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)

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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: 
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.

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.

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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

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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

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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)

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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 length (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.

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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)

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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 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

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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.

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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)
N = 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)

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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.

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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)

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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)

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)

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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  

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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)

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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. 

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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.

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.

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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.

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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 :

(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)

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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.

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.

Q2: Observe the following diagram showing an image formation in an eye:

(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)

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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)

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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.

Q5: The colour of the clear sky from the Earth appears blue but from space, it appears black. Why?   (2023)

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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:

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)

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Ans: 

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)

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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)

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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)

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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.  

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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.  

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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)

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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)

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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 

⇒ f = -x
The ray diagram of the defective eye and its rectification are shown here.

Normal Eye 

Myopic Eye

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.
(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.

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.
An 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.

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:

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.

(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.

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.

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.

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.

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 λ_b < λ_r, only red colour having a longer wavelength and is least scattered, reaches our eyes. Hence, the sun appears red at sunrise or sunset.

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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

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
Cause 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.

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,

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.
(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
Whenever 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.

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.

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.
The 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.

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 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

For a concave lens take u =−60 cm (object on left) and f = −30 cm. Substitute:

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.

• 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.

• 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):

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.

(a) (iii):
Focal lengths: f_A = 10 cm = 0.10 m, f_B = 20 cm = 0.20 m.
Power P = 1/f
 so

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) 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

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: 

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
For a concave mirror take u = -10 cm (object to the left) and f = -15 cm. Substitute:
so, 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
 

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 Take u = -30 cm (object to the left) and v = -60 cm (real image in front of mirror, same side as object).

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: Step 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), thenand
Image distance v =−24 cm indicates image is on the same side as object (virtual image).
Step 4: 
Substitute v=−24 cm and u=6 cm:

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)

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_o = 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:

Solve for u: (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 Mirror formula:

Substitute v = -2u

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:

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):

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: 

– Refractive index of medium B: 

– Refractive index of medium C: 

From the problem, if medium B has a higher optical density than A and C, it implies:
This means that the speed of light in medium B is less than that in A and C:

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: v_A(higher speed)
• Medium B: v_B(lowest speed)
• Medium C: v_C (medium speed)

2. Optical Densities Comparison:

• Medium A: lowest optical density
• Medium B: highest optical density
• Medium C: medium optical density

– Speed of Light: 

– Optical Densities: 

(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: i_A 
• Angle of refraction in medium B: r_B

Using Snell’s law:

⇒ 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: 
= 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.

OR
(iii) (B) Here ƒ = –12 cm, u = –18 cm, n = ?

Mirror formula:

v = -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; n_m = 1.5; v_m = ?

Absolute refractive index of a medium (n_m):

n_m = speed of light in vacuum (c)speed of light in medium (v_m)

n_m = cv_m

v_m = cn_m = 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

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Ans: (a)

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)

(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.  

(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. 

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
(2) is directed towards its principal focus(ii) Given u = –16 cm, f = + 24 cm, h = 4 cm
Lens formula:

Magnification:

Image height:

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:

Hide Answer  

• 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:The image is at  v=+10cm, virtual and behind the mirror.
  Find Focal Length:
  • Use the mirror formula
    The positive focal length indicates a convex mirror, consistent with the positive magnification.
    • New Magnification Requirement:
      • New magnification: m′=+1/3
      • Magnification formula:Mirror Formula: The new object distance is 40 cm from the mirror.
    • Verification:
      Calculate v′
      Magnification:
      The 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.

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: 

• 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.

Hide Answer  

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) r_water < r_glass < r_{carbon 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_g = 4/3

Speed of light in vacuum 
Refractive index of water 

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, η_g = 32

Refractive index of water, η_w = 43

Refractive index of glass, η_g = 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, η_w = 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? 

Hide Answer  

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

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 erect, magnified, and on the same side as the object.
• Concave Lens: Always forms a virtual image. The image is erect, diminished, 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)

(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.

Hide Answer  

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)

(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)

Hide Answer  

Ans: (A)
(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.        

Hide Answer  

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.  

Hide Answer  

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                

Hide Answer  

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              

Hide Answer  

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  

Hide Answer  

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. 

Hide Answer  

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.

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     

Hide Answer  

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.  

Hide Answer  

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. 

Hide Answer  

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) n₁ < n₂ < n₃
(b) n₁ > n₂ > n₃
(c) n₁ < n₂ = n₃
(d) n₁ = n₃ < n₂     

Hide Answer  

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.

Hide Answer  

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 

Hide Answer  

Ans: (c)
Sol: Given, combined power = +1 D
Focal length of one lens, f₁ = 20 cm
Combined focal length, f_combined = 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)

(a) Concave; between O’ and Y
(b) Concave : between O and Y
(c) Convex; between O’ and Y
(d) Convex; between O and Y  

Hide Answer  

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  

Hide Answer  

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.

Hide Answer  

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.

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)

(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)

(a) n₁ = n₂
(b) n₁ > n₂
(c) n₂ > n₃
(d) n₃ > n₁  

Hide Answer  

Ans: (d)
Sol: 

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  

Hide Answer  

Ans: (d)
Sol: Given, refractive index of medium A, μ_A = 1.5
Refractive index of medium B, μ_B = 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:

μ_A = cv_A ⇒ 1.5 = 3 × 10⁸v_A m/s

v_A = 2 × 10⁸ m/s

For medium B:

μ_B = cv_B ⇒ 1.33 = 3 × 10⁸v_B m/s

V_B = 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)

(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)

Hide Answer  

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)

Hide Answer  

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)

Hide Answer  

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 = h_ih_o = vu

Substitute the known values:

m = -15-30 = 12

Also, To find the height the value of h_i

Using the magnification formula:

m = h_ih_o

Substitute the values of m and h_o:

12 = h_i4

Solve for h_i:

h_i = 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, vμ_x = 23

Refractive index of medium y with respect to z, zμ_y = 43

∴ Refractive index of medium x with respect to z, zμ_x = v_{μx x} zμ_y = 23 × 43 = 89

∴ Refractive index of medium z with respect to x, xμ_z = 1zμ_z = 98

Now, speed of light in x = 3 × 10⁸ m/s

Speed of light in y, v_y = ?

yμ_x = Speed of light in xSpeed of light in y
⇒ v_y = 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

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

(C) An incident ray coming obliquely to the principal axis towards a point (Pole of the mirror) on the convex mirror is reflected obliquely.

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)

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).

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.

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.

The 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
  

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.

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.

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)

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

1f = 1v – 1u ⇒ 120 = 1v – 1(-30)

1v = 120 – 130 ⇒ 1v = 360 – 260

Thus, v = 60 cm

m = h_ih_o = vu ⇒ h_i4 = 60(-30)h_i = – 8 cm
Ratio = h_i/h_o 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.

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 = h_ih_o = -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:

So, 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:
The 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.
The 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.

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 n_m = 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.

(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.(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.

(b) m=−3, 
u=−20 cm , 
v=? 

Image Position

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.

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

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.

(ii) a magnified and inverted image when an object is placed in the following positions: Between F₁ and 2F₁

(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.

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).

b. Concave mirror when object is in between P and F its enlarged, erected and virtual image is formed.

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.

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:

(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.

(ii) For object distance 20 cm.

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.


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

b. Convex mirror

Concave mirror form magnified virtual image of an object.

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.

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 = ?

• 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.

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.Hence, 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.

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

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.When 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:1 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)

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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 pair. During 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)

​(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.
(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.

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.(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 —

(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.

​​​​​​​​​​​​​​​​​

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:

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 generation, 25% of the plants bear white flowers.
(ii) The ratio of gene combinations in F₂ generation is 1 WW : 2 Ww : 1 ww.

(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 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)(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)

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:
 75% 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 F₂ progeny 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 F₁ progeny?  
(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 F₂ progeny 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 F₂ generation?
(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 F₂ generation 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 F₂ generation (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:

OR

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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.

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 F₁ progeny 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%.

(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

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.

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 F₂ progeny 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:

The colour in the F₁ progeny is green stemmed as green stem colour is dominant.

(ii) F₁ progeny on self pollination:

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 :So, 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:

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:

(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 

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.

(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.

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 F₁ generation 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.

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.
Skeleton of forelimbs of (a) Human (b) Dog  (c) Bird (d) Whale showing homologous features.

 (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.

Comparison of early Development stages (a) Fish (b) Salamander (c) Tortoise (d) Chicken (e) Human

(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.

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.

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 F₁ generation 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.

Independent 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:

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:
Hence, 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 – 1

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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:
(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 F₁ progeny 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.

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)

A 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:

(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.

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.

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.

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)

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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)

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 vesicles, add 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:

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:

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 = Anther(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:

(b) Uterus Changes:

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)

(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 

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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.

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)
(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)

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.

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.                  

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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:
Fertilisation, 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:

Q4: What are testes? List two functions performed by testes in human beings. (2019 C)

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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:
(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)

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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.
The 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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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:

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)

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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.

In asexual reproduction, only mitotic divisions are involved and the chromosome number remains the same.

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:

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.