: The separation of the right side and the left side of heart is useful to  (2022)
(a) Keep oxygenated blood from mixing with deoxygenated blood
(b) Allow a slow supply of oxygen in the body
(c) Supply energy to animals with low energy needs
(d) Often change their body temperature.                  

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Ans: (a)

  • The right side handles deoxygenated blood, sending it to the lungs for oxygen.
  • The left side manages oxygenated blood, distributing it to the body.
  • This separation prevents mixing of the two blood types, ensuring a more efficient supply of oxygen.
  • It is particularly beneficial for animals with high energy needs, such as birds and mammals, which require constant energy to maintain their body temperature.


Q17: In spring, sugar stored in root or stem tissue of plants is transported to the buds for  (2022)
(a) The energy needs of the buds to grow
(b) Temperature regulation
(c) Balancing the storage in different organs
(d) Diffusion process.                 

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Ans: (a)
 In spring, new buds are formed. These buds need more energy to grow than other parts of the plant. This energy comes from the sugar stored in root or stem tissue and is transported to the buds by phloem.


Q18: Upward movement of water in tall trees is due to  (2022)
(a) Translocation
(b) Excretion
(c) Photosynthesis
(d) Transpiration.               

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Ans: (d)
The upward movement of water in tall trees is primarily driven by transpiration. This process involves:

  • Water evaporating from the stomata in the leaves.
  • This evaporation creates a negative pressure in the leaves and xylem tissues.
  • As a result, water is pulled up from the roots through the xylem.
    In summary, transpiration not only aids in the movement of water but also helps regulate the plant’s temperature.


Q19: Thin walled blood vessels are called  (2022)
(a) Aorta
(b) Capillaries
(c) Arteries 
(d) Vena cava       

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Ans: (b)

Capillaries are tiny blood vessels with very thin walls, just one cell thick. They play a crucial role in the body by:

  • Allowing the exchange of materials between blood and surrounding cells.
  • Facilitating the transfer of oxygen and nutrients to tissues.
  • Collecting waste products from cells for removal.

After the exchange, capillaries merge to form veins, which carry blood away from the tissues.


Q20: Consider the following statements in connection with the functions of the blood vessels marked A and B in the diagram of a human heart as shown.  (2022)

Previous Year Questions 2022

(i) Blood vessel A – It carries carbon dioxide rich blood to the lungs.
(ii) Blood vessel B – It carries oxygen rich blood from the lungs.
(iii) Blood vessel B – Left atrium relaxes as it receives blood from this blood vessel.
(iv) Blood vessel A – Right atrium has thick muscular wall as it has to pump blood to this blood vessel.
The correct statements are
(a) (i) and (ii) only 
(b) (ii) and (iii) only
(c) (ii), (iii) and (iv)
(d) (i), (ii) and (iii).              

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Ans: (d)
 Blood vessel A represents the pulmonary artery, while blood vessel B signifies the pulmonary vein.
The key functions of these blood vessels are:

  • Pulmonary artery: Carries deoxygenated blood from the right ventricle of the heart to the lungs for oxygenation.
  • Pulmonary vein: Transports oxygen-rich blood from the lungs to the left atrium of the heart.

Thus, oxygen-rich blood from the lungs enters the heart via the pulmonary vein, ensuring efficient oxygen supply to the body.


Q21: Identify the two components of phloem tissue that help in transportation of food in plants.  (2022)
(a) Phloem parenchyma and sieve tubes
(b) Sieve tubes and companion cells
(c) Phloem parenchyma and companion cells
(d) Phloem fibres and sieve tubes  

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Ans: (b)
 The two components of phloem tissue that assist in the transportation of food in plants are:

  • Sieve tubes
  • Companion cells

These components work together to move nutrients, such as sugars, throughout the plant.


Q22: Read the following and answer the questions from (i) to (iv).  (2022)
The figure shown below represents a common type of dialysis called as haemodialysis- It removes waste products from the blood, such as excess salts, and urea which are insufficiently removed by the kidney in patients with kidney failure. During the procedure, the patient’s blood is cleaned by filtration through a series of semi-permeable membranes before being returned to the blood of the patient. On the basis of this answer the following questions.

Previous Year Questions 2022

(i) The haemodialyser has semi-permeable lining c tubes which help
(a) To maintain osmotic pressure of blood
(b) To filter nitrogenous wastes from the dialysin solution
(c) In passing the waste products in the dialysing solution
(d) To pump purified blood back into the body of the patient.          

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Ans: (c) 

  • The haemodialyser contains tubes with a semi-permeable lining that serve several important functions:
  • They help to filter waste products like urea and creatinine from the blood.
  • They allow the passage of these waste products into the dialysing fluid.
  • This process mimics the kidney’s function, although it does not involve re-absorption of substances.
  • Finally, the purified blood is pumped back into the patient’s body.

(ii) Which one of the following is not a function of artificial kidney?
(a) To remove nitrogenous wastes from the blood
(b) To remove excess fluids from the blood.
(c) To reabsorb essential nutrients from the blood.
(d) To filter and purify the blood.               

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Ans: (c)

  • An artificial kidney does not reabsorb essential nutrients from the blood.
  • Its primary functions include:
  • Removing nitrogenous wastes from the blood.
  • Eliminating excess fluids from the blood.
  • Filtering and purifying the blood.

(iii) The ‘used dialysing’ solution is rich in
(a) Urea and excess salts
(b) Blood cells
(c) Lymph    
(d) Proteins.                

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Ans: (a)

  • The used dialysing solution is rich in waste products.
  • It contains high levels of urea and excess salts.
  • During dialysis, the patient’s blood flows through tubes with a semi-permeable lining.
  • As blood passes through, waste products diffuse into the dialysing fluid.
  • This process helps to purify the blood before it is returned to the patient.

(iv) Which part of the nephron in human kidney, serves the function of reabsorption of certain substances
(a) Glomerulus
(b) Bowman’s Capsule
(c) Tubules
(d) Collecting Duct                  

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Ans: (c)

  • The tubules of the nephron are responsible for the selective reabsorption of essential substances.
  •  These substances include: Glucose, Amino acids, Salts, Water
  • This process allows these useful substances to return to the blood capillaries.

Previous Year Questions 2021

Q1: (i) Plants absorb water from the soil. Explain how it is taken up and transported from the soil.  (2021)
(ii) “When we are injured and start bleeding, it requires the loss of blood from the system to be minimized.” What will happen if the blood loss is not stopped? Is there anything the system could do on its own to prevent the loss?

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Ans: (i)  

  • Minerals and water needed by the plants are absorbed by roots. 
  •  The root hairs absorb water from the soil by the process of osmosis and take in minerals by the process of diffusion. 
  •  Thus, a difference in concentration of ions is created between the roots and the soil, which enables the water to enter into the roots to compensate for the difference in concentration. 
  •  This water, along with dissolved minerals from the root hairs, passes into xylem vessels through cells of the cortex, endodermis, and pericycle, and then ascent of sap takes place from the xylem of roots into the xylem of stem and finally leaf veins through vessels and tracheids.

(ii) Bleeding leads to a loss of pressure, which reduces the efficiency of the pumping system. When an injury is caused, the blood platelets release certain chemicals which are called the platelet factors (e.g., thromboplastin). These platelet factors help in the clotting of blood.

Also watch: What are Life Processes

Previous Year Questions 2020

Q1: In the human body, the site of absorption of digested food is the small intestine. How is the process of absorption carried out, and why is the absorption of digested food necessary?  (2020)

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Ans:  

  •  The small intestine is the main region for the absorption of digested food. 
  •  It consists of the following parts: 
  •  The inner surface of the small intestine has millions of tiny finger-like projections called villi. 
  •  These villi increase the surface area for efficient food absorption. 
  •  Within these villi, many blood vessels are present that absorb the digested food and carry it to the bloodstream. 
  •  The small intestine in human beings is the site of the complete digestion of food like carbohydrates, proteins, and fats. 
  •  The absorption of digested food is necessary because an organism takes the complex organic materials by the process of absorption and then utilizes absorbed nutrients for various metabolic processes. 
  •  With the help of these metabolic processes, the body uses these food nutrients for energy, growth, and cell repair. 


Q2: Complete the following flow chart as per the given instructions.   (2020)
Previous Year Questions 2020

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Ans: a – Hydrochloric acid (HCI)
b – Pepsin
c – Mucus
d – HCI makes medium acidic for the activation of an enzyme pepsin.
e – Pepsin acts in an acidic medium, which breaks down proteins into peptones.
f – The mucus protects the inner lining of the stomach from the corroding action of HCI.


Q3: (a) State the role played by the following in the process of digestion:  (2020)
(i) Enzyme trypsin
(ii) Enzyme lipase
(b) List two functions of finger-like projections present in the small intestine. 

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Ans: (a) (i) Enzyme trypsin: This enzyme is produced by the pancreas in an inactive form called trypsinogen.
Trypsin converts the remaining proteins into peptones, and the peptones into peptides and amino acids.
(ii) Enzyme lipase: It is secreted by the pancreas and small intestine. Lipase converts fats into fatty acids and glycerol.
(b) Internally, the wall of the small intestine is provided with long finger-like projections called villi. Two functions of villi are:
(i) The villi greatly increase the absorptive surface area of the inner lining of the small intestine.
(ii) Villi are richly supplied with blood vessels that carry absorbed food to all cells of the body, where it is utilized for obtaining energy.


Q4: (a) In the process of respiration, state the function of alveoli.  (2020)
(b) The rate of breathing in aquatic organisms is much faster than that in terrestrial organisms. Give reasons.
(c) Complete the following pathway showing the breakdown of glucose.

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Ans: (a) Functions of alveoli are:
(i) They increase the surface area for the exchange of gases.
(ii) The thin walls of alveoli facilitate the rapid exchange of oxygen and carbon dioxide between alveolar air and blood.
(b) Aquatic animals like fishes obtain dissolved oxygen from water present through their gills. The amount of dissolved oxygen is quite small compared to the amount of oxygen in the air. Therefore, to obtain the required amount of oxygen from water, aquatic animals have to breathe much faster than terrestrial organisms.
(c)


Q5: Give reasons:  (2020)
(a) Ventricles have thicker muscular walls than atria.
(b) The transport system in plants is slow.
(c) Circulation of blood in aquatic vertebrates differs from that in terrestrial vertebrates.
(d) During the daytime, water and minerals travel faster through the xylem as compared to the night.
(e) Veins have valves, whereas arteries do not.

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Ans: (a) Since ventricles have to pump blood into various organs with high pressure, they have thicker walls than atria.
(b) The transport system in plants is slow due to:

  • A less elaborate system compared to animals.
  • Plants are generally less active, so their cells do not require rapid material supply.

(c) 

  •  Circulation of blood in aquatic vertebrates differs from terrestrial vertebrates because: 
  • Aquatic vertebrates, like fish, have gills for oxygenation and exhibit single circulation, where blood passes through the heart once per cycle. 
  • Terrestrial vertebrates, like birds and humans, have lungs and use double circulation, with blood passing through the heart twice per cycle. 

(d) During the daytime, water and minerals travel faster through the xylem because the rate of transpiration is higher during the daytime.
(e) The lumen of veins has valves, which allow the blood to flow in only one direction. Thus preventing the backflow of blood.


Q6: (a) “Blood circulation in fishes is different from the blood circulation in human beings”. Justify the statement. 
(b) Describe “blood circulation” in human beings.  (2020)

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Ans: (a) 

  • Fishes have only two chambers in their heart. The blood is pumped to the gills for oxygenation, and from there, it passes directly to the rest of the body. Thus, the blood goes only once through the heart during one cycle of passage through the body. This type of circulation is termed as single circulation. 
  •  In human beings, during circulation, blood travels twice through the heart in one complete cycle of the body, which is called double circulation. 
  •  The pathway of blood from the heart to the lungs and back to the heart is called pulmonary circulation, and the pathway of blood from the heart to the rest of the body and back to the heart is called systemic circulation. 

(b)

  • In human beings, deoxygenated blood from the body tissues is poured into the right atrium.
  • Contraction of the heart forces it into the right ventricle.
  • From the right ventricle, deoxygenated blood flows to the lungs through the pulmonary artery.
  • Oxygenated blood from the lungs is returned to the left atrium and then into the left ventricle.
  • The left ventricle forces the oxygenated blood to the whole body.
  • Thus, to make one complete round or circulation circuit around all body parts, the blood passes through the heart twice.
  • This is known as the double circulation of blood.


Q7: (a) Describe the structure and function of the basic filtering unit of the kidney.  (2020)
(b) List two factors on which the reabsorption of water from urine depends.

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Ans: (a) Nephron is the structural and functional unit of the kidney. Its main role is to regulate the concentration of water and soluble substances like sodium and other salts by filtering the blood, reabsorbing the excess water that is needed, and excreting the rest as urine.
(b) Two factors on which the reabsorption of water from urine depends are:

  • The amount of excess water which is present in the body.
  • The amount of dissolved waste that needs to be excreted out of the body.


Q8: (a) Name the organs that form the excretory system in human beings.  (2020)
(b) Describe in brief how urine is produced in the human body. 

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Ans: (a) The Excretory system (urinary system) in human beings consists of a pair of kidneys, a pair of ureters, a urinary bladder and a urethra.
(b) 

  •  In the kidney, the wastes are converted to urine by three processes: 
  • (i) Ultrafiltration: A large amount of water, along with certain harmful substances like ureauric acidK, ammonium salts, creatinine, etc., and certain useful substances like glucose, amino acids, Na, etc., pass through glomerular capillaries and glomerular membrane into the cavity of Bowman’s capsule of nephrons under pressure. The filtrate so formed is called nephric filtrate, which is moved towards the ureter. 
  • (ii) Selective reabsorption: A large amount of water and sodium, a whole of glucose and amino acids, and a small amount of urea are passed back from the nephric filtrate into the blood capillaries. It occurs either by back diffusion (i.e., water and urea) or active transport (i.e., Na, glucose, and amino acids). It generally occurs in the PCT (Proximal convoluted tubule) of nephrons. 
  • (iii) Tubular secretion: Certain harmful chemicals like uric acid, creatinine, K, etc., are passed from blood capillaries surrounding the nephron into nephric filtrate by active transport. It generally occurs in the DCT (Distal convoluted tubule) of nephrons. Now, the fluid is termed as urine and is excreted out of the excretory organs. 


Q9: (a) Why is nutrition necessary for the human body?  (CBSE 2020)
(b) What causes the movement of food inside the alimentary canal?
(c) Why is the small intestine in herbivores longer than in carnivores?
(d) What will happen if mucus is not secreted by the gastric glands?

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Ans: (a) 

  • Nutrition is necessary for the human body because the human body continuously requires energy for life activities like respiration, circulation, excretion, etc.
  • Energy is required even when we are sleeping because a number of biological processes keep on occurring.
  • All these processes require energy, and this energy is obtained from nutrition.
  • It is also needed for the growth and repair of the human body.

(b)

  • The wall of the alimentary tract contains muscles that can contract and expand alternately.
  • The contraction and expansion movement of the wall of the food pipe is called peristaltic movement.
  • The peristaltic movement moves the partially digested food in all the digestive organs throughout the alimentary canal.

(c)

  • The small intestine in herbivores is longer than in carnivores because:
  • Herbivores consume plants rich in cellulose, which takes longer to digest.
  • They rely on enzymes from symbiotic bacteria to break down cellulose.
  • Carnivores eat flesh, which is easier to digest, resulting in a shorter small intestine.

(d) If mucus is not secreted by the gastric glands:

  • The inner lining of the stomach would be exposed to hydrochloric acid and digestive enzymes.
  • This could lead to erosion of the stomach lining, causing acidity and potentially ulcers.


Q10: (a) Why is there a difference in the rate of breathing between aquatic organisms and terrestrial organisms? Explain.  (2020)
(b) Draw a diagram of the human respiratory system and label it – the pharynx, trachea, lungs, diaphragm and alveolar sac on it. 

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Ans: (a) 

  •  The difference in the rate of breathing between aquatic and terrestrial organisms is primarily due to their oxygen sources: 
  •  Terrestrial organisms breathe in oxygen from the atmosphere, which contains about 21% oxygen. 
  •  Aquatic organisms rely on oxygen that is dissolved in water, which is less than 1%. 
  • Oxygen diffuses through water much slower than in air. 
  •  As a result, terrestrial organisms can obtain oxygen more easily and quickly. 
  •  Aquatic organisms must work harder to extract the same amount of oxygen, leading to a faster breathing rate. 

(b) The labelled diagram of the human respiratory system is as follows
Previous Year Questions 2020


Q11: (A) Write the correct sequence of steps followed during journey of oxygen rich blood from lungs to various organs of human body. (B) What happens when the system of blood vessels develop a leak? (CBSE 2020)

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Ans: (A) Journey of oxygen rich blood from lungs to various parts of the body is as follows: 
(i) The pulmonary vein brings the oxygenated blood from the lungs in the left atrium of the heart. 
(ii) The left atrium contracts and pumps blood into left ventricle through the valve. 
(iii) When the left ventricle contracts, the oxygenated blood enters the largest artery known as aorta. 
(iv) The blood travels from the aorta to larger and smaller arteries into the capillaries. 
(v) The aorta transports the blood to all the organs of the body except the lungs. 
(vi) The oxygenated blood releases oxygen, nutrients and other substances and takes away carbon dioxide and waste substances.
(vii) The deoxygenated blood enters the vena cava, which carries it to the right atrium of the heart
Previous Year Questions 2020

(B) When the system of blood vessels develop a leak, the platelets will increase, which will minimise the leakage. Whenever there is a leak in the blood vessels, the blood flows, through them. If not checked, that may cause an excessive loss of blood.

Previous Year Questions 2019

Q1: Name an enzyme present in pancreatic juice.  (2019)

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Ans: Pancreatic juice contains the digestive enzyme trypsin.


Q2: What causes the movement of food inside the alimentary canal in human beings?  (2019)

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Ans:

  • The movement of food in the alimentary canal is primarily due to peristalsis. This is the rhythmic contraction and relaxation of the muscles in the canal, which helps to push food downwards.
  • Peristalsis occurs throughout the digestive tract.
  • Food travels from the mouth to the stomach via the oesophagus.
  • The stomach’s muscular walls mix food with digestive juices.


Q3: (a) Write the function of the following in the human alimentary canal:  (CBSE 2019)
(i) Saliva
(ii) HCI in the stomach
(iii) Bile juice
(iv) Villi
(b) Write one function for each of the following enzymes:
(i) Pepsin
(ii) Lipase   

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Ans: (a) (i) Saliva contains salivary amylase and is released in our mouth. It breaks down starch into sugar (complex carbohydrates into simpler ones).
(ii) Acid (HCI) plays an important role in the process of digestion. These are: 

  • Hydrochloric acid (HCl) creates an acidic medium in the stomach, which is essential for the activation and functioning of the gastric enzyme pepsin.
  • HCI kills the harmful bacteria present in the food. 
  • Bile brings about the emulsification of fat (i.e., breaks fat molecules into small globules). 
  • Internally, the wall of the small intestine is provided with long finger-like projections called villi. Two functions of villi are: 
  • The villi greatly increase the absorptive surface area of the inner lining of the small intestine. 
  • Villi are richly supplied with blood vessels that carry absorbed food to all cells of the body, where it is utilized for obtaining energy. 

(b) (i) Pepsin is a protein-digesting enzyme present in gastric juice. Pepsin gets activated in an acidic medium and splits proteins into peptones and peptides.
(ii) Lipase is secreted by the pancreas and small intestine. Lipase converts fats into fatty acids and glycerol.


Q4: Write two different ways in which glucose is oxidized to provide energy in the human body. Write the products formed in each case.  (2019)

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Ans: The two different ways by which glucose is oxidized to provide energy in the human body are:
(i) Aerobic respiration (In the presence of oxygen): The end products in aerobic respiration are carbon dioxide, water, and energy.
Previous Year Questions 2019
(ii) Anaerobic respiration (In the lack of oxygen): The end products are lactic acid and energy.
Previous Year Questions 2019


Q5: Explain the ways in which glucose is broken down in the absence or shortage of oxygen.  (2019)

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Ans: In the absence or shortage of oxygen, glucose is oxidized in the following ways:
(i) In cytoplasm, glucose is oxidized to pyruvate and release energy.
(ii) In the muscles of humans, i.e., under anaerobic conditions, pyruvate breaks down to lactic acid and releases energy.
(iii) In yeast, i.e., during the fermentation process, pyruvate converts to ethanol while carbon dioxide releases a certain amount of energy.


Q6: (a) List in tabular form two differentiating features between xylem and phloem.   (2019)
(b) Write two advantages of transpiration in plants. 

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Ans: (a) Differences between transport in the xylem and transport in the phloem are as follows:
Previous Year Questions 2019
(b) Significance of transpiration in plants: 
(i) The absorbed water is transported from roots to leaves through xylem vessels, which is greatly influenced by transpiration pull.
(ii) The water stream moving upwards carries dissolved minerals with it. Transpiration also helps in distributing these minerals throughout the plant.
(iii) The evaporation of water during transpiration provides a cooling effect to the leaves.


Q7: Write three types of blood vessels. Give one important feature of each.  (2019)

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Ans: The three types of blood vessels in the human body are:
(i) Arteries are the blood vessels that carry blood from the heart to various parts of the body. They are thick-walled, elastic, and muscular, which enables them to dilate but do not rupture when the heart contracts and forces blood into them.
(ii) Veins are thin-walled blood vessels that bring blood from the body back to the heart. They are larger and hold more blood than the arteries. The lumen of veins has valves which prevent the backflow of blood.
(iii) Capillaries are thin-walled and extremely narrow blood vessels that occur at the terminals of arteries and veins. The walls of capillaries are permeable to water and dissolved substances so that the exchange of materials between the blood and body cells can take place.


Q8: (a) Write two water-conducting tissues present in plants. How does water enter continuously into the root xylem?   (2019)
(b) Explain why plants have low energy needs as compared to animals.

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Ans: (a)  

  • Xylem tracheids and vessels are two water-conducting tissues present in plants that help in the rapid movement of water. 
  •  In xylem tissue, vessels and tracheids of the roots, stems, and leaves are interconnected to form a continuous system of water-conducting channels reaching all parts of the plant. 
  • Minerals and water needed by the plants are absorbed by root hairs from the soil by the process of osmosis and take in minerals by the process of diffusion. 
  •  Thus, a difference in concentration of ions is created between the roots and the soil, which enables the water to enter into the roots to compensate for the difference in concentration. 
  •  The water, along with dissolved minerals from root hairs, passes into xylem vessels through cells of the cortex, endodermis, and pericycle and then ascent of sap (i.e., upward movement of water and mineral salts from roots to the aerial parts of the plant against the gravitational force) takes place from xylem of the roots to the xylem of stem and leaves through vessels and tracheids. 
  • Evaporation of water molecules from the cells of leaves creates a suction pressure that pulls the water from the xylem cells. 

(b) Plants are autotrophic and do not have to move from one place to another in search of their food. Movements in a plant occur at the cellular level, and hence, they require less energy. Animals are heterotrophic and move in search of food and other activities; hence, they require a higher amount of energy than plants.


Q9: List four functions of the human heart. Why is double circulation necessary in the human body?  (2019)

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Ans: The functions of the human heart include:

  • Maintaining blood pressure: The heart ensures that blood is pumped throughout the body effectively.
  • Pumping vital substances: It delivers essential nutrients and substances to various body parts.
  • Circulating oxygen-rich blood: The heart sends oxygenated blood from the lungs to the entire body and carries carbon dioxide-rich blood back to the lungs for purification.
  • Removing metabolic waste: It helps eliminate waste products, such as carbon dioxide, from body tissues.

The human heart consists of two sides: the right side and the left side. The right side receives deoxygenated blood and sends it to the lungs for purification. The left side receives oxygenated blood from the lungs and pumps it to the rest of the body. This process is known as double circulation.
Double circulation is necessary because:

  • It separates oxygenated and deoxygenated blood, which is crucial for meeting the high energy demands of the body.
  • This separation allows for efficient oxygen delivery to tissues, supporting the body’s metabolic needs.


Q10: (a) Define the term excretion. Why should animals excrete waste matter?  (2019)
(b) Name the main excretory organ of human beings and state the form in which the excretory matter is thrown out of the body.

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Ans: (a)  Excretion is a biological process that removes harmful metabolic wastes from the body. Key points include:

  • It eliminates substances like nitrogenous waste, which includes urea, salts, and water.
  • Accumulation of these wastes can be harmful to the body.
  • The main excretory organs in humans are the kidneys.
  • Kidneys filter blood to produce urine, which carries away waste.

Overall, excretion is essential for maintaining the body’s health by preventing the buildup of toxic substances.

(b) A pair of kidneys are the main excretory organs in a human being. It helps excrete nitrogenous waste in the form of urea, toxins, excess salts, water-soluble vitamins, etc. and then eliminates it in the form of urine.


Q11: In the experimental set-up to show that “CO2 is given out during respiration” name the substance taken in the small test tube kept in the conical flask. State its function and the consequence of its use.  (2019)

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Ans: In the experiment in which carbon dioxide is given out during respiration, KOH (potassium hydroxide) solution or pellets are taken in a tube and placed within the conical flask. KOH absorbs the carbon dioxide, and it prevents the carbon dioxide from being utilized by the plant for the process of photosynthesis. Function of KOH
KOH: KOH solution kept in the flask absorbs carbon dioxide and creates a partial vacuum in the flask. The air present in the bent tube moves into a conical flask, and this pulls the water level up in the tube. Consequences:

  1. The whole amount of CO2 liberated gets absorbed by KOH to prevent it from being utilized by the plant for the process of photosynthesis.
  2. This CO2 absorption leads to creating a vacuum in a flask that causes an observable rise in the water level of the connected U tube.
  3. Any gas produced can neither escape nor can get outside air in.
  4. Germinating seed produces CO2 via the following reactions: Aerobic respiration: C6H12O+ 6O2 → 6CO2 + 6H2O + 38ATP


Q12: In the experimental setup shown: 

(A) Name the material X filled in the small test tube and the material Y placed at the bottom of the conical flask. 
(B) Why is there a rise in water level in the delivery tube?   (CBSE 2019)

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Ans: (A) X – KOH pellets 
Y – Wet germinating seeds
(B) Seeds use oxygen present in the flask and release carbon dioxide which is absorbed by potassium hydroxide. Thus, partial vacuum is created in the conical flask, as a result, water from the beaker rises in the delivery tube


Q13: Write the function of the following in the human alimentary canal:  (CBSE 2019)
(i) Saliva
(ii) HCI in the stomach
(iii) Bile juice
(iv) Villi

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Ans: (i) Saliva contains salivary amylase and is released in our mouth. It breaks down starch into sugar (complex carbohydrates into simpler ones).
(ii) Acid (HCI) plays an important role in the process of digestion. These are: 

  • Hydrochloric acid (HCl) creates an acidic medium in the stomach, which is essential for the activation and functioning of the gastric enzyme pepsin.
  • HCI kills the harmful bacteria present in the food. 

(iii) Bile brings about the emulsification of fat (i.e., breaks fat molecules into small globules). 
(iv) Internally, the wall of the small intestine is provided with long finger-like projections called villi. Two functions of villi are: 

  • The villi greatly increase the absorptive surface area of the inner lining of the small intestine. 
  • Villi are richly supplied with blood vessels that carry absorbed food to all cells of the body, where it is utilized for obtaining energy. 


Q14: Give reasons for the following: 
(A) There is a difference in the rate of breathing between aquatic organisms and terrestrial organisms. 
(B) Plants have low energy needs as compared to animals. (CBSE 2019)

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Ans: (A) The rate of breathing differs between aquatic and terrestrial organisms due to their environments:

  • Aquatic organisms breathe faster because they extract dissolved oxygen from water.
  • Water contains less oxygen than air, necessitating a quicker breathing rate.
  • Terrestrial organisms breathe more slowly as they obtain oxygen from the air, which is about 21% oxygen.

(B)  Plants have lower energy needs compared to animals for several reasons:

  • Plants are stationary and do not move, resulting in less energy expenditure.
  • They are generally less active than animals.
  • Many plant tissues contain a large amount of dead cells that provide structural support.
  • As a result, plant cells do not require rapid supply of materials, leading to lower energy needs and a slower respiration rate.

Previous Year Questions 2018

Q1: (a) Mention any two components of blood.  (2018)
(b) Trace the movement of oxygenated blood in the body.
(c) Write the function of valves present in between atria and ventricles.
(d) Write one structural difference between the composition of arteries and veins. 

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Ans: (a) Two components of blood are blood plasma and blood corpuscles.
(b) Oxygenated blood from lungs → Pulmonary veins → Left atrium of heart → Left ventricle → Aorta → Arteries → Body parts
(c) When blood is pumped, valves prevent the backflow of blood between ventricles and atria. They open and allow the right amount of blood to flow from one chamber to the other.
(d) Structural difference between veins and arteries is as follows:

Previous Year Questions 2018


Q2: (a) Name the process and explain the type of nutrition found in green plants. List the raw materials required for this process. Give a Chemical equation for the mentioned process.  (2018)
(b) Write three events that occur during this process.  

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Ans: (a) Photosynthesis is the process where green plants show autotrophic nutrition. The raw materials required for photosynthesis are carbon dioxide water, and energy in the form of sunlight.
The equation is as follows:
Previous Year Questions 2018

(b) The three events in photosynthesis are:

  • Absorption of light energy by chlorophyll.
  • Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen.
  • Reduction of carbon dioxide to carbohydrates.


Q3: (a) Write the reaction that occurs when glucose breaks down anaerobically in yeast.   (2018)
(b) Write the mechanism by which fishes breathe in water. 
(c) Name the balloon-like structures present in the lungs. List its two functions. 
(d) Name the respiratory pigment and write its role in human beings. 

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Ans: (a) Previous Year Questions 2018

(b) Fishes breathe with the gills by diffusion.
(c) Alveoli are the balloon-like structures:

  • They provide a surface for the exchange of gases.
  • They contain a residual volume of air so that there is sufficient time for the exchange of gases.

(d) Haemoglobin is the respiratory pigment in humans. It transports a major part of oxygen and some amount of carbon dioxide through the blood.


Q4: (a) Mention any two components of blood.  (2018)
(b) Trace the movement of oxygenated blood in the body.
(c) Write the function of valves present in between atria and ventricles.
(d) Write one structural difference between the composition of arteries and veins.

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Ans: (а) Blood is composed of plasma and three types of cells: red blood cells, white blood cells, and platelets.
(b) 

  • Oxygenated blood from the lungs is brought to the left atrium by pulmonary veins.
  • When the atrium contracts, blood is transferred to the left ventricle.
  • When the ventricle contracts, blood is pushed into the aorta and through arteries to all parts of the body.

(c) The function of valves between the atria and ventricles is to:

  • Prevent backflow of blood during pumping
  • Ensure the correct amount of blood flows from one chamber to another

(d) 


Q5: (а) Define excretion.  (2018)
(b) Name the basic filtration unit present in the kidney.
(c) Draw the excretory system in human beings and label the following organs of the excretory system that perform the following functions:
(i) form urine.
(ii) is a long tube that collects urine from the kidney.
(iii) store urine until it is passed out.

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Ans:
(a) Excretion is the biological process of removal of harmful metabolic wastes from the body.
(b) Nephrons
(c) (i) Kidney, (ii) Ureter, (iii) Urinary bladder.

Excretory System of Human BeingsExcretory System of Human Beings

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Previous Year Questions 2017

Q1: Diffusion is insufficient to meet the oxygen requirement of multicellular organisms like humans. State reason.  (2017)

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Ans:  Due to the higher metabolic rate and the large volume of the body, oxygen cannot diffuse into all cells of the human body quickly as oxygen will have to travel large distances to reach each and every cell. So, diffusion is insufficient to meet the oxygen demand of multicellular organisms.


Q2: Name the vein that brings blood to the left atrium from the lungs.  (2017)

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Ans: The pulmonary vein carries oxygenated blood from the lungs to the left atrium of the heart.


Q3: Explain the significance of photosynthesis. Write the balanced chemical equation involved in the process.   (2017)

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Ans: Photosynthesis is important for a number of reasons:
(i) Food: By photosynthesis, green plants synthesize food from simple raw materials like CO2 and H2O. Thus, it sustains life on earth.
(ii) Oxygen: Oxygen released during the process of photosynthesis is needed by animals and humans for respiration. Oxygen also supports the combustion of fuels. The balanced chemical equation involved in the process of photosynthesis is given as:
Previous Year Questions 2017


Q4: Differentiate between autotrophs and heterotrophs and give one example of each.  (2017)

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Ans: Differences between autotrophs and heterotrophs are as follows:
Previous Year Questions 2017


Q5: Explain how the translocation of materials in phloem tissue in plants is achieved by utilizing energy.  (2017)

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Ans:

  • The phloem cells transport the soluble food materials to all parts of the plant.
  • The transport of food from leaves to different parts of the plant is termed translocation.
  • Components of phloem are sieve tubes, companion cells, phloem parenchyma, and phloem fibres.
  • The food is manufactured in the mesophyll cells (or photosynthetic cells) of a leaf.
  • The manufactured food enters into sieve tubes of the phloem and is transported as a dilute aqueous solution either in an upward or downward direction.
  • Food is transported to all non-green parts of the plant for their growth and metabolic activities.
  • Besides food molecules, the phloem also transports amino acids, hormones synthesized in the shoot tips and root tips, and other metabolites.
  • In this process, glucose is transferred to phloem tissue using energy from ATP.
  • This increases the osmotic pressure of the tissue, causing the water to move into it (endosmosis).
  • Soluble material is then transferred from phloem tissue to other tissues which have less pressure than in the phloem.
  • Thus, according to the plant’s requirement, the material is translocated from higher osmotic pressure areas to lower osmotic pressure areas.


Q6: Draw a diagram of the human excretory system and label kidney ureters on it.  (2017)

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Ans: The diagram of the human excretory system is as follows:
Previous Year Questions 2017


Q7: (a) Write the reaction that occurs when glucose breaks down anaerobically in yeast.   (2017)
(b) Write the mechanism by which fishes breathe in water. 
(c) Name the balloon-like structures present in the lungs. List its two functions. 
(d) Name the respiratory pigment and write its role in human beings.

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Ans:  (a) Glucose breaks down anaerobically i.e., in the absence of O2; in yeast in the following manner:

Previous Year Questions 2017

(b) Since fishes are aquatic, they take in O2 dissolved in water. They take water through their mouth, which goes to their gills, where the dissolved oxygen is taken up by the blood and transported. The CO2 is given out through the same passage.

(c) The balloon-like structures in our lungs are the alveoli.
Their functions are:

  • They are very thin-walled and hence help in gaseous exchange.
  • They contain an extensive network of blood vessels that help in the transport of respiratory gases.
  •  They increase the surface area for the absorption of gases.

(d) The respiratory pigment is Haemoglobin.
Its role is:

  • It has an affinity for oxygen and hence helps in the transport of oxygen from alveoli to body tissues.
  • It also has an affinity for CO2 and hence helps in the transport of CO2 from body tissues to alveoli.


Q8: (a) Name the process and explain the type of nutrition found in green plants. List the raw materials required for this process. Give the chemical equation for the mentioned process.   (2017)
(b) Write three events that occur during this process. 

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Ans: (a) The process of nutrition in green plants is Photosynthesis. The type of nutrition is autotrophic nutrition. Autotrophic nutrition requires sunlight, chlorophyll, water, and carbon dioxide.
Plants prepare their own food in the form of glucose, which gets converted to starch. This is done using COand H2O, which are fixed using energy from sunlight from the chlorophyll molecules present in green plants.
Equation – 

Previous Year Questions 2017

(b) The three events involved are:

  • The absorption of sunlight occurs through chlorophyll.
  • Conversion of this light energy into chemical energy and splitting of water into H+ and OH using this energy.
  • Reduction of CO2 to carbohydrate by using H+ produced due to splitting of water.


Q9: (a) Write the name of different components of the transport system in human beings and state their functions in brief.   (2017)
(b) How does a blood clot form if a leak develops in the system of blood vessels?

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Ans: (a) Various components of the transport system and their functions:Previous Year Questions 2017

(b) If there is a leak in the blood vessels due to injury, then the blood platelets help plug these leaks. The platelets make the clot by forming a mesh-like structure over the leak in which other blood cells get entangled, ultimately plugging the leak.

Previous Year Questions 2016

Q1: Define translocation in reference to plants.  (2016)

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Ans: Translocation is the process by which plants transport the products of photosynthesis from the leaves to various parts of the plant, including:

  • Roots
  • Stems
  • Branches

This transport occurs through the phloem, a type of vascular tissue, and involves:

  • Movement of sucrose and other nutrients.
  • Delivery to storage organs like roots, fruits, and seeds.
  • Support for growing organs.

Translocation is an energy-dependent process, using ATP to increase osmotic pressure, which facilitates the movement of substances in the phloem.

This allows the plant to distribute nutrients according to its needs, such as transporting stored sugars to new buds in spring.


Q2: Mention the raw materials required for photosynthesis.  (2016)

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Ans: Raw materials required for photosynthesis are carbon dioxide (CO2), water, light, and chloroplast.


Q3:  Write three points of difference between breathing and respiration.  (2016)

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Ans: Differences between breathing and respiration are as follows:
Previous Year Questions 2016


Q4: Draw a flow chart to show the breakdown of glucose by various pathways.  (2016)

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Ans: Glucose is first broken down in the cell cytoplasm into a three-carbon molecule called pyruvate. Pyruvate is further broken down in the following ways to provide energy.
Previous Year Questions 2016


Q5: Draw a diagram of the human alimentary canal and label the following:  (2016)
(i) Part in which starch digestion is initiated.
(ii) Organ in which bile is stored.
(iii) The gland that secretes digestive enzymes as well as hormones.
(iv) Part of the alimentary canal where water is reabsorbed.
(v) Part of the gut where finger-like projections are present to facilitate absorption of digested food.

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Ans: (i) Mouth, (ii) Gall Bladder, (iii) Pancreas, (iv) Large Intestine, (v) Small intestine.

Previous Year Questions 2016


Q6: (a) State the form in which the following are stored :   (2016)
(i) Unused carbohydrates in plants. 
(ii) Energy derived from food in humans. 
(b) Describe the process of nutrition in the amoeba with the help of a diagram.

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Ans: 
(a) 
(i) As starch in fruits/storage roots tubers etc.
(ii) As glycogen.
(b) Amoeba is a unicellular eukaryotic organism that exhibits a holozoic mode of nutrition.
Five steps are involved in this mode of nutrition:

  • Ingestion: Amoeba feeds on microscopic organisms. It takes in food using finger-like projections called pseudopodia. It encircles the food and engulfs or ingests it by a process called phagocytosis. The food comes to lie in a vesicle called a phagosome or food vacuole.
  • Digestion: A lysosome fuses with a phagosome, and complex substances are broken down into simpler ones. Such a type of digestion that occurs within a cell is referred to as Intracellular digestion. The digested food then diffuses into the cytoplasm.
  • Absorption and Assimilation: The digested food that has diffused into the cytoplasm is quickly absorbed by the organism and converted into various constituents of protoplasm.
  • Egestion: The undigested food material reaches the rear end of the organism and is thrown out by the process of exocytosis. The membrane of the vesicle fuses with the surface membrane.
    Previous Year Questions 2016
Also watch: What are Life Processes

Previous Year Questions 2015

Q1: Give reason:   (2015)
(l) Fine hair and mucus are present in the nasal passage. 
(ii) Rings of cartilage are present in the throat.

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Ans: (i) Fine hair and mucus are present in the nasal passage so that any dust particles, pathogens, etc., can be trapped in here, and only clean air will enter the lungs. This is one of the defence mechanisms of our body.
(ii) Rings of cartilage are present in the throat so that the trachea does not collapse on respiration.


Q2: Define the term parasite. Name one plant parasite and one animal parasite. Some organisms break down the food material outside the body and then absorb it. Give two examples.  (2015)

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Ans: Parasites are organisms that obtain nutrients from the bodies of other living organisms and harm them as a result.
(i) Plant parasite – cuscuta
(ii) Animal parasite – Tapeworm
Organisms that break down food outside the body and then absorb it are saprophytes.
Example: fungi, bacteria.


Q3: What do the following transport?  (2015)
(i) Xylem
(ii) Phloem
(iii) Pulmonary vein
(iv) Vena cava
(v) Pulmonary artery
(vi) Aorta. 

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Ans: 
Material transported by:
(i) Xylem: Water and minerals.
(ii) Phloem: Organic molecules like sucrose.
(iii) Pulmonary vein: Oxygenated blood from lungs to heart.
(iv) Vena cava: Impure and deoxygenated blood from the body to the heart.
(v) Pulmonary artery: Deoxygenated blood from the heart to the lungs.
(vi) Aorta: Oxygenated blood from the heart to body parts.


Q4: In single-celled organisms, diffusion is sufficient to meet all their requirements for food, gas exchange, or removal of waste, but it is not in the case of multicellular organisms. Explain the reason for this difference.   (2015)

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Ans: In the case of a single-celled organism, the entire surface of the organism is in contact with the environment, and hence, no specific organ for taking in food, exchanging gases, or removing waste may be needed. In multicellular organisms, only the cells of the skin are in direct contact with the environment. Diffusion is a very slow process, and it will take a very long time to reach all the cells of the body parts. Diffusion is insufficient to melt oxygen requirements.


Q5: State the role of the following in the human digestive system:   (2015)
(a) Digestive enzymes 
(b) Hydrochloric acid 
(c) Villi 

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Ans: (a) Digestive enzymes digest the food we eat.
(b) Hydrochloric acid creates an acidic medium to facilitate the action of the enzyme pepsin.
(c) Villi increases the surface area inside the small intestine to facilitate the absorption of food.

Previous Year Questions 2014

Q1: State the location and function of gastric glands.  (2014)

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Ans: Gastric glands are present in the wall of the stomach. They secrete gastric juices containing mucus, protein-digesting enzymes pepsin, rennin, and hydrochloric acid (HCI).


Q2: How does nutrition take place in Amoeba? How is it different in Paramoecium?  (2014)

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Ans: Nutrition in amoeba: 

  • Nutrition occurs through phagocytosis.
  • Amoeba captures food using pseudopodia, which are temporary extensions of the cell.
  • The ingested food is enclosed in a food vacuole.
  • Inside the food vacuole, enzymes break down the food into simpler substances.
  • Undigested food is expelled through the cell membrane.

Nutrition in Paramoecium:

  • Paramoecium has a definite shape and takes in food at a specific location.
  • Food is directed to this location by the movement of cilia that cover the cell’s surface.


Q3: Why is it advisable to breathe through nose?  (2014)

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Ans: Breathing through the nose is advisable for several reasons:

  • The inner lining of the nose contains fine hairs and mucus glands that filter incoming air, trapping germs and dust.
  • This process helps to warm and humidify the air before it reaches the lungs.
  • Breathing through the nose also promotes better oxygen absorption in the body.


Q4: Define transpiration. How does transpiration help in upward movement of water from roots to leaves?  (2014)

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Ans: Transpiration is the process through which water vapour is lost from the aerial parts of a plant, primarily via the stomata in the leaves.

This process plays a crucial role in the upward movement of water from the roots to the leaves:

  • As water evaporates from the leaf cells, it creates a suction force.
  • This suction pulls water from the xylem cells in the roots.
  • Transpiration helps in the absorption of water and minerals, facilitating their movement to the leaves.
  • It also aids in regulating the plant’s temperature.

During the day, when the stomata are open, the transpiration pull is the main driving force for water movement in the xylem.

04. Previous Year Questions: Carbon & its Compounds – 2

Previous Year Questions 2020

Q1: Several factories were pouring their wastes in rivers A and B. Water samples were collected from these two rivers. It was observed that sample collected from river A was acidic while that of river B was basic. The factories located near A and B are    (2020)
(a) Soaps and detergents factories near A and alcohol distillery near B.
(b) Soaps and detergents factories near B and alcohol distillery near A.
(c) Lead storage battery manufacturing factories near A and soaps and detergents factories near B.
(d) Lead storage battery manufacturing factories near B and soaps and detergents factories near A.

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Ans: (c)


Q2: (a) Carry out following conversions:
(i) Ethanol to ethene
(ii) Ethanol to ethanoic acid
(b) Differentiate between addition reaction and substitution reaction. Give one example of each.
    (2020)

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Ans: (a) (i) When ethanol is heated with cone. H2SO4 at 443 K, ethene is obtained due to dehydration of ethanol.Previous Year Questions 2020
(a) (ii) When 5 % alkaline KMnO4 solution is added drop by drop to warm ethanol then it gets oxidized to ethanoic acid.Previous Year Questions 2020

(b) Addition reactions : Those reactions in which atoms or group of atoms are simply added to a double or triple bond without the elimination of any atom or molecule, are known as addition reactions.
Previous Year Questions 2020
Substitution reactions: The reactions which involve the displacement or substitution of an atom or a group of atoms in an organic compound by another atom or group of atoms, are known as substitution reactions.

Saturated hydrocarbons are fairly unreactive and inert in the presence of most of the reagents. However, in presence of sunlight, hydrocarbons undergo rapid substitution reactions, e.g.,


Q3: (a) What is a homologous series? Explain with an example.
(b) Define the following terms giving one example of each.
(i) Esterification
(ii) Addition reaction    (2020)

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Ans: (a) A homologous series is the family of organic compounds having the same functional group, similar chemical properties but the successive (adjacent) members of the series differ by a -CH2 unit or 14 mass units.
For example, alkane series has general formula CnH2n + 2.
First member of homologous series of alkane is .methane, i.e., CH4.
Second member of homologous series of alkane is ethane, i.e., C2H6.
Third member of homologous series of alkane is propane i.e., C3H8.  
(b) (i) Carboxylic acids react with alcohols in the presence of a little concentrated sulphuric acid to form pleasant smelling esters. This reaction is called esterification reaction.

(ii) Addition reactions : Those reactions in which atoms or group of atoms are simply added to a double or triple bond without the elimination of any atom or molecule, are known as addition reactions.


Q4: Assertion (A) : Ethanoic acid is also known as glacial acetic acid.
Reason (R) : The melting point of pure ethanoic acid is 290 K and hence it often freezes during winters in cold climates.    (2020)
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

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Ans: (a)
Pure ethanoic acid or acetic acid freezes below room temperature into white crystals that resemble glaciers. 


Q5: Assertion (A) : Esterification is a process in which a sweet smelling substance is produced.
Reason (R): When esters react with sodium hydroxide, an alcohol and sodium salt of carboxylic acid are obtained.    (2020)
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

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Ans: (b)
When an ester reacts with the base saponification reaction occurs.  


Q6: 3 mL of ethanol is taken in a test tube and warmed gently in a water bath. A 5% solution of alkaline potassium permanganate is added first drop by drop to this solution, then in excess. 
(i) How is 5% solution of KMnO4 prepared?
(ii) State the role of alkaline potassium permanganate in this reaction. What happens when one adds it in excess?
(iii) Write chemical equation of this reaction.    (2020)

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Ans: (i) 5% solution of KMnO4 is prepared by adding 5 g of KMnO4 in 95 g of water.
(ii) Here alkaline KMnO4 acts as an oxidising agent. It oxidises ethanol to ethanoic acid by donating nascent oxygen. If excess of KMnO4 is added the purple colour will persist indicating no more alcohol is left and there is no reaction.
Previous Year Questions 2020


Q7: Carbon, a member of group 14, forms a large number of carbon compounds estimated to be about three million. Why is this property not exhibited by other elements of this group? Explain.    (2020)

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Ans: Carbon has the unique ability to form bonds with other atoms of carbon, giving rise to large molecules. This property is called catenation. Carbon shows catenation due to its small size and Stronger carbon-carbon bond strength. As we move down the group, the element-element bond energies decrease rapidly. For this reason other elements of this group show little or no catenation property.


Q8: Assertion (A) : Following are the members of a homologous series
CH3OH, CH3CH2OH, CH3CH2CH2OH
Reason (R) : A series of compounds with same functional group but differing by -CH2 unit is called homologous series.    (2020)
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true. 

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Ans: (a)
The given compounds are members of homologous series of alcohol.



Q9: Name a cyclic unsaturated carbon compound.    (2020)

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Ans: Benzene is a cyclic unsaturated carbon compound.

Ans: Benzene is a cyclic unsaturated carbon compound.


Q10: Covalent compounds have low melting and boiling point. Why? [2020]

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Ans: Covalent compounds have low melting and boiling points because the forces of attraction between molecules of covalent compounds are very weak. On applying a small amount of heat these molecular forces break.  

Previous Year Questions 2019

Q11: (a) Define the term isomer.
(b) Two compounds have same molecular formula C3H6O. Write the name of these compounds and their structural formula.
(c) How would you bring the following conversions:
(i) Ethanol to ethene
(ii) Propanol to propanoic acid?    (2019)

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Ans: (a) Isomers are those molecules which have the same molecular formula but different structural formula i.e., show different properties.

(b) 

(c)  (i) When ethanol is heated with conc. H2SO4 at 443 K, ethene is obtained due to dehydration of ethanol.

(c) (ii) Propanol to propanoic acid
Previous Year Questions 2019


Q12: Write the chemical formula and name of the compound which is the active ingredient of all alcoholic drinks. List its two uses. Write chemical equation and name of the product formed when this compound reacts with
(i) sodium metal
(ii) hot concentrated sulphuric acid.    (2019)

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Ans: 
Ethanol having chemical formula C2H5OH is the active ingredient of all alcoholic drinks.
Uses of ethanol:
1. Ethanol is widely used in industry as a solvent.
2. Ethanol is used as an antiseptic for wounds in the form of rectified spirit.
Chemical equations: 

(i) 2C2H5OH + 2Na → 2C2H5ONa+ + H2 ↑ 

(ii)Previous Year Questions 2019


Q13: (a) State the reason why carbon can neither form C4+ cations nor C4- anions, but forms covalent bonds. Also state reasons to explain why covalent compounds
(i) are bad conductors of electricity.
(ii) have low melting and boiling points.
(b) Write the structural formula of benzene, C6H6    (2019)

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Ans: 
(a)  Ionic compounds are formed either by gaining or losing electrons from the outermost shells, but carbon which has four electrons in its outermost shell cannot form ionic bonds because
1. If carbon forms ionic bonds by gaining four electrons to attain a noble gas configuration then it would be difficult for six protons in the nucleus to hold ten electrons.
2. If carbon forms ionic bonds by loss of four electrons then it would require a lot of energy to remove these electrons from outermost shell.
Due to these reasons carbon forms covalent bonds by sharing the valence electrons.
Type of bonds formed in ionic compounds are called electrovalent bonds and the type of bonds formed in carbon compounds are called covalent bonds. 

(i) Covalent bonds are those bonds which are formed by sharing of the valence electrons between two atoms. 

 Electron dot structure of Methane

Covalent compounds are generally poor conductors of electricity because they do not have tree electrons or ions. 

(ii) Covalent compounds have low melting and boiling points because the forces of attraction between molecules of covalent compounds are very weak. On applying a small amount of heat these molecular forces break.

(b) Structural formula of Benzene, C6H6 is shown belowPrevious Year Questions 2019



Q14: What happens when this compound burns with oxygen?    (2019)

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Ans: 
(i) Methane is the first member of alkane series having formula CH4.
Covalent bonds are those bonds which are formed by sharing of the valence electrons between two atoms. Electron dot structure of methane is shown in the figure. 

Previous Year Questions 2019 Electron dot structure of Methane

Covalent compounds are generally poor conductors of electricity because they do not have tree electrons or ions.
(ii)  When methane is burnt in presence of oxygen then carbon dioxide will be produced.
CH4 + O2 → CO2 + H2O + heat + light 

Covalent compounds have low melting and boiling points because the forces of attraction between molecules of covalent compounds are very weak. On applying a small amount of heat these molecular forces break.


Q15: What is methane? Draw its electron dot structure. Name the type of bonds formed in this compound. Why are such compounds:    (Delhi 2019)
(i) poor conductors of electricity? and
(ii) have low melting and boiling points? What happens when this compound in oxygen?

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Ans: (i) Methane is the simplest hydrocarbon which is produced when carbon atom shares its 4 valence electrons with 4 hydrogen atoms.
(ii) Methane achieves noble gas electronic configuration by sharing its four electrons with other elements, i.e. it forms covalent compounds.

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Previous Year Questions 2018

Q16: Which type of compounds can be formed by carbon?    (CBSE 2018)

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Ans: Covalent compounds can be formed by carbon.

Previous Year Questions 2018


Q17: What happens when hydrogen is added to a vegetable oil in the presence of nickel? Name the reaction and write one difference between the physical property of the vegetable oil and the product obtained in this reaction. Write the role of nickel in this reaction.    (CBSE 2018C)

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Ans:

  • Vegetable oil undergoes hydrogenation to form saturated product known as vegetable ghee or vanaspati ghee. This reaction is known as hydrogenation of oils.
  • Vegetable oil is liquid at room temperature whereas vegetable ghee is solid at room temperature.
  • Nickel acts as catalyst and forms saturated hydrocarbons.


Q18: (a) Why are most carbon compounds poor conductors of electricity?
(b) Write the name and structure of a saturated compound in which the carbon atoms are arranged in a ring. Give the number of single bonds present in this compound.    (CBSE 2018)

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Ans: 
(a) Carbon compounds are covalent in nature and they do not dissociate to form ions because of which they are poor conductors of electricity.
(b) Cyclohexane is the saturated compound in which the carbon atoms are arranged in a ring.

Previous Year Questions 2018Cyclohexane

Previous Year Questions 2018

It has 18 single bonds.


Q19: A compound A (C2H4O2) reacts with Na metal to form a compound B and evolves a gas which burns with a pop sound. Compound A on treatment with an alcohol C in the presence of an acid forms a sweet smelling compound, D (C4H8O2). On addition of NaOH to D gives back B and C. Identify A, B, C and D. Write the reactions involved.    (CBSE Sample Questions Paper 2018)

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Ans: The compounds A, B, C and D are given as under:
A = CH3COOH
B = CH3COONa
C = C2H5OH
D = CH3COOC2H5
The reactions are explained as under:

Previous Year Questions 2018
Also read: PPT: Carbon and its Compounds

Previous Year Questions 2017

Q20: Two carbon compounds X and Y have the molecular formula C4H8 and C5H12 respectively. Which one of these is most likely to show addition reaction? Justify your answer. Also give the chemical equation to explain the process of addition reaction in this case.    (Delhi 2017)

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Ans:  ‘X’ (C4H8) will show addition reaction because it is unsaturated, i.e. it has a double bond e.g.

Previous Year Questions 2017
 When a molecule or a compound is added to unsaturated compound to form a saturated compound is called addition reaction.


Q21: Complete the following chemical equations:    (Delhi 2017)
(i) CH3COOC2H5 + NaOH →
(ii) CH3COOH + NaOH →
(iii) 

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Ans: (i) CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
(ii) CH3COOH + NaOH → CH3COONa + H2O
(iiiC2H5OH + CH3COOH → CH3COOC2H+ H2O


Q22: Write the structural formula of ethanol and list its two physical properties. What happens when it is heated with excess of cone. H2SO4 at 443 K? State the role of conc. H2SO4 in this reaction.     (DoE, AI 2017, Foreign 2013)

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Ans: CH3—CH2OH is structural formula of ethanol
Physical properties:
(i) It is liquid and has sweet smell.
(ii) It is soluble in water in all proportions.
When ethanol is heated with cone. H2SO4 at 443 K, ethene is formed.

Previous Year Questions 2017
Conc. H2SO4 acts as dehydrating agent.


Q23: Ethanoic acid reacts with absolute ethanol in the presence of Conc. H2SO4 to form a compound:    [AI 2017 C]
(i) Write the smell and class of compounds to which this compound belongs.
(ii) Write the chemical equation for the reaction and state the role of Conc. H2SO4 in the reaction.
(iii) Write one use of the product of this reaction.

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Ans:
(i) The compound formed is ‘Esters’ It has a pleasant fruity smell.

(ii) Previous Year Questions 2017
Conc. H2SO4 acts as dehydrating agent.
(iii) It is used as a solvent in glass, nail polish removers and decaffeinating tea and coffee. It is used is solvent in printing to modify its drying rate.
Uses of esters:

  • Esters are used as flavouring agents in ice creams, cold drinks etc.
  • They are used in perfumes.


Q24: Distinguish between esterification and saponification reactions of organic compounds with the help of the chemical equation for each. What is the use of (i) esters and (ii) saponification process?    (AI 2017, Foreign 2012)

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Ans:


Uses:
(i) Esters are used in cold drinks, ice creams, perfumes and as artificial flavouring agents.
(ii) Saponification process is used in manufacture of soaps.


Q25: Name two oxidising agents that are used for the conversion of alcohols to acids. Distinguish between ethanol and ethanoic acid on the basis of (a) litmus test, and (b) reaction with sodium hydrogen carbonate.    (Delhi 2013; AI 2012; Foreign 2017)

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Ans: Alkaline KMnO4/NaOH (Potassium permanganate) and Acidified K2Cr2O7 / H2SO4 (Potassium dichromate).Previous Year Questions 2017


Q26: What is the homologous series of carbon compounds? List any two characteristics. Write the name and formula of next higher homologue of HCOOH.   (CBSE 2017)

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Ans: A series of compounds whose two successive members differ by – CH2 in the molecule formula and which show similar chemical properties are called homologous series.

Characteristics:
1. Melting and boiling points increase regularly with increase in molecular mass.
2. Solubility in a particular solvent shows a regular gradation.
3. Chemical properties remain the same in the homologous series.
Next higher homologue of HCOOH is CH3COOH. Its name is ethanoic acid or acetic acid.


Q27: When ethanol reacts with ethanoic acid in the presence of conc. H2SO4, a substance with fruity smell is produced. Answer the following:    (CBSE 2016)
(i) State the class of compounds to which the fruity smelling compounds belong. Write the chemical equation for the reaction and write the chemical name of the product formed.
(ii) State the role of cone. H2SO4 in this reaction.

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Ans:

(i) Fruity smelling compounds belong to the class of esters.

Previous Year Questions 2017
(ii) Conc. H2SO4 acts as a dehydrating agent and removes the water formed in the reaction.


Q28: Soaps and detergents are both types of salts. State the difference between the two. Write the mechanism of the cleansing action of soaps. Why do soaps not form lather (foam) with hard water? Mention any two problems that arise due to the use of detergents instead of soaps.   (Delhi 2017)

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Ans: Soaps are sodium or potassium salts of fatty acids having —COONa group. Detergents are sodium or potassium salts of sulphonic acids having — SO3Na and —SO4Na group.
Cleansing Action of Soap. Soaps consist of a large hydrocarbon tail which is hydrophobic (water-hating or water repelling) with a negatively charged head which is hydrophilic (water-loving) as shown in figure.

Previous Year Questions 2017Previous Year Questions 2017Previous Year Questions 2017When a soap is dissolved in water, the molecules associate together as clusters called micelles in which, water molecules being polar in nature, surround the ions and the hydrocarbon part of the molecule attracts grease, oil and dirt.Hard water has Ca2+ and Mg2+ ions when they react with soap to form insoluble compound and soap goes waste.
Disadvantages of Detergents
(i) Detergents are expensive.
(ii) Many detergents are branched chain hydrocarbon which are not biodegradable and create water pollution.

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Previous Year Questions 2016

Q29: What are covalent compounds? Why are they different from ionic compounds? List their three characteristic properties.    (CBSE 2016)

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Ans:

Covalent compounds are the compounds that contain covalent bonds in the molecule. A covalent bond is formed by sharing of electrons between the atoms.
While a covalent compound contains covalent bonds, the ionic compound is made up of ionic bond which is formed by transference of electrons from one atom to the other.
Characteristics of covalent compounds:
(i) These compounds show low melting points and boiling points because the intermolecular forces are weak.
(ii) Such compounds are poor conductors of electricity.
(iii) Generally insoluble or less soluble in water but soluble in organic solvents.


Q30: Write chemical equation of the reaction of ethanoic acid with the following:
(а) Sodium; 
(b) Sodium hydroxide; (c) Ethanol.
Write the name of one main product of each reaction.    (CBSE 2016)

Hide Answer  

Ans:

Previous Year Questions 2016


Q31: An aldehyde as well as a ketone can be represented by the same molecular formula, say C3H6O. Write their structures and name them. State the relationship between the two in the language of science.    (CBSE 2016)

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Ans: The structures of the aldehyde and ketone with the molecular formula C3H6O are

Previous Year Questions 2016

Both these compounds are isomers (because they have the same molecular formula but different structural formula).


Q32: (i) The chemical properties of ethanol are different from methyl ethanoate. Justify the statement with a proper reason.
(ii) Methyl ethanoate is used in making perfumes. Justify.
(iii) Ethanol is converted into ethene with excess of hot concentrated H2SO4. Justify with the help of chemical equation.   (CBSE 2016)

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Ans:
(i) Ethanol reacts with sodium metal whereas methyl ethanoate does not. They differ in functional group, and in chemical properties.
(ii) It is because it has pleasant fruity smell.
(iii) Previous Year Questions 2016

Ethanol gets dehydrated with conc. H2SO4 to form ethene and H2O.


Q33: (a) Define the term functional group. Identify the functional group present in

(i)

Previous Year Questions 2016

(ii)

Previous Year Questions 2016

(b) What happens when 5% alkaline KMnO4 solution is added drop by drop to warm ethanol taken in a test tube? State the role of alkaline KMnO4 solution in this reaction.    (Foreign 2016)

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Ans:

(a) It is an atom or group of atoms or reactive part of compound which largely determine the chemical properties of compound (i) Methanol (ii) Ethanoic acid.

(b) Previous Year Questions 2016

Alkaline KMnO4 acts as an oxidizing agent.


Q34: On dropping a small piece of sodium in a test tube containing carbon compound ‘X ’ With molecular formula C2H6O, a brisk effervescence is observed and a gas ‘Y’ is produced, On bringing a burning splinter at the mouth of the test tube the gas evolved burns with a pop sound. Identify ‘X ’ and ‘Y’. Also write the chemical equation for the reaction. Write the name and structure of the product formed, when you heat ‘X’ with excess cone, sulphuric acid.    (AI 2016)

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Ans:
‘X’ is C2H5OH

2C2H5OH + 2Na → C2H5Na + H2
‘Y’ is H2(g) which burns with ‘pop’ sound.

Previous Year Questions 2016

Ethene is the product.


Q35: An organic compound ‘P’ is a constituent of wine. ‘P ’ on reacting with acidified K2Cr2O7 forms another compound ‘Q’. When a piece of sodium is added to ‘Q’ a gas ‘R’ evolves which burns with a pop sound. Identify P, Q and R and write the chemical equations of the reactions involved.    (Foreign 2016)

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Ans: 

‘P ’ is C2H5OH, Ethanol

Previous Year Questions 2016

Q is Acetic Acid.
2CH3COOH + 2Na → 2CH3COONa + H2
‘R’ is hydrogen gas which burns with ‘pop’ sound.


Q36: (a) Give chemical tests to detect the presence of
(i) Ethanol 

(ii) Ethanoic acid
(b) Why ethanoic acid is also called glacial acetic acid?   (CBSE 2016)

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Ans:

(a) Add sodium hydrogen carbonate. Ethanol will not react. Ethanoic acid will give brisk effervescence due to carbon dioxide.
(b) Pure ethanoic acid exists as solid like glaciers at 291 K, therefore, called glacial acetic acid.


Q37: Write a chemical equation of the action of ethanoic acid with the following: (a) Sodium; (b) Sodium hydroxide; (c) Ethanol.
Write the name of one main product of each reaction.   (Al 2016)

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Ans:

(a) 2CH3COOH + 2Na → 2CH3COONa + H2
Sodium ethoxide is the main product formed.
(b) CH3COOH + NaOH→ CH3COONa + H2O
Sodium ethoxide is the main product formed.

(c)
Previous Year Questions 2016

Ethyl ethanoate (ester) is the main product formed.


Q38: Under what conditions an oxidation reaction can be called a combustion reaction? Illustrate your answer with examples.   (CBSE 2016)

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Ans: When complete oxidation of fuel takes place with the release of high amount of heat and light, it is called combustion reaction. It is highly exothermic and generally accompanied by evolution of heat and light.
Examples – Burning of coal
C + O2 → CO2 + heat + light
Burning of the rocket fuel (liquid H2)
2H2O + O2 → 2H2O + heat + light
Burning of acetylene used in welding torches.
2C2H2 + 5O2 → 4CO2 + 2H2O + heat + light


Q39: C3H6, C4H8 and C5H10 belong to the same homologous series.    (CBSE 2016)
(i) Define homologous series.
(ii) Why are the melting and boiling points of C5H10 is higher than C4H8?
(iii) Arrange these hydrocarbons in order of increasing boiling points.

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Ans:  (i) The series of organic compounds which have similar chemical properties, same functional group and same general formula is called homologous series.
(ii) It is because C5H10 has higher molecular weight, more Vander Wall’s force of attraction and higher boiling points and melting points.
(iii) C3H6 < C4H8 < C5H10 is increasing order of boiling point.


Q40: An aldehyde as well as ketone can be represented by the same molecular formula, say C3H6O. Write their structures and name them. State the relation between the two in the language of science.    (Al 2016)

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Ans:


They are functional isomers.


Q41: Write the name and molecular formula of an organic compound having its name suffixed with ‘ol’ and having two carbon atoms in its molecule. Write the balanced chemical equation to indicate what happens when this compound is heated with excess conc. H2SO4 and the name of main product formed. Also state the role of conc. H2SO4 in the reaction.   (2016)

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Ans:
Ethanol, C2H5OH

Previous Year Questions 2016

Conc. H2SO4 acts as dehydrating agent.


Q42: What is an oxidising agent? What happens when an oxidising agent is added to propanol? Explain with the help of a chemical equation.    (Delhi 2016)

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Ans:
Those substances which add oxygen are called oxidising agent and propanoic acid is formed.

Previous Year Questions 2016

Reaction of Propanol with Oxidizing Agent
Propanol will get oxidised to propanoic acid by acidified KMnO4.

Q43: (a) You have three unlabelled test tubes containing ethanol, ethanoic acid and soap solution. Explain the method you would use to identify the compounds in different test tubes by chemical tests using litmus paper and sodium metal.
(b) Give the reason of formation of scum when soaps are used with hard water.   (Foreign 2016)

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Ans:
(a) Ethanol will not be affected by blue litmus as well as red litmus paper. Ethanoic acid will turn blue litmus red whereas red litmus will remain as it is. Soap solution will turn red litmus blue but blue litmus will remain as it is. Sodium metal will liberate hydrogen gas with ethanol as well as ethanoic acid. Soap solution will not react with sodium metal.
(b) It is because soap will react with Ca2+ and Mg2+ ions present in hand water to form calcium salts of fatty acids which are insoluble and called scum.


Q44: (a) Give a chemical test to distinguish between saturated and unsaturated hydrocarbon.
(b) Name the products formed when ethane burns in air. Write the balanced chemical equation for the reaction showing the types of energies liberated.
(c) Why is reaction between methane and chlorine in the presence of sunlight considered a substitution reaction?    (Delhi 2016)

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Ans:
(a) Add bromine water. Unsaturated hydrocarbon will decolourise bromine water whereas saturated hydrocarbon will not.
(b) Carbon dioxide and Water are formed.

2C2H6 (g) + 7O2 (g) → 4CO(g) + 6H2O +Heat + Light
(c) It is because hydrogen atom of methane gets substituted by chlorine atom to form chloromethane, therefore, it is called a substitution reaction.


Q45: What are micelles? Why does it form when soap is added to water? Will a micelle be formed in other solvents such as ethanol also? State briefly how the formation of micelles help to clean the clothes having oily spots.    (Foreign 2016)

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Ans:  Micelles: When molecular ions in soaps and detergents aggregate, they form micelles.
It is because a large number of molecular ions of soaps get aggregated and form colloidal solution. Soap has hydrophobic tail (hydrocarbon) which dissolves in hydrocarbon part and hydrophilic part dissolves in water. Ethanol is non-polar solvent therefore micelles are not formed because hydrocarbon part gets attracted towards ethanol and the ionic end will not dissolve in alcohol.


Q46: A carbon compound ‘P ’ on heating with excess cone. H2SO4 forms another carbon compound ‘Q’ which on addition of hydrogen in the presence of nickel catalyst forms a saturated carbon compound ‘R’. One molecule of ‘R’ on combustion forms two molecules of carbon dioxide and three molecules of water. Identify P, Q and R and write chemical equations for the reactions involved.   (CBSE 2016)

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Ans: The compounds are identified as under:
P = Ethanol (CH3CH2OH)
Q = Ethene (CH2 = CH2)
R = Ethane (CH3 – CH3)
Reactions are given as under:

Previous Year Questions 2016
Also read: PPT: Carbon and its Compounds

Previous Year Questions 2015

Q47: Draw the electron-dot structure for ethyne. A mixture of ethyne and oxygen is burnt for welding. In your opinion, why cannot we use a mixture of ethyne and air for this purpose?   (AI 2015)

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Ans:

Previous Year Questions 2015

Electron-dot structure of Ethyne

Ethyne and air will not produce enough heat due to incomplete combustion needed for welding purpose. Ethyne and oxygen will produce a lot of heat due to complete combustion which can be used for welding purposes.


Q48: That is meant by isomers? Draw the structures of two isomers of butane, C4H10. Explain why we cannot have isomers of the first three members of alkane series.   (DoE, Delhi 2015)

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Ans: Isomers are those compounds which have the same molecular formula but different structural formula.

Previous Year Questions 2015

Butane & Iso-Butane are two isomers of C4H10. Isomers are not possible for the first three members because branching is not possible.


Q49: Write the name and general formula of a chain of hydrocarbons in which an addition reaction with hydrogen is possible. State the essential conditions for an addition reaction. Stating this condition, write a chemical equation giving the name of the reactant and the product of the reaction.   (AI 2015)

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Ans:  CnH2n and CnH2n- 2 are the general formula of alkene and alkynes in which addition reaction with hydrogen is possible. Hydrogen is added to unsaturated hydrocarbon (having double or triple bond) in presence of heated nickel as catalyst.

Previous Year Questions 2015


Q50: Write the molecular formula of the following compounds and draw their electron-dot structures:    (Foreign 2015)
(i) Ethane 
(ii) Ethene 
(iii) Ethyne

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Ans:
(i) Ethane (C2H6)
Previous Year Questions 2015

(ii) Ethene (C2H4)

Previous Year Questions 2015

(iii) Ethyne (C2H2)

Previous Year Questions 2015


Q51: Write the respective chemical equations to show what happens when    (Foreign 2015)
(i) Methane is burned in presence of oxygen?
(ii) Ethanol is heated with concentrated sulphuric acid at 443 K?
(iii) Ethanol reacts with ethanoic acid in the presence of an acid acting as a catalyst?

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Ans:
(i) CH4 + 2O2 → 2H2O + CO2
Water and carbon dioxide formed.

(ii)

Previous Year Questions 2015

The concentrated sulphuric acid can be regarded as a dehydrating agent which removes water from ethanol.

(iii) 


 Esters are sweet smelling substances.


Q52: Write the chemical equation to explain what happens when ethanol is heated with alkaline solution of potassium permanganate. Mention two physical properties and two uses of ethanol.    (Foreign 2015)

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Ans:
Alkaline KMnO4 is dark pink in colour. So when it is added to ethanol and heated, the pink colour of the solution disappears. When excess of KMnO4 is added, the pink colour does not disappear, indicating that all the ethanol has been converted to ethanoic acid.
• Physical properties of ethanol:
(a) It is a colourless liquid with pleasant smell and burning taste.
(b) It is a volatile liquid with low boiling point.
• Uses of ethanol:
(a) It is used in the manufacture of medicines, varnished, paints, dyes, soap, etc.
(b) It is a good solvent. Many organic compounds which are insoluble in water are soluble in ethanol.


Q53: What is the difference between the molecules of soaps and detergents? Explain the cleansing action of soaps.    (CBSE 2015)

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Ans:
Soaps are sodium or potassium salts of long chain carboxylic acids.
Detergents are ammonium or sulphonate salts of long chain carboxylic acids.

Previous Year Questions 2015

Cleansing action of soap: The ionic end of soap dissolves in water while the carbon chain dissolves in oil. Soap molecules form micelles where one end of the soap molecule is linked to oil droplet while the ionic part faces water. This forms an emulsion in water. On adding water, these micelles containing dirt are washed away and our clothes are washed clean. This is represented in the figure.


Q54: With the help of an example, explain the process of hydrogenation. Mention the essential conditions for the reaction and state the change in physical property with the formation of the product.    (CBSE 2015)

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Ans: Conversion of oil into fat by passing hydrogen gas in the presence of nickel as catalyst is called hydrogenation. This is how hydrogenation is carried out.
Conditions: The presence of a catalyst like nickel is essential.
Change in physical property: Oils are liquids while the fats obtained on hydrogenation are solids i.e., a change in the physical state takes place on hydrogenation.

Q55: Explain why carbon forms compounds mainly by covalent bond. Explain in brief two main reasons for carbon forming a large number of compounds. Why does carbon form strong bonds with most other elements?    (CBSE 2015)

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Ans: Carbon has no tendency to lose electrons easily. Similarly it has no tendency to gain electrons. It completes its octet by sharing the electrons with other atoms. Therefore, carbon forms compounds mainly by covalent bond.
Two main reasons for carbon forming a large number of compounds are given below:
Catenation: It is the property of carbon to link with other carbon atoms to form straight chain, branched chain or cyclic compounds. Thus, carbon forms a large number of compounds containing 2, 3, 4, 5,….etc., carbon atoms.
Tetravalency: Carbon has four electrons in the outermost shell. It shares these four electrons with four electrons from four other monovalent atoms. Carbon has the tendency to form bonds with hydrogen, oxygen, nitrogen and halogens. This increases the number of compounds of carbon.
Carbon forms strong bonds with most other elements: This is because carbon has a small atomic size. It can hold strongly the four pairs of electrons that it shares with other atoms. Therefore, it forms strong bonds with other elements.

Previous Year Questions: Carbon & its Compounds – 2

Table of contents
Previous Year Questions 2020
Previous Year Questions 2019
Previous Year Questions 2018
Previous Year Questions 2017
Previous Year Questions 2016
Previous Year Questions 2015

Previous Year Questions 2020

Q1: Several factories were pouring their wastes in rivers A and B. Water samples were collected from these two rivers. It was observed that sample collected from river A was acidic while that of river B was basic. The factories located near A and B are    (2020)
(a) Soaps and detergents factories near A and alcohol distillery near B.
(b) Soaps and detergents factories near B and alcohol distillery near A.
(c) Lead storage battery manufacturing factories near A and soaps and detergents factories near B.
(d) Lead storage battery manufacturing factories near B and soaps and detergents factories near A.

Hide Answer  

Ans: (c)


Q2: (a) Carry out following conversions:
(i) Ethanol to ethene
(ii) Ethanol to ethanoic acid
(b) Differentiate between addition reaction and substitution reaction. Give one example of each.
    (2020)

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Ans: (a) (i) When ethanol is heated with cone. H2SO4 at 443 K, ethene is obtained due to dehydration of ethanol.Previous Year Questions 2020
(a) (ii) When 5 % alkaline KMnO4 solution is added drop by drop to warm ethanol then it gets oxidized to ethanoic acid.Previous Year Questions 2020

(b) Addition reactions : Those reactions in which atoms or group of atoms are simply added to a double or triple bond without the elimination of any atom or molecule, are known as addition reactions.
Previous Year Questions 2020
Substitution reactions: The reactions which involve the displacement or substitution of an atom or a group of atoms in an organic compound by another atom or group of atoms, are known as substitution reactions.

Saturated hydrocarbons are fairly unreactive and inert in the presence of most of the reagents. However, in presence of sunlight, hydrocarbons undergo rapid substitution reactions, e.g.,


Q3: (a) What is a homologous series? Explain with an example.
(b) Define the following terms giving one example of each.
(i) Esterification
(ii) Addition reaction    (2020)

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Ans: (a) A homologous series is the family of organic compounds having the same functional group, similar chemical properties but the successive (adjacent) members of the series differ by a -CH2 unit or 14 mass units.
For example, alkane series has general formula CnH2n + 2.
First member of homologous series of alkane is .methane, i.e., CH4.
Second member of homologous series of alkane is ethane, i.e., C2H6.
Third member of homologous series of alkane is propane i.e., C3H8.  
(b) (i) Carboxylic acids react with alcohols in the presence of a little concentrated sulphuric acid to form pleasant smelling esters. This reaction is called esterification reaction.

(ii) Addition reactions : Those reactions in which atoms or group of atoms are simply added to a double or triple bond without the elimination of any atom or molecule, are known as addition reactions.


Q4: Assertion (A) : Ethanoic acid is also known as glacial acetic acid.
Reason (R) : The melting point of pure ethanoic acid is 290 K and hence it often freezes during winters in cold climates.    (2020)
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

Hide Answer  

Ans: (a)
Pure ethanoic acid or acetic acid freezes below room temperature into white crystals that resemble glaciers. 


Q5: Assertion (A) : Esterification is a process in which a sweet smelling substance is produced.
Reason (R): When esters react with sodium hydroxide, an alcohol and sodium salt of carboxylic acid are obtained.    (2020)
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

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Ans: (b)
When an ester reacts with the base saponification reaction occurs.  


Q6: 3 mL of ethanol is taken in a test tube and warmed gently in a water bath. A 5% solution of alkaline potassium permanganate is added first drop by drop to this solution, then in excess. 
(i) How is 5% solution of KMnO4 prepared?
(ii) State the role of alkaline potassium permanganate in this reaction. What happens when one adds it in excess?
(iii) Write chemical equation of this reaction.    (2020)

Hide Answer  

Ans: (i) 5% solution of KMnO4 is prepared by adding 5 g of KMnO4 in 95 g of water.
(ii) Here alkaline KMnO4 acts as an oxidising agent. It oxidises ethanol to ethanoic acid by donating nascent oxygen. If excess of KMnO4 is added the purple colour will persist indicating no more alcohol is left and there is no reaction.
Previous Year Questions 2020


Q7: Carbon, a member of group 14, forms a large number of carbon compounds estimated to be about three million. Why is this property not exhibited by other elements of this group? Explain.    (2020)

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Ans: Carbon has the unique ability to form bonds with other atoms of carbon, giving rise to large molecules. This property is called catenation. Carbon shows catenation due to its small size and Stronger carbon-carbon bond strength. As we move down the group, the element-element bond energies decrease rapidly. For this reason other elements of this group show little or no catenation property.


Q8: Assertion (A) : Following are the members of a homologous series
CH3OH, CH3CH2OH, CH3CH2CH2OH
Reason (R) : A series of compounds with same functional group but differing by -CH2 unit is called homologous series.    (2020)
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true. 

Hide Answer  

Ans: (a)
The given compounds are members of homologous series of alcohol.



Q9: Name a cyclic unsaturated carbon compound.    (2020)

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Ans: Benzene is a cyclic unsaturated carbon compound.

Ans: Benzene is a cyclic unsaturated carbon compound.


Q10: Covalent compounds have low melting and boiling point. Why? [2020]

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Ans: Covalent compounds have low melting and boiling points because the forces of attraction between molecules of covalent compounds are very weak. On applying a small amount of heat these molecular forces break.  

Previous Year Questions 2019

Q11: (a) Define the term isomer.
(b) Two compounds have same molecular formula C3H6O. Write the name of these compounds and their structural formula.
(c) How would you bring the following conversions:
(i) Ethanol to ethene
(ii) Propanol to propanoic acid?    (2019)

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Ans: (a) Isomers are those molecules which have the same molecular formula but different structural formula i.e., show different properties.

(b) 

(c)  (i) When ethanol is heated with conc. H2SO4 at 443 K, ethene is obtained due to dehydration of ethanol.

(c) (ii) Propanol to propanoic acid
Previous Year Questions 2019


Q12: Write the chemical formula and name of the compound which is the active ingredient of all alcoholic drinks. List its two uses. Write chemical equation and name of the product formed when this compound reacts with
(i) sodium metal
(ii) hot concentrated sulphuric acid.    (2019)

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Ans: 
Ethanol having chemical formula C2H5OH is the active ingredient of all alcoholic drinks.
Uses of ethanol:
1. Ethanol is widely used in industry as a solvent.
2. Ethanol is used as an antiseptic for wounds in the form of rectified spirit.
Chemical equations: 

(i) 2C2H5OH + 2Na → 2C2H5ONa+ + H2 ↑ 

(ii)Previous Year Questions 2019


Q13: (a) State the reason why carbon can neither form C4+ cations nor C4- anions, but forms covalent bonds. Also state reasons to explain why covalent compounds
(i) are bad conductors of electricity.
(ii) have low melting and boiling points.
(b) Write the structural formula of benzene, C6H6    (2019)

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Ans: 
(a)  Ionic compounds are formed either by gaining or losing electrons from the outermost shells, but carbon which has four electrons in its outermost shell cannot form ionic bonds because
1. If carbon forms ionic bonds by gaining four electrons to attain a noble gas configuration then it would be difficult for six protons in the nucleus to hold ten electrons.
2. If carbon forms ionic bonds by loss of four electrons then it would require a lot of energy to remove these electrons from outermost shell.
Due to these reasons carbon forms covalent bonds by sharing the valence electrons.
Type of bonds formed in ionic compounds are called electrovalent bonds and the type of bonds formed in carbon compounds are called covalent bonds. 

(i) Covalent bonds are those bonds which are formed by sharing of the valence electrons between two atoms. 

 Electron dot structure of Methane

Covalent compounds are generally poor conductors of electricity because they do not have tree electrons or ions. 

(ii) Covalent compounds have low melting and boiling points because the forces of attraction between molecules of covalent compounds are very weak. On applying a small amount of heat these molecular forces break.

(b) Structural formula of Benzene, C6H6 is shown belowPrevious Year Questions 2019



Q14: What happens when this compound burns with oxygen?    (2019)

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Ans: 
(i) Methane is the first member of alkane series having formula CH4.
Covalent bonds are those bonds which are formed by sharing of the valence electrons between two atoms. Electron dot structure of methane is shown in the figure. 

Previous Year Questions 2019 Electron dot structure of Methane

Covalent compounds are generally poor conductors of electricity because they do not have tree electrons or ions.
(ii)  When methane is burnt in presence of oxygen then carbon dioxide will be produced.
CH4 + O2 → CO2 + H2O + heat + light 

Covalent compounds have low melting and boiling points because the forces of attraction between molecules of covalent compounds are very weak. On applying a small amount of heat these molecular forces break.


Q15: What is methane? Draw its electron dot structure. Name the type of bonds formed in this compound. Why are such compounds:    (Delhi 2019)
(i) poor conductors of electricity? and
(ii) have low melting and boiling points? What happens when this compound in oxygen?

Hide Answer  

Ans: (i) Methane is the simplest hydrocarbon which is produced when carbon atom shares its 4 valence electrons with 4 hydrogen atoms.
(ii) Methane achieves noble gas electronic configuration by sharing its four electrons with other elements, i.e. it forms covalent compounds.

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Previous Year Questions 2018

Q16: Which type of compounds can be formed by carbon?    (CBSE 2018)

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Ans: Covalent compounds can be formed by carbon.

Previous Year Questions 2018


Q17: What happens when hydrogen is added to a vegetable oil in the presence of nickel? Name the reaction and write one difference between the physical property of the vegetable oil and the product obtained in this reaction. Write the role of nickel in this reaction.    (CBSE 2018C)

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Ans:

  • Vegetable oil undergoes hydrogenation to form saturated product known as vegetable ghee or vanaspati ghee. This reaction is known as hydrogenation of oils.
  • Vegetable oil is liquid at room temperature whereas vegetable ghee is solid at room temperature.
  • Nickel acts as catalyst and forms saturated hydrocarbons.


Q18: (a) Why are most carbon compounds poor conductors of electricity?
(b) Write the name and structure of a saturated compound in which the carbon atoms are arranged in a ring. Give the number of single bonds present in this compound.    (CBSE 2018)

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Ans: 
(a) Carbon compounds are covalent in nature and they do not dissociate to form ions because of which they are poor conductors of electricity.
(b) Cyclohexane is the saturated compound in which the carbon atoms are arranged in a ring.

Previous Year Questions 2018Cyclohexane

Previous Year Questions 2018

It has 18 single bonds.


Q19: A compound A (C2H4O2) reacts with Na metal to form a compound B and evolves a gas which burns with a pop sound. Compound A on treatment with an alcohol C in the presence of an acid forms a sweet smelling compound, D (C4H8O2). On addition of NaOH to D gives back B and C. Identify A, B, C and D. Write the reactions involved.    (CBSE Sample Questions Paper 2018)

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Ans: The compounds A, B, C and D are given as under:
A = CH3COOH
B = CH3COONa
C = C2H5OH
D = CH3COOC2H5
The reactions are explained as under:

Previous Year Questions 2018
Also read: PPT: Carbon and its Compounds

Previous Year Questions 2017

Q20: Two carbon compounds X and Y have the molecular formula C4H8 and C5H12 respectively. Which one of these is most likely to show addition reaction? Justify your answer. Also give the chemical equation to explain the process of addition reaction in this case.    (Delhi 2017)

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Ans:  ‘X’ (C4H8) will show addition reaction because it is unsaturated, i.e. it has a double bond e.g.

Previous Year Questions 2017
 When a molecule or a compound is added to unsaturated compound to form a saturated compound is called addition reaction.


Q21: Complete the following chemical equations:    (Delhi 2017)
(i) CH3COOC2H5 + NaOH →
(ii) CH3COOH + NaOH →
(iii) 

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Ans: (i) CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
(ii) CH3COOH + NaOH → CH3COONa + H2O
(iiiC2H5OH + CH3COOH → CH3COOC2H+ H2O


Q22: Write the structural formula of ethanol and list its two physical properties. What happens when it is heated with excess of cone. H2SO4 at 443 K? State the role of conc. H2SO4 in this reaction.     (DoE, AI 2017, Foreign 2013)

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Ans: CH3—CH2OH is structural formula of ethanol
Physical properties:
(i) It is liquid and has sweet smell.
(ii) It is soluble in water in all proportions.
When ethanol is heated with cone. H2SO4 at 443 K, ethene is formed.

Previous Year Questions 2017
Conc. H2SO4 acts as dehydrating agent.


Q23: Ethanoic acid reacts with absolute ethanol in the presence of Conc. H2SO4 to form a compound:    [AI 2017 C]
(i) Write the smell and class of compounds to which this compound belongs.
(ii) Write the chemical equation for the reaction and state the role of Conc. H2SO4 in the reaction.
(iii) Write one use of the product of this reaction.

Hide Answer  

Ans:
(i) The compound formed is ‘Esters’ It has a pleasant fruity smell.

(ii) Previous Year Questions 2017
Conc. H2SO4 acts as dehydrating agent.
(iii) It is used as a solvent in glass, nail polish removers and decaffeinating tea and coffee. It is used is solvent in printing to modify its drying rate.
Uses of esters:

  • Esters are used as flavouring agents in ice creams, cold drinks etc.
  • They are used in perfumes.


Q24: Distinguish between esterification and saponification reactions of organic compounds with the help of the chemical equation for each. What is the use of (i) esters and (ii) saponification process?    (AI 2017, Foreign 2012)

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Ans:


Uses:
(i) Esters are used in cold drinks, ice creams, perfumes and as artificial flavouring agents.
(ii) Saponification process is used in manufacture of soaps.


Q25: Name two oxidising agents that are used for the conversion of alcohols to acids. Distinguish between ethanol and ethanoic acid on the basis of (a) litmus test, and (b) reaction with sodium hydrogen carbonate.    (Delhi 2013; AI 2012; Foreign 2017)

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Ans: Alkaline KMnO4/NaOH (Potassium permanganate) and Acidified K2Cr2O7 / H2SO4 (Potassium dichromate).Previous Year Questions 2017


Q26: What is the homologous series of carbon compounds? List any two characteristics. Write the name and formula of next higher homologue of HCOOH.   (CBSE 2017)

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Ans: A series of compounds whose two successive members differ by – CH2 in the molecule formula and which show similar chemical properties are called homologous series.

Characteristics:
1. Melting and boiling points increase regularly with increase in molecular mass.
2. Solubility in a particular solvent shows a regular gradation.
3. Chemical properties remain the same in the homologous series.
Next higher homologue of HCOOH is CH3COOH. Its name is ethanoic acid or acetic acid.


Q27: When ethanol reacts with ethanoic acid in the presence of conc. H2SO4, a substance with fruity smell is produced. Answer the following:    (CBSE 2016)
(i) State the class of compounds to which the fruity smelling compounds belong. Write the chemical equation for the reaction and write the chemical name of the product formed.
(ii) State the role of cone. H2SO4 in this reaction.

Hide Answer  

Ans:

(i) Fruity smelling compounds belong to the class of esters.

Previous Year Questions 2017
(ii) Conc. H2SO4 acts as a dehydrating agent and removes the water formed in the reaction.


Q28: Soaps and detergents are both types of salts. State the difference between the two. Write the mechanism of the cleansing action of soaps. Why do soaps not form lather (foam) with hard water? Mention any two problems that arise due to the use of detergents instead of soaps.   (Delhi 2017)

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Ans: Soaps are sodium or potassium salts of fatty acids having —COONa group. Detergents are sodium or potassium salts of sulphonic acids having — SO3Na and —SO4Na group.
Cleansing Action of Soap. Soaps consist of a large hydrocarbon tail which is hydrophobic (water-hating or water repelling) with a negatively charged head which is hydrophilic (water-loving) as shown in figure.

Previous Year Questions 2017Previous Year Questions 2017Previous Year Questions 2017When a soap is dissolved in water, the molecules associate together as clusters called micelles in which, water molecules being polar in nature, surround the ions and the hydrocarbon part of the molecule attracts grease, oil and dirt.Hard water has Ca2+ and Mg2+ ions when they react with soap to form insoluble compound and soap goes waste.
Disadvantages of Detergents
(i) Detergents are expensive.
(ii) Many detergents are branched chain hydrocarbon which are not biodegradable and create water pollution.

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Previous Year Questions 2016

Q29: What are covalent compounds? Why are they different from ionic compounds? List their three characteristic properties.    (CBSE 2016)

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Ans:

Covalent compounds are the compounds that contain covalent bonds in the molecule. A covalent bond is formed by sharing of electrons between the atoms.
While a covalent compound contains covalent bonds, the ionic compound is made up of ionic bond which is formed by transference of electrons from one atom to the other.
Characteristics of covalent compounds:
(i) These compounds show low melting points and boiling points because the intermolecular forces are weak.
(ii) Such compounds are poor conductors of electricity.
(iii) Generally insoluble or less soluble in water but soluble in organic solvents.


Q30: Write chemical equation of the reaction of ethanoic acid with the following:
(а) Sodium; 
(b) Sodium hydroxide; (c) Ethanol.
Write the name of one main product of each reaction.    (CBSE 2016)

Hide Answer  

Ans:

Previous Year Questions 2016


Q31: An aldehyde as well as a ketone can be represented by the same molecular formula, say C3H6O. Write their structures and name them. State the relationship between the two in the language of science.    (CBSE 2016)

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Ans: The structures of the aldehyde and ketone with the molecular formula C3H6O are

Previous Year Questions 2016

Both these compounds are isomers (because they have the same molecular formula but different structural formula).


Q32: (i) The chemical properties of ethanol are different from methyl ethanoate. Justify the statement with a proper reason.
(ii) Methyl ethanoate is used in making perfumes. Justify.
(iii) Ethanol is converted into ethene with excess of hot concentrated H2SO4. Justify with the help of chemical equation.   (CBSE 2016)

Hide Answer  

Ans:
(i) Ethanol reacts with sodium metal whereas methyl ethanoate does not. They differ in functional group, and in chemical properties.
(ii) It is because it has pleasant fruity smell.
(iii) Previous Year Questions 2016

Ethanol gets dehydrated with conc. H2SO4 to form ethene and H2O.


Q33: (a) Define the term functional group. Identify the functional group present in

(i)

Previous Year Questions 2016

(ii)

Previous Year Questions 2016

(b) What happens when 5% alkaline KMnO4 solution is added drop by drop to warm ethanol taken in a test tube? State the role of alkaline KMnO4 solution in this reaction.    (Foreign 2016)

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Ans:

(a) It is an atom or group of atoms or reactive part of compound which largely determine the chemical properties of compound (i) Methanol (ii) Ethanoic acid.

(b) Previous Year Questions 2016

Alkaline KMnO4 acts as an oxidizing agent.


Q34: On dropping a small piece of sodium in a test tube containing carbon compound ‘X ’ With molecular formula C2H6O, a brisk effervescence is observed and a gas ‘Y’ is produced, On bringing a burning splinter at the mouth of the test tube the gas evolved burns with a pop sound. Identify ‘X ’ and ‘Y’. Also write the chemical equation for the reaction. Write the name and structure of the product formed, when you heat ‘X’ with excess cone, sulphuric acid.    (AI 2016)

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Ans:
‘X’ is C2H5OH

2C2H5OH + 2Na → C2H5Na + H2
‘Y’ is H2(g) which burns with ‘pop’ sound.

Previous Year Questions 2016

Ethene is the product.


Q35: An organic compound ‘P’ is a constituent of wine. ‘P ’ on reacting with acidified K2Cr2O7 forms another compound ‘Q’. When a piece of sodium is added to ‘Q’ a gas ‘R’ evolves which burns with a pop sound. Identify P, Q and R and write the chemical equations of the reactions involved.    (Foreign 2016)

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Ans: 

‘P ’ is C2H5OH, Ethanol

Previous Year Questions 2016

Q is Acetic Acid.
2CH3COOH + 2Na → 2CH3COONa + H2
‘R’ is hydrogen gas which burns with ‘pop’ sound.


Q36: (a) Give chemical tests to detect the presence of
(i) Ethanol 

(ii) Ethanoic acid
(b) Why ethanoic acid is also called glacial acetic acid?   (CBSE 2016)

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Ans:

(a) Add sodium hydrogen carbonate. Ethanol will not react. Ethanoic acid will give brisk effervescence due to carbon dioxide.
(b) Pure ethanoic acid exists as solid like glaciers at 291 K, therefore, called glacial acetic acid.


Q37: Write a chemical equation of the action of ethanoic acid with the following: (a) Sodium; (b) Sodium hydroxide; (c) Ethanol.
Write the name of one main product of each reaction.   (Al 2016)

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Ans:

(a) 2CH3COOH + 2Na → 2CH3COONa + H2
Sodium ethoxide is the main product formed.
(b) CH3COOH + NaOH→ CH3COONa + H2O
Sodium ethoxide is the main product formed.

(c)
Previous Year Questions 2016

Ethyl ethanoate (ester) is the main product formed.


Q38: Under what conditions an oxidation reaction can be called a combustion reaction? Illustrate your answer with examples.   (CBSE 2016)

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Ans: When complete oxidation of fuel takes place with the release of high amount of heat and light, it is called combustion reaction. It is highly exothermic and generally accompanied by evolution of heat and light.
Examples – Burning of coal
C + O2 → CO2 + heat + light
Burning of the rocket fuel (liquid H2)
2H2O + O2 → 2H2O + heat + light
Burning of acetylene used in welding torches.
2C2H2 + 5O2 → 4CO2 + 2H2O + heat + light


Q39: C3H6, C4H8 and C5H10 belong to the same homologous series.    (CBSE 2016)
(i) Define homologous series.
(ii) Why are the melting and boiling points of C5H10 is higher than C4H8?
(iii) Arrange these hydrocarbons in order of increasing boiling points.

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Ans:  (i) The series of organic compounds which have similar chemical properties, same functional group and same general formula is called homologous series.
(ii) It is because C5H10 has higher molecular weight, more Vander Wall’s force of attraction and higher boiling points and melting points.
(iii) C3H6 < C4H8 < C5H10 is increasing order of boiling point.


Q40: An aldehyde as well as ketone can be represented by the same molecular formula, say C3H6O. Write their structures and name them. State the relation between the two in the language of science.    (Al 2016)

Hide Answer  

Ans:


They are functional isomers.


Q41: Write the name and molecular formula of an organic compound having its name suffixed with ‘ol’ and having two carbon atoms in its molecule. Write the balanced chemical equation to indicate what happens when this compound is heated with excess conc. H2SO4 and the name of main product formed. Also state the role of conc. H2SO4 in the reaction.   (2016)

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Ans:
Ethanol, C2H5OH

Previous Year Questions 2016

Conc. H2SO4 acts as dehydrating agent.


Q42: What is an oxidising agent? What happens when an oxidising agent is added to propanol? Explain with the help of a chemical equation.    (Delhi 2016)

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Ans:
Those substances which add oxygen are called oxidising agent and propanoic acid is formed.

Previous Year Questions 2016

Reaction of Propanol with Oxidizing Agent
Propanol will get oxidised to propanoic acid by acidified KMnO4.

Q43: (a) You have three unlabelled test tubes containing ethanol, ethanoic acid and soap solution. Explain the method you would use to identify the compounds in different test tubes by chemical tests using litmus paper and sodium metal.
(b) Give the reason of formation of scum when soaps are used with hard water.   (Foreign 2016)

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Ans:
(a) Ethanol will not be affected by blue litmus as well as red litmus paper. Ethanoic acid will turn blue litmus red whereas red litmus will remain as it is. Soap solution will turn red litmus blue but blue litmus will remain as it is. Sodium metal will liberate hydrogen gas with ethanol as well as ethanoic acid. Soap solution will not react with sodium metal.
(b) It is because soap will react with Ca2+ and Mg2+ ions present in hand water to form calcium salts of fatty acids which are insoluble and called scum.


Q44: (a) Give a chemical test to distinguish between saturated and unsaturated hydrocarbon.
(b) Name the products formed when ethane burns in air. Write the balanced chemical equation for the reaction showing the types of energies liberated.
(c) Why is reaction between methane and chlorine in the presence of sunlight considered a substitution reaction?    (Delhi 2016)

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Ans:
(a) Add bromine water. Unsaturated hydrocarbon will decolourise bromine water whereas saturated hydrocarbon will not.
(b) Carbon dioxide and Water are formed.

2C2H6 (g) + 7O2 (g) → 4CO(g) + 6H2O +Heat + Light
(c) It is because hydrogen atom of methane gets substituted by chlorine atom to form chloromethane, therefore, it is called a substitution reaction.


Q45: What are micelles? Why does it form when soap is added to water? Will a micelle be formed in other solvents such as ethanol also? State briefly how the formation of micelles help to clean the clothes having oily spots.    (Foreign 2016)

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Ans:  Micelles: When molecular ions in soaps and detergents aggregate, they form micelles.
It is because a large number of molecular ions of soaps get aggregated and form colloidal solution. Soap has hydrophobic tail (hydrocarbon) which dissolves in hydrocarbon part and hydrophilic part dissolves in water. Ethanol is non-polar solvent therefore micelles are not formed because hydrocarbon part gets attracted towards ethanol and the ionic end will not dissolve in alcohol.


Q46: A carbon compound ‘P ’ on heating with excess cone. H2SO4 forms another carbon compound ‘Q’ which on addition of hydrogen in the presence of nickel catalyst forms a saturated carbon compound ‘R’. One molecule of ‘R’ on combustion forms two molecules of carbon dioxide and three molecules of water. Identify P, Q and R and write chemical equations for the reactions involved.   (CBSE 2016)

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Ans: The compounds are identified as under:
P = Ethanol (CH3CH2OH)
Q = Ethene (CH2 = CH2)
R = Ethane (CH3 – CH3)
Reactions are given as under:

Previous Year Questions 2016
Also read: PPT: Carbon and its Compounds

Previous Year Questions 2015

Q47: Draw the electron-dot structure for ethyne. A mixture of ethyne and oxygen is burnt for welding. In your opinion, why cannot we use a mixture of ethyne and air for this purpose?   (AI 2015)

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Ans:

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Electron-dot structure of Ethyne

Ethyne and air will not produce enough heat due to incomplete combustion needed for welding purpose. Ethyne and oxygen will produce a lot of heat due to complete combustion which can be used for welding purposes.


Q48: That is meant by isomers? Draw the structures of two isomers of butane, C4H10. Explain why we cannot have isomers of the first three members of alkane series.   (DoE, Delhi 2015)

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Ans: Isomers are those compounds which have the same molecular formula but different structural formula.

Previous Year Questions 2015

Butane & Iso-Butane are two isomers of C4H10. Isomers are not possible for the first three members because branching is not possible.


Q49: Write the name and general formula of a chain of hydrocarbons in which an addition reaction with hydrogen is possible. State the essential conditions for an addition reaction. Stating this condition, write a chemical equation giving the name of the reactant and the product of the reaction.   (AI 2015)

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Ans:  CnH2n and CnH2n- 2 are the general formula of alkene and alkynes in which addition reaction with hydrogen is possible. Hydrogen is added to unsaturated hydrocarbon (having double or triple bond) in presence of heated nickel as catalyst.

Previous Year Questions 2015


Q50: Write the molecular formula of the following compounds and draw their electron-dot structures:    (Foreign 2015)
(i) Ethane 
(ii) Ethene 
(iii) Ethyne

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Ans:
(i) Ethane (C2H6)
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(ii) Ethene (C2H4)

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(iii) Ethyne (C2H2)

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Q51: Write the respective chemical equations to show what happens when    (Foreign 2015)
(i) Methane is burned in presence of oxygen?
(ii) Ethanol is heated with concentrated sulphuric acid at 443 K?
(iii) Ethanol reacts with ethanoic acid in the presence of an acid acting as a catalyst?

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Ans:
(i) CH4 + 2O2 → 2H2O + CO2
Water and carbon dioxide formed.

(ii)

Previous Year Questions 2015

The concentrated sulphuric acid can be regarded as a dehydrating agent which removes water from ethanol.

(iii) 


 Esters are sweet smelling substances.


Q52: Write the chemical equation to explain what happens when ethanol is heated with alkaline solution of potassium permanganate. Mention two physical properties and two uses of ethanol.    (Foreign 2015)

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Ans:
Alkaline KMnO4 is dark pink in colour. So when it is added to ethanol and heated, the pink colour of the solution disappears. When excess of KMnO4 is added, the pink colour does not disappear, indicating that all the ethanol has been converted to ethanoic acid.
• Physical properties of ethanol:
(a) It is a colourless liquid with pleasant smell and burning taste.
(b) It is a volatile liquid with low boiling point.
• Uses of ethanol:
(a) It is used in the manufacture of medicines, varnished, paints, dyes, soap, etc.
(b) It is a good solvent. Many organic compounds which are insoluble in water are soluble in ethanol.


Q53: What is the difference between the molecules of soaps and detergents? Explain the cleansing action of soaps.    (CBSE 2015)

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Ans:
Soaps are sodium or potassium salts of long chain carboxylic acids.
Detergents are ammonium or sulphonate salts of long chain carboxylic acids.

Previous Year Questions 2015

Cleansing action of soap: The ionic end of soap dissolves in water while the carbon chain dissolves in oil. Soap molecules form micelles where one end of the soap molecule is linked to oil droplet while the ionic part faces water. This forms an emulsion in water. On adding water, these micelles containing dirt are washed away and our clothes are washed clean. This is represented in the figure.


Q54: With the help of an example, explain the process of hydrogenation. Mention the essential conditions for the reaction and state the change in physical property with the formation of the product.    (CBSE 2015)

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Ans: Conversion of oil into fat by passing hydrogen gas in the presence of nickel as catalyst is called hydrogenation. This is how hydrogenation is carried out.
Conditions: The presence of a catalyst like nickel is essential.
Change in physical property: Oils are liquids while the fats obtained on hydrogenation are solids i.e., a change in the physical state takes place on hydrogenation.

Q55: Explain why carbon forms compounds mainly by covalent bond. Explain in brief two main reasons for carbon forming a large number of compounds. Why does carbon form strong bonds with most other elements?    (CBSE 2015)

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Ans: Carbon has no tendency to lose electrons easily. Similarly it has no tendency to gain electrons. It completes its octet by sharing the electrons with other atoms. Therefore, carbon forms compounds mainly by covalent bond.
Two main reasons for carbon forming a large number of compounds are given below:
Catenation: It is the property of carbon to link with other carbon atoms to form straight chain, branched chain or cyclic compounds. Thus, carbon forms a large number of compounds containing 2, 3, 4, 5,….etc., carbon atoms.
Tetravalency: Carbon has four electrons in the outermost shell. It shares these four electrons with four electrons from four other monovalent atoms. Carbon has the tendency to form bonds with hydrogen, oxygen, nitrogen and halogens. This increases the number of compounds of carbon.
Carbon forms strong bonds with most other elements: This is because carbon has a small atomic size. It can hold strongly the four pairs of electrons that it shares with other atoms. Therefore, it forms strong bonds with other elements.