Q1: What are trophic levels? Give an example of a food chain and state the different trophic levels in it. Ans:
A trophic level is a level of species in an ecosystem based on the source of nutrition, such as producers, primary consumers, secondary consumers, etc.
The producers form the first trophic level as they manufacture food; primary consumers form the second trophic level, the secondary consumers form the third, and the tertiary consumers form the fourth trophic level.
Various trophic levels are connected through food chains. Example: In an aquatic food chain, phytoplankton are the producers, zooplankton are the primary consumers, small fish are the secondary consumers, and so on.
Q2: What is the role of decomposers in the ecosystem? Ans:
Decomposers include micro-organisms such as bacteria and fungi that obtain nutrients by breaking down the remains of dead plants and animals.
They help in the breakdown of organic matter or biomass from the body of dead plants and animals into simple inorganic raw materials, such as CO2, H2O, and some nutrients.
Page No. 214
Q1: Why are some substances biodegradable and some non-biodegradable? Ans:
Some substances, such as metal, glass, plastic, etc., which cannot be decomposed by living organisms, are non-biodegradable wastes.
These substances are non-biodegradable because the microorganisms do not have enzymes that can digest these substances.
Therefore, we classify them as non-biodegradable wastes.
Other substances, such as paper, vegetable wastes, etc. that can be easily broken down by enzymes are biodegradable wastes.
Q2: Give any two ways in which biodegradable substances would affect the environment. Ans: Biodegradable substances affect the environment in the following ways:
Biodegradable substances such as tree leaves, plant parts, and kitchen wastes can be used as humus after composting. This will enhance soil fertility.
Biodegradable substances mainly contain carbon. These substances, after decomposition, release that carbon back into the atmosphere.
Q3: Give any two ways in which non-biodegradable substances would affect the environment. Ans: Non-biodegradable substances affect the environment in the following ways:
They contaminate soil and water resources as they cannot be decomposed by microorganisms.
These substances, when accidentally eaten by stray animals, can harm them and can even cause their death.
Page No. 216
Q1: What is ozone, and how does it affect any ecosystem? Ans:
Ozone is a colorless gas that acts as a screen for ultraviolet radiation.
It is continuously formed at the higher levels of the atmosphere due to the action of UV rays on molecular oxygen.
The high-energy UV radiations break down O2 molecules into nascent oxygen.
Then, this free oxygen atom combines with an oxygen molecule to form ozone.
In recent years, the amount of ozone in the atmosphere has been depleted.
This ozone depletion causes a greater amount of ultraviolet radiation to enter the Earth’s atmosphere.
This has an indirect effect on the ecosystem. (Ecosystem includes both the biological community and the non-living components of an area). It results in the death of many phytoplankton, thereby affecting the process of photosynthesis.
Plants utilize atmospheric CO2 to make their food. In the absence of plants, the levels of CO2 in the atmosphere will increase, which would, in turn, lead to an increase in global warming.
The depletion of the ozone layer also increases the frequency of infectious diseases as it suppresses the immune systems of both human beings and animals.
The frequency of skin cancer also increases in human beings because of direct exposure to ultraviolet radiation.
Q2: How can you help in reducing the problem of waste disposal? Give any two methods. Ans:
Proper waste management can help solve the issue of waste disposal.
By following these steps, we can manage waste more effectively and reduce its impact on our surroundings. (i) Use separate bins (blue and green) to dispose of non-biodegradable and biodegradable wastes. (ii) Different garbage bins for disposing of biodegradable waste and non-biodegradable waste (iii) Reduce the usage of non-biodegradable products such as plastics.
Exercise
Q1: Which of the following groups contain only biodegradable items? (a) Grass, flowers, and leather (b) Grass, wood, and plastic (c) Fruit peels, cake, and lime-juice (d) Cake, wood, and grass Ans: (c) Fruit peels, cake, and lime-juice (d)Cake, wood, and grass
Q2: Which of the following constitutes a food chain? (a) Grass, wheat, and mango (b) Grass, goat, and human (c) Goat, cow, and elephant (d) Grass, fish, and goat Ans: (b) Grass, goat, and human
Food Chain
Q3: Which of the following are environment-friendly practices? (a) Carrying cloth bags to put purchases in while shopping (b) Switching off unnecessary lights and fans (c) Walking to school instead of getting your mother to drop you on her scooter (d) All of the above Ans: (d) All of the above
Q4: What will happen if we kill all the organisms at one trophic level?
Ans:
Various trophic levels are connected through the food chains. If all the organisms of any one trophic level are killed, it will disrupt the entire food chain. Example: In a food chain, if all the plants are killed, then all the deer will die due to lack of food. If all the deer are dead, then the tigers will die soon. Due to the death of these animals, the decomposer population will rise in that area.
This is just an example of one food chain. However, in nature, food chains are not isolated.
They are interconnected in the form of food webs.
Therefore, killing all the plants in an area will not only affect the deer, but it will also affect other herbivores such as goats, cattle, sheep, etc.
Q5: Will the impact of removing all the organisms in a trophic level be different for different trophic levels? Can the organisms of any trophic level be removed without causing any damage to the ecosystem? Ans:
Organisms of all trophic levels are equally important and are an integral part of the ecosystem.
If all the producers are removed, then it will affect all the herbivores as it is their primary food source.
The death of herbivores will soon affect the primary carnivores and so on.
Now let us suppose that all the deer (herbivores) are killed in a region. This can lead to an increase in the number of producers.
Simultaneously, there will be an increase in the number of other herbivores, such as rabbits, goats, sheep, etc., due to less competition.
This will also lead to an increase in the population of only consumers of these increased herbivores. Thus, the balance in the ecosystem gets disturbed if any of its component organisms are removed.
Q6: What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem? Ans:
Biomagnification is the increase in the concentration of pollutants or harmful chemicals within each step of the food chain.
The levels of biomagnification will be different at different trophic levels. Example: In a pond of water, DDT was sprayed, and the producers were found to have 0.04 ppm concentration of DDT. Since many types of plankton are eaten by some fishes and clams, their body accumulates 0.23 ppm of DDT. Seagulls that feed on clams accumulate more DDT as one seagull eats many clams. Hawk, the top carnivore, has the highest concentration of DDT.
Biomagnification
Q7: What are the problems caused by the non-biodegradable wastes that we generate? Ans: Non-biodegradable substances affect the environment in the following ways:
Since the non-biodegradable substances cannot be broken down, they accumulate and thus contaminate the soil and the water resources.
These substances, when accidentally eaten by some stray animals, can harm them and can even cause their death.
These substances occupy more space in landfills and require special disposal techniques.
These materials can accumulate in the environment and can also enter the food chain.
Q8: If all the waste we generate is biodegradable, will this have no impact on the environment?
Ans:
The generation of only biodegradable waste will have a positive impact on the environment.
There will not be any pollution caused by the non-biodegradable wastes.
The problem associated with waste management and disposal will also not occur.
The population of decomposers will increase to break down the extra biodegradable waste generated.
Q9: Why is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage? Ans:
Ozone depletion occurs widely in the stratosphere.
However, it is more prominent over the Antarctic region and is known as the ozone hole.
Diagram Representing Ozone Hole
Consequences of ozone depletion:
It causes skin darkening, skin cancer, aging, and corneal cataracts in human beings.
It can result in many phytoplankton deaths, which leads to increased global warming.
The steps taken to limit the damage:
To limit the damage to the ozone layer, the release of CFCs into the atmosphere must be reduced.
CFCs used as refrigerants and in fire extinguishers should be replaced with environmentally safe alternatives.
Also, the release of CFCs through industrial activities should be controlled.
Q1: Why does a compass needle get deflected when brought near a bar magnet? Ans: The needle of a compass is a small magnet. That’s why when a compass needle is brought near a bar magnet, its magnetic field lines interact with that of the bar magnet. Hence, a compass needle gets deflected.
Page No: 200
Q1: Draw magnetic field lines around a bar magnet. Ans: Magnetic field lines of a bar magnet emerge from the north pole and terminate at the south pole. Inside the magnet, the field lines emerge from the south pole and terminate at the north pole, as shown in the given figure. Magnetic Field Lines Q2: List the properties of magnetic field lines. Ans: Properties of magnetic lines of force –
The magnetic field lines originate from the north pole of a magnet and end at its south pole.
The magnetic field lines become closer to each other near the poles of a magnet but they are widely separated at other places.
Two magnetic field lines do not intersect one another.
Q3: Why don’t two magnetic field lines intersect each other? Ans: If two magnetic field lines of force intersect each other then at the point of intersection, the compass needle would show two different directions which is not possible.
Page No: 201
Q1: Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right-hand rule to find out the direction of the magnetic field inside and outside the loop. Ans: Inside the loop = Pierce inside the table Outside the loop = Appear to emerge out from the table For downward direction of current flowing in the circular loop, the direction of magnetic field lines will be as if they are emerging from the table outside the loop and merging in the table inside the loop. Similarly, for upward direction of current flowing in the circular loop, the direction of magnetic field lines will be as if they are emerging from the table outside the loop and merging in the table inside the loop, as shown in the given figure.
Q2: The magnetic field in a given region is uniform. Draw a diagram to represent it. Ans:
Magnetic Field The magnetic field lines inside a current-carrying long straight solenoid are uniform.
Page No: 202
Q1: Choose the correct option. The magnetic field inside a long straight solenoid-carrying current (a) is zero (b) decreases as we move towards its end (c) increases as we move towards its end (d) is the same at all points Ans: (d) Sol: The magnetic field inside a long straight current-carrying solenoid is uniform therefore it is the same at all points.Page No: 203
Q1: Which of the following property of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.) (a) mass (b) speed (c) velocity (d) momentum Ans: (c) Sol: When a proton enters the region of magnetic field, it experiences magnetic force. Due to which the path of the proton becomes circular. As a result, the velocity and the momentum change.
Page No: 204
Q1: In Activity 12.7, How do we think the displacement of rod AB will be affected if, (i) – current in rod AB is increased; (ii) – a stronger horse-shoe magnet is used; and (iii) – length of the rod AB is increased? Ans: (i) When the current in the rod AB is increased, force exerted on the conductor increases, so the displacement of the rod increases. (ii) When a stronger horse-shoe magnet is used, the magnitude of the magnetic field increases. This increases the force exerted on the rod and the displacement of the rod. (iii) When the length of the rod AB is increased, force exerted on the conductor increases, so the displacement of the rod increases. Displacement of Rod Q2: A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is (a) towards south (b) towards east (c) downward (d) upward Ans: (d) Sol: The direction of the magnetic field can be determined using Fleming’s Left-hand rule. According to the rule, if we arrange our thumb, forefinger and the middle finger of the left-hand right perpendicular to each other, then the thumb points towards the direction of the magnetic force, the middle finger the direction of current and the forefinger the direction of the magnetic field. Since the direction of positively charged particle is towards west, the direction of the current will also be towards the west. The direction of the magnetic force is towards the north, hence the direction of the magnetic field will be upward according to Fleming’s Left-hand rule.
Left – Hand Rule
Page No: 205
Q1: Name two safety measures commonly used in electric circuits and appliances. Ans: Two safety measures commonly used in electric circuits and appliances are
Electric Fuse: An electric fuse is connected in series it protects the circuit from overloading and prevents it from short-circuiting.
Proper earthing of all electric circuit in which any leakage of current in an electric appliance is transferred to the ground and people using the appliance do not get the shock.
Q2: An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain. Ans: The current drawn by the electric oven, I = P/V Substituting the values, I = 2000/220 = 9.09 A The current drawn by the electric oven is 9.09 A which exceeds the safe limit of the circuit. Due to excessive current, the fuse wire will blow and the circuit will break.
Q3: What precaution should be taken to avoid the overloading of domestic electric circuits? Ans:The precautions that should be taken to avoid the overloading of domestic circuits are as follows:
Too many appliances should not be connected to a single socket.
Too many appliances should not be used at the same time.
Faulty appliances should not be connected in the circuit.
Fuse should be connected in the circuit.
Page No: 207
Exercise
Q1: Which of the following correctly describes the magnetic field near a long straight wire? (a) The field consists of straight lines perpendicular to the wire (b) The field consists of straight lines parallel to the wire (c) The field consists of radial lines originating from the wire (d) The field consists of concentric circles centred on the wire Ans: (d) Sol: The magnetic field near a long straight wire are concentric circles. Their centers lie on the wire.
Q2: At the time of short circuit, the current in the circuit (a) reduces substantially (b) does not change (c) increases heavily (d) vary continuously Ans: (c) increases heavily Sol: When two naked wires in the circuit come in contact with each other, the amount of current flowing in the circuit increase abruptly resulting in short circuit.
Q3. State whether the following statements are true or false. (a) The field at the centre of a long circular coil carrying current will be parallel straight lines. Ans: True
A long circular coil is a solenoid. The magnetic field lines inside a solenoid are parallel straight lines.
(b) A wire with a green insulation is usually the live wire of an electric supply. Ans: False
Live wire has red insulation cover, whereas earth wire has green insulation colour in the domestic circuits.
Q4: List two methods of producing magnetic fields. Ans: Two methods of producing magnetic fields are: (i) The permanent magnet produces a magnetic field around it. (ii) Passing current through a conductor also produces the magnetic field.
Q5: When is the force experienced by a current–carrying conductor placed in a magnetic field largest? Ans: The force experienced by a current-carrying conductor is largest when the direction of the current is perpendicular to the direction of the magnetic field.
Q6: Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field? Ans: The direction of the magnetic field is vertically downwards. The direction of current is from the front wall to the back wall because negatively charged electrons are moving from the back wall to the front wall. The direction of magnetic force is rightward. Hence, using Fleming’s left hand rule, it can be concluded that the direction of magnetic field inside the chamber is downward.
Q7: State the rule to determine the direction of a (i) magnetic field produced around a straight conductor-carrying current, (ii) force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and (iii) current induced in a coil due to its rotation in a magnetic field. Ans: (i) The rule used to determine the direction of the magnetic field produced around a straight conductor-carrying current is Maxwell’s right-hand thumb rule. (ii) The rule used to determine the force experienced by a current-carrying straight conductor placed in a magnetic field that is perpendicular to it is Fleming’s left-hand rule. (iii) The rule used to determine the current induced in a coil due to its rotation in a magnetic field is Fleming’s right-hand rule.
Q8: When does an electric short circuit occur? Ans: Listed below are two instances of when a short-circuit can occur: (i) When too many appliances are connected to a single socket or when high power rating appliances are connected to a light circuit, the resistance of the circuit becomes low as a result the current flowing through the circuit becomes very high. This condition results in a short circuit. (ii) When live wires whose insulation have worn off come in contact with each other, the current flowing in the circuit increases abruptly which results in a short circuit.
Q9: What is the function of an earth wire? Why is it necessary to earth metallic appliances? Ans: It is necessary to earth metallic appliances because it ensures that if there is any current leakage in the metallic cover, the potential of the appliance becomes equal to that of the earth. The potential of the earth is zero. As a result, the person handling the appliance will not get an electric shock.
Q1: What does an electric circuit mean? Ans: A continuous and closed path of an electric current is called an electric circuit. An electric circuit consists of electric devices, source of electricity and wires that are connected with the help of a switch.
Q2: Define the unit of current. Ans: The unit of electric current is ampere(A). 1 A is defined as the flow of 1 C of charge through a wire in 1 s.
Q3: Calculate the number of electrons constituting one coulomb of charge. Ans: One electron possesses a charge of 1.6 ×10-19C, i.e., 1.6 ×10-19C of charge is contained in 1 electron. ∴ 1 C of charge is contained in Therefore, 6 x 1018 electrons constitute one coulomb of charge.
Page No. 174
Q1: Name a device that helps to maintain a potential difference across a conductor. Ans: Any source of electricity like a battery, cell, power supply, etc. helps to maintain a potential difference across a conductor.Simple Battery CircuitQ2: What is meant by saying that the potential difference between two points is 1 V? Ans: If 1 J of work is required to move a charge of amount 1 C from one point to another, then it is said that the potential difference between the two points is 1 V.
Q3: How much energy is given to each coulomb of charge passing through a 6V battery? Ans: The energy given to each coulomb of charge is equal to the amount of work which is done in moving it. Now we know that, Potential difference = ∴ Work done = Potential difference × charge Where, Charge = 1 C and Potential difference = 6 V ∴ Work done = 6×1 = 6 Joule.
Page No. 181
Q1: On what factors does the resistance of a conductor depend? Ans: The resistance of a conductor depends upon the following factors: (a) Length of the conductor (b) Cross-sectional area of the conductor (c) Material of the conductor (d) Temperature of the conductor
Q2: Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why? Ans: Resistance of a wire, Where, ρ = Resistivity of the material of the wire l = Length of the wire A = Area of cross-section of the wire Resistance is inversely proportional to the area of cross-section of the wire. Thicker the wire, lower is the resistance of the wire and vice-versa. Therefore, current can flow more easily through a thick wire than a thin wire.
Q3: Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it? Ans: According to Ohm’s law V = IR I= V/R … (1) Now Potential difference is decreased to half ∴ New potential difference Vʹ=V/2 Resistance remains constant So the new current Therefore, the amount of current flowing through the electrical component is reduced by half.
Q4: Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal? Ans: The resistivity of an alloy is higher than the pure metal. Moreover, at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.
Q5: Use the data in Table 12.2 to answer the following Table 12.2 Electrical resistivity of some substances at 20°C
–
Material
Resistivity (Ω m)
Conductors
Silver
1.60 x 10-8
Copper
1.62 x 10-8
Aluminium
2.63 x 10-8
Tungsten
5.20 x 10–8
Nickel
6.84 x 10-8
Iron
10.0 x 10-8
Chromium
12.9 x 10-8
Mercury
94.0 x 10-8
Manganese
1.84 x 10-6
Constantan (alloy of Cu and Ni)
49 x 10-6
Alloys
Manganin
(alloy of Cu, Mn and Ni}
44 x 10-6
Nichronie
(alloy of Ni, Cr, Mn and Fe)
100 x 10-6
Glass
1010 – 1014
Insulators
Hard rubber
1013 – 1016
Ebonite
Diamond
1015 – 1017
1012 – 1013
Paper (dry)
1012
Ans: (a) Resistivity of iron = 10.0 x 10-8 Ω Resistivity of mercury = 94.0 x 10-8 Ω Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury. (b) It can be observed from Table 12.2 that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.
Page No. 185
Q1: Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series. Ans: Three cells of potential 2 V, each connected in series, therefore, the potential difference of the battery will be 2 V + 2 V + 2 V = 6V. The following circuit diagram shows three resistors of resistances 5 Ω, 8 Ω and 12 Ω respectively connected in series and a battery of potential 6 V and a plug key which is closed means the current is flowing in the circuit.
Q2: Redraw the circuit of question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter? Ans: An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel, as shown in the following figure.
The resistances are connected in series. Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According to Ohm’s law, V = IR, Where, Potential difference, V = 6 V Current flowing through the circuit/resistors = I Resistance of the circuit, R = 5 + 8 + 12 = 25Ω I = V/R = 6/25 = 0.24 A Potential difference across 12 Ω resistor = V1 Current flowing through the 12 Ω resistor, I = 0.24 A Therefore, using Ohm’s law, we obtain V1 = IR = 0.24 x 12 = 2.88 V Therefore, the reading of the ammeter will be 0.24 A. The reading of the voltmeter will be 2.88 V.
Page No. 188
Q1: Judge the equivalent resistance when the following are connected in parallel − (a) 1 Ω and 106Ω, (b) 1 Ω and 103Ω and 106Ω. Ans: (a) When 1 Ω and 106 Ω are connected in parallel: Let R be the equivalent resistance.
Therefore, equivalent resistance ≈ 1 Ω
(b) When 1Ω, 103 Ω and 106 Ω are connected in parallel: Let R be the equivalent resistance.
Therefore, equivalent resistance = 0.999 Ω
Q2: An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it? Ans: Resistance of electric lamp, R1 = 100 Ω Resistance of toaster, R2 = 50 Ω Resistance of water filter, R3 = 500 Ω Potential difference of the source, V = 220 V These are connected in parallel, as shown in the following figure.
Let R be the equivalent resistance of the circuit.
7.04 A of current is drawn by all the three given appliances. Therefore, current drawn by an electric iron connected to the same source of potential 220 V= 7,04 A Let R’ be the resistance of the electric iron. According to Ohm’s law, V= IR’
Therefore, the resistance of the electric iron is 31.25 Ω and the current flowing through it is 7.04 A,
Q3: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series? Ans: Advantages of connecting electrical devices in parallel: (i) When the appliances are connected in parallel with the battery, each appliance gets the same potential difference as that of battery which is not possible in series connection. (ii) Each appliance has different resistances and requires different currents to operate properly. This is possible only in parallel connection, as in series connection, the same current flows through all devices, irrespective of their resistances. (iii) If one appliance fails to work, others will continue to work properly. If they are connected in parallel.
Q4: How can three resistors of resistances 2Ω, 3Ω and 6Ω be connected to give a total resistance of (a) 4Ω, (b) 1Ω? Ans: There are three resistors of resistances 2Ω, 3Ω, and 6Ω respectively. (a) The following circuit diagram shows the connection of the three resistors.
Here, 6 Ω and 3 Ω resistors are connected in parallel. Therefore, their equivalent resistance will be given by
This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series. Therefore, the equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω Hence the total resistance of the circuit is 4 Ω. (b) The following circuit diagram shows the connection of the three resistors.
All the resistors are connected in series. Therefore, their equivalent resistance will be given as
Therefore, the total resistance of the circuit is 1 Ω.
Q5: What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω? Ans: There are four coils of resistances 4 Ω, 8 Ω, 12 Ω and 24 Ω respectively. (a) If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum 4 + 8 + 12 + 24 = 48 Ω (b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by
Therefore, 2 Ω is the lowest total resistance.
Page No. 190
Q1: Why does the cord of an electric heater not glow while the heating element does? Ans: The cord of an electric heater is made up of metallic wire such as copper or aluminium which has low resistance while the heating element is made up of an alloy which has more resistance than its constituent metals. Also, heat produced ‘H’ is H = I2Rt Thus, for the same current H ∝ R, more heat is produced by heating element as it has more resistance, and it glows.
Q2: Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V. Ans: Given Charge, Q = 96000C Time, t= 1hr = 60 x 60= 3600s Potential difference, V= 50volts Now we know that H= VIt So we have to calculate I first As I= Q/t ∴ I = 96000/3600 = 80/3 A
Therefore, the heat generated is 4.8 x 106 J.
Q3: An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30s. Ans: The amount of heat (H) produced is given by the joule’s law of heating as H= Vlt Where, Current, I = 5 A Time, t = 30 s Voltage, V = Current x Resistance = 5 x 20 = 100 V H= 100 x 5 x 30 = 1.5 x 104 J. Therefore, the amount of heat developed in the electric iron is 1.5 x 104 J.
Page No. 192
Q1: What determines the rate at which energy is delivered by a current? Ans: The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.
Q2: An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h. Ans: Power (P) is given by the expression,P = VI Where, Voltage,V = 220 V Current, I = 5 A P= 220 x 5 = 1100 W Energy consumed by the motor = Pt Where, Time, t = 2 h = 2 x 60 x 60 = 7200 s ∴ P = 1100 x 7200 = 7.92 x 106 J Therefore, power of the motor = 1100 W Energy consumed by the motor = 7.92 x 106 J
Page No. 193
Exercises
Q1: A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is – (a) 1/25 (b) 1/5 (c) 5 (d) 25 Ans: (d) Resistance of a piece of wire is proportional to its length. A piece of wire has a resistance R. The wire is cut into five equal parts.
Therefore, resistance of each part = (R/5)
All the five parts are connected in parallel. Hence, equivalent resistance (R’) is given as
Q2: Which of the following terms does not represent electrical power in a circuit? (a) I2R (b) IR2 (c) VI (d) V2/R Ans: (b)
(b) Electrical power is given by the expression, P = VI … (i) According to Ohm’s law, V = IR … (ii) Where, V = Potential difference I = Current R = Resistance
From equation (i), it can be written P = (IR) × I
From equation (ii), it can be written
Q3: An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W (b) 75 W (c) 50 W (d) 25 W Ans : (d) Energy consumed by an appliance is given by the expression,
Where, Power rating, P = 100 W Voltage, V = 220V
Resistance,
The resistance of the bulb remains constant if the supply voltage is reduced to 110 V. If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power as
Therefore, the power consumed will be 25 W.
Q4: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be – (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1 Ans: (c) The Joule heating is given by, H = i2Rt
Let, R be the resistance of the two wires.
The equivalent resistance of the series connection is RS = R + R = 2R
If V is the applied potential difference, then it is the voltage across the equivalent resistance.
The heat dissipated in time f is,
The equivalent resistance of the parallel connection is
y is the applied potential difference across this Rp.
The heat dissipated in time f is,
So; the ratio of heat produced is,
Note: H ∝ R also H ∝ i2 and H ∝ t. In this question, f is same for both the circuit. But the current through the equivalent resistance of both the circuit is different. We could have solved the question directly using H α R if in case the current was also same. As we know the voltage and resistance of the circuits, we have calculated I in terms of voltage and resistance and used in the equation H = I2Rt to find the ratio.
Q5: How is a voltmeter connected in the circuit to measure the potential difference between two points? Ans: To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.
Q6: A copper wire has a diameter 0.5 mm and resistivity of 1.6 × 10−8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled? Ans: Area of cross-section of the wire, A =π (d/2)2 Diameter= 0.5 mm = 0.0005 m Resistance, R = 10 Ω We know that
if the diameter of the wire is doubled , new diameter = 2 x 0.5=1 mm = 0.001 m
let new Resistance be R´
Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 Ω.
Q7: The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below −
/ (amperes)
0.5
1.0
2,0
3.0
4.0
V (volts)
1.6
3.4
6.7
10.2
13.2
Ans: Plot a graph between V and I and calculate the resistance of that resistor. The plot between voltage and current is called IV characteristic. The voltage is plotted on x-axis and current is plotted on y-axis. The values of the current for different values of the voltage are shown in the given table.
V (volts)
1.6
3.4
6.7
10.2
13.2
/ (amperes)
0.5
1.0
2.0
3.0
4.0
The IV characteristic of the given resistor is plotted in the following figure.
The slope of the line gives the value of resistance (R) as,
Therefore, the resistance of the resistor is 3.4 Ω.
Q8: When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor. Ans: Resistance (R) of a resistor is given by Ohm’s law as, V= IR R= V/I
Where,
Potential difference, V= 12 V
Current in the circuit, I= 2.5 mA = 2.5 x 10-3 A
Therefore,
the resistance of the resistor is 4.8 kΩ
Q9: A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor? Ans: There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as
V= IR (I = (V/ R)) Where, R is the equivalent resistance of resistances 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. These are connected in series. Hence, the sum of the resistances will give the value of R. R= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω Potential difference, V= 9 V (I = { 9/13.4 })= 0.671 A
Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.
Q10: How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? Ans: For x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is given by Ohm’s law asV= IR (R = V/I)) Where, Supply voltage, V= 220 V
Current, I = 5 A Equivalent resistance of the combination = R,given as
Therefore, four resistors of 176 Ω are required to draw the given amount of current.
Q11: Show how you would connect three resistors, each of resistance 6 Ω so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω. Ans: (i) When two 6 Ω resistances are in parallel and the third is in combination with this, the equivalent resistance will be 9Ω.
(if) When two 6 Ω resistances are in series and the third is in parallel to them, then it will be 4 Ω.
Q12: Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A? Ans: Resistance R1 of the bulb is given by the expression, Supply voltage, V = 220 V Maximum allowable current, I = 5 A Rating of an electric bulb P=10watts
Because of R=V2/P
According to ohm’s law
V=IR
Let R is the Total Resistance of the circuit for x number of electric bulbs
Resistance of each electric bulb, R1 = 4840 Ω
∴ Number of electric bulbs connected in parallel is 110.
Q13: A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases? Ans: Supply voltage, V= 220 V Resistance of one coil, R= 24 Ω (i) Coils are used separately According to Ohm’s law, V= I1R1 Where, I1 is the current flowing through the coil I1 = V/R1 = 220/24 = 9.166 A Therefore, 9.16 A current will flow through the coil when used separately.
(ii) Coils are connected in series Total resistance, R2 = 24 Ω + 24 Ω = 48 Ω According to Ohm’s law,V = I2R2 Where, I2 is the current flowing through the series circuit I2 = V/R2 = 220/48 = 4.58 A Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in parallel Total resistance, R3 is given as =
According to Ohm’s law, V= I3R3 Where, I3 is the current flowing through the circuit I3 = V/R3 = 220/12 = 18.33 A Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.
Q14: Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors. Ans: (i) Potential difference, V = 6 V 1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω According to Ohm’s law, V = IR Where, I is the current through the circuit I= 6/3 = 2 A This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 Ω resistor is 2 A. Power is given by the expression, P= (I)2R = (2)2 x 2 = 8 W (ii) Potential difference, V = 4 V 12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 Ω resistor will be 4 V. Power consumed by 2 Ω resistor is given by P= V2/R = 42/2 = 8 W Therefore, the power used by 2 Ω resistor is 8 W.
Page No. 194
Exercises
Q15: Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V? Ans: Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit. Current drawn by the bulb of rating 100 W is given by, Power = Voltage × Current (Current = (Power/ Voltage)=(100/ 220}A) Similarly, current drawn by the bulb of rating 60 W is given by, Power = Voltage × Current (Current = (Power/Voltage)={60/220}A) Hence, current drawn from the line
Q16: Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes? Ans: Energy consumed by an electrical appliance is given by the expression, H= Pt Where, Power of the appliance = P Time = t Energy consumed by a TV set of power 250 W in 1 h = 250 ×3600 = 9 ×105 J Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600 Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600 = 7.2×105 J Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.
Q17: An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater. Ans: Rate of heat produced by a device is given by the expression for power as, P= I2R Where, Resistance of the electric heater, R= 8 Ω Current drawn, I = 15 A P= (15)2 x 8 = 1800 J/s Therefore, heat is produced by the heater at the rate of 1800 J/s.
Q18: Explain the following. (a) Why is tungsten used almost exclusively for filament of electric lamps? (b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal? (c) Why is the series arrangement not used for domestic circuits? (d) How does the resistance of a wire vary with its area of cross-section? (e) Why are copper and aluminium wires usually employed for electricity transmission? Ans: (a) Tungsten is a metal that has a high melting point and is resistant to heat, making it ideal for use in electric lamps where the filament needs to be heated to produce light. The filament is the part of the lamp that gets hot and produces light when an electric current is passed through it. (b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of metals which produces large amount of heat. (c) In series circuits voltage is divided. Each component of a series circuit receives a small voltage so the amount of current decreases and the device becomes hot and does not work properly. Hence, series arrangement is not used in domestic circuits. (d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A), i.e.
(e) Copper and aluminium are good conductors of electricity also they have low resistivity. So they are usually used for electricity transmission.
Q1. What is meant by power of accommodation of the eye?
Ans: When the ciliary muscles are relaxed, the eye lens becomes thin, the focal length increases, and the distant objects are clearly visible to the eyes. To see the nearby objects clearly, the ciliary muscles contract making the eye lens thicker. Thus, the focal length of the eye lens decreases and the nearby objects become visible to the eyes. Hence, the human eye lens is able to adjust its focal length to view both distant and nearby objects on the retina. This ability is called the power of accommodation of the eyes.
Q2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision? Ans: The person is able to see nearby objects clearly, but he is unable to see objects beyond 1.2 m. This happens because the image of an object beyond 1.2 m is formed in front of the retina and not on the retina, as shown in the given figure.
Myopic Eye
To correct this defect of vision, he must use a concave lens. The concave lens will bring the image back to the retina as shown in the given figure.
Correction of Myopic Eye
Q3. What is the far point and near point of the human eye with normal vision? Ans: The near point of the eye is the minimum distance of the object from the eye, which can be seen distinctly without strain. For a normal human eye, this distance is 25 cm. The far point of the eye is the maximum distance to which the eye can see the objects clearly. The far point of the normal human eye is infinity.
Q4. A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected? Ans: A student has difficulty reading the blackboard while sitting in the last row. It shows that he is unable to see distant objects clearly. He is suffering from myopia. This defect can be corrected by using a concave lens.
Page No. 170 Exercises
Q1. The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to (a) presbyopia (b) accommodation (c) near-sightedness (d) far-sightedness Ans: (b) accommodation
Human eye can change the focal length of the eye lens to see the objects situated at various distances from the eye. This is possible due to the power of accommodation of the eye lens.
Q2. The human eye forms the image of an object at its (a) cornea (b) iris (c) pupil (d) retina Ans: (d) retina
The retina is the layer of nerve cells lining the back wall inside the eye. This layer senses light and sends signals to the brain so you can see.
Q3. The least distance of distinct vision for a young adult with normal vision is about (a) 25 m (b) 2.5 cm (c) 25 cm (d) 2.5 m Ans: (c) 25 cm
The least distance of distinct vision is the minimum distance of an object to see a clear and distinct image. It is 25 cm for a young adult with normal vision.
Q4. The change in focal length of an eye lens is caused by the action of the (a) pupil (b) retina (c) ciliary muscles (d) iris Ans: (c) ciliary muscles
The relaxation or contraction of ciliary muscles changes the curvature of the eye lens. The change in curvature of the eye lens changes the focal length of the eyes. Hence, the change in focal length of an eye lens is caused by the action of ciliary muscles.
Q5. A person needs a lens of power −5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision? Ans: For distant vision = −0.181 m, for near vision = 0.667 m The power P of a lens of focal length f is given by the relation Power (P) = 1 / f(in metres)
(i) Power of the lens used for correcting distant vision = −5.5 D Focal length of the required lens, f = 1 / p f = 1 / -5.5 => f= -0.181 m The focal length of the lens for correcting distant vision is −0.181 m. (ii) Power of the lens used for correcting near vision = +1.5 D Focal length of the required lens, f = 1 / p
f = 1 / 1.5 = +0.667 m The focal length of the lens for correcting near vision is 0.667 m.
Q6. The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem? Ans: The person is suffering from an eye defect called myopia. In this defect, the image is formed in front of the retina. Hence, a concave lens is used to correct this defect of vision. Object distance, u = infinity = ∞ Image distance, v = −80 cm Focal length = f According to the lens formula,
We know,
A concave lens of power −1.25 D is required by the person to correct his defect.
Q7. Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm. Ans: A person suffering from hypermetropia can see distinct objects clearly but faces difficulty in seeing nearby objects clearly. It happens because the eye lens focuses the incoming divergent rays beyond the retina. This defect of vision is corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following figure.
Correction of Hypermetropic EyeThe convex lens actually creates a virtual image of a nearby object (N’ in the figure) at the near point of vision (N) of the person suffering from hypermetropia. The person will be able to clearly see the object kept at 25 cm (near point of the normal eye) if the image of the object is formed at his near point, which is given as 1 m. Object distance, u = −25 cm Image distance, v = −1 m = −100 cm Focal length, f Using the lens formula,
A convex lens of power +3.0 D is required to correct the defect.
Q8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm? Ans: 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. If the object is placed at a distance less than 25 cm from the eye, then the object appears blurred and produces strain in the eyes.
Q9. What happens to the image distance in the eye when we increase the distance of an object from the eye? Ans: Since the size of the eyes cannot increase or decrease, the image distance remains constant. When we increase the distance of an object from the eye, the image distance in the eye does not change. The increase in the object distance is compensated by the change in the focal length of the eye lens. The focal length of the eyes changes in such a way that the image is always formed on the retina of the eye.
Q10. Why do stars twinkle? Ans: Stars twinkle due to atmospheric refraction of starlight. As the stars are very away they behave as almost point sources of light. As on account of atmospheric refraction, the path of rays of light coming from the star goes on varying slightly, the apparent position of the star fluctuates and the amount of light entering the eye flickers, so sometimes the star appear brighter and at some other time, fainter. Thus the stars twinkle.
Q11. Explain why the planets do not twinkle? Ans: Planets are much closer to the earth and are seen as extended source. So, a planet may be considered as a collection of a large number of point-sized light sources. Although light coming from individual point-sized sources flickers but the total amount of light entering our eye from all the individual point-sized sources average out to be constant. Thereby, planets appear equally brighter and there is no twinkling of planets.
Q12. Why does the sky appear dark instead of blue to an astronaut? Ans: Blue colour of the sky is on account of scattering of light of shorter wavelength by particles in the atmosphere of earth. If the earth had no atmosphere, there would not have been any scattering and sky would have looked dark. When astronaut in his spacecraft goes above the atmosphere of earth, sky appears dark to him because there is no scattering of light.
Q1. Define the principal focus of a concave mirror. Ans: Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of the concave mirror.
Q2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length? Ans: Radius of curvature, R = 20 cm. The radius of curvature of a spherical mirror = 2 × Focal length (f) i.e. R = 2 f
Hence, the focal length of the given spherical mirror is 10 cm.
Q3. Name the mirror that can give an erect and enlarged image of an object. Ans: When an object is placed between the pole and the principal focus of a concave mirror, the image formed is virtual, erect, and enlarged. Therefore, the mirror is concave.
Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Ans: Convex mirrors give a virtual, erect, and diminished image of the objects placed in front of them. They are preferred as a rear-view mirror in vehicles because they give a wider field of view, which allows the driver to see most of the traffic behind him.
Page No. 145
Q1. Find the focal length of a convex mirror whose radius of curvature is 32 cm. Ans: Radius of curvature, R = 32 cm Radius of curvature = 2 × Focal length ( f ) i.e. R = 2 f
Hence, the focal length of the given convex mirror is 16 cm.
Q2. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located? Ans: Magnification produced by a spherical mirror is given by the relation,
Let the height of the object, ho = h Then, height of the image, hi = −3h (Image formed is real)
∵ Object distance, u = −10 cm ⇒ v = 3 × (−10) = −30 cm Here, the negative sign indicates that an inverted image is formed at a distance of 30 cm in front of the given concave mirror.
Page No. 150
Q1. A ray of light travelling in the air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why? Ans: The light ray bends towards the normal. When a ray of light travels from an optically rarer medium to an optically denser medium, it gets bent towards the normal. Since water is optically denser than air, a ray of light travelling from air into the water will bend towards the normal.
Q2. Light enters from air to glass having a refractive index of 1.50. What is the speed of light in the glass? The speed of light in a vacuum is 3 × 108 ms−1. Ans: Refractive index of a medium nm is given by,
Speed of light in vacuum, c = 3 × 108 m s−1 Refractive index of glass, ng = 1.50 Q3. Find out, from Table, the medium having the highest optical density. Also, find the medium with the lowest optical density.
Ans: Highest optical density = Diamond, Lowest optical density = Air
The optical density of a medium is directly related to the refractive index of that medium. A medium that has the highest refractive index will have the highest optical density and vice-versa.
It can be observed from the table that diamond and air respectively have the highest and lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density.
Q4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table.
Ans: The speed of light in a medium is given by the relation for refractive index (nm). The relation is given as:
It can be inferred from the relation that light will travel the slowest in the material which has the highest refractive index and travel the fastest in the material which has the lowest refractive index. It can be observed from table 10.3 that the refractive indices of kerosene, turpentine, and water are 1.44, 1.47, and 1.33 respectively. Therefore, light travels the fastest in water.
Q5. The refractive index of diamond is 2.42. What is the meaning of this statement? Ans: The Refractive index of a medium nm is related to the speed of light in that medium v by the relation:
Where c is the speed of light in vaccum/air. The refractive index of diamond is 2.42. This suggests that the speed of light in a diamond will reduce by a factor of 2.42 compared to its speed in air.
Page No. 158
Q1. Define 1 dioptre of power of a lens. Ans: The power of the lens is defined as the reciprocal of its focal length. If P is the power of a lens of focal length f in metres, then
P = 1 / f(in metres) The S.I. unit of power of a lens is Dioptre. It is denoted by D. 1 dioptre is defined as the power of a lens of focal length 1 metre i.e. 1 D = 1 m−1.
Q2. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens. Ans: When an object is placed at the centre of curvature, 2F1, of a convex lens, its image is formed at the centre of curvature, 2F2, on the other side of the lens. The image formed is inverted and of the same size as the object, as shown in the given figure.
It is given that the image of the needle is formed at a distance of 50 cm from the convex lens. Hence, the needle is placed in front of the lens at a distance of 50 cm. Object distance, u = −50 cm, Image distance, v = 50 cm According to the lens formula,
Hence, the power of the given lens is +4 D.
Q3. Find the power of a concave lens of focal length 2 m. Ans: Focal length of a concave lens, f=−2m (negative due to the divergent nature of the concave lens) Power of a lens,
Hence, the power of the given concave lens is −0.5D.
Page No. 159
Exercises
Q1. Which one of the following materials cannot be used to make a lens? (a) Water (b) Glass (c) Plastic (d) Clay
Ans: (d)
Explanation: A lens allows light to pass through it. Since clay does not show such property, it cannot be used to make a lens.
Q2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object? (a) Between the principal focus and the centre of curvature (b) At the centre of curvature (c) Beyond the centre of curvature (d) Between the pole of the mirror and its principal focus
Ans: (d) Explanation: When an object is placed between the pole and principal focus of a concave mirror, the image formed is virtual, erect, and larger than the object.
Q3. Where should an object be placed in front of a convex lens to get a real image of the size of the object? (a) At the principal focus of the lens (b) At twice the focal length (c) At infinity (d) Between the optical centre of the lens and its principal focus.
Ans: (b) Explanation: When an object is placed at the centre of curvature in front of a convex lens, its image is formed at the centre of curvature on the other side of the lens. The image formed is real, inverted, and of the same size as the object.
Q4. A spherical mirror and a thin spherical lens have each a focal length of −15 cm. The mirror and the lens are likely to be (a) both concave (b) both convex (c) the mirror is concave and the lens is convex (d) the mirror is convex, but the lens is concave
Ans: (a) Explanation: By convention, the focal length of a concave mirror and a concave lens are taken as negative. Hence, both the spherical mirror and the thin spherical lens are concave in nature.
Page No. 160
Q5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be (a) plane (b) concave (c) convex (d) either plane or convex
Ans: (d) Explanation: A convex mirror always gives a virtual and erect image of the smaller size of the object placed in front of it. Similarly, a plane mirror will always give a virtual and erect image of same size as that of the object placed in front of it. Therefore, the given mirror could be either plane or convex.
Q6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary? (a) A convex lens of focal length 50 cm (b) A concave lens of focal length 50 cm (c) A convex lens of focal length 5 cm (d) A concave lens of focal length 5 cm
Ans: (c) Explanation: A convex lens gives a magnified image of an object when it is placed between the radius of curvature and focal length. Also, magnification is more for convex lenses having a shorter focal length. Therefore, for reading small letters, a convex lens of focal length 5 cm should be used.
Q7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case. Ans: Range of object distance = 0 cm to 15 cm A concave mirror gives an erect image when an object is placed between its pole (P) and the principal focus (F). Hence, to obtain an erect image of an object from a concave mirror of focal length 15 cm, the object must be placed anywhere between the pole and the focus. The image formed will be virtual, erect, and magnified in nature, as shown in the given figure.
Q8. Name the type of mirror used in the following situations. (a) Headlights of a car (b) Side/rear-view mirror of a vehicle (c) Solar furnace Support your answer with reason. Ans: (a) Concave (b) Convex (c) Concave Explanation:(a) Concave mirror is used in the headlights of a car. This is because concave mirrors can produce a powerful parallel beam of light when the light source is placed at their principal focus. (b) Convex mirror is used inside/rearview mirror of a vehicle. Convex mirrors give a virtual, erect, and diminished image of the objects placed in front of it. Because of this, they have a wide field of view. It enables the driver to see most of the traffic behind him/her. (c) Concave mirrors are convergent mirrors. That is why they are used to construct solar furnaces. Concave mirrors converge the light incident on them at a single point known as the principal focus. Hence, they can be used to produce a large amount of heat at that point.
Q9. One-half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations. Ans: The convex lens will form a complete image of an object, even if its one half is covered with black paper. It can be understood by the following two cases. Case I: When the upper half of the lens is covered In this case, a ray of light coming from the object will be refracted by the lower 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 following figure.
Case II: When the lower half of the lens is covered In this case, a ray of light coming from the object is 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 following figure.
Q10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and nature of the image formed. Ans: We know that a converging lens forms the image of an object placed in front of it by converging the light rays falling on it. We can find the image distance very easily using the lens’ formula when the focal length and the object distance of the system is given
According to the lens’ formula –
Where, f is the focal length of the lens,
v is the image distance,
u is the object distance.
It is given that the object is 25 cm away from the lens and the focal length of the lens is 10 cm. We can find the image distance by substituting the values as –
We can find the magnification of the image formed by finding the ratio between the image distance to the object distance. This is given by –
From this we can conclude that the image is two-third of the object size. The image size can be given as –
The image size has a negative sign, this implies that the image formed is real and inverted
The image is formed 16.67 cm away from the lens. The image size is 3.33 cm. The image is real and inverted.
Q11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram. Ans: Focal length of concave lens (OF1), f = −15 cm, Image distance, v = −10 cm
According to the lens formula,
The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown in the following ray diagram.
Q12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image. Ans: Focal length of convex mirror, f = +15 cm, Object distance, u = −10 cm
According to the mirror formula,
The positive value of ‘v’ indicates that the image is formed behind the mirror.
The positive value of magnification indicates that the image formed is virtual and erect.
Q13. The magnification produced by a plane mirror is +1. What does this mean? Ans: Magnification produced by a mirror is given by the relation:
The magnification produced by a plane mirror is +1. It shows that the image formed by the plane mirror is of the same size as that of the object. The positive sign shows that the image formed is virtual and erect.
Q14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size: Ans. Object distance, u = −20 cm, Object height, h = 5 cm, Radius of curvature, R = 30 cm
Radius of curvature = 2 × Focal length i.e. R = 2f ⇒ f = 15 cm
According to the mirror formula,
The positive value of ‘v’ indicates that the image is formed behind the mirror.
The positive value of magnification indicates that the image formed is virtual
The positive value of image height indicates that the image formed is erect. Therefore, the image formed is virtual, erect, and smaller in size.
Q15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharply focused image can be obtained? Find the size and nature of the image. Ans: Object distance, u = −27 cm, Object height, h = 7 cm, Focal length, f = −18 cm
According to the mirror formula,
The screen should be placed at a distance of 54 cm in front of the given mirror.
The negative value of magnification indicates that the image formed is real.
The negative value of image height indicates that the image formed is inverted.
Q16. Find the focal length of a lens of power −2.0 D. What type of lens is this?
Ans:
A concave lens has a negative focal length. Hence, it is a concave lens.
Q17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging? Ans:
As the power of the lens is +ve, the lens is a converging (convex) lens.
Q1: If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier? Ans.
In asexual reproduction, the reproducing cells produce a copy of their DNA through some chemical reactions.
However, this copying of DNA is not accurate and therefore, the newly formed DNA has some variations.
It can be easily observed in the above figure that in asexual reproduction, very few variations are allowed.
Therefore, if a trait is present in only 10% of the population, it is more likely that the trait has arisen recently.
Hence, it can be concluded that trait B that exists in 60% of the same population has arisen earlier than trait A.
Q2: How does the creation of variations in a species promote survival? Ans. Variations within a species can significantly enhance survival in changing environments.
When environmental conditions shift dramatically, such as a sudden increase in water temperature, many organisms may struggle to survive.
In such cases, only a few heat-resistant variants may endure, while others perish.
If these resistant variants did not exist, the entire species could face extinction.
While these variations aid in survival, not all variations are advantageous; some may be neutral or even detrimental.
Page No. 133
Q1: How do Mendel’s experiments show that traits may be dominant or recessive? Ans. Mendel’s experiments with pea plants demonstrated that traits can be dominant or recessive through the following steps:
Mendel selected true breeding tall (TT) and dwarf (tt) pea plants. Then, he crossed these two plants.
The seeds formed after fertilization were grown and these plants that were formed represent the first filialor F1 generation.
All the F1 plants obtained were tall.
Then, Mendel self-pollinated the F1 plants and observed that all plants obtained in the F2 generation were not tall.
Instead, one-fourth of the F2 plants were short.
From this experiment, Mendel concluded that the F1 tall plants were not true-breeding.
They were carrying traits of both short height and tall height. They appeared tall only because the tall trait is dominant over the dwarf trait.
Q2: How do Mendel’s experiments show that traits are inherited independently? Ans. Mendel’s experiments demonstrated that traits are inherited independently through careful cross-breeding of pea plants.
Mendel crossed pea plants having round green seeds (RRyy) with pea plants having wrinkled yellow seeds (rrYY).
Since the F1 plants are formed after crossing pea plants having green round seeds and pea plants having yellow wrinkled seeds, F1 generation will have both these characters in them.
However, as we know that yellow seed colour and round seeds are dominant characters, therefore, the F1 plants will have yellow round seeds.
The first generation (F1) plants displayed the dominant traits, resulting in yellow round seeds.
When F1 plants self-pollinated, the second generation (F2) exhibited a variety of combinations:
Yellow round seeds (9/16)
Green round seeds (3/16)
Yellow wrinkled seeds (3/16)
Green wrinkled seeds (1/16)
The ratio of these traits in F2 was found to be 9:3:3:1.
This indicates that the traits for seed shape and colour are inherited independently.
Q3: A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits − blood group A or O − is dominant? Why or why not? Ans. No, this information is not enough to determine which trait is dominant between blood group A and O. Here’s why:
The blood group of the man with group A can be either AA or AO.
Since their daughter has blood group O, she must have inherited an O allele from both parents.
This means the father must have the A allele as AO to pass on the O allele.
However, without knowing the blood groups of all offspring, we cannot definitively conclude which trait is dominant.
Q4: How is the sex of the child determined in human beings? Ans. In human beings, the sex of a child is determined by the combination of sex chromosomes inherited from the parents:
Females have two X chromosomes (XX).
Males have one X and one Y chromosome (XY).
During reproduction, each parent contributes one sex chromosome:
The mother can only provide an X chromosome.
The father can provide either an X or a Y chromosome.
Thus, the sex of the child is determined as follows:
If the child inherits an X chromosome from the father, the child will be a girl (XX).
If the child inherits a Y chromosome from the father, the child will be a boy (XY).
Thus, the mother provides only X chromosomes. The sex of the baby is determined by the type of male gamete (X or Y) that fuses with the X chromosome of the female.
Exercises
Q1: A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as (a) TTWW (b) TTww (c) TtWW (d) TtWw Ans. (c)
Explanation:
The genetic make-up of the tall parent can be depicted as TtWW
Since all the progeny bore violet flowers, it means that the tall plant having violet flowers has WW genotype for violet flower colour.
Since the progeny is both tall and short, the parent plant was not a pure tall plant.
Its genotype must be Tt.
Therefore, the cross involved in the given question is
Therefore, half the progeny is tall, but all of them have violet flowers.
Q2: A study found that children with light-coloured eyes are likely to have parents with light coloured eyes. On this basis, can we say anything about whether the light eye colour is dominant or recessive? Why or why not? Ans: No. From the given statement, we cannot say with certainty whether light eye colour is dominant or recessive. However, since both children and their parents have light eye colour, the possibility is that light eye colour is a recessive trait. Had the light eye colour been a dominant trait, the homozygous light eyed parents would have only light eyed children but heterozygous light eyed parent might had some recessive dark eyed children (3 : 1 ratio).
Q3: Outline a project which aims to find the dominant coat colour in dogs? Ans: To determine the dominant coat colour in dogs, follow these steps:
Select a homozygous black male dog (BB) and a homozygous white female dog (bb).
Crossbreed the two dogs to produce the F1 generation.
If all offspring are black, it indicates that the black coat colour is dominant over white.
Q4: How is the equal genetic contribution of male and female parents ensured in the progeny? Ans: The equal genetic contribution of male and female parents to their offspring is ensured through the following mechanisms:
Each parent produces gametes that contain half the genetic information needed for the offspring.
The father contributes 23 chromosomes and the mother contributes another 23 chromosomes.
This results in a total of 46 chromosomes in the offspring, ensuring equal genetic contribution from both parents.
Q1: What is the importance of DNA copying in reproduction? Ans:
DNA (Deoxyribonucleic acid) is the genetic material found in the chromosomes, which are present in the nucleus of a cell.
The DNA is the information site for making proteins, and each specific type of protein leads to a specific type of body design.
Thus, it is the DNA molecule that determines the body design of an individual.
Therefore, it can be concluded that it is the DNA which gets transferred from parents to offspring and makes them look similar.
DNA determines the body’s structure.
Q2: Why is variation beneficial to the species but not necessarily for the individual? Ans:
Variations are more beneficial to the species than the individual because sometimes, for a species, the environmental conditions change so drastically that their survival becomes difficult. Example: If the temperature of water increases suddenly, then most of the bacteria living in that water will die. Only a few variants that are resistant to heat would be able to survive.
However, if these variants had not been present, then the entire species of bacteria would have been destroyed.
Thus, these variants help in the survival of the species. However, not all variations are necessarily beneficial for individual organisms.
Page No. 133
Q1: How does binary fission differ from multiple fission? Ans:
In binary fission, a single cell divides into two equal halves. Amoeba and Bacteria divide by binary fission.
Binary Fission in Bacteria
In multiple fission, a single cell divides into many daughter cells simultaneously. Amoeba and Plasmodium divide by multiple fission.
Multiple Fission in Plasmodium
Q2: How will an organism be benefited if it reproduces through spores? Ans: There are many advantages if an organism reproduces through spores.
Advantages of spore formation:
Large numbers of spores are produced in one sporangium.
Spores are distributed easily by air to far-off places to avoid competition in one place.
Spores are covered by thick walls to prevent dehydration under unfavourable conditions.
Q3: Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration? Ans:
Simple organisms such as Hydra and Planaria are capable of producing new individuals through the process of regeneration.
The process of regeneration involves the formation of new organisms from their body parts.
Simple organisms can utilize this method of reproduction as their entire body is made of similar kind of cells in which any part of their body can be formed by growth and development.
However, complex organisms have an organ-system level of organization.
All the organ systems of the body work together as an interconnected unit.
They can regenerate their lost body parts such as skin, muscles, blood, etc. However, they cannot give rise to new individuals through regeneration.
Q4: Why is vegetative propagation practised for growing some types of plants? Ans:
Plants raised by vegetative propagation can bear flowers and fruits earlier than those produced from seeds.
It also helps in the propagation of plants such as bananas, oranges, roses and jasmine that have lost the capacity to produce seeds.
All plants produced by this method are genetically similar to the parent plant in that they have all characteristics.
Thus, desirable features of the parent plant can be retained and expressed without any change in future generations.
Q5: Why is DNA copying an essential part of the process of reproduction? Ans:
DNA (Deoxyribonucleic acid) copying is an essential part of reproduction as it passes genetic information from parents to offspring.
It determines the body design of an individual.
The reproducing cells produce a copy of their DNA through chemical reactions that result in two copies of DNA.
The copying of DNA always takes place along with the creation of the additional cellular structure.
This process is then followed by the division of a cell to form two cells.
Page No. 140
Q1: How is the process of pollination different from fertilization? Ans:
Q2: What is the role of the seminal vesicles and the prostate gland? Ans: The secretions from seminal vesicles and prostate glands lubricate the sperms and provide a fluid medium for easy transport of sperms. Their secretion also provides nutrients in the form of fructose, calcium, and some enzymes.
Q3: What are the changes seen in girls at the time of puberty? Ans: Secondary sexual characteristics in girls:
Increase in breast size and darkening of the skin of the nipples present at the tips of the breasts.
The appearance of hair in the genital area.
The appearance of hair in other areas of skin like underarms, face, hands, and legs.
Increase in the size of the uterus and ovary.
Beginning of menstrual cycle.
More secretion of oil from the skin, which results in the appearance of pimples.
Q4: How does the embryo get nourishment inside the mother’s body? Ans:
The embryo develops inside the mother’s body for about nine months. Inside the uterus, the outer tissue surrounding the embryo develops finger-like projections called villi.
These villi are surrounded by uterine tissue and maternal blood.
They provide a large surface area for the exchange of oxygen and nutrients.
Also, there is a special tissue called the placenta, which is embedded in the uterine wall.
The embryo receives the oxygen and nutrients from the mother’s blood via the placenta.
The waste materials produced by the embryo are also removed through the placenta.
Q5: If a woman is using a copper−T, will it help in protecting her from sexually transmitted diseases? Ans: No. Using a copper-T will not provide protection from sexually transmitted diseases, as it does not prevent the entry of semen. It only prevents the implantation of the embryo in the uterus.
Page No. 141
Exercises
Q1: Asexual reproduction takes place through budding in (a) amoeba (b) yeast (c) plasmodium (d) leishmania
Ans:(b) Asexual reproduction takes place through budding in yeast. Yeast is an example of asexual reproduction taking place through budding. A small protuberance is produced on the parent cell that grows in full size, forming a bud. In the parent cell, the daughter nucleus splits and migrates to the daughter cell. By forming a constriction, the bud detaches from the mother’s body at the base. This process of budding continues to form a chain of bud cells. The mother cell is smaller than the daughter cell.
Q2: Which of the following is not a part of the female reproductive system in human beings? (a) Ovary (b) Uterus (c) Vas deferens (d) Fallopian tube Ans:(c) Vas deferens is not a part of the female reproductive system in human beings.
Vas deferens is a part of the male reproductive system. It is a long, muscular tube travelling from the epididymis into the pelvic cavity. It is behind the bladder. Its function is to transport the mature sperm to the urethra. It also carries urine outside of the body.
Q3: The anther contains (a) sepals (b) ovules (c) pistil (d) pollen grains Ans:(d) The anther contains pollen grains.
(a) Stamen (b) 3-D Cut-section of an Anther
The stamen are male reproductive organs. They contain anther, which is a site of pollen development. Inside the anther, the male sporogenous cell differentiates and undergoes meiosis to produce microspores that develop into pollen grains.
Q4: What are the advantages of sexual reproduction over asexual reproduction? Ans: Advantages of sexual reproduction:
In sexual reproduction, more variations are produced. Thus, it ensures the survival of species in a population.
The newly formed individual has the characteristics of both parents.
Variations are more viable in the sexual mode than in the asexual one. This is because, in asexual reproduction, DNA has to function inside the inherited cellular apparatus.
Q5: What are the functions performed by the testes in human beings? Ans: The testes are the male reproductive organs that are located outside the abdominal cavity within a pouch called the scrotum. Functions of testes:
Produce sperms.
Produce a hormone called testosterone, which brings about secondary sexual characteristics in boys.
Q6: Why does menstruation occur? Ans:
Menstruation is a process in which blood and mucous flows out every month through the vagina.
This process occurs every month because one egg is released from the ovary every month and at the same time, the uterus (womb) prepares itself to receive the fertilized egg.
Thus, the inner lining of the uterus gets thickened and is supplied with blood to nourish the embryo.
If the egg does not get fertilized, then the lining of the uterus breaks down slowly and gets released in the form of blood and mucus from the vagina.
Q7: Draw a labelled diagram of the longitudinal section of a flower. Ans:
Longitudinal Section of a Flower
Q8: What are the different methods of contraception? Ans: The contraceptive methods can be broadly divided into the following types: 1. Natural Method
It involves avoiding the chance of meeting the sperm and ovum.
In this method, the sexual act is avoided from the 10th to the 17th of the menstrual cycle because, during this period, ovulation is expected, and therefore, the chances of fertilization are very high.
2. Barrier Method
In this method, the fertilization of the ovum and sperm is prevented with the help of barriers.
Barriers are available for both males and females.
Condoms are barriers made of thin rubber that are used to cover the penis in males and the vagina in females.
3. Oral Contraceptives
In this method, tablets or drugs are taken orally.
These contain small doses of hormones that prevent the release of eggs, and thus, fertilization cannot occur.
4. Implants and Surgical Methods
Contraceptive devices such as the loop or Copper-T are placed in the uterus to prevent pregnancy.
Some surgical methods can also be used to block the gamete transfer. It includes the blocking of the vas deferens to prevent the transfer of sperm, known as vasectomy.
Similarly, the fallopian tubes of the females can be blocked so that the egg will not reach the uterus, known as tubectomy.
Q9: How are the modes of reproduction different in unicellular and multicellular organisms? Ans:
In unicellular organisms, reproduction occurs by the division of the entire cell.
The modes of reproduction in unicellular organisms can be fission, budding, etc.
Meanwhile, specialised reproductive organs are present in multicellular organisms. Therefore, they can reproduce by complex reproductive methods such as vegetative propagation, spore formation, etc.
In more complex multicellular organisms such as human beings and plants, the mode of reproduction is sexual reproduction.
Q10: How does reproduction help in providing stability to populations of species? Ans:
Living organisms reproduce for the continuation of a particular species.
It helps in providing stability to the population of species by producing a new individual that resembles the parents.
This is the reason why cats give birth to only cats or dogs give birth to only dogs.
Therefore, reproduction provides stability to populations of dogs or cats or any other species.
Q11: What could be the reasons for adopting contraceptive methods? Ans: Contraceptive methods are mainly adopted because of the following reasons:
To prevent unwanted pregnancies.
To control population rise or birth rate.
To prevent the transfer of sexually transmitted diseases.
Q1: What is the difference between a reflex action and walking? Ans:
Reflex action is the involuntary action that occurs in response to stimuli.
They occur without the involvement of conscious areas of the brain.
All the reflex actions are unconscious.
Reflex action occurs brain and spinal cord of the central nervous systems.
This kind of response occurs within a fraction of a second.
On the other hand, voluntary actions are those which occur under the control of the cerebellum of the brain.
Walking is learnt as we grow. Walking is controlled by the brain. This kind of response takes a longer time.
Q2: What happens at the synapse between two neurons? Ans: A synapse is a tiny gap between the end of one neuron’s axon and the dendrite of another neuron. It functions as a one-way valve for transmitting signals. Here’s how it works:
The axon of the first neuron releases chemicals into the synapse.
These chemicals cross the synapse and trigger an electrical impulse in the dendrite of the next neuron.
This process ensures that signals travel in only one direction.
This mechanism is essential for the rapid transmission of nervous impulses throughout the body.
DendriteQ3: Which part of the brain maintains the posture and equilibrium of the body? Ans: Cerebellum, a part of the hindbrain is responsible for maintaining posture and equilibrium of the body.
Q4: How do we detect the smell of an agarbatti (incense stick)? Ans: The thinking part of our brain is the forebrain. It has separate areas that are specialized for hearing, smelling, sight, taste, touch, etc. The fore-brain also has regions that collect information or impulses from various receptors.
When the smell of an incense stick reaches us, our forebrain detects it. Then, the forebrain interprets it by putting it together with the information received from other receptors and also with the information already stored in the brain.
Q5: What is the role of the brain in reflex action? Ans: Reflex actions are sudden responses, which do not involve any thinking.
For example, when we touch a hot object, we withdraw our hand immediately without thinking as thinking may take time which would be enough to get us burnt.
The sensory nerves that detect heat are connected to the nerves that move the muscles of the hand.
Such a connection of detecting the signal from the nerves (input) and responding to it quickly (output) is called a reflex arc.
The reflex arcs (connections present between the input and output nerves) meet in a bundle in the spinal cord.
Reflex arcs are formed in the spinal cord and the information (input) reaches the brain. The brain is only aware of the signal and the response that has taken place. However, the brain has no role to play in the creation of the response. Page No. 108
Q1: What are plant hormones? Ans: Plant hormones are the fluid that are secreted within the plant also known as phytohormones. Plant hormones regulate the growth and development of the plant. Examples of plant hormones include:
Auxin – promotes cell elongation and growth.
Gibberellins – stimulate stem growth.
Cytokinins – encourage cell division.
Q2: How is the movement of leaves of the sensitive plant different from the movement of a shoot towards light? Ans:Movement of leaves of sensitive plantMovement of shoot towards light(i) Growth independent movement. (ii) Movement is non-directional i.e., it is neither towards nor away from the stimulus. (iii) Such movements are referred to d as nastic movements. (iv) Such movements are reversible.(I) Growth-dependent movement. (ii) Movement is towards the source of stimulus (light). So it is a directional movement. (in) Such movements are referred d as tropic movements. (iv) Such movements are irreversible.
Q3: Give an example of a plant hormone that promotes growth. Ans:Auxin is an example of a growth-promoting plant hormone. Auxins are responsible for cell elongation in the shoot and accelerates growth.
Q4: How do auxins promote the growth of a tendril around a support? Ans: Auxins are hormones produced at the shoot tip of plants. They play a crucial role in growth, particularly in tendrils. Here’s how they work:
When a tendril touches a support, auxins promote faster growth on the side away from the support.
This uneven growth causes the tendril to coil around the support.
The result is a structure that resembles a watch spring, allowing the plant to cling effectively.
Q5: Design an experiment to demonstrate hydrotropism. Ans: To demonstrate hydrotropism in plants. Procedure:
Plant a seedling in a vessel containing soil.
Adjacent to the seedling put a porous pot containing water.
Leave thset-upup for few days.
Observation: On examining the roots it is observed that the roots bend towards the source of water and do not grow straight.
Result: It confirms that the plant shows hydrotropism as the roots bend towards the porous pot of water. As hydrotropism is a plant growth response in which the direction of growth is determined by a stimulus of a gradient in water concentration.
Experiment of Hydrotropism
Page No. 111
Q1: How does chemical coordination take place in animals? Ans: Chemical coordination in animals occurs through hormones, A hormone is the chemical messenger that regulates the physiological processes in living organisms.
It is secreted by glands.
The regulation of physiological processes and control and coordination by hormones comes under the endocrine system.
The nervous system along with the endocrine system in our body controls and coordinates the physiological processes.
Q2: Why is the use of iodized salt advisable? Ans: The use of iodized salt is important for several reasons:
Iodine is essential for the thyroid gland to produce the hormone thyroxin.
Thyroxin helps regulate the metabolism of carbohydrates, fats, and proteins in the body.
A deficiency in iodine can lead to an enlarged thyroid gland, known as goitre.
Goitre is often characterised by a swollen neck.
Therefore, incorporating iodized salt into our diet supports the normal functioning of the thyroid gland.
Q3: How does our body respond when adrenaline is secreted into the blood? Ans: Adrenalineis a hormone released by the adrenal glands during times of danger or stress. When it enters the bloodstream, it triggers several responses in the body:
It is secreted directly into the blood and is transported to different parts of the body.
When secreted in large amounts, it speeds up the heartbeat and hence supplies more oxygen to the muscles.
The breathing rate also increases due to contractions of the diaphragm and rib muscles.
It also increases the blood pressure.
All these responses enable the body to deal with any stress or emergency.
Q4: Why are some patient of diabetes treated by giving injections of insulin? Ans: Diabetes is caused due to less or no secretion of the hormone insulin by the pancreas. In such a person, blood sugar level is high. Insulin converts extra sugar present in the blood into glycogen. Thus, patients suffering from diabetes are given insulin injections on to control their blood sugar levels.
Page No. 112 Exercises Q1. Which of the following is a plant hormone? (a) Insulin (b) Thyroxin (c) Oestrogen (d) Cytokinin Ans: (d) Cytokinin Cytokinin is a plant hormone whereas Insulin, Thyroxin and, Oestrogen are the hormones produced by animals.
Q2. The gap between two neurons is called a (a) dendrite. (b) synapse. (c) axon. (d) impulse. Ans: (b) Synapse The gap between two neurons is called a synapse. At the synapse, chemicals are released that help transmit signals to the next neuron.
Q3. The brain is responsible for (a) thinking. (b) regulating the heartbeat. (c) balancing the body. (d) all of the above. Ans: (d) all of the above The brain is responsible for thinking, regulating the heartbeat and balancing the body.
Q4. What is the function of receptors in our body? Think of situations where receptors do not work properly. What problems are likely to arise? Ans: Receptors are sensory structures (organs/tissues or cells) present all over the body. The receptors are either grouped in the case of the eye or ear or scattered in the case of the skin. Functions of receptors:
They sense external stimuli such as heat or pain.
They also trigger an impulse in the sensory neuron which sends a message to the spinal cord.
When the receptors are damaged, the external stimuli transferring signals to the brain are not felt. For example, in the case of damaged receptors, if we accidentally touch any hot object, then our hands might get burnt as damaged receptors cannot perceive the external stimuli of heat and pain.
Q5. Draw the structure of a neuron and explain its function. Ans:
Structure of Neuron
Functions of the three parts of a neuron:
Axon: It conducts messages away from the cell body.
Dendrite: It receives information from the axon of another cell and conducts the messages towards the cell body.
Cell body: It contains a nucleus, mitochondria, and other organelles. It is mainly concerned with maintenance and growth.
Q6. How does phototropism occur in plants? Ans: The growth movement in plants in response to light stimulus is known as phototropism. The shoots show positive phototropism and the roots show negative phototropism. This means that the shoots bend towards the source of light whereas the roots bend away from the light source.
Some examples of phototropism are as follows:
The flower head of the sunflower is positively phototropic and hence it moves from east to west along with the sun.
The ovary stalk of groundnut is positively phototropic before fertilization and becomes negatively phototropic after fertilization so that the fruit is formed underground.
Q7. Which signals will get disrupted in case of a spinal cord injury? Ans: The reflex arc connections between the input and output nerves meet in a bundle in the spinal cord. Nerves from all over the body meet in a bundle in the spinal cord on their way to the brain. In case of any injury to the spinal cord, the signals coming from the nerves as well as the signals coming to the receptors will be disrupted.
Q8. How does chemical coordination occur in plants? Ans:
In animals, control and coordination occur with the help of the nervous system.
However, plants do not have a nervous system.
Plants respond to stimuli by showing movements.
The growth, development, and responses to the environment in plants are controlled and coordinated by a special class of chemical substances known as hormones.
These hormones are produced in one part of the plant body and are translocated to other needy parts.
For example, a hormone produced in roots is translocated to other parts when required.
The five major types of phytohormone are auxins, gibberellins, cytokinins, abscisic acid, and ethylene.
These phytohormones are either growth promoters (such as auxins, gibberellins, cytokinins, and ethylene) or growth inhibitors such as abscisic acid.
Q9. What is the need for a system of control and coordination in an organism? Ans: The need for a system of control and coordination in an organism is crucial for maintaining body functions. This system allows various body systems to work together in response to changes.
All the movements that occur in response to stimuli are carefully coordinated and controlled.
In animals, the control and coordination movements are provided by the nervous and muscular systems.
The nervous system sends messages to and away from the brain.
The spinal cord plays an important role in the relay of messages.
In the absence of this system of control and coordination, our body will not be able to function properly.
For example, when we accidentally touch a hot utensil, we immediately withdraw our hand. In the absence of nerve transmission, we will not withdraw our hands and may get burnt.
Q10. How are involuntary actions and reflex actions different from each other? Ans: Involuntary actions cannot be consciously controlled. For example, we cannot consciously control the movement of food in the alimentary canal. These actions are however directly under the control of the brain. On the other hand, the reflex actions such as the closing of eyes immediately when bright light is focused show sudden responses and do not involve any thinking. This means that, unlike involuntary actions, reflex actions are not under the control of the brain.
Q11. Compare and contrast nervous and hormonal mechanisms for control and coordination in animals. Ans:
Q12. What is the difference between the manner in which movement takes place in a sensitive plant and the movement in our legs? Ans:
Q1: Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms like humans? Ans: Multicellular organisms such as humans possess complex body designs. They have specialized cells and tissues for performing various necessary functions of the body, such as intake of food and oxygen. Unlike unicellular organisms, multicellular cells are not in direct contact with the outside environment. Therefore, diffusion cannot meet their oxygen requirements.
Q2: What criteria do we use to decide whether something is alive? Ans: If something shows the following characteristics then it is a requirement a living being: (i) It can move on its own. (ii) It needs food to get energy and nutrients. (iii) It respires. (iv) It responds to changes that take place in its surroundings. (v) It exhibits growth and development. (vi) It removes its metabolic wastes.
Characteristics of lifeQ3: What are outside raw materials used for by an organism? Ans: An organism uses outside raw materials, mostly in the form of food and oxygen. The raw materials required by an organism can be quite varied depending on the complexity of the organism and its environment.
Q4: What processes would you consider essential for maintaining life? Ans: Life processes such as nutrition, respiration, transportation, excretion, etc. are essential for maintaining life.
Page No. 87
Q1: What are the differences between autotrophic nutrition and heterotrophic nutrition? Ans:
Q2: Where do plants get each of the raw materials required for photosynthesis? Ans: The following raw materials are required for photosynthesis:
Carbon Dioxide: Plants get CO2 from the atmosphere through stomata.
Water: Plants absorb water from the soil through roots and transport it to leaves.
Sunlight: Sunlight, which is absorbed by the chlorophyll and other green parts of the plant.
Q3: What is the role of the acid in our stomach? Ans: Following are the roles of acid in our stomach: (i) It creates an acidic medium for the action of the proteolytic enzyme pepsin. (ii) Kills harmful bacteria present in the food. (iii) Inactivates salivary amylase. (iv) Prevents fermentation of food. (v) HCl converts inactive pepsinogen into active pepsin.
Q4: What is the function of digestive enzymes? Ans: Digestive enzymes such as amylase, lipase, pepsin, trypsin, etc. help in the breakdown of complex food particles into simple ones. These simple particles can be easily absorbed by the blood and thus transported to all the cells of the body.
Q5: How is the small intestine designed to absorb digested food? Ans: The small intestine has millions of tiny finger-like projections called villi. These villi increase the surface area for more efficient food absorption. Within these villi, many blood vessels are present that absorb the digested food and carry it to the bloodstream. From the bloodstream, the absorbed food is delivered to each and every cell of the body.
Fig: Longitudinal section of villi
Page No. 91
Q1: What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration? Ans: Terrestrial organisms take up oxygen from the atmosphere, whereas aquatic animals obtain oxygen from water. Air contains more O2 as compared to water. Since the content of O2 in the air is high, terrestrial animals do not have to breathe faster to get more oxygen. Therefore, unlike aquatic animals, terrestrial animals do not need adaptations for gaseous exchange.
Q2: What are the different ways in which glucose is oxidized to provide energy in various organisms? Ans: At first, glucose (6 carbon molecules) is broken in the cytoplasm of cells of all organisms. This process yields a 3-carbon molecule compound called pyruvate. Further breakdown of pyruvate takes place in different manners in different organisms.
Anaerobic Respiration: This process takes place in the absence of oxygen, e.g., in yeast during fermentation. In this case, pyruvate is converted into ethanol and carbon dioxide.
Aerobic Respiration: In aerobic respiration, the breakdown of pyruvate takes place in the presence of oxygen to give rise to 3 molecules of carbon dioxide and water. The release of energy in aerobic respiration is much more than anaerobic respiration.
Lack of Oxygen: Sometimes, when there is a lack of oxygen, especially during vigorous activity, in our muscles, pyruvate is converted into lactic acid (3-carbon molecule compounds). The formation of lactic acid in muscles causes cramps.
Q3: How is oxygen and carbon dioxide transported in human beings? Ans: Haemoglobin transports oxygen molecules to all the body cells for cellular respiration. The hemoglobin pigment present in the blood gets attached to four O2 molecules that are obtained from breathing.
It thus forms oxyhemoglobin and the blood becomes oxygenated. This oxygenated blood is then distributed to all the body cells by the heart. After giving away O2 to the body cells, blood takes away CO2, which is the end product of cellular respiration. Now, the blood becomes deoxygenated.
Since hemoglobin pigment has less affinity for CO2, CO2 is mainly transported in the dissolved form. This deoxygenated blood gives CO2 to lung alveoli and takes O2 in return.
Q4: How are the lungs designed in human beings to maximize the area for the exchange of gases? Ans: The exchange of gases takes place between the blood of the capillaries that surround the alveoli and the gases present in the alveoli. Thus, alveoli are the site for exchange of gases. The lungs get filled up with air during the process of inhalation as ribs are lifted up and the diaphragm is flattened. The air that is rushed inside the lungs fills the numerous alveoli present in the lungs. Each lung contains 300-350 million alveoli. These numerous alveoli increase the surface area for gaseous exchange, making the process of respiration more efficient.
Page No. 96
Q1: What are the components of the transport system in human beings? What are the functions of these components? Ans: The main components of the transport system in human beings are the heart, blood, and blood vessels.
Heart: The heart pumps oxygenated blood throughout the body. It receives deoxygenated blood from the various body parts and sends this impure blood to the lungs for oxygenation. Blood: Blood helps in the transport of oxygen, nutrients, CO2, and nitrogenous wastes. Blood Vessels: The blood vessels (arteries, veins, and capillaries) carry blood either away from the heart to various organs or from various organs back to the heart.
Q2: Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds? Ans: Warm-blooded animals such as birds and mammals maintain constant body temperature by cooling themselves when they are in a hotter environment and by warming their bodies when they are in a cooler environment. Hence, these animals require more oxygen (O2) for more cellular respiration so that they can produce more energy to maintain their body temperature.
Thus, it is necessary for them to separate oxygenated and deoxygenated blood so that their circulatory system is more efficient and can maintain their constant body temperature.
Q3: What are the components of the transport system in highly organized plants? Ans: In highly organized plants, there are two different types of conducting tissues – xylem and phloem. Xylem conducts water and minerals obtained from the soil (via roots) to the rest of the plant. The phloem transports food materials from the leaves to different parts of the plant body.
Q4: How are water and minerals transported in plants? Ans: The components of xylem tissue (tracheids and vessels) of roots, stems, and leaves are interconnected to form a continuous system of water-conducting channels that reaches all parts of the plant. Transpiration creates a suction pressure, as a result of which water is forced into the xylem cells of the roots. Then, there is a steady movement of water from the root xylem to all the plant parts through the interconnected water-conducting channels.
Fig: Xylem tissue
Q5: How is food transported in plants? Ans: Phloem transports food materials from the leaves to different parts of the plant body. The transportation of food in phloem is achieved by utilizing energy from ATP. As a result of this, the osmotic pressure in the tissue increases, causing water to move into it. This pressure moves the material in the phloem to the tissues which have less pressure. This is helpful in moving materials according to the needs of the plant. For example, the food material, such as sucrose, is transported into the phloem tissue using ATP energy.
Fig: Phloem tissue
Page No. 98
Q1: Describe the structure and functioning of nephrons. Ans: Nephrons are the basic filtering units of the kidneys. Each kidney possesses a large number of nephrons, approximately 1-1.5 million. The main components of the nephron are the glomerulus, Bowman’s capsule, and a long renal tubule.
Fig: Structure of Nephron
Functioning of a nephron: The blood enters the kidney through the renal artery, which branches into many capillaries associated with the glomerulus. The water and solute are transferred to the nephron at Bowman’s capsule. In the proximal tubule, some substances such as amino acids, glucose, and salts are selectively reabsorbed, and unwanted molecules are added in the urine. The filtrate then moves down into the loop of Henle, where more water is absorbed. From here, the filtrate moves upwards into the distal tubule and finally to the collecting duct. Collecting duct collects urine from many nephrons. The urine formed in each kidney enters a long tube called a ureter. From the ureter, it gets transported to the urinary bladder and then into the urethra.
Q2: What are the methods used by plants to get rid of excretory products? Ans: Plants can get rid of excess of water by transpiration. Waste materials may be stored in the cell vacuoles or as gum and resin, especially in old xylem. It is also stored in the leaves that later fall off.
Q3: How is the amount of urine produced regulated? Ans: The amount of urine produced depends on the amount of excess water and dissolved wastes present in the body. Some other factors, such as habitat of an organism and hormone such as Antidiuretic hormone (ADH), also regulates the amount of urine produced.
Page No. 99
Exercise
Q1: The kidneys in human beings are a part of the system for (a) nutrition (b) respiration (c) excretion (d) transportation Ans: (c) In human beings, the kidneys are a part of the system for excretion.
Q2: The xylem in plants are responsible for (a) transport of water (b) transport of food (c) transport of amino acids (d) transport of oxygen Ans: (a) In a plant, the xylem is responsible for transport of water.
Q3: The autotrophic mode of nutrition requires (a) carbon dioxide and water (b) chlorophyll (c) sunlight (d) all of the above Ans: (d) The autotrophic mode of nutrition requires carbon dioxide, water, chlorophyll and sunlight.
Q4: The breakdown of pyruvate to give carbon dioxide, water and energy takes place in (a) cytoplasm (b) mitochondria (c) chloroplast (d) nucleus Ans: (b) The breakdown of pyruvate to give carbon dioxide, water and energy takes place in mitochondria.
Q5: How are fats digested in our bodies? Where does this process take place? Ans: Fats are present in the form of large globules in the small intestine. The small intestine receives the secretions from the liver and the pancreas. The bile salts (from the liver) break down the large fat globules into smaller globules so that the pancreatic enzyme lipase can easily act on them. This is referred to as emulsification of fats. This process takes place in the small intestine.
Q6: What is the role of saliva in the digestion of food? Ans: Saliva is secreted by the salivary glands, located under the tongue. It moistens the food for easy swallowing. It contains a digestive enzyme called salivary amylase, which breaks down starch into sugar.
Q7: What are the necessary conditions for autotrophic nutrition and what are its by-products? Ans: Autotrophic nutrition takes place through the process of photosynthesis. Carbon dioxide, water, chlorophyll pigment, and sunlight are the necessary conditions required for autotrophic nutrition. Carbohydrates (food) and O2 are the by-products of photosynthesis.
Q8: What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration. Ans:
Q9: How are the alveoli designed to maximize the exchange of gases? Ans: The alveoli are the small balloon-like structures present in the lungs. The walls of the alveoli consist of an extensive network of blood vessels. Each lung contains 300−350 million alveoli, making it a total of approximately 700 million in both lungs. The alveolar surface, when spread out, covers about 80 m2 area. This large surface area makes the gaseous exchange more efficient.
Fig: Alveoli and capillaries
Q10: What would be the consequences of a deficiency of hemoglobin in our bodies? Ans: Haemoglobin is the respiratory pigment that transports oxygen to the body cells for cellular respiration. Therefore, a deficiency of hemoglobin in the blood can affect the oxygen-supplying capacity of blood. This can lead to a deficiency of oxygen in the body cells. It can also lead to a disease called anemia.
Q11: Describe double circulation in human beings. Why is it necessary? Ans: The human heart is divided into four chambers − the right atrium, the right ventricle, the left atrium, and the left ventricle.
Flow of blood in the heart:
The heart has superior and inferior vena cava, which carries deoxygenated blood from the upper and lower regions of the body respectively and supplies this deoxygenated blood to the right atrium of the heart.
Fig: Structure of heart
The flow of blood in the human heart
The right atrium then contracts and passes the deoxygenated blood to the right ventricle, through an auriculo-ventricular aperture.
Then, the right ventricle contracts and passes the de-oxygenated blood into the two pulmonary arteries, which pump it to the lungs, where the blood becomes oxygenated. From the lungs, the pulmonary veins transport the oxygenated blood to the left atrium of the heart.
Then, the left atrium contracts and through the auriculo-ventricular aperture, the oxygenated blood enters the left ventricle.
The blood passes to the aorta from the left ventricle. The aorta gives rise to many arteries that distribute the oxygenated blood to all the regions of the body.
Fig: Flow of blood
Schematic diagram of blood circulation in humans
Therefore, the blood goes twice through the heart. This is known as double circulation.
Importance of double circulation:
The separation of oxygenated and de-oxygenated blood allows a more efficient supply of oxygen to the body cells. This efficient system of oxygen supply is very useful in warm-blooded animals such as human beings.
As we know, warm-blooded animals have to maintain a constant body temperature by cooling themselves when they are in a hotter environment and by warming their bodies when they are in a cooler environment.
Hence, they require more O2 for more respiration so that they can produce more energy to maintain their body temperature. Thus, the circulatory system of humans is more efficient because of the double circulatory heart.
Q12: What are the differences between the transport of materials in xylem and phloem? Ans:
Q13: Compare the functioning of alveoli in the lungs and nephrons in the kidneys with respect to their structure and functioning. Ans:
Alveoli
Nephrons
Structure
Structure
Alveoli are tiny balloon-like structures present inside the lungs.
Nephrons are tubular structures present inside the kidneys.
The walls of the alveoli are one cell thick and it contains an extensive network of blood capillaries.
Nephrons are made of glomerulus, bowman’s capsule, and a long renal tube.
Function
Function
The exchange of 02 and C02 takes place between the blood of the capillaries that surround the alveoli and the gases present in the alveoli.
The blood enters the kidneys through the renal artery. The blood is entered here and the nitrogenous waste in the form of urine is collected by collecting duct.
Q1: What would be the electron dot structure of carbon dioxide, which has the formula CO2? Ans:
A molecule of CO2 consists of one atom of carbon and two atoms of oxygen.
The electronic configuration of carbon is 2,4 while that of oxygen is 2, 6. Each of the two atoms of oxygen shares two electrons with the carbon atom to complete the octet of both the elements, thereby forming a double covalent bond. Electron Dot Structure of CO2
Q2: What would be the electron dot structure of a molecule of sulphur, which is made up of eight atoms of sulphur? (Hint – the eight atoms of sulphur are joined together in the form of a ring.) Ans: In S8, the atoms are joined together in the form of a ring. Electron Dot Structure of S8
Page No. 68
Q1: How many structural isomers can you draw for pentane? Ans: We can draw three structural isomers of pentane (C5H12).
Q2: What are the two properties of carbon which lead to the huge number of carbon compounds we see around us? Ans: The two features of carbon that give rise to a large number of compounds are as follows:
Catenation: The self linking ability of carbon to form a long straight-chain, branched-chain, and closed ring structures.
Tetravalency: With the valency of four, carbon is capable of bonding with four other atoms.
Q3: What will be the formula and electron dot structure of cyclopentane? Ans: The formula of cyclopentane is C5H10.
Page No. 69
Q4: Draw the structures for the following compounds. (a) Ethanoic acid (b) Bromopentane (c) Butanone (d) Hexanal Are structural isomers possible for bromopentane? Ans:
(a)
(b)
(c)
(d)
Yes, the structural isomers of bromopentane are possible by changing the bromine position and branching in the parent carbon chain.
Structural Isomers of Bromopentane
Q5: How would you name the following compounds?
(i) CH3—CH2—Br
(ii)
(iii)
Ans:
(i) Bromoethane (ii) Methanal (formaldehyde) (iii) Hex-1-yne
Page No. 71
Q1: Why is the conversion of ethanol to ethanoic acid an oxidation reaction? Ans:
Since the conversion of ethanol to ethanoic acid involves the addition of oxygen to ethanol, therefore it is an oxidation reaction.
Q2: A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used? Ans:
2HC ≡ CH + 5O2 → 4CO2 + 2H2O + Heat
When ethyne is burnt in air, it gives a sooty flame.
This is due to incomplete combustion caused by a limited supply of air.
However, if ethyne is burnt with oxygen, it gives a clean flame with a temperature of 3000°C because of complete combustion.
This oxy-acetylene flame is used for welding. It is not possible to attain such a high temperature without mixing oxygen.
This is the reason why a mixture of ethyne and air is not used.
Page No. 74
Q1: How would you distinguish experimentally between alcohol and carboxylic acid? Ans:
Q2: What are oxidising agents? Give an example. Ans: Those substances that give oxygen or replace hydrogen on reaction with other compounds are known as oxidising agents, such as potassium permanganate (KMnO4).
Page No. 76
Q1: Would you be able to check if the water is hard by using a detergent? Ans:
Detergents are ammonium or sulphonate salts of long-chain carboxylic acids. Unlike soap, they do not react with calcium and magnesium ions present in hard water to form scum.
They give a good amount of lather irrespective of whether the water is hard or soft.
This means that detergents can be used in both soft and hard water. Therefore, it cannot be used to check whether the water is hard or not.
Q2: People use a variety of methods to wash clothes. Usually, after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush, or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes? Ans:
A soap molecule has two parts, namely hydrophobic and hydrophilic. With the help of these, it attaches to the grease or dirt particles and forms a cluster called a micelle.
These micelles remain suspended in solution as a colloid.
When water is agitated, the oily dirt tends to lift off from the dirty surface and dissociates into fragments.
This gives an opportunity to other tails to stick to oil. This results in the formation of an emulsion in water.
This emulsion now contains small globules of oil surrounded by soap or detergent molecules.
The negatively charged heads present in water prevent the small globules from coming together and form clusters.
Thus, the oily dirt is removed from the object.
Exercise (Page 77)
Q1: Ethane, with the molecular formula C2H6, has (a) 6 covalent bonds (b) 7 covalent bonds (c)8 covalent bonds (d) 9 covalent bonds
Ans: (b)
Sol:
The structural formula for ethane is:
So, it has 7 covalent bonds.
Q2: Butanone is a four-carbon compound with the functional group (a) Carboxylic acid (b) Aldehyde (c) Ketone (d) Alcohol Ans: (c) Sol: Butanone has the formula CH3COCH2CH3. Thus, it has ketone as a functional group. Option (c) is correct.
Q3: While cooking, if the bottom of the vessel is getting blackened on the outside, it means that (a) The food is not cooked completely. (b) The fuel is not burning completely. (c) The fuel is wet. (d) The fuel is burning completely. Ans: (b) Sol: This means that the fuel is not burning completely, and unburnt carbon particles get deposited on the bottom of the vessel, making it black.
Q4: Explain the nature of the covalent bond using the bond formation in CH3Cl. Ans: Carbon can neither lose 4 electrons nor do gain four electrons as these process make the system unstable due to requirement of extra energy. Therefore CH3Cl completes its octet configuration by sharing its 4 electrons with carbon atoms or with atoms of other elements. Hence the bonding that exists in CH3Cl is a covalent bonding.
Here, carbon requires 4 electrons to complete its octet, while each hydrogen atom requires one electron to complete its duplet. Also, chlorine requires an electron to complete the octet. Therefore, all of these share the electrons and as a result, carbon forms 3 bonds with hydrogen and one with chlorine.
Q5: Draw the electron dot structures for (i) Ethanoic acid (ii) H2S (iii) Propanone (iv) F2 Ans:
Q6: What is a homologous series? Explain with an example. Ans:
It is a series of organic compounds having the same general formula, same functional group, same general methods of preparation, similar chemical properties, and gradation in physical properties where the adjacent members differ by a CH2 group. Example: The general formula for the homologous series of alkanes is CnH2n+2. CH4 (Methane), C2H6 (Ethane), C3H8 (Propane), and C4H10 (Butane).
Every homologous series have a general formula.
In alkane, a single bond is a functional group, an alkene double bond, and in alkyne triple bond.
Q7: How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties? Ans:
(i) Difference on the physical basis
(ii) Difference on the chemical basis
Q8: Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also? Ans:
A soap molecule has both a hydrophilic and a hydrophobic end. The hydrophilic end is soluble in water, whereas the hydrophobic end is insoluble in water.
When soap is added to water, the hydrophilic part gets dissolved in water, but the hydrocarbon tail being the hydrophobic part, forms clusters called micelles.
As soap is soluble in ethanol, micelle formation will not take place in it.
Q9: Why are carbon and its compounds used as fuels for most applications? Ans: During the process of combustion of carbon and its compounds, a large amount of heat and light is released, i.e., They have high calorific value, and because of this, carbon and its compounds are used as fuels.
Q10: Explain the formation of scum when hard water is treated with soap. Ans:
Soaps are sodium or potassium salts of fatty acids having cleansing action in the water.
Hard water contains Ca2+ and Mg2+ ions, which react with soap to form calcium and magnesium salts of fatty acids, which are insoluble and are called scum.
Q11: What change will you observe if you test soap with litmus paper (red and blue)? Ans: As a soap solution is basic in nature, it will turn red litmus paper into blue, but it will not affect blue litmus paper.
Q12: What is hydrogenation? What is its industrial application? Ans:
Hydrogenation is a process in which Hydrogen is added in the presence of nickel or palladium as a catalyst.
Industrial application: The process of hydrogenation is used to prepare vegetable ghee from vegetable oil.
Q13: Which of the following hydrocarbons undergo addition reactions: C2, H6, C3H8, C3H6, C2H2, and CH4. Ans:Unsaturated hydrocarbons undergo addition reactions. Being unsaturated hydrocarbons, C3H6, and C2H2 will undergo addition reactions.
Q14: Give a test that can be used to differentiate chemically between butter and cooking oil. Ans:
Add bromine water to each of them.
Cooking oil will decolourize bromine water showing that it is unsaturated, whereas butter will not decolourize bromine water showing that it is saturated.
Q15: Explain the mechanism of the cleaning action of soaps. Ans:
A soap molecule has hydrophobic and hydrophilic ends. In water, hydrophobic ends of soap, which consists of hydrocarbon chains, cluster together to form micelles.
The oily dirt collects in the centre of the micelle.
These micelles stay in solution as a colloid and will not come together to form precipitate because of ion-ion repulsion. Formation of Micelle
Thus the dirt suspended in the micelles can be easily rinsed away, and hence, soaps are effective in cleaning.
(ii) Different garbage bins for disposing of biodegradable waste and non-biodegradable waste
Electron Dot Structure of CO2
Formation of Micelle