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Heat Class 10 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Bhagya

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 5 Heat Notes, Textbook Exercise Important Questions and Answers.

Std 10 Science Part 1 Chapter 5 Heat Question Answer Maharashtra Board

Class 10 Science Part 1 Chapter 5 Heat Question Answer Maharashtra Board

Question 1.
Fill in the blanks and rewrite the sentences:
a. The amount of water vapour in air is determined in terms of its………..
Answer:
The amount of water vapour in air is determined in terms of its absolute humidity.

b. If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their………….
Answer:
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacities.

c. When a liquid is getting converted into solid, the latent heat is………. (Practical Activity Sheet – 1 and 2)
Answer:
When a liquid is getting converted into solid, the latent heat is released.

Question 2.
Observe the following graph. Considering the change in volume of water as its temperature is raised from 0 °C, discuss the difference in the behaviour of water and other substances. What is this behaviour of water called?

Answer:
If the temperature of water is raised from 0 °C to 10 °C, its volume goes on decreasing in the range 0 °C to 4 °C. It is minimum at 4 °C. The volume of water goes on increasing in the range 4 °C to 10 °C.

In general, when a substance is heated, its volume goes on increasing with temperature. Thus, in the range 0 °C to 4 °C, behaviour of water is different from other substances. It is called anomalous behaviour of water.

Question 3.
What is meant by specific heat capacity?
How will you prove experimentally that different substances have different specific heat capacities?
Answer:
The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C is called the specific heat capacity of that object.

Question 4.
While deciding the unit for heat, which temperatures interval is chosen? why?
Answer:
While deciding the unit for heat, the temperature interval chosen is 14.5 °C to 15.5 °C. For the reason, see the information given in the following box.

Question 5.
Explain the following temperature vs time graph:

(Practice Activity Sheet – 1 and 4; March 2019)
Answer:
The graph shows what happens when a mixture of ice and water is heated continuously. The temperature of the mixture remains constant (0 °C) till all the ice melts as shown by the line AB. This temperature is the melting point of ice. On further heating, the temperature rises steadily from 0 °C to 100 °C as shown by the line BC, At 100 °C water starts converting into steam. This temperature is the boiling point of water. Further heating does not change the temperature and the conversion waters steam continues as shown by the line CD.

Question 6.
Explain the following:
a. the role of anomalous behaviour of water in preserving aquatic life in regions of cold climate?
Answer:

In cold regions, during winter, the temperature of the atmosphere falls well below 0 °C. As the temperature decreases, the water at the surfaces of lakes and ponds starts contracting. Hence, its density increases and it sinks to the bottom. This process continues till the temperature of all the water in a lake falls to 4 °C. As the water at the surface cools further, i.e., its temperature falls below 4 °C, it starts expanding instead of contracting. Therefore, its density decreases and it remains at the surface.

The temperature of the water at the surface continues to fall to 0 °C. Finally, the water at the surface is converted into ice, but the water below the layer of ice is at 4 °C. Ice is a bad conductor of heat. Hence, the layer of the ice at the surface does not allow transfer of heat from the water to the atmosphere. As the water below the layer of ice remains at 4 °C, fish and other aquatic animals and plants can survive in it.

b. How can you relate the formation of water droplets on the outer surface of a bottle taken out of a refrigerator with formation of dew?
Answer:
At a given temperature, there is a limit on how much water vapour the given volume of air can hold. The lower the temperature, the lower is the capacity of air to hold water vapour.

The temperature of a bottle kept in a refrigerator is lower than room temperature. Hence, when the bottle is taken out of the refrigerator, the temperature of the air surrounding the bottle is lowered. Therefore, the capacity of the air to hold water vapour becomes less. Hence, the excess water vapour condenses to form water droplets (like dew) on the outer surface of the bottle.

c. In cold regions in winter, the rocks crack due to anomalous expansion of water.
Answer:
Sometimes water enters into crevices of the rocks. When the temperature of the atmosphere falls below 4 °C, water expands. Even when water freezes to form ice, there is increase in its volume. As there is no room for expansion, it exerts a tremendous pressure on the rocks which crack and break up into small pieces.

Question 7.
a. What is meant by latent heat? How will the state of matter transform if latent heat is given off?
Answer:
When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

When a liquid is converted into the gaseous phase (gaseous state), at the boiling point, the heat absorbed is used for breaking the bonds between the atoms or molecules. This heat is called the latent heat of vaporization. Some solids, under certain conditions, are directly transformed into the gaseous phase. Here the heat is absorbed but the temperature remains constant. The absorbed heat is used for breaking the bonds between atoms or molecules. This heat is called the latent heat of sublimation.

In general, latent heat is the heat absorbed or given out by a substance during a change of state at constant temperature.
In transformations from liquid to solid, gas to liquid and gas to solid, latent heat is given out by the body (substance).
(Note: change of state = change of phase)

b. Which principle is used to measure the specific heat capacity of a substance?
Answer:
The principle of heat exchange is used to measure the specific heat capacity of a substance. This principle is as follows: If a system of two objects is isolated from the environment by keeping it inside a heat resistant box, then no energy can leave the box or enter the box. In this situation, heat energy lost by the hot object = heat energy gained by the cold object.

c. Explain the role of latent heat in the change of state of a substance.
Answer:
When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

d. what basis and how will you determine whether air is saturated with vapour or not?
Answer:
Whether the air is saturated with water vapour or not is determined on the basis of the extent of water vapour present in the air. If the relative humidity is 100%, the air is saturated with water vapour. In that case, we can see the formation of water droplets on the leaves of plants/grass.
If the relative humidity is less than 100%, the air is not saturated with water vapour.

Question 8.
Read the following paragraph and answer the questions:
If heat is exchanged between a hot and cold object, the temperature of the cold object goes on increasing due to gain of energy and the temperature of the hot object goes on decreasing due to loss of energy.

The change in temperature continues till the temperatures of both the objects attain the same value. In this process, the cold object gains heat energy and the hot object loses’ heat energy. If the system of both the objects is isolated from the environment by keeping it inside a heat resistant box (meaning that the energy exchange takes place between the two objects only), then no energy can flow from inside the box or come into the box.
(1) Heat is transferred from where to where?
(2) Which principle do we learn about from this process?
(3) How will you state the principle briefly?
(4) Which property of the substance is measured using this principle?
Answer:
(1) Heat is transferred from a hot object to a cold object.
(2) This process shows the principle of heat exchange.
(3) In this process, the cold object gains heat energy and the hot object loses energy. If a system of two objects is isolated from the surroundings, heat energy lost by the hot object = heat energy gained by the cold object.
(4) This principle is used to measure the specific heat capacity of a substance.

Question 9.
Solve the following problems:
a. Equal heat is given to two objects A and B of mass 1 g. The temperature of A increases by 3 °C and B by 5°C. Which object has more specific heat? And by what factor?
Solution:
Data: m = 1 g, Δ T₁ = 3 °C, Δ T₂ = 5 °C,
Q same
Here, Q = mc₁ ΔT₁ = mc₂ΔT₂

Thus, c₁ > c₂
The specific heat of A is more than that of B and

b. Liquid ammonia is used in ice factory for making ice from water. If water at 20 °C is to be converted into 2 kg, ice at 0 °C, how many grams of ammonia is to be evaporated?
(Given: The latent heat of vaporization of 1 ammonia = 341 cal/g)
Solution:
Data : m₁ = 2kg, ΔT₁=20 °C – 0 °C
= 20 °C, c₁ = 1 kcal/kg·°C, L₁ (ice) = 80 kcal/kg,
L₂ (vaporization of ammonia) = 341 cal/g = 341 kcal/kg, m₂ =?
Q₁ (heat lost by water) = m₁c₁ΔT₁ + m₁L₁
= 2kg × 1 kcal/kg·°C × 20 °C + 2 kg × 80 kcal/kg
=40 kcal + 160 kcal = 200 kcal
Q₂ (heat absorbed by ammonia) = m₂L₂
= m₂ × 34l kcal/kg
According to the principle of heat exchange, Q₁ = Q₂
∴ 200 kcal = m₂ × 341 kcal/kg
∴ m₂ = \(\frac{200}{341}\) kg = 0.5864 kg = 586.4 g
586.4 g of ammonia are to be evaporated.

c. A thermally insulated pot has 150 g ice at temperature 0 °C. How much steam of 100 °C has to he mixed to it, so that water of temperature 50 °C will be obtained?
(Given: Latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g °C)
Solution:
Data: m₁ = 150 g, ΔT₁ = 50 °C – 0 °C
= 50 °C, c_w = 1 cal/g.°C, L₁ = 80 cal/g, L₂ = 540 cal/g,
Δ T₂ = 100°C – 50 °C = 50 °C, m₂ = ?
Q₁ (heat absorbed by ice) = m₁L₁
= 150 g × 80 cal/g = 12000 cal
Q₂ (heat absorbed by water formed on melting of ice) =m₁ c_w ΔT₁
= 150 g × 1 cal/g·°C × 50 °C = 7500 cal
Q₃ (heat given out by steam) = m₂L₂
= m₂ × 540 cal/g
Q₄ (heat given out by water formed on condensation of steam)
= m₂ c_w ΔT₂ = m₂ × 1 cal/g·°C × 50 °C
According to the principle of heat exchange,
Q₁ + Q₂ = Q₃ + Q₄
∴ 12000 cal + 7500 cal = m₂ × 540 cal/g + m₂ × 50 cal/g
∴ 19500 cal = m₂ (540 + 50) cal/g
∴ m₂ = \(\frac{19500}{590}\) g
33.5 g of steam is to be mixed.

d. A calorimeter has mass 100 g and specific heat 0.1 kcal/kg ·°C. It contains 250 g of liquid at 30 °C having specific heat of 0.4 kcal/kg·°C. If we drop a piece of ice of mass 10 g at 0 °C into the liquid, what will be the temperature of the mixture?
Solution:
Data: m₁ = 100 g, c₁ = 0.1 kcal/kg·°C,
= 0.1 cal/g·°C, T₁ = 30 °C, m₂ = 250 g,
c₂ = 0.4 kcal/kg·°C = 0.4 cal/g·°C, T₂ = 30 °C,
m₃ = 10 g, T₃ = 0 °C, L = 80 cal/g,
c (water) = 1 cal/g·°C, T = ?
Q₁ (heat lost by calorimeter) = m₁c₁(T- T₁),
Q₂ (heat lost by liquid) = m₂c₂ (T – T₂),
Q₃ (heat absorbed by ice) = m₃ L,
Q₄ (heat absorbed by water formed on melting of ice) = m₃c (T – 0 °C)
According to the principle of heat exchange,
Q₁ + Q₂ = Q₃ + Q₄
∴ m₁c₁ (T₁ – T) + m₂c₂ (T₂ – T) = m₃L + m₃c (T – 0 °C)
∴ m₁c₁T₁ – m₁c₁T + m₂c₂T₂ – m₂c₂T = m₃L + m₃c (T – 0°C)
∴ m₁c₁T₁ + m₂c₂T₂ = m₃L + (m₁c₁ + m₂c₂ + m₃c)T
∴ 100g × 0.1 cal/g°C × 30 °C + 250g × 0.4 cal/g.°C × 30 °C J
= 10 g x× 80 cal/g + (100 g × 0.1 cal/g.°C + 250 g × 0.4 cal/g.°C + 10 g × 1 cal/g.°C) T
∴ (10 + 100 + 10) T = (300 + 3000 – 800)°C
∴ 120 T = 2500 °C
∴ T = \(\frac{2500}{120}\) °C = \(\frac{125}{6}\) °C = 20.83 °C
This is the temperature of the mixture.

Project:
Take help of your teachers to make a working model of Hope’s apparatus and perform the experiment. Verify the results you obtain. [Do it your self]

Can you recall? (Text Book Page No. 62)

Question 1.
What is the difference between heat and temperature?
Answer:
Heat is a form of energy. Particles of matter (atoms, molecules, etc.) possess potential energy and kinetic energy. Total energy (potential energy + kinetic energy) of all particles of matter in a given sample is called it’s thermal energy. When two bodies at different temperatures are in thermal contact with each other, there is transfer of thermal energy from a body at higher temperature to a body at lower temperature. This energy in transfer is called heat. It is expressed in joule, calorie and erg.

Temperature is a quantitative measure of degree of hotness or coldness of a body. It is expressed in °C, °F or K (kelvin). Temperature determines the direction of energy transfer.

Question 2.
What are the different ways of heat transfer?
Answer:
Ways of heat transfer: conduction, convection and radiation.
[Note: heat ≡ heat energy. In the textbook, both the terms are used.]

Use your brain power! (Text Book Page No. 63)

Question 1.
Is the concept of latent heat applicable during transformation of gaseous phase to liquid phase and from liquid phase to solid phase?
Answer:
Yes.

Question 2.
Where does the latent heat go during these transformations?
Answer:
During these transformations, the latent heat is given out by the substance to the surroundings.

Use your brain power! (Text Book Page No. 64)

Question 1.
In the above experiment, the wire moves through the ice slab. However, the ice slab does not break. Why?
Answer:
When the thin wire with two equal weights attached to its ends is hung over the block of ice, it exerts pressure on the ice below it. Due to this, the melting point of the ice below the wire is lowered and some ice melts. The wire passes through the water so formed.

The water above the wire is no longer under pressure and, therefore, refreezes. Once again the ice below the wire melts, and the wire passes through it, and the process continues. In this way, due to alternate melting of ice and refreezing of water, the wire cuts right through the block of ice leaving the block intact.

Question 2.
Is there any relationship of latent heat with regelation?
Answer:
Yes. when the ice melts, heat is absorbed, but the temperature does not change. Also, when water refreezes, heat is given out, but the temperature does not change. This heat absorbed or given out is the latent heat.

Question 3.
You know that as we go higher than the sea level, the boiling point of water decreases. What would be the effect on the melting point of a solid?
Answer:
As we go higher than the sea level, the melting point of solids (i) that expand on melting is lowered due to a decrease in pressure (ii) that contract on melting is raised due to a decrease in pressure.

[The wire used in the experiment is made of a metal (usually copper). Metals are good conductors of heat. Hence, exchange of heat between the portion of the ice above the wire and that below the wire takes place readily.]

Can you tell? (Text Book Page No. 64)

Question 1.
We feel that some objects are cold, and some are hot. Is this feeling related in some way to our body temperature?
Answer:
Yes. If the temperature of the object is lower than our body temperature, e.g., ice, we feel the object is cold. If the temperature of the object is higher than our body temperature, e.g., hot water, we feel the object is hot.

Use your brain power! (Text Book Page No. 66)

Question 1.
How will you explain the following statements with the help of the anomalous behaviour of water?
(1) In regions with cold climate, the aquatic plants and animals can survive even when the atmospheric temperature goes below 0 °C.
(2) In cold regions in winter the pipes for water supply break and even rocks crack.
Answer:

In cold regions, during winter, the temperature of the atmosphere falls well below 0 °C. As the temperature decreases, the water at the surfaces of lakes and ponds starts contraeting. Hence, its density increases and it sinks to the bottom. This process continues till the temperature of all the water in a lake falls to 4 °C. As the water at the surface cools further, i.e., its temperature falls below 4 °C, it starts expanding instead of contracting. Therefore, its density decreases and it remains at the surface.

The temperature or the water at the surface continues to fall to 0 °c. Finally, the water at the surface is converted Into ice, but the water below the layer of ice is at 4 °C. Ice is a bad conductor of heat. Hence, the layer of the ice at the surface does not allow transfer of heat from the water to the atmosphere. As the water below the layer of ice remains at 4 °C, fish and other aquatic animals and plants can survive in it.

(2) Sometimes water enters into crevices of the rocks. when the temperature of the atmosphere falls below 4 °C, water expands. Even when water freezes to form ice, there is increase in its volume. As there is no room for expansion, it exerts tremendous pressure on the rocks which crack and break up Into small pieces.

In cold countries, in winter, the temperature of the atmosphere falls below 0 °C. when the temperature of water falls below 4 °C, it expands. Hence, the water in pipes expands. Even if ice is rormed, there is an increase in the volume.

As there is no room for expansion, water (or ice) exerts a large pressure on the pipes. Hence, the pipelines carrying water burst.

Fill in the blanks and rewrite the sentences:

Question 1.
The amount of water vapour in air is determined in terms of its…………..
Answer:
The amount of water vapour in air is determined in terms of its absolute humidity.

Question 2.
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their……………
Answer:
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacities.

Question 3.
When a liquid is getting converted into solid, the latent heat is…………. (Practical Activity Sheet – 1 and 2)
Answer:
When a liquid is getting converted into solid, the latent heat is released.

Rewrite the following statements by selecting the correct options:

Question 1.
……….is used to study the anomalous behaviour of water.
(a) Calorimeter
(b) Joule’s apparatus
(c) Hope’s apparatus
(d) Thermos flask
Answer:
(c) Hope’s apparatus

Question 2.
When water boils and is converted into steam, then………..
(a) heat is taken in and temperature remains constant
(b) heat is taken in and temperatures rises
(c) heat is given out and temperature lowers
(d) heat is given out and temperature remains constant
Answer:
(a) heat is taken in and temperature remains constant

Question 3.
When steam condenses to form water,………..
(a) heat is absorbed and temperature increases
(b) heat is absorbed and temperature remains the same
(c) heat is given out and temperature decreases
(d) heat is given out and temperature remains the same
Answer:
(d) heat is given out and temperature remains the same

Question 4.
The temperature of ice can be decreased below 0 °C by mixing………..in it. (Practice Activity Sheet – 3)
(a) saw dust
(b) sand
(c) salt
(d) coal
Answer:
(c) salt

Question 5.
Ice/water is a substance that………..
(a) expands on melting and contracts on freezing
(b) contracts on melting and does not undergo change in volume on freezing
(c) contracts on melting and expands on freezing
(d) does not undergo any change in volume on melting or freezing
Answer:
(c) contracts on melting and expands on freezing

Question 6.
Heat absorbed when 1 g of ice melts at 0 °C to form 1 g of water at the same temperature is………..cal.
(a) 80
(b) 800
(c) 540
(d) 54
Answer:
(a) 80

Question 7.
The latent heat of vaporization of water is………..
(a) 540 cal/g
(b) 800 cal/g
(c) 80 cal/g
(d) 54 cal/g
Answer:
(a) 540 cal/g

Question 8.
The latent heat of fusion of ice is………..
(a) 540 cal/g
(b) 80 cal/g
(c) 800 cal/g
(d) 4cal/g
Answer:
(b) 80 cal/g

Question 9.
If the temperature of water is decreased from 4 °C to 10 °C, then its………..
(a) volume decreases and density increases
(b) volume increases and density decreases
(c) volume decreases and density decreases
(d) volume increases and density increases
Answer:
(b) volume increases and density decreases

Question 10.
At 4 °C, the density of water is………..
(a) 10 g/cm³
(b) 4g/cm³
(c) 4 × 10³ kg/m³
(d) 1 × 10³ kg/m³
Answer:
(d) 1 × 10³ kg/m³

Question 11.
The density of water is maximum at………..
(a) 0 °C
(b) – 4 °C
(c) 100 °C
(d) 4 °C
Answer:
(d) 4 °C

Question 12.
………..heat is needed to raise the temperature of 1 kg of water from 14.5 °C to 15.5 °C.
(a) 4180 J
(b) 103 J
(c) 1 cal
(d) 4180 cal
Answer:
(a) 4180 J

Question 13.
………..heat is needed to convert 1 g of water at 0 °C and at a pressure of one atmosphere into 1 g of steam under the same conditions.
(a) 80 cal
(b) 540 cal
(c) 89 J
(d) 540 J
Answer:
(b) 540 cal

Question 14.
Water expands on reducing its temperature below………..°C. (March 2019)
(a) 0
(b) 4
(c) 8
(d) 12
Answer:
(b) 4

State whether the following statements are true or false. (If a statement is false, correct it and rewrite it.):

Question 1.
Specific latent heat of fusion is expressed in g/cal.
Answer:
False. (Specific latent heat of fusion is expressed in cal/g.)

Question 2.
If the temperature of water is raised from 0 °C to 10 °C, its volume goes on increasing.
Answer:
False. (If the temperature of water is raised from 0 °C to 10 °C, its volume goes on decreasing in the range 0 °C to 4 °C and then goes on increasing in the range 4 °C to 10 °C.)

Question 3.
At dew point relative humidity is 100%.
Answer:
True.

Question 4.
1 kcal = 4.18 joules.
Answer:
False. (1 kcal = 4180 joules.)

Question 5.
Specific heat capacity is expressed in cal/g·°C
Answer:
True.

Question 6.
Latent heat of fusion, Q = mL.
Answer:
True.

Question 7.
If the relative humidity is more than 60%, we feel that the air is humid.
Answer:
True.

Question 8.
If the relative humidity is less than 60%, we feel that the air is dry.
Answer:
True.

Question 9.
Relative humidity has no unit.
Answer:
True.

Question 10.
Absolute humidity is expressed in kg/m3.
Answer:
True.

Identify the odd one and give the reason:

Question 1.
Temperature, conduction, convection, radiation.
Answer:
Temperature. It is a physical quantity. Others are modes of transfer of heat.

Question 2.
The joule, The erg, The calorie, The newton.
Answer:
The newton. It is a unit of force. Others are units of energy (as well as work.)

Question 3.
cal/g, cal/g·°C, k cal/kg·°C, erg/g·°C.
Answer:
cal/g. It is a unit of specific latent heat. Others are units of specific heat capacity.

Match the columns:

Column A	Column B
1. Latent heat	a. Q = mc ΔT
2. Specific heat capacity	b. Q = mL
3. Heat absorbed or given out by a body when its temperature changes.	c. kcal
	d. cal/g·°C

Answer:
(1) Latent heat – Q = mL
(2) Specific heat capacity – cal/g·°C
(3) Heat absorbed or given out by a body when its temperature changes – Q = mc ΔT.

Answer the following questions in one sentence each:

Question 1.
State units of temperature.
Answer:
Units of temperature: °C, °F and K (kelvin).

Question 2.
State units of energy.
Answer:
Units of energy: the erg, the joule, the calorie.

Question 3.
State the relation between the joule and the calorie.
Answer:
1 calorie = 4.18 joules.

Question 4.
State the relation between the erg and the joule.
Answer:
1 joule = 10⁷ ergs.

Question 5.
State the relation between the erg and the kilocalorie.
Answer:
1 kilocalorie = 4.18 × 10¹⁰ ergs.

Question 6.
State the relation between the joule and the kilocalorie.
Answer:
1 kilocalorie = 4.18 × 10³ joules.

Question 7.
When heat energy is absorbed by an object, ΔT represents the rise in temperature. What would ΔT represent if the object loses heat energy? (Practice Activity Sheet – 4)
Answer:
If the object loses heat energy, ΔT would represent the decrease in temperature.

Answer the following questions:

Question 1.
Define latent heat of fusion.
(OR)
What is latent heat of fusion? State its units.
Answer:
When a solid is converted into liquid at constant temperature (melting point of the substance) the amount of heat absorbed by it is called the latent heat of fusion.
Heat is a form of energy. Hence, latent heat is expressed in units joule, erg, calorie or kilocalorie.

Question 2.
Define specific latent heat of fusion.
(OR)
What is specific latent heat of fusion? State its units.
Answer:
The amount of heat energy absorbed at constant temperature by unit mass of a solid to convert into liquid phase is called the specific latent heat of fusion.
It is expressed in units J/kg, erg/g, cal/g, kJ/ kg and kcal/kg.

[Note: Specific latent heat (L) = \(\frac{\text { latent heat }(Q)}{\text { mass of the substance }(m)}\)
:. SI unit of specific latent heat = SI unit of energy / SI unit of mass = J/kg]

Question 3.
Explain the term latent heat of vaporization.
Answer:
When a liquid is heated continuously, initially, its temperature increases. Later, at a certain stage, its temperature does not increase even when heat is supplied to it. At this temperature, heat absorbed by the liquid is used for breaking the bonds between its atoms or molecules, i.e., for doing work against the forces of attraction between the atoms or molecules and conversion into gaseous phase.

This heat is called the latent heat of vaporization and the constant temperature at which this change of state occurs is called the boiling point of the liquid.

Question 4.
Define boiling point of a liquid.
(OR)
What is boiling point of a liquid?
Answer:
The constant temperature at which a liquid transforms into gaseous state is called the boiling point of the liquid.
[Note: On application of pressure, the boiling point of a liquid is raised. On reducting the pressure, the boiling point is lowered.]

Question 5.
Define specific latent heat of vaporization.
OR
What is specific latent heat of vaporization?
Answer:
The amount of heat energy absorbed at constant temperature by unit mass of a liquid to convert into gaseous phase is called the specific latent heat of vaporization.

Question 6.
The specific latent heat of fusion of ice is 80 cal/g. Explain this statement.
Answer:
When 1 g of ice at a pressure of one atmosphere and at a temperature 0 °C is converted into 1 g of water, heat absorbed by the ice is 80 cal.

Question 7.
The specific latent heat of fusion of silver is 88.2 kJ/kg. Explain this statement.
Answer:
When 1 kg of silver at a pressure of one atmosphere and at a temperature of 962 °C (melting point of silver) is converted into 1 kg of silver in liquid phase, heat absorbed by the silver is 88.2 kJ.

Question 8.
The specific latent heat of vaporization of water is 540 cal/g. Explain this statement.
Answer:
When 1 g of water at a pressure of one atmosphere and at a temperature of 100 °C is converted into 1 g of steam, heat absorbed by the water is 540 cal.

Question 9.
Define regelation.
(OR)
What is regelation?
Answer:
The phenomenon in which the ice converts to liquid due to applied pressure and then re-converts to ice once the pressure is removed is called regelation.

Question 10.
The terms hot and cold are used in relative context. Explain.
Answer:

(1) Take three large bowls, P, Q and R. Fill bowl P with cold water, bowl Q with lukewarm water, and bowl R with hot water.
(2) Immerse your right hand in bowl P, and left hand in bowl R for about five seconds.
(3) Now, immerse both the hands in bowl Q at the same time.
(4) You will find that the water in bowl Q appears warm to your right hand, and cold to your left hand. Thus, the hand immersed in cold water for some time finds the lukewarm water hot while the one immersed in hot water finds the same lukewarm water cold. This experiment shows that the terms hot and cold are relative.

Question 11.
Draw a neat labelled diagram of Hope’s apparatus. Explain how this apparatus can be used to demonstrate anomalous behaviour of water. Draw a graph of temperature of water against time.
Answer:

The figure shows Hope’s apparatus. Initially, the cylindrical container in Hope’s apparatus is filled with water at about 12 °C and the flat bowl is filled with a freezing mixture of ice and salt.

The temperature of water in the upper part of the container (T₂) is recorded by thermometer T₂ and that of water in the lower part of the container (T₁) is recorded by thermometer T₁. Figure shows variation of temperature of water with time.

Initially, both the thermometers show the same temperature (say, 12 °C). In a short time, the temperature shown by the lower thermometer starts decreasing, while the temperature shown by the upper thermometer does not change very much.

This process continues till the temperature shown by the lower thermometer falls to 4 °C and remains constant thereafter. This shows that in the temperature range 12 °C to 4 °C, the density of the water in the central part of the container goes on increasing and hence the water sinks to the bottom. It means that water contracts, i.e., its volume decreases as its temperature falls from 12 °C to 4 °C.

As the temperature of the water in the central part of the container becomes less than 4 °C, the temperature shown by the upper thermometer begins to fall rapidly to 0 °C. But the temperature shown by the lower thermometer remains constant (4 °C). Later, the heading shown by the lower thermometer decreases to 0 °C.

In the temperature range 4 °C to 0 °C, the water moves upward. This shows that the density of water goes on decreasing in this range. It means that water expands, i.e. its volume increases as its temperature falls from 4 °C to 0 °C.

Thus, the volume of a given mass of water is minimum at 4 °C, i.e., the density of water is maximum at 4 °C.

In the above figure, the point of intersection of the two curves shows the temperature at which the density of water is maximum. This temperature is 4 °C.

Question 12.
A mountaineer climbing on the Everest, experienced the following facts. Explain each fact with the scientific reason : (1) He found j fishes alive below the ice (2) Time required for cooking was more as he went higher (3) He saw many times cliffs falling suddenly (4) He saw tubes carrying water broken.
Answer:
Explanation:
(1) Water expands as its temperature decreases from 4 °C to 0 °C. Water is converted into ice at 0 °C. The density of water is more than that of ice. Fishes can remain alive in the water (at 4 °C) below the ice.
(2) At high altitudes, atmospheric pressure is low and hence water boils at a temperature lower than its normal boiling point. Therefore, the time required for cooking food is more at higher altitudes.
(3) Water expands while freezing. Hence, the water present in the crevices of the rocks exerts a tremendous pressure on the rocks, while freezing. Therefore, the cliffs fall.
(4) Water expands while freezing. Hence, the water in the tube exerts a large pressure on the tube, while freezing. Therefore, the tube carrying water breaks.

Question 13.
What is humidity?
Answer:
The moisture, i.e., the presence of water vapour, in the atmosphere is called humidity.

Question 14.
When is air said to be saturated with water vapour?
Answer:
When air contains maximum possible water vapour, it is said to be saturated with water vapour at that temperature.

Question 15.
What does the amount of water vapour needed to saturate air depend on?
Answer:
The extent of water vapour needed to saturate air depends on the temperature. The greater the temperature, the greater is the amount of water vapour needed to saturate air.

Question 16.
When is air said to be unsaturated with water vapour?
Answer:
When air contains water vapour less than its capacity to hold water vapour at that temperature, it is said to be unsaturated with water’vapour.

Question 17.
What is dew point temperature?
(OR)
Define dew point temperature.
Answer:
If the temperature of unsaturated air is decreased, a temperature is reached at which the air becomes saturated with water vapour. This temperature is called the dew point temperature.

Question 18.
Name the physical quantity used to express the amount of water vapour present in air.
Answer:
Absolute humidity.

Question 19.
Define absolute humidity.
(OR)
What is absolute humidity? State its unit.
Answer:
The mass of water vapour present in a unit volume of air is called absolute humidity. Generally it is expressed in kg/m³.

Question 20.
Define relative humidity.
(OR)
What is relative humidity? Write the formula for % relative humidity.
Answer:
The ratio of the actual mass of water vapour content in the air for a given volume and temperature to that required to make the same volume of air saturated with water vapour at the same temperature is called the relative humidity.

% Relative humidity = [the actual mass of water vapour content in the air for a given volume and temperature ÷ the mass of water vapour required to make the same volume of air saturated with water vapour at the same temperature] × 100%.

Question 21.
What is the value of relative humidity at the dew point temperature?
Answer:
At the dew point temperature, relative humidity is 100%.

Question 22.
The mass of water vapour in air enclosed in a certain space is 60 g and the mass of water vapour needed to saturate the same air with water vapour under the same conditions is 100 g. What is the corresponding % relative humidity?
Answer:
Here, % relative humidity = (\(\frac{60 \mathrm{g}}{100 \mathrm{g}}\)) × 100% = 60%

Question 23.
During winter, sometimes we see a white trail at the back of a flying aeroplane in a clear sky. Explain why.
Answer:
In winter, air temperature is low. Hence, when an aeroplane flies, the vapour released by its engine condenses and forms white clouds. If the relative humidity of the air surrounding the plane is high, we see this white trail at the back of the plane for a long time before it disappears. If.the relative humidity is low, the white trail is short and disappears quickly. If the relative humidity is very low, there is no formation of the white trail.

Question 24.
State two effects of humidity present in atmosphere.
Answer:
Effects of humidity present in atmosphere: When the temperature of air falls below the dew point, dew and fog are formed.

Question 25.
Explain how dew and fog are formed.
(OR)
Write a short note on formation of dew and fog.
Answer:
At a particular temperature, a given volume of air can contain a certain maximum amount of water vapour. Normally, the temperature of air during the day is such that air is not saturated with water vapour present in it.

As the temperature falls, the capacity of air to hold water vapour becomes less. During a cold night, the temperature of air may fall to the dew point, or even below the dew point. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed. If the water vapour condenses on the fine dust particles present in the atmosphere, mist or fog is formed.

Question 26.
State the units of heat.
Answer:
Units of heat: joule, erg, calorie, kilocalorie.

Question 27.
Define the kilocalorie.
Answer:
The amount of heat necessary to raise the temperature of 1 kg of water by 1 °C from 14.5 °C to 15.5 °C is called one kilocalorie.

Question 28.
Define the calorie.
Answer:
The amount of heat necessary to raise the temperature of 1 g of water by 1 °C from 14.5 °C to 15.5 °C is called one calorie.

Question 29.
State the relation between the kilocalorie and the calorie.
Answer:
1 kilocalorie = 10³ calories.

Question 30.
Study the following procedure and answer the questions below: (Practice Activity Sheet – 2)
1. Take 3 spheres of iron, copper and lead of equal mass.
2. Put all the 3 spheres in boiling water in a beaker for some time.
3. Take 3 spheres out of the water. Put them immediately on a thick slab of wax.
4. Note the depth that each sphere goes into the wax.
(i) Which property of a substance can be studied with this procedure?
(ii) Describe that property in minimum words.
(iii) Explain the rule of heat exchange with this property.
Answer:
(i) Specific heat.
(ii) Specific heat: The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C.
(iii) According to the rule/principle of heat exchange, heat energy lost by the hot object = heat energy gained by the cold object.

In this activity, heat absorbed by the iron sphere is transmitted more in the wax, hence the sphere goes deepest into the wax, while the lead sphere absorbs less heat, resulting in less transmission of heat in the wax, hence, the sphere goes the least depth into the wax.

Question 31.
Write the symbol for specific heat capacity. State the units of specific heat capacity.
Answer:
Symbol for specific heat capacity: c.
Units of specific heat capacity: J/kg·°C,
erg/g·°C, cal/g·°C, kcal/kg·°C.
[ Notes: (1) Specific heat capacity

In SI, heat is expressed in joule (J), mass in kg and temperature in kelvin(K).
∴ SI unit of specific heat capacity = \(\frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}\). (2) The specific heat capacity of a substance depends upon its constituent particles (atoms, molecules, etc.), interaction between them, structure of the substance (atomic/molecular arrangement), temperature of the substance, etc.]

Question 32.
Explain the principle of heat exchange. Ans. Suppose two objects A and B at different temperature T₁ and T₂ respectively are enclosed in a box of heat resistant material as shown in figure.

Let m₁ = mass of A, m₂ = mass of B, c₁ = specific heat capacity of A, c₂ = specific heat capacity of B and T = common temperature attained by A and B after the heat exchange between A and B. Here, no heat leaves the box or enters the box from outside. Hence, if T₁ > T₂, heat energy lost by A (Q₁) = heat energy gained by B (Q₂).
∴ m₁c₁ (T₁ – T) = m₂c₂(T – T₂)
[Note: If m₁, c₁, T₁, T, m₂ and T₂ are known, c₂ can be determined.]

Question 33.
The specific heat capacity of silver is 0.056 kcal/kg·°C. Explain this statement.
Answer:
The amount of heat needed to raise the temperature of 1 kg of silver by 1 °C is 0.056 kcal.

Question 34.
Explain how the specific heat capacity of a solid can be determined (measured) by the method of mixture.
Answer:
A hot solid is put in water in a calorimeter. The mixture is stirred continuously and the maximum temperature of the mixture is measured with a thermometer. Heat exchange between the hot solid, water and calorimeter results in sill bodies attaining the same temperature after some time. Hence, according to the principle of heat exchange, heat lost by the solid = heat gained by the water in the calorimeter + heat gained by the calorimeter.

Now, heat lost by the solid (Q) = mass of the solid × its specific heat capacity × decrease in its temperature, heat gained by the water (Q₁) = mass of the water × its specific heat capacity × increase in its temperature and heat gained by the calorimeter (Q₂) = mass of the calorimeter × its specific heat capacity × increase in its temperature.

Heat lost by the hot object = heat gained by the calorimeter + heat gained by the water. Q = Q₂ + Q₁
Using this equation, the specific heat capacity of the solid can be determined (measured) when the other quantities are known.

Give scientific reasons:

Question 1.
Even though heat is supplied to boiling water, there is no increase in its temperature.
Answer:
Once water starts boiling, all the heat supplied to it is used in conversion of water into steam at the boiling point of water. Hence, there is no rise in its temperature.

Question 2.
Burns from steam are worse those from boiling water at the same temperature.
Answer:
1. A given quantity of steam contains more heat than the same quantity of boiling water at the same temperature.
2. When steam comes in contact with one’s body, it releases extra heat of 540 calories per gram and causes a more serious burn than that caused by boiling water.

Question 3.
In winter, the pipelines carrying water burst in cold countries.
Answer:
1. In cold countries, in winter, the temperature of the atmosphere falls below 0 °C. When the temperature of water falls below 4 °C, it expands. Hence, the water in pipes expands. Even if ice is formed, there is an increase in the volume.

2. As there is no room for expansion, water (or ice) exerts a large pressure on the pipes. Hence, the pipelines carrying water burst.

Question 4.
If crushed ice is pressed and then the pressure is released, a lump of ice is formed.
Answer:
1. When crushed ice is pressed, its melting point is lowered and some ice melts to form water.
2. When pressure is released, the melting point becomes normal and the water freezes to form ice forming a lump.

Question 5.
In cold countries, in winter, even when the water of lakes freezes, aquatic animals and plants can survive.
Answer:
1. In cold countries, in winter, a layer of ice is formed on the surface of lakes when the atmospheric temperature falls below 0 °C. However, below this layer, there is water at 4 °C.
2. Ice, being a bad conductor of heat, does not allow transfer of heat from this water to the atmosphere. Hence, aquatic animals and plants can survive in this water.

Question 6.
Water droplets are seen on’ the outer surface of a cold drink bottle.
Answer:
1. The temperature of the outer surface of a cold drink bottle is less than that of the atmosphere.
2. Therefore, the excess of water vapour from the air condenses to form droplets on the outer surface of the cold drink bottle.

Question 7.
During cold nights, sometimes dew is formed.
Answer:
1. During a cold night, the temperature of air may fall to the dew point, or even below the dew point. 2. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed.

Question 8.
When you enter a warm room after being outside on a frosty early morning, your spectacles ‘steam up’.
Answer:
1. On a frosty early morning, the temperature of air outside a warm room is lower than the dew point.
2. Hence, when you enter the room from outside, some water vapour in the room condenses on the glass of your spectacles, i.e., the spectacles ‘steam up’.

Question 9.
A plastic bottle, completely filled with water, when kept in a freezer, is likely to break.
Answer:
The temperature of air in the freezer (deep freeze) compartment of a refrigerator is less than 0 °C. 2. When a plastic bottle, completely filled with water, is kept in this compartment, the temperature of water falls below 4 °C and the water expands. Even when water freezes and ice is formed, there is an increase in the volume. It exerts a large pressure on the sides of the bottle and hence the bottle is likely to break.

Question 10.
The outer surface of a beaker containing ice cubes becomes wet in a short while.
Answer:
1. When ice cubes are placed in a beaker, ice starts melting. The heat required for melting is absorbed from the surrounding air and also from the beaker to some extent.
Hence, the temperature of the air and beaker falls.

2. The capacity of air to hold water vapour depends upon the temperature of the air, and this capacity decreases as the temperature decreases. At a certain low temperature, the surrounding air becomes saturated with water vapour present in it. As the temperature falls further, the air is unable to hold all the water vapour.

Hence, the extra water vapour starts condensing on the cold outer surface of the beaker in the form of minute drops. Therefore, the outer surface of the beaker containing ice cubes becomes wet in a short while.

Distinguish between the following:

Question 1.
Absolute humidity and Relative humidity.
Answer:
Absolute humidity:

Absolute humidity is the mass of water vapour present in a unit volume of air.
It is commonly expressed in kg/m³.

Relative humidity:

Relative humidity is the ratio of the mass of water vapour in a given volume of air at a given temperature to the mass of water vapour required to saturate the same volume of air at the same temperature.
It does not have unit.

Solve the following examples/Numerical problems:
[Use the data given in the Tables on pages 130 and 131.]

Problem 1.
Calculate the amount of heat required to convert 5 g of ice of 0 °C into water at 0 °C. (Specific latent heat of fusion of ice = 80 cal/g)
Solution: Here, m = 5 g, L = 80 cal/g; Q = ?
Amount of heat required, Q = mL
= 5 g × 80 cal/g
= 400 calories.

Problem 2.
Find the amount of heat required to convert 10 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution: Here, m = 10 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 10 g × 540 cal/g
= 5400 calories.

Problem 3.
Calculate the amount of heat required to convert 15 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
m = 15 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 15 g × 540 cal/g
= 8100 calories.

Problem 4.
How many calories of heat will be absorbed when 3 kg of ice at 0 °C melts?
Solution:
m = 3 kg = 3000 g; L = 80 cal/g; Q = ?
Quantity of heat absorbed, Q = mL
= 3000 g × 80 cal/g
∴ Q = 240000 calories.

Problem 5.
Calculate the amount of heat required to convert 10 g of water at 30 °C into steam at 100 °C. (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
Here, m = 10 g; c = 1 cal/g·°C
T₂ – T₁ = 100 °C – 30 °C = 70 °C; L = 540 cal/g; Q = ?
Amount of heat required, Q = mc (T₂ – T₁) + mL
= 10 g × 1 cal/g·°C × 70 °C + 10 g × 540 cal/g
= 700 cal + 5400 cal
∴ Q = 6100 calories.

Problem 6.
If water of mass 80 g and temperature 45 °C is mixed with water of mass 20 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution:
Data : m₁ = 80 g, T₁ = 45 °C, m₂ = 20 g,
T₂ = 30 °C, T = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
∴ m₁c (T1 – T) = m₂c (T – T₂)
∴ m₁T₁ – m₁T = m₂T – m₂T₂
∴ m₁T₁ + m₂T₂ = (m1 + m₂)T
∴ Maximum temperature of the mixture,

= (36 + 6) °C
= 42°C.

Problem 7.
When water of mass 70 g and temperature 50 °C is added to water of mass 30 g, the maximum temperature of the mixture is found to be 41 °C. Find the temperature of water of mass 30 g before hot water was added to it.
Solution:
Data : m₁ = 70 g, T₁ = 50 °C, m₂ = 30 g, T = 41 °C, T₂ = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
∴ m₁c (T₁ – T) = m₂c (T – T₂)
∴ m₁T₁ – m₁T = m₂T – m₂T₂
∴ m₂T₂ = (m₁ + m₂) T – m₁T₁

This is the required temperature.

Problem 8.
Find the heat needed to raise the temperature of a silver container of mass 100 g by 10 °C. (c = 0.056 cal/g.°C)
Solution:
Data: m = 100 g, ΔT = 10 °C, c = 0.056 cal/g·°C
Heat needed to raise the temperature of the container = mc ΔT
= 100 g × 0.056 cal/g·°C × 10 °C .
= 56 calories.

Problem 9.
If steam of mass 100 g and temperature 100 °C is released on an ice slab of temperature 0 °C, how much ice will melt?
Solution:
Data: m₁ = 100 g, L₁ = 540 cal/g,
T₁ = 100 °C, mass of ice, m = ?, L₂ = 80 cal/g, c (water) = 1 cal/g·°C
According to the principle of heat exchange, heat lost by hot body = heat gained by cold body. Conversion of steam into water:
Q₁ = m₁L₁ = 100 g × 540 cal/g = 54000 cal
Decrease in the temperature of this water to 0 °C:
Q₂ = m₁c × (T₁ – 0 °C) = 100 g × 1 cal/g·°C × (100 °C – 0 °C) = 10000 Cal
Melting of ice: Q₃ = mL₂
= m × 80 cal/g
Now, Q₁ + Q₂ = Q₃
∴ (54000 + 10000) cal = m × 80 cal/g
∴ m = \(\frac{64000}{80}\) = 800 g
800 g of ice will melt.

Numerical problems for practice:

Problem 1.
Calculate the amount of heat required to convert 80 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
6400 cal

Problem 2.
Find the heat required to convert 20 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
1600 cal

Problem 3.
Calculate the quantity of heat released during the conversion of 10 g of ice cold water (temperature 0 °C) into ice at the same temperature. (Specific latent heat of freezing of water = 80 cal/g)
Solution:
800 cal

Problem 4.
How many calories of heat will be absorbed when 2 kg of ice at 0 °C melts? (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
160000 cal

Problem 5.
How much heat will be required to convert 20 g of water at 100 °C into steam at 100 °C? (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
10800 cal

Problem 6.
Find the heat absorbed by 25 g of water at 100 °C to convert into steam at the same temperature. (Specific latent heat of vaporization of water = 540 cal/g.)
Solution:
13500 cal

Problem 7.
If water of mass 60 g and temperature 50 °C is mixed with water of mass 40 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution:
42 °C

Problem 8.
If water of mass 60 g and temperature 60 °C is mixed with water of mass 60 g and temperature 40 °C, what will be the maximum temperature of the mixture?
Solution:
50 °C

Problem 9.
Find the heat needed to raise the temperature of a piece of iron of mass 500 g by 20 °C. (c = 0.110 cal/g·°C)
Solution:
1100 cal

Problem 10.
Water of mass 200 g and temperature 30 °C is taken in a copper calorimeter of mass 50 g and temperature 30 °C. A copper sphere of mass 100 g and temperature 100 °C is released into it. What will be the maximum temperature of the mixture? [c (water) = 1 cal/g·°C, c (copper) =0.1 cal/g·°C]
Solution:
33.26 °C

Problem 11.
A copper calorimeter of mass 100 g and temperature 30 °C contains water of mass 200 g and temperature 30 °C. If a piece of ice of mass 40 g and temperature 0 °C is added to it, what will be the maximum temperature of the mixture? [c (copper) = 0.1 cal/g·°C, c (water) = 1 cal/g·°C, L = 80 cal/g]
Solution:
12.4 °C

10th Std Science Part 1 Questions And Answers:

Carbon Compounds Class 10 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Bhagya

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds Notes, Textbook Exercise Important Questions and Answers.

Std 10 Science Part 1 Chapter 9 Carbon Compounds Question Answer Maharashtra Board

Class 10 Science Part 1 Chapter 9 Carbon Compounds Question Answer Maharashtra Board

Question 1.
Match the pairs.

Group A	Group B
a. C₂H₆	1. Unsaturated hydrocarbon
b. C₂H₂	2. Molecular formula of an alcohol
c. CH₄O	3. Saturated hydrocarbon
d. C₃H₆	4. Triple bond

Question 2.
Draw an electron dot structure of the following molecules. (Without showing the circles)
a. Methane.
Answer:
Molecular formula: CH₄

b. Ethene.
Answer:
Molecular formula: H₂C = CH₂

c. Methanol.
Answer:
Molecular formula: H₃C – OH

d. Water.
Answer:
Molecular formula: H₂O

Question 3.
Draw all possible structural formulae of compounds from their molecular formula given below.
a. C₃H₈
b. C₄H₁₀
c. C₃H₄
Answer:
a. C₃H₈ Propane:

b. C₄H₁₀ Butane:

c. C₃H₄ Propyne:

Question 4.
Explain the following terms with example.
a. Structural isomerism.
Answer:
The phenomenon in which compounds having different structural formulae have the same molecular formula is called structural isomerism. Butane is represented by two different compounds as their structural formulae are different. The first compound is a straight chain compound and the second compound is a branched chain compound. These two different structural formulae have the same molecular formula i.e. C₄H₁₀.

b. Covalent bond.
Answer:
The chemical bond formed by sharing of two valence electrons between the two atoms is called covalent bond.
Example:
1. Hydrogen molecule formation: The atomic number of hydrogen is 1, its atom contains 1 electron in K shell. It requires one more electron to complete the K shell and attain the configuration of helium (He). To meet this requirement two hydrogen atoms share their electrons with each other to form H₂ molecule. One covalent bond, i.e. a single bond is formed between two hydrogen atoms by sharing of two electrons.

2. Formation of oxygen molecule:
(1) The atomic number of oxygen is 8. The electronic configuration of oxygen is (2, 6). Oxygen has 6 electrons in the outermost shell.
(2) It requires 2 electrons to complete the L shell and attain the configuration of neon (Ne).
(3) Each oxygen atom shares its valence electron with the valence electron of another oxygen atom to give two shared pairs of electrons which results in the formation of oxygen molecule.
(4) Thus, two electron pairs are shared between two oxygen atoms, forming double covalent bond (=).

c. Hetero atom in a carbon compound.
Answer:
Carbon compounds are formed by formation of bonds of carbon with other elements such as halogens, oxygen, nitrogen, sulfur. The atoms of these elements substitute one or more hydrogen atoms in the hydrocarbon chain and thereby the tetravalency of carbon is satisfied. The atom of the element which is substitute for hydrogen is referred to as a heteroatom. Sometimes hetero atoms are not alone but exist in the form of certain groups of atoms.

d. Functional group.
Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the functional groups.

e. Alkane.
Answer:
In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called alkane.
Example: In methane, four hydrogen atoms are bonded to carbon atom by four single covalent bonds.

f. Unsaturated hydrocarbons.
Answer:
The carbon compounds having a double bond or triple bond between two carbon atoms are called unsaturated hydrocarbons. The unsaturated hydrocarbons containing a carbon-carbon double bond are called alkenes.
e.g. Ethene (CH₂ = CH₂), Propene (CH₃– CH = CH₂).
The unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes e.g. Ethyne (CH ≡ CH).

g. Homopolymer.
Answer:
The polymers formed by repetition of single monomer are called homopolymer. e.g. polyethylene (CH₂– CH₂)n.

h. Monomer.
Answer:
The small unit that repeats regularly to form a polymer is called monomer.
Example: Ethylene.

i. Reduction.
Answer:
In a chemical reaction, removal of oxygen from a compound or addition of hydrogen to a compound is called a reduction.

j. Oxidant.
Answer:
An oxidant is a reactant that oxidizes or removes electrons from other reactants during a redox reaction. An oxidant may also be called an oxidizer or oxidizing agent. When the oxidant includes oxygen, it may be called an oxygenation reagent or oxygen-atom transfer (OT) agent.
Examples of oxidants include:

Hydrogen peroxide
Ozone
Nitric acid
Sulfuric acid

Question 5.
Write the IUPAC names of the following structural formulae.
a. CH₃ – CH₂ – CH₂ – CH₃
Answer:
The number of carbon atopic in the longest chain: 4
Parent alkane: Butane IUPAC name: n-Butane

b. CH₃ – CHOH – CH₃ (Practice Activity Sheet – 3)
Answer:
The number of carbon atoms in the longest chain: 3

Parent alkane: Propane
Functional group: -OH (ol)
Assign the number: 2
The carbon atom to which the -OH group is attached is numbered as C₂. If the carbon chain of the compound contains a -OH group then change the ending of the parent name, i.e., ‘e’ of propane is replaced by ‘ol’. (ol stands for alcohol)
Parent suffix: Propan-2-ol
IUPAC name: Propan-2-ol

c. CH₃– CH₂– COOH (Practice Activity Sheet – 3)
Answer:
The number of carbon atoms in the longest chain: 3
Parent alkane: Propane
Functional group: -COOH (-oic acid)
If the carbon chain of the compound contains a -COOH group then change the ending of the parent name, i.e., ‘e’ of propane is replaced by ‘oic acid’.
Parent suffix: Propanoic acid
IUPAC name: Propanoic acid

d. CH₃ – CH₂ – NH₂
Answer:
Number of carbon atoms: 2
Parent alkane: Ethane
Functional group: -NH₂ (amine)
If the carbon chain of the compound contains a -NH₂ group, then change the ending of the parent name, i.e., ‘e’ of ethane is replaced by ‘amine’.
Parent suffix: Ethanamine
IUPAC name: Ethanamine.

e. CH₃ – CHO
Answer:
Number of carbon atoms: 2
Parent alkane: Ethane
Functional group: -CHO (al)
If the carbon chain of the compound contains a -CHO group, then change the ending of the parent name, i.e., ‘e’ of ethane is replaced by ‘al’.
Parent suffix: Ethanal
IUPAC name: Ethanal

f. CH₃ – CO – CH₂ – CH₃
Answer:
Number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -CO- (one)
Assign the number:

In the longest chain, the numbering of carbon atom starts from the carbon atom nearest to the function group.
If the carbon chain of the compound contains a (-CO-) group, then change the ending of the parent name, i.e., ‘e’ of butane is replaced by ‘one’.
Parent suffix: Butan-2-one
IUPAC name: Butan-2-one

Question 6.
Identify the type of the following reaction of carbon compounds.
1. CH₃ – CH₂ – CH₂ – OH + (O) → CH₃ – CH₂ – COOH
2. CH₃ – CH₂ – CH₃ + O₂ → 3CO₂ + 4H₂O
3. CH₃ – CH = CH – CH₃ + Br₂ → CH₃ – CHBr – CHBr – CH₃
4. CH₃– CH₃+ Cl₂ → CH₃– CH₂– Cl + HCl
5. CH₃– CH₂– CH₂– CH₂– OH → CH₃– CH₂– CH = CH₂+ H₂O
6. CH₃– CH₂– COOH + NaOH → CH₃– CH₂– COONa⁺ + H₂O
7. CH₃– COOH + CH₃– OH → CH₃– COO – CH₃ + H₂O
Answer:

Question 7.
Write the structural formulae for the following IUPAC names:
a. Pent-2-one
Answer:
Pent-2-one.
(1) Pent stands for 5 carbon atoms in a chain.
Number the carbon atoms in a chain as 1, 2, 3,…..

(2) ‘one’ stands for functional group

ketone. The number assigned for the ketone group is 2. Show the ketone group at C₂.

(3) Now satisfy the valencies of each carbon atom

b. 2-Chlorobutane
Answer:
(1) In 2-chlorobutane, butane is parent alkane stands for 4 carbon atoms and number the carbon atoms in a chain as 1, 2, 3,….

(2) Chloro (Halo) is the prefix and the number assigned for prefix (chloro) is 2. Show the chloro atom at C₂

(3) Now satisfy the valencies of each carbon atom

c. Propan-2-ol
Answer:
(1) Propan stands for 3 carbon atoms in a chain. Number the carbon atom in a chain as 1, 2, 3.

(2) ‘-ol’ stands for (-OH) hydroxyl group. The number assigned for the hydroxyl group is 2. Show the -OH group at C₂.

(3) Now satisfy the valencies of each carbon atom

d. Methanal
Answer:
(1) Meth – stands for one carbon atom and assigned the number ‘1’ to carbon in the functional group -CHO.
(2) ‘-al’ stands for functional group (-CHO) aldehyde.
(3) Now satisfy the valencies of carbon in -CHO.

e. Butanoic acid
Answer:
(1) But stands for 4 carbon atoms in a chain. Number the carbon atoms in a chain as 1, 2, 3,….

‘-oic acid’ stands for functional group -COOH. Assign the number 1 to carbon in the functional group -COOH.

Now satisfy the valencies of each carbon atom

f. 1-Bromopropane.
Answer:
(1) In 1-bromopropane, propane is parent alkane stands for 3 carbon atoms and number the carbon atoms in a chain as 1, 2, 3…….

(2) Bromo (Halo) is the prefix and the number assigned for prefix (bromo) is 1, show the bromine atom at C₁.

(3) Now satisfy the valencies of each carbon atom

g. Ethanamine
Answer:
(1) Eth stands for 2 carbon atoms in a chain and the parent alkane is ethane.
– C – C –
(2) ‘amine’ stands for (- NH₂) amino group. Show the amino (-NH₂) at any carbon atom.
– C – C – NH₂
(3) Now satisfy the valencies of each carbon atom

h. Butanone.
Answer:
(1) But stands for 4 carbon atoms in a chain and the parent alkane is butane. Number the carbon atoms in a chain 1, 2, 3,….

(2) ‘one’ stands for functional group

ketone. The number assigned for the ketone group is 2. Show the ketone group at C₂.

(3) Now satisfy the valencies of each carbon atom

Question 8.
a. What causes the existance of very large number of carbon compound?
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms; this results in formation of big molecules. This property of carbon is called catenation power. The carbon compounds contain open chains or closed chains of carbon atoms. An open chain can be a straight chain or a branched chain. A closed chain is a ring structure. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

(2) One, two or three covalent bonds can bond together two carbon atoms. These bonds are called single covalent bond, double covalent bond and triple covalent bond respectively. Due to the ability of carbon atoms to form multiple bonds as well as single bonds, the number of carbon compounds increases. For example, there are three compounds, namely, ethane (CH₃– CH₃), ethene (CH₂ = CH₂) and ethyne (CH = CH) which contain two carbon atoms.

(3) Carbon being tetravalent, one carbon atom can form bonds with four other atoms (carbon or any other). This results in formation of many compounds. These compounds possess different properties as per the atoms to which carbon is bonded. For example, five different compounds are formed using one carbon atom and two monovalent elements hydrogen and chlorine: CH₄, CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄. Similarly carbon atoms form covalent bonds with atoms of elements like O, N, S, halogen and P to form different types of carbon compounds in large number.

(4) Isomerism is one more characteristic of carbon compound which is responsible for large number of carbon compounds.

b. Saturated hydrocarbons are classified into three types. Write these names giving one example each.
Answer:
In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called saturated hydrocarbons. Methane molecule contains only one carbon atom. In methane, four hydrogen atoms are bonded to carbon atom by four covalent bonds.

c. Give any four functional groups containing oxygen as the heteroatom in it. write name and structural formula of one example each.
Answer:

d. Give names of three functional groups containing three different heteroatoms. Write name and structural formula of one example each.
Answer:

e. Give names of three natural polymers. write the place of their occurance and names of monomers from which they are formed.
Answer:

Poly saccharide is a natural polymer. It occurs in starch/carbohydrates. It is formed from monomer glucose.
Protein is a natural polymer. It occurs in muscles, hair, enzymes, skin, egg. It is formed from alpha amino acids.
Rubber is a natural polymer. It occurs in latex of rubber tree. It is formed from monomer isoprene.

f. What is meant by vinegar and gasohol? What are their uses?
Answer:
(1) Vinegar is a 5 – 8% aqueous solution of acetic acid. It is used as preservative in pickles. It is used to cook meat. 1t is used as a salad dressing.
(2) To increase the efficiency of petrol, it is mixed with 10% anhydrous ethanol, such a fuel is called gasohol. It is used as a fuel in cars and other vehicles.

g. what is a catalyst ? write any one reaction which is brought about by use of catalyst?
Answer:
Catalyst is a substance, which changes the rate of reaction, without causing any disturbance to it. Vegetable oil (unsaturated compound) undergoes addition reaction with hydrogen in the presence of nickel catalyst to form vanaspati ghee (saturated compound).

Project:
Prepare a chart giving detailed information of carbon compounds in everyday use. Display it in the cluss and discuss.

Can you recall? (Text Book Page No. 110)

Question 1.
What are the types of compounds?
Answer:
Organic and inorganic compounds are the two important types of compounds.

Question 2.
Objects in everyday use such as foodstuff, fibres, paper, medicines, wood, fuels are made of various compounds. Which constituent elements are common in these compounds?
Answer:
The constituent elements common in these compounds are carbon (C), hydrogen (H) and oxygen (O).

Question 3.
To which group in the periodic table does the element carbon belongs? Write down the electronic configuration of carbon and deduce the valency of carbon.
Answer:
The element carbon belongs to group 14 and its electronic configuration is 2, 4. The valency of carbon is 4.

Use your brain Power! (Text Book Page No. 115)

Question 1.
The molecular formula of ethyne is C₂H₂. From this draw its structural formula and electron-dot structure.
Answer:
Ethyne: Molecular formula: C₂H₂

Question 2.
How many bonds have to be there in between the carbon atoms in ethyne so as to satisfy their tetra valency?
Answer:
To satisfy their tetravalency, three double bonds have to be there in between two carbon atoms in ethyne.

Use your brain power! (Text Book Page No. 116)

Question 1.
Draw the electron-dot structure of cyclohexane.
Answer:
Cyclohexane: Molecular formula: C₆H₁₂

Use your brain power! (Text Book Page No. 112)

Question 1.
Atomic number of chlorine is 17. What is the number of electrons in the valence shell of the chlorine?
Answer:
There are seven electrons in the valence shell of the chlorine.

Question 2.
Molecular formula of chlorine is Cl₂. Draw electron-dot and line structure of a chlorine molecule.
Answer:

Question 3.
The molecular formula of water is H₂O. Draw electron-dot and line structures for triatomic molecule. (Use dots for electron of oxygen atom and crosses for electrons of hydrogen atoms.)
Answer:

Question 4.
The molecular formula of ammonia is NH₃. Draw electron-dot and line structures for ammonia molecule.
Answer:

Question 5.
The molecular formula of carbon dioxide is CO₂. Draw the electron-dot structure (without showing circle) and line structure for CO₂.
Answer:

Question 6.
With which bond C atom in CO₂ is bonded to each of the O atoms ?
Answer:
In CO₂, carbon atom is bonded to each of the O atoms by double bond.

Question 7.
The molecular formula of sulfur is S₈ in which eight sulphur atoms are bonded to each other to form one ring. Draw electron-dot structure for S₈ without showing the circles.
Answer:

The above S₈ molecule of sulphur has crown shaped structure. One molecule of sulpur is made up of eight atoms of sulphur.

Use your brain power! (Text Book Page No. 113)

Question 1.
Hydrogen peroxide decomposes of its own by the following reaction:
2H – O – O – H → 2H – O – H + O₂
From this, what will be your inference about the strength O – O covalent bond ?
Tell from the above example whether oxygen has catenation power or not.
Answer:
In hydrogen peroxide (H₂O₂), the O – O covalent bond is not strong as oxygen has no catenation power.

Name	Molecular formula	Condensed Structural formula	Number of carbon atoms	Number of -CH₂– units	Boiling point ° C
Ethene	C₂H₄	CH₂ = CH₂	2	0	-102
Propone	C₃H₆	CH₃ – CH = CH₂	3	1	-48
1-Butene	C₄H₈	CH₃ – CH₂ – CH = CH₂	…	…	-6.5
1-Pentene	C₅H₁₀	CH₃ – CH₂ – CH₂ – CH = CH₂	…	…	30

Use your brain power! (Text Book Page No. 120)

Question 1.
The above table shows the homologous series of alkenes. Inspect the molecular formulae of the members of this series. Do you find any relationship, in the number of carbon atoms and the number of hydrogen atoms in the molecular formulae?
Answer:
In the above homologous series, if we observe the molecular formulae of alkenes then the number of carbon atoms are half the number of hydrogen atoms.

Question 2.
If the number of carbon atoms in the molecular formulae of alkenes is denoted by ‘n’ what will be the number of hydrogen atoms?
Answer:
If the number of carbon atoms in the molecular formulae of alkenes is denoted by ‘n’ then the number of hydrogen atoms would be 2n.

Question 3.
What would be the general formula for the molecular formulae of the members of the homologous series of alkanes? What would be the value of ‘n’ for the first member of this series?
Answer:
The general formula for the homologous series of alkane is C_nH_{2n + 2}. The value of ‘n’ for the first member of homologous series is 1.
C_nH_2n+2 = C₁H_{2 × 1} + 2 = CH₄

Question 4.
The general molecular formula for the homologous series of alkynes is C_nH_{2n – 2}. Write down the individual molecular formulae of the first, second and third members by substituting the values 2, 3 and 4 respectively for ‘n’ in this formula.
Answer:
The general molecular formula for the homologous series of alkynes is C_nH_{2n – 2}
n = 2 C₂H_{2 × 2 – 2} = C₂H₂ Ethyne
n = 3 C₃H_{2 × 3 – 2} = C₃H₄ Propyne
n = 4 C₄H_{2 × 4 – 2} = C₄H₆ Butyne

Question 5.
Write down structural formulae of the first four members of the various homologous series formed by making use of the functional groups.
Answer:

Functional group Halo – X (Cl, Br, -I)	Functional group Aldehyde – CHO	Functional group Carboxylic acid – COOH	Functional group Amine -NH₂
CH₃Cl Chloromethane	HCHO Methanal	HCOOH Methanoic acid	CH₃NH₂Methenamine
CH₃ – CH₂ – Cl Chloroethane	CH₃CHO Ethanal	CH₃COOH Ethanoic acid	CH₃CH₂NH₂Ethanamine
CH₃ – CH₂ – CH₂ – Cl 1-Chloropropane	CH₃CH₂CHO Propanal	CH₃CH₂COOH Propanoic acid	CH₃CH₂CH₂NH₂Propanamine
CH₃ – CH₂ – CH₂ CH₂ – Cl 1-Chlorobutane	CH₃CH₂CH₂ CHO Butanal	CH₃CH₂CH₂COOH Butanoic acid	CH₃CH₂CH₂CH₂NH₂Butanamine

Question 6.
General formula of the homologous series of alkanes is C_nH_{2n + 2}. Write down the molecular formula of the 8th and 12th member using this.
Answer:
General formula of alkanes is C_nH_{2n + 2}
n = 8 C₈H_{2 × 8 + 2}= C₆H₁₈ Octane
n = 12 C₁₂H_{2 × 12 + 2} = C₁₂H₂₆ Dodecane

Use your brain power! (Text Book Page No. 121)

Question 1.
Draw three structural formulae having molecular formula C₅H₁₂. Give the names n-pentane, i-pentane and neo-pentane to the above structural formulae.
Answer:

Question 2.
Draw all possible structural formulae having molecular formula C₆H₁₄. Give names to all the isomers.
Answer:

Try This! (Text Book Page No. 124)

Question 1.
Light a bunsen burner. Open and close the air hole at the bottom of the burner by means of the movable ring around it. When do you get yellow sooty flame? When do you get blue flame?
Answer:
When the air hole at the bottom of the burner is open, sufficient oxygen is mixed gaseous fuel for complete combustion and a clean blue flame is obtained. When the air hole is partially blocked by means of the movable ring around it, the air supply is limited which results in incomplete combustion. Hence, yellow sooty flame is produced.

(Text Book Page No. 126)

Question 1.
The names of four fatty acids separated from vegetable oils are given in the table. Identify the number of carbon – carbon double bonds from their structure and molecular formula from the below fatty acids which one when reacts with iodine will make the colour of iodine disappear.
Answer:

Name	Molecular Formula	Number of C = C double bonds	Will it decolorise I₂?
Stearic acid	C₁₇H₃₅COOH	———————–	~~yes~~/no
Oleic acid	C₁₇H₃₃COOH	One double bond	yes/no
Plamitic acid	C₁₅H₃₁COOH	———————–	~~yes~~/no
Linoleic acid	C₁₇H₃₁COOH	Two double bonds	yes/no

Use your brain power! (Text Book Page No. 128)

Question 1.
Explain by writing a reaction, what will happen when pieces of sodium metal are put in n-propyl alcohol.
Answer:
n-Propyl alcohol reacts with pieces of sodium metal, sodium propoxide and hydrogen gas are obtained.

Question 2.
Explain by writing a reaction, which product will be formed on heating n-butyl alcohol with concentrated sulphuric acid.
Answer:
When n-butyl alcohol is heated with concentrated sulphuric acid, one molecule of water is removed from its molecule to form 1-butene.

Use your brain power! (Text Book Page No. 129)

Question 1.
Which one of ethanoic acid and hydrochloric acid is stronger?
Answer:
Hydrochloric acid is stronger acid.

Question 2.
Which indicator paper out of blue litmus paper and pH paper is useful to distinguish between ethanoic acid and hydrochloric acid ?
Answer:
pH paper is useful to distinguish between ethanoic acid and hydrochloric acid.

Use your brain power! (Text Book Page No. 130)

Question 1.
Explain why does the lime water turns milky in the reaction of acetic acid with sodium carbonate.
Answer:
In the reaction of acetic acid with sodium carbonate, carbon dioxide gas is evolved which turns lime water milky resulting in the formation of insoluble calcium carbonate.

Question 2.
Explain the reaction that would take place when a piece of sodium metal is dropped in ethanoic acid.
Answer:
When a piece of sodium metal is dropped in ethanoic acid, sodium acetate and hydrogen gas is formed.

Question 3.
Two test tubes contain two colourless liquids ethanol and ethanoic acid. Explain by writing reaction which chemical test you would perform to tell which substance is present in which test tube.
Answer:
Ethanol does not react with sodium bicarbonate, while ethanoic acid reacts with sodium bicarbonate to form carbon dioxide gas.

Use your brain power! (Text Book Page No. 131)

Question 1.
When fat is heated with sodium hydroxide solution, soap and glycerin are formed. Which functional groups might be present in fat and glycerin? What do you think?
Answer:
The functional group carboxylic acid (-COOH) is present in fat whereas the functioned group hydroxyl group (-OH) is present in glycerin.

Can you tell? (Text Book Page No. 131)

Question 1.
What are the chemical names of the nutrients that we get from the foodstuff, namely, cereals, pulses and meat?
Answer:
The nutrients that we get from the foodstuff, namely cereals, pulses and meat are alpha amino acids.

Question 2.
What are the chemical substances that make cloth, furniture and elastic objects?
Answer:
The chemical substances that make cloth, furniture and elastic objects are cellulose and rubber.

Use your brain power! (Text Book Page No. 133)

Question 1.
Structural formulae of some monomers are given below. Write the structural formula of the homopolymer formed from them.

Answer:

Question 2.
From the given structural formula of polyvinyl acetate, that is used in paints and glues, deduce the name and structural formula of the corresponding monomer.

Answer:

(Text Book Page No. 133)

Question 1.
Complete the following table by writing their Structural formulae and Molecular formulae.
Answer:
(Answer is given directly in bold letters.)

Fill in the gaps in the table: (Text Book Page No. 119)
(Answer is given directly in bold letters.)
a. Homologous series of alkanes.
Answer:

b. Homologous series of alcohol.
Answer:

c. Homologous series of alkenes.
Answer:

(Text Book Page No. 123)

Question 1.
Complete the table by writing the IUPAC names in the third column.
(Answer is directly given with underline.)
Answer:

Common name	Structural formula	IUPAC name
Ethylene	CH₂ = CH₂	Ethene
Acetylene	HC = CH	Ethyne
Acetic acid	CH₃ -COOH	Ethanoic acid
Methyl alcohol	CH₃ – OH	Methanol
Ethyl alcohol	CH₃ – CH₂ – OH	Ethanol

Use your brain power! (Text Book Page No. 119)

Question 1.
By how many -CH₂– (methylene) units do the formulae of the first two members of homologous series of alkanes, methane (CH₄) and ethane (C₂H₆) differ? Similarly, by how many -CH₂– units do the neighbouring members ethane (C₂H₆) and propane (C₃H₈) differ from each other?
Answer:
The first two members of homologous series of alkanes, methane (CH₄) and ethane (C₂H₆) differed by one -CH₂– unit. Similarly, ethane (C₂H₆) and propane (C₃H₈) differed by -CH₂– unit.

Question 2.
How many methylene units are extra in the formula of the fourth member than the third member of the homologous series of alcohols?
Answer:
There is only one, methylene unit extra in the formula of the fourth member and the third member of the homologous series of alcohols.

Question 3.
How many methylene units are less in the formula of the second member than the third member of the homologous series of alkenes?
Answer:
There is only one methylene unit less in the formula of the second member of and the third member of the homologous series of alkenes.

Fill in the blanks and rewrite the completed statements:

Question 1.
The organic compounds having double or triple bonds in them are termed as …………
Answer:
The organic compounds having double or triple bonds in them are termed as unsaturated hydrocarbons.

Question 2.
The general formula of alkane is ……………
Answer:
The general formula of alkane is C_nH_{2n + 2}.

Question 3.
The compounds of homologous series have the same ………….. group.
Answer:
The compounds of homologous series have the same functional group.

Question 4.
A double bond is formed between carbon atoms by ………… pairs of electrons.
Answer:
A double bond is formed between carbon atoms by two pairs of electrons.

Question 5.
The compounds having different structural formulae having the same molecular formula is called ……….
Answer:
The compounds having different structural formulae having the same molecular formula is called structural isomerism.

Question 6.
The functional group of ether is …………..
Answer:
The functional group of ether is -O-.

Question 7.
The general formula of alkene is …………
Answer:
The general formula of alkene is C_nH_2n.

Question 8.
The bond between two atoms of nitrogen is a ………… bond.
Answer:
The bond between two atoms of nitrogen is a triple bond.

Question 9.
Benzene ring is made up of ………….. carbon atoms.
Answer:
Benzene ring is made up of six carbon atoms.

Question 10.
Due to …………., vegetable oil is converted into vanaspati ghee.
Answer:
Due to hydrogenation, vegetable oil is converted into vanatspati ghee.

Question 11.
………….. control the heredity at molecular level.
Answer:
Nucleic acids control the heredity at molecular level.

Question 12.
The regular repetition of a small unit is called …………..
Answer:
The regular repetition of a small unit is called polymer.

Question 13.
The structural formula of polypropylene is ……………….
Answer:
The structural formula of polypropylene is

Question 14.
The monomers of proteins are ……………..
Answer:
The monomers of proteins are alpha amino acids.

Question 15.
The monomer of cellulose is …………
Answer:
The monomer of cellulose is glucose.

Question 16.
…………. have sweet odour.
Answer:
Esters have sweet odour.

Choose the correct alternative and rewrite the statement:

Question 1.
The property of direct bonding between atoms of the same element to form a chain is called ………..
(a) catenation
(b) isomerism
(c) dehydration
(d) polymerization
Answer:
The property of direct bonding between atoms of the same element to form a chain is called catenation.

Question 2.
The molecular weight of two adjacent members in homologous series of an alkane differ by ………. units.
(a) 16
(b) 20
(c) 14
(d) 12
Answer:
The molecular weight of two adjacent members in homologous series of an alkane differ by 14 units.

Question 3.
Consecutive members of a homologous series differ by ………. group.
(a) -CH
(b) -CH2
(C) -CH3
(d) -CH4
Answer:
Consecutive members of a homologous series differ by CH₂ group.

Question 4.
……….. is used to prepare carbon black.
(a) Methane
(b) Ethene
(c) Propane
(d) Butane
Answer:
Methane is used to prepare carbon black.

Question 5.
……….. is the general formula of alkene.
(a) C_nH_2n
(b) C_nH_{2n + 2}
(c) C_nH_{2n – 2}
(d) C_nH_{n – 2}
Answer:
C_nH_2n is the general formula of alkene.

Question 6.
The reaction of methane with chlorine in the presence of sunlight is called ………..
(a) pyrolysis
(b) an elimination reaction
(c) a substitution reaction
(d) an addition reaction
Answer:
The reaction of methane with chlorine in the presence of sunlight is called a substitution reaction.

Question 7.
The general formula for alkynes is ………….
(a) C_nH_2n
(b) C_nH_{2n + 2}
(c) C_nH_{2n – 2}
(d) C_nH_{2n – 1}
Answer:
The general formula for alkynes is C_nH_{2n – 2}

Question 8.
The reaction of ………… with ethanol is a fast reaction.
(a) calcium
(b) magnesium
(c) sodium
(d) aluminum
Answer:
The reaction of sodium with ethanol is a fast reaction.

Question 9.
Ethylene has …………. bond between two carbon atoms.
(a) a single
(b) a double
(c) a triple
(d) an ionic
Answer:
Ethylene has a double bond between two carbon atoms.

Question 10.
The saturated hydrocarbons are those in which carbon atom are linked by ………….
(a) a single bond
(b) a double bond
(c) a triple bond
(d) an ionic bond
Answer:
The saturated hydrocarbons are those in which carbon atom are linked by a single bond.

Question 11.
C₇H₁₆ is ………….
(a) hexane
(b) octane
(c) methane
(d) heptane
Answer:
C₇H₁₆ is heptane.

Question 12.

(a) carboxylic acid group
(b) aldehyde group
(c) ketonic group
(d) alcohol group
Answer:
(a) carboxylic acid group

Question 13.
The possible isomers for C₅H₁₂ are ……………
(a) 2
(b) 4
(c) 1
(d) 3
Answer:
The possible isomers for C₅H₁₂ are 3.

Question 14.
………. contains alcoholic functional group.

Answer:
(d) all of these

Question 15.
Oxygen molecule has ………… bond between two oxygen atoms.
(a) a double
(b) a single
(c) a triple
(d) an ionic
Answer:
Oxygen molecule has a double bond between two oxygen atoms.

Question 16.
Some acetic acid is treated with solid NaHCO₃. The resulting solution will be ………..
(a) colourless
(b) blue
(c) green
(d) yellow
Answer:
Some acetic acid is treated with solid NaHCO₃. The resulting solution will be colourless.

Question 17.
Ethanoic acid has a ……… odour.
(a) rotten eggs
(b) pungent
(c) mild
(d) vinegar-like
Answer:
Ethanoic acid has a vinegar-like odour.

Question 18.
Acetic acid ………..
(a) turns red litmus blue
(b) has pungent odour
(c) is red in colour
(d) is odourless
Answer:
Acetic acid has pungent odour.

Question 19.
When acetic acid reacts with sodium metal ……….. gas is formed.
(a) oxygen
(b) hydrogen
(c) chlorine
(d) nitrogen
Answer:
When acetic acid reacts with sodium metal hydrogen gas is formed.

Question 20.
The molecular formula of acetic acid (ethanoic acid) is …………
(a) HCOOH
(b) CH₃COOH
(c) C₂H₅COOH
(d) C₃H₇COOH
Answer:
The molecular formula of acetic acid (ethanoic acid) is CH₃COOH.

Question 21.
When sodium bicarbonate solution is added to dilute acetic acid …………
(a) a gas is evolved
(b) a solid settles at the bottom
(c) the mixture becomes warm
(d) the colour of the mixture becomes yellow
Answer:
When sodium bicarbonate solution is added to dilute acetic acid a gas-is evolved.

Question 22.
2 ml of ethanoic acid was taken in each of test tubes A, B, C and 2 ml, 4 ml, 6 ml of water was added respectively to them. A clear solution is obtained in ………..
(a) test tube A
(b) test tube B
(c) test tube C
(d) all the test tubes
Answer:
2 ml of ethanoic acid was taken in each of test tubes A, B, C and 2 ml, 4 ml, 6 ml of water was added respectively to them. A clear solution is obtained in all the test tubes.

Question 23.
In the presence of acid catalyst, ethanoic acid reacts with ethanol and ……….. ester is produced.
(a) ethanol
(b) ethanoic
(c) ethyl ethanoate
(d) ethyl ethanol (Practice Activity Sheet – 1)
Answer:
In the presence of acid catalyst, ethanoic acid reacts with ethanol and ethyl ethanoate ester is produced.

Question 24.
The following structural formula belongs to which carbon compound?

(a) Camphor
(b) Benzene
(c) Starch
(d) Glucose (Practical activity sheet- 2)
Answer:
(b) Benzene

Question 25.
What type of reaction is shown below?
\(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \stackrel{\text { Sunlight }}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{Cl}+\mathrm{HCl}\)
(a) Addition
(b) Substitution
(c) Decomposition
(d) Reduction (Practice Activity Sheet – 3)
Answer:
(b) substitution

Question 26.
The carbon compound is used in daily life is ………..
(a) edible oil
(b) salt
(c) carbon dioxide
(d) baking soda (March 2019)
Answer:
The carbon compound is used in daily life is edible oil

State whether the following statements are true or false. (If a statement is false, correct it and rewrite it.):

Question 1.
Generally the melting and boiling points of carbon compounds are high.
Answer:
False. (Generally the melting and boiling points of carbon compounds are low.)

Question 2.
Till now the number of known carbon compounds is about 10 million.
Answer:
True.

Question 3.
Unsaturated hydrocarbons are less reactive than saturated hydrocarbons.
Answer:
False. (Unsatured hydrocarbons are more reactive than saturated hydrocarbons.)

Question 4.
Benzene is an aromatic compound.
Answer:
True.

Question 5.
The carbon-carbon double and triple bonds are also recognised as functional groups.
Answer:
True.

Question 6.
The general formula of alkyne is C_nH_2n.
Answer:
False. (The general formula of alkyne is C_nH_{2n – 2})

Question 7.
Naphthalene burns with a yellow flame.
Answer:
True.

Question 8.
When vegetable oil and tincture iodine react, the color of iodine does not change.
Answer:
False. (When vegetable oil and tincture iodine react, the colour of iodine changes.)

Question 9.
Saturated fats are healthy.
Answer:
False. (Saturated fats are harmful to health.)

Question 10.
Aqueous solution of ethanol is found to be neutral.
Answer:
True.

Question 11.
Denatured ethanol is used as industrial solvent.
Answer:
True.

Question 12.
Vinegar is a 12-15 % aqueous solution of acetic acid.
Answer:
False. (Vinegar is a 5-8 % aqueous solution of acetic acid.)

Question 13.
The functional group of ethanoic acid is a carboxylic group.
Answer:
True.

Question 14.
Sodium hydroxide is used in the preparation of soap from fats and oils.
Answer:
True.

Question 15.
Rubber is a manmade macromolecule.
Answer:
False. (Rubber is a natural macromolecule.)

Question 16.
Polyvinyl chloride is used in the manufacture of P.V.C. pipes and bags.
Answer:
True.

Question 17.
Polyethylene is a homopolymer.
Answer:
True.

Question 18.
The chemical bonds in carbon compounds do not produce ions.
Answer:
True.

Find the odd one out:

Question 1.
Propane, methane, ethene, pentane
Answer:
Ethene. (Others are saturated hydrocarbons.)

Question 2.
Methane, butane, benzene, sodium chloride
Answer:
Sodium chloride. (Others are organic compounds.)

Question 3.
CH₄, C₂H₆, C₃H₈, CaCO₃
Answer:
CaCO₃. (Others are organic compounds.)

Question 4.
C₂H₂, C₃H₈, C₂H₆, CH₄
Answer:
C₂H₂. (Others are saturated hydrocarbons.)

Question 5.
C₂H₄, C₄H₁₀, C₃H₈, CH₄
Answer:
C₂H₄. (Others are saturated hydrocarbons.)

Question 6.
Polyethylene, Polysaccharide, Polystyrene, Polypropylene
Answer:
Polysaccharide (Others are manmade polymers.)

Question 7.
-NH₂, -COOH,-SO₄, -Br
Answer:
-SO₄ (Others are functional groups.)

Question 8.
Methane, Ethane, Propene, Propane, Butane
Answer:
Propene (Others are members of homologous series of alkanes.)

Match the columns:

Question 1.

Column I	Column II
(1) CH₄	(a) CH₂ = CH₂
(2) Ethane	(b) C_nH_{2n –}₂
(3) Alkene	(c) Methane
(4) Alkyne	(d) C₂H₆
	(e) C₃H₈

Answer:
(1) CH₄ – Methane
(2) Ethane – C₂H₆
(3) Alkene – CH₂ = CH₂
(4) Alkyne – C_nH_{2n –}₂.

Question 2.

Column I	Column II
(1) Aromatic hydrocarbon	(a) Propyne
(2) Alkane	(b) Benzene
(3) Alkyne	(c) Saturated hydrocarbon
(4) Alkene	(d) C_nH_2n
	(e) C_n H_{2n – 1}OH

Answer:
(1) Aromatic hydrocarbon – Benzene
(2) Alkane – Saturated hydrocarbon
(3) Alkyne – Propyne
(4) Alkene – C_nH_2n.

Question 3.

Column I	Column II
(1) Cyclohexane	(a) CH₃COOH
(2) Methanol	(b) CH₃Cl
(3) Acetaldehyde	(c) CH₂Cl₂
(4) Ethanoic acid	(d) CH₃OH
	(e) C₆H₁₂
	(f) CH₃CHO

Answer:
(1) Cyclohexane – C₆H₁₂
(2) Methanol – CH₃OH
(3) Acetaldehyde – CH₃CHO
(4) Ethanoic acid – CH₃COOH.

Question 4.

Answer:
(1) (-OH) – Alcohol
(2) (-COOH) – Carboxylic acid
(3) (-CHO) – Aldehyde

Question 5.

Column I	Column II
(1) Ethyne	(a) C₂H₆
(2) Ethene	(b) C₂H₂
(3) Ethane	(c) C₃H₆
(4) Propyne	(d) C₂H₄
	(e) C₃H₄

Answer:
(1) Ethyne – C₂H₂
(2) Ethene – C₂H₄
(3) Ethane – C₂H₆
(4) Propyne – C₃H₄.

Question 6.

Column I	Column II
(1) Cellulose	(a) P.V.C. pipes, bags
(2) R.N.A	(b) Blankets
(3) Polyacrylonitrile	(c) Wood
(4) Polyvinyl chloride	(d) Chromosomes of plants

Answer:
(1) Cellulose – Wood
(2) R.N.A. – Chromosomes of plants
(3) Polyacrylonitrile – Blankets
(4) Polyvinyl chloride – P.V.C. pipes, bags.

Consider the relation between Column I and II. Fill in Column IV to match Column III.

Column I	Column II	Column III	Column IV
(1) Ethylene	Polyethylene	Tetrafluoroethylene	—————–
(2) Polypropylene	Propylene	Polystyrene	—————–
(3) Polysaccharide	Glucose	Proteins	—————–
(4) Rubber	Isoprene	D.N.A.	—————–
(5) Wood	Cellulose	Chromosomes of plants	—————–

Answer:
(1) Teflon
(2) Styrene
(3) Alpha aminoacid
(4) Nucleotide
(5) R.N.A.

Define the following:

Question 1.
Define Alkane
Answer:
Alkane: In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called alkane.
Example: Methane (CH₄), Ethane (C₂H₆)

Question 2.
Define Alkene.
Answer:
Alkene: The unsaturated hydrocarbons containing a carbon-carbon double bond are called alkenes.
Example : Ethene (CH₂ = CH₂)

Question 3.
Define Alkyne.
Answer:
Alkyne: The unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes.
Example: Ethyne C₂H₂ (CH ≡ CH).

Question 4.
Define Addition reaction.
Answer:
Addition reaction: When a carbon compound combines with another compound to form a product that contains all the atoms in both the reactants; it is called an addition reaction.

Question 5.
Define Substitution reaction.
Answer:
Substitution reaction: The reaction in which the place of one type of atom/group in a reactant is taken by another atom/group of atoms, is called substitution reaction.

Question 6.
Define Esterification.
Answer:
Esterification: A carboxylic acid reacts with an alcohol in presence of an acid catalyst, an ester is formed. The reaction is known as esterification.

Question 7.
Define Saponification.
Answer:
Saponification: When an ester reacts with the alkali, i.e. sodium hydroxide, the corresponding alcohol and sodium salt of carboxylic acid are obtained. This reaction is called saponification reaction. It is used in the preparation of soap.
Ester + Sodium hydroxide → Sodium salt of carboxylic acid + Alcohol.

Question 8.
Define Polymerization.
Answer:
Polymerization: The reaction by which monomer molecules are converted into a polymer is called polymerization.

Name the following:

Question 1.
The higher homologue of hexane.
Answer:
Keptane.

Question 2.
The number of double bonds in benzene.
Answer:
Three.

Question 3.
The functional group in ether and halogen.
Answer:
Functional groups:
Ether: – O –
Halogen: – X (-Cl, -Br, -I).

Question 4.
Polymer of tetrafluoroethylene.
Answer:
Teflon.

Question 5.
The monomer of polysaccharide.
Answer:
Glucose.

Question 6.
The Polymer of nucleotide.
Answer:
D.N.A./R.N.A.

Question 7.
The Monomer of rubber.
Answer:
Isoprene.

Question 8.
Two oxidising compounds.
Answer:
Potassium permanganate, Potassium dichromate.

Question 9.
IUPAC name of sodium acetate.
Answer:
Sodium ethanoate.

Question 10.
The main component of natural gas.
Answer:
Methane.

Question 11.
Two isomers of butane.
Answer:
n-butane and i-butane.

Question 12.
A nomenclature system based on the structure of the compounds and it was accepted all over the world.
Answer:
International Union of Pure and Applied Chemistry (IUPAC).

Answer the following questions in one sentence each:

Question 1.
State the atomic number and electronic configuration of carbon.
Answer:
The atomic number of carbon is 6 and the electronic configuration of carbon is (2, 4).

Question 2.
State number of electrons in the outermost orbit of carbon and valency of carbon.
Answer:
Four electrons are present in the outermost orbit of carbon and the valency of carbon is 4.

Question 3.
What are hydrocarbons? Give one example.
Answer:
The compounds containing only carbon and hydrogen are called hydrocarbons. These compounds are known as organic compounds. E.g. Methane, Ethane.

Question 4.
What is the molecular formula and structural of methane?
Answer:
The molecular formula of methane is CH₄.

Question 5.
How many atoms of carbon and hydrogen are present in methane?
Answer:
The molecule of methane has one carbon atom and four hydrogen atoms.

Question 6.
State the general formula of alkane.
Answer:
The general formula of an alkane is C_nH_{2n + 2}.

Question 7.
Give two examples of alkanes.
Answer:
Methane (CH₄) and ethane (C₂H₆) are alkanes.

Question 8.
Give two examples of alkenes.
Answer:
Ethene (CH₂ = CH₂) and propene (CH₃ – CH = CH₂) are alkenes.

Question 9.
Give two examples of alkynes.
Answer:
Ethyne (HC ≡ CH) and propyne (CH₃ – C ≡ CH) are alkynes.

Question 10.
Write the name and molecular formula of a higher homologue of propane.
Answer:
Butane (C₄H₁₀) is a higher homologue of propane.

Question 11.
Write the structure and molecular formula of ethane.
Answer:
Structure of ethane:

Molecular formula of ethane: C₂H₆

Question 12.
What is meant by catenation power?
Answer:
Carbon has a unique ability to form strong covalent bonds with other carbon atoms, this result in formation of big molecules. This property of carbon is called catenation power.

Question 13.
State the structural and molecular formula of benzene.
Answer:

Question 14.
Which functional groups are present in aldehyde and ketone?
Answer:
The functional group -CHO is present in aldehyde and the functional group

is present in ketone.

Question 15.
Which functional group is present in

Answer:

Question 16.
Compare: The proportions of carbon atoms in ethanol (C₂H₅OH) and naphthalene (C₁₀H₈).
Answer:
Ethanol contains two carbon atoms while naphthalene contains 10 carbon atoms. Ethanol is a saturated hydrocarbon and naphthalene is an unsaturated hydrocarbon.

Question 17.
What are the products of combustion of methane?
Answer:
Carbon dioxide (CO₂) and water (H₂O) are the products of combustion of methane.

Question 18.
Which gas is evolved when ethanol reacts with sodium?
Answer:
Hydrogen gas (H₂) is evolved when ethanol reacts with sodium.

Question 19.
Compare: How is the transformation of ethanol into ethanoic acid on oxidation reaction?
Answer:
The transformation of ethanol into ethanoic acid is an oxidation process, in which ethanol accepts oxygen.

Question 20.
Complete the following:

Answer:

Question 21.
Which of the following hydrocarbons undergo addition reactions:
C₂H₆, C₃H₈, C₂H₄, C₃H₈?
Answer:
C₂H₄ (ethene) and C₃H₆ (propene) undergo addition reactions.

Question 22.
How many covalent bonds are there in a molecule of cyclohexane?
Answer:
A molecule of cyclohexane contains 18 covalent bonds.

Question 23.
Give the IUPAC name for CH₃COOH.
Answer:
The IUPAC name for CH₃COOH is ethanoic acid.

Question 24.
Write the IUPAC name of CH₃COONa.
Answer:
IUPAC name of CH₃COONa is sodium ethanoate.

Question 25.
What is meant by denatured alcohol?
Answer:
Ethanol is the important commercial solvent. To prevent the misuse of this solvent, it is mixed with the poisonous methanol. Such ethanol is called denatured spirit.

Question 26.
What is meant by glacial acetic acid ?
Answer:
The melting point of pure acetic acid is 17 °C. Therefore, during winter in old countries acetic acid freezes at room temperature itself and looks like ice. Therefore it is named glacial acetic acid.

Question 26.
Which useful components of hydro¬carbon are obtained by fractional distillation of crude oil?
Answer:
Various useful components of hydrocarbon such as CNG, LPG, petrol (gasoline), rockel, diesel, engine oil, lubricant, etc. are obtained by separation of crude oil using fractional distillation.

Question 28.
Which functional groups are present in ester and amine?
Answer:
Ester: -COO-
Amine: -NH₂–

Question 29.
Give two examples of natural macromolecules.
Answer:
Examples: Polysaccharide, protein and nucleic acid.

Question 30.
Write the structure of polystyrene and give its uses.
Answer:
Structure:

Polystyrene is used to make thermocoal articles.

Question 31.
Write the name and the structure of monomer of polyacrylonitrile.
Answer:
The name and structure of monomer: Acrylonitrile CH₂ = CH – CN

Question 32.
Write the name and the structure of monomer of teflon and its uses.
Answer:
The name and structure of monomer: Tetrafluro ethylene CF₂ = CF₂
Teflon is used to make nonstick cookware.

Question 33.
What is meant by copolymers?
Answer:
The polymers formed from two or more monomers are called copolymers.
Examples: Poly ethylene terephthalate.

Answer the following questions:

Question 1.
How is hydrogen molecule formed?
Answer:

The atomic number of hydrogen is 1, its atom contains 1 electron in K shell. It requires one more electron to complete the K shell and attain the configuration of helium (He). To meet this requirement two hydrogen atoms share their electrons with each other to form H₂ molecule. One covalent bond, i.e. a single bond is formed between two hydrogen atoms by sharing of two electrons.

Question 2.
Describe the formation of oxygen molecule (O₂).
Answer:

(1) The atomic number of oxygen is 8. The electronic configuration of oxygen is (2, 6). Oxygen has 6 electrons in the outermost shell.
(2) It requires 2 electrons to complete the L shell and attain the configuration of neon (Ne).
(3) Each oxygen atom shares its valence electron with the valence electron of another oxygen atom to give two shared pairs of electrons which results in the formation of oxygen molecule.
(4) Thus, two electron pairs are shared between two oxygen atoms, forming double covalent bond (=).

Question 3.
Describe the formation of nitrogen molecule.
Answer:

(1) The atomic number of nitrogen is 7. The electronic configuration of nitrogen is (2, 5). Nitrogen has 5 electrons in the outermost shell.
(2) It requires three more electrons to complete the L shell and attain the configuration of neon (Ne).

(3) Two nitrogen atoms come close together and share three pairs of electrons with each other, resulting in the formation of a triple bond.
(4) Thus, two nitrogen atoms are bound with a triple bond (=) to form a nitrogen molecule.

Question 4.
How is the methane molecule formed?
Answer:
(1) The electronic configuration of carbon is (2, 4). Carbon has four electrons in the outermost shell, hence it is tetravalent.
(2) The electronic configuration of hydrogen is 1, hence it is monovalent.
(3) Carbon needs four electrons to complete the L shell and attain the configuration of neon (Ne).
(4) Four atoms of hydrogen share 1 electron each with 4 electrons of carbon.
(5) A single covalent bond is formed by sharing of two electrons.

Thus, the methane molecule contains four single bonds between the carbon and hydrogen atoms.

Question 5.
State the various compounds and its formulae formed by a single atom of carbon with monovalent hydrogen and chlorine.
Answer:

Compounds	Names
CH₄	Methane
CH₃Cl	Methyl chloride
CH₂Cl₂	Methylene dichloride
CHCl₃	Methylene trichloride
CCl₄	Carbon tetrachloride

Question 6.
Observe the straight chain hydrocarbons given below and answer the following questions:

(i) Which of the straight chain compounds from A and B is saturated and unsaturated straight chains?
(ii) Name these straight chains.
(iii) Write their chemical formulae and number of -CH₂ units. (Practice Activity Sheet – 2)
Answer:
(i) A is a saturated hydrocarbon, B is an unsaturated hydrocarbon.
(ii) A = Propane, B = Propene
(iii) The chemical formula of A = C₃H₈ and number of -CH₂ units are 3.
The chemical formula of B = C₃H₆ and number of -CH₂ unit is 1.

Question 7.
Draw electron-dot and line structure of an ethane molecule.
Answer:
The molecular formula of ethane is C₂H₆.

Question 8.
The molecular formula of propane is C₃H₈. From this draw its structural formula. (Practice Activity Sheet – 3)
Answer:
The structural formula of propane:

Question 9.
Draw the structure and carbon skeleton for cyclohexane.
Answer:

Question 10.
Classify into saturated and unsaturated hydrocarbons: (1) Methane (2) Ethene (3) Ethane (4) Ethyne (5) Propene (6) Propyne (7) Butane (8) Cyclohexene (9) Cyclopentane (10) Heptane.
Answer:
(i) Saturated hydrocarbons: (1) Methane (2) Ethane (3) Butane (4) Cyclopentane (5) Heptane.
(ii) Unsaturated hydrocarbons: (1) Ethene (2) Ethyne (3) Propene (4) Propyne (5) Cyclohexene.

Question 11.
Classify into alkanes, alkenes and alkynes: (1) Ethane (2) Ethene (3) Methane (4) Propene (5) Ethyne (6) Propyne (7) Butane (8) Pentane.
Answer:
Alkanes: (1) Ethane (2) Methane (3) Butane. (4) Pentane
Alkenes: (1) Ethene (2) Propene
Alkynes: (1) Ethyne (2) Propyne

Question 12.
Classify into straight chain carbon compounds, branched chain carbon compounds and ring carbon compounds:
(1) Propene (2) Butane (3) Iso-butane (4) Cyclopentane (5) Benzene (6) Isobutylene.
Answer:
Straight chain carbon compounds:

Propene
Butane.

Branched chain carbon compounds:

Iso-butane
Isobutylene.

Ring carbon compounds:

Cyclopentane
Benzene.

Question 13.
Draw chain and ring structures of organic compound having six carbon atoms in it.
Answer:
Chain structures of an organic compound having six carbon atoms:

Ring structures of an organic compound having six carbon atoms:

Question 14.
Explain the structure of benzene.
Answer:
The molecular formula of benzene is C₆H₆. It is a cyclic unsaturated hydrocarbon. Benzene ring is made of six carbon atoms. In benzene, each carbon atom is linked to two other carbon atoms, on one side by a single bond and on the other side by a double bond, i.e. three alternate single bonds and double bonds in the six membered ring structure of benzene. The compound having this characteristic unit in their structure are called aromatic compounds.

Question 15.
Draw the structures of isomers of pentane (C₅H₁₂).
Answer:

Question 16.
Recognize the carbon chain type for each of the following:

(Practice Activity Sheet – 1)
Answer:
In the reaction of acetic acid with sodium carbonate, carbon dioxide gas is evolved which turns lime water milky resulting in the formation of insoluble calcium carbonate.

Question 17.
What is meant by functional group? Give examples.
(OR)
Explain the term functional group with example.
Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the functional groups.
Example: Methyl alcohol, acetic acid.

In methane (CH₄), when one hydrogen atom is replaced by an -OH group, methyl alcohol (CH₃OH), is formed. The -OH is known as the alcoholic functional group.
Similarly, from methane (CH₄) when one hydrogen atom is replaced by -COOH group, acetic acid (CH₃COOH) is formed. The -COOH group is known as the carboxylic acid functional group.

Question 18.
Define functional group and complete the following table:

Functional group	Compound	Formula
——————–	Ethyl alcohol	——————–
——————–	Acetaldehyde	——————–

Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the
functional groups.

Functional group	Compound	Formula
-OH	Ethyl alcohol	C₂H₅OH
-CHO	Acetaldehyde	CH₃CHO

Question 19.
What is meant by homologous series?
Answer:
The length of the carbon chains in carbon compounds is different their chemical properties are very much similar due to the presence of the same functional group in them. The series of compounds formed by joining the same functional group in place of a particular hydrogen atom on the chains having sequentially increasing length is called homologous series. Two adjacent members of the series differ by only one -CH₂– (methylene) unit and their mass differ by 14 units.

The homologous series of straight chain alkanes can be represented by the general formula CnH_{2n + 2}. The members of this series are as follows:

Methane – CH₄Ethane – C₂H₆	– These differ by – CH₂ units
Ethane – C₂H₆Propane – C₃H₈	– These differ by – CH₂ units
Butane – C₄H₁₀Pentane – C₅H₁₂	– These differ by – CH₂ units

Question 20.
State the four characteristics of homologous series.
Answer:
Characteristics of Homologous series:
(1) In homologous series while going in an increasing order of the length of carbon chain
(a) one methylene unit ( -CH₂– ) gets added
(b) molecular mass increases by 14 u (c) number of carbon atoms increases by one.
(2) Chemical properties of members of a homologous series show similarity due to the presence of the same functional group in them.
(3) Each member of the homologous series can be represented by the same general molecular formula.
(4) While going in an increasing order of the length there is gradation in the physical properties i.e. the boiling and melting points.

Question 21.
Write names of first four homologous series of alcohols: (March 2019)

Answer:
First four homologous series of alcohols are

Methanol CH₃ – OH
Ethanol C₂H₅ – OH
Propanol C₃H₇ – OH
Butanol C₄H₉ – OH

Question 22.
Describe the IUPAC rules of naming organic compounds.
Answer:
IUPAC nomenclature system: International Union for Pure and Applied Chemistry (IUPAC) put forth a nomenclature system based on the structure of the compounds and it was accepted all over the world. There are three units in the IUPAC name of any carbon compound: parent, suffix and prefix. These are arranged in the name as follows:

Prefix-parent-suffix:
An IUPAC name is given to a compound on the basis of the name of its parent alkane. The name of the compound is constructed by attaching appropriate suffix and prefix to the name of the parent-alkane. The steps in the IUPAC nomenclature of straight chain compounds are as follows:

Step 1: Draw the structural formula of the straight chain compound and count the number of carbon atoms in it. The alkane with the same number of carbon atoms is the parent alkane of the concerned compound. Write the name of this alkane.

In case the carbon chain of concerned compound contains a double bond, change the ending of the parent name from ‘ane’ to ‘ene’. If the carbon chain in the concerned compound contains a triple bond, change the ending of the parent name from ‘ane’ to ‘yne’.

Sr. No.	Structural formula	Straight chain	Parent name
1.	CH₃– CH₂– CH₃	C – C – C	propane
2.	CH₃– CH₂– OH	C – C	ethane
3.	CH₃– CH₂– COOH	C – C – C	propane
4.	CH₃– CH₂– CH₂– CHO	C – C – C – C	butane
5.	CH₃– CH = CH₂	C – C = C	propene
6.	CH₃– C ≡ CH	C – C ≡ C	propyne

Step 2: If the structural formula contains a functional group, replace the last letter ‘e’ from the parent name by the condensed name of the functional group as the suffix. (Exception: The condensed name of the functional group ‘halogen’ is always attached as the prefix.)

Step 3: Number the carbon atoms in the carbon chain from one end to the other. Assign the number T to carbon in the functional group -CHO or -COOH, if present. Otherwise, the chain can be numbered in two directions. Accept that numbering which gives smaller number to the carbon carrying the functional group. In the final name, a digit (number) and a character (letter) should be separated by a small horizontal line.

Question 23.
Write the IUPAC names of the following structural formulae.
a. CH₃ – CH₂ – CH = CH₂
Answer:
Number of carbon atoms in the longest chain: 4
Parent alkane: Butene
Functional group: double bond
Assign the number:

In the longest chain, the numbering of carbon atom starts from the carbon atom nearest to the double bond and the other c-atoms are numbered accordingly.
Parent suffix: But-1-ene
IUPAC name: But-1-ene

b. CH₃ – C ≡ C – H
Answer:
Number of carbon atoms in the longest chain: 3
Parent alkane: Propyne
Functional group: triple bond
Parent suffix: Propyne
IUPAC name: Propyne

c.

Answer:
The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Prefix functional group: Chloro
Assign the number: 2

The carbon atom to which the -Cl atom is attached is numbered as C₂ and the other C atoms are numbered accordingly.
Prefix parent: 2-Chloropentane
IUPAC name: 2-Chloropentane

d. CH₃ – CH₂ – CH₂ – Br
Answer:
The number of carbon atoms in the longest chain: 3
Parent alkane : Propane Prefix functional group: Bromo
Assign the number:

The carbon atom to which the -Br atom is attached is numbered as C₁ and the other C atoms are numbered accordingly.
Prefix parent: 1-Bromopropane
IUPAC name: 1-Bromopropane

e.

Answer:
The number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -OH (ol)
Assign the number:

The carbon atom to which the -OH group is attached is numbered as C₂.
If the carbon chain of the compound contains a -OH group, then change the ending ‘e’ of the parent name, i.e. ,‘e’ of butane is replaced by ‘ol’ (ol for alcohol).
Parent suffix: Butan-2-ol
IUPAC name: Butan-2-ol

f.

The number of carbon atoms: 3
Parent alkane: Propane
Functional group: -NH₂ (amine)
Assign the number:

If the carbon chain of the compound contains a -NH2 group then change the ending of the parent name, i.e. ‘e’ of propane is replaced by ‘amine’.
Parent suffix: 2-Propanamine
IUPAC name: 2-Propanamine

g. HCOOH
Answer:
The number of carbon atoms: 1
Parent alkane: Methane
Functional group: -COOH (-oic cid)
If the carbon chain of the compound contains a -COOH group, then change the ending of the parent name, i.e. ‘e’ of methane is replaced by ‘-oic acid’.
Parent suffix: Methanoic acid
IUPAC name: Methanoic acid

h. CH₃ – CH₂ – CH₂ – CHO
Answer:
The number of carbon atoms in the longest chain: 4
Parent alkane : Butane Functional group: -CHO (al)
Assign the number: 1
Assign the number ‘1’ to carbon in the functional group

If the carbon chain of the compound contains a -CHO group then change the ending of the parent name, i.e. ‘e’ of the butane is replaced by ‘al’.
Parent suffix: Butanal
IUPAC name: Butanal

i.

Answer:
The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Functional group:

Assign the numbering:

In the longest chain, the numbering of carbon atom starts from the carbon nearest to the functional group (both the numbering equivalent).
If the carbon chain of compound contains a

group, then change the ending of the parent name i.e. ‘e’ of pentane is replaced by ‘one’.
Parent suffix: Pentan-3-one
IUPAC name: Pentan-3-one.

Question 24.
What happens when methane is burnt in air? Write the balanced chemical equation for the same.
Answer:
When methane burns in air, carbon dioxide and water are formed. The reaction is exothermic with release of large amount of heat and light.

Question 25.
What happens when ethanol is burnt in air?
Answer:
When ethanol is burnt in air, it burns with a clean blue flame, carbon dioxide and water are formed. In this reaction, release of large amount of heat and light takes place.

Question 26.
What happens when ethanol is treated with alkaline potassium permanganate? Write the balanced chemical equation for the same.
Answer:
When ethanol is treated with alkaline potassium permanganate, ethanol gets oxidised by alkaline potassium permanganate to’ form ethanoic acid.

Question 27.
What happens when vegetable oil is hydrogenated? Write the balanced chemical equation.
Answer:
When vegetable oil (unsaturated compound) is hydrogenated in the presence of nickel catalyst, vanaspati ghee (saturated) compound is formed.

Question 28.
What happens when chlorine is treated with methane?
(OR)
Describe the action of chlorine on methane.
(OR)
Write a note on chlorination of methane.
Answer:
Methane reacts rapidly with chlorine in the presence of sunlight to form four products. In this reaction, chlorine atoms replace, one by one, all the hydrogen atoms in the methane.

The reaction in which the place of one type of atom/group in a reactant is taken by another atom/group of atoms is called substitution reaction. Chlorination of methane is a substitution reaction.

Question 29.
What happens when ethanol is reacted with sodium?
Answer:
When ethanol is reacted with sodium at room temperature, sodium ethoxide is formed and hydrogen gas is liberated.

Question 30.
What happens when ethanol is heated at 170 °C with excess of conc. sulphuric acid?
Answer:
When ethanol is heated at 170 °C with excess of conc. sulphuric acid, one molecule of water is removed from its molecule to form ethene (unsaturated compound).

Question 31.
What happens when ethanoic acid is treated with sodium hydroxide? Write the balanced equation for the same.
Answer:
When ethanoic acid is treated with sodium hydroxide, neutralization takes place to form sodium acetate (sodium ethanoate) and water.

Question 32.
What happens when ethanoic acid is treated with sodium carbonate?
Answer:
When ethanoic acid is treated with sodium carbonate, sodium ethanoate, carbon dioxide and water is formed.

Question 33.
What happens when ethanoic acid is treated with sodium bicarbonate?
Answer:
When ethanoic acid is treated with sodium bicarbonate, sodium ethanoate, water and carbon dioxide is formed.

Question 34.
What happens when ethanoic acid is treated with ethanol? Write the balanced equation for the same.
Answer:
When ethanoic acid is treated with ethanol in the presence of an acid catalyst, an ester, i.e., ethyl ethanoate is formed.

Question 35.
What happens when ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst?
Ans. When ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst, it polymerizes to form polyethylene or polythene (plastic).

Question 36.
State the physical properties of ethyl alcohol ethanol.
Answer:

Ethanol is a colourless liquid and it is soluble in water in all proportions and has pleasant odour.
The boiling point of ethanol is 78 °C and the freezing point is -114 °C.
It is combustible and burns with a blue flame.
An aqueous solution of ethanol is neutral to litmus paper.

Question 37.
State the properties of ethanoic acid.
Answer:

Ethanoic acid is a colourless liquid with boiling point 118 °C and melting point 17 °C. It has a pungent odour.
Its aqueous solution is acidic and turns blue litmus red.
A 5-8% aqueous solution of acetic acid is used as vinegar.
It is a weak acid.

Write short notes:

Question 1.
Catenation power.
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms; this results in formation of big molecules. This property of carbon is called catenation power.

(2) Carbon shows catenation. Two or more carbon atoms can share their valence electrons and bond with each other. Thus, carbon chains can be straight or branched or closed chain ring structure forming large molecules. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

(3) Hence, carbon atoms can form an unlimited number of compounds.

Question 2.
Characteristics of Carbon.
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms: this results in formation of big molecules. This property of carbon is called catenation power. The carbon compounds contain open chains or closed chains of carbon atoms. An open chain can be a straight chain or a branched chain. A closed chain is a ring structure. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

Question 3.
Functional group.
Answer:
(1) The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of the length and nature of the carbon chain in that compound. Therefore these hetero atoms or the groups of atoms containing hetero atoms are called functional groups.

All organic compounds are derivatives of hydrocarbons. The derivatives are formed by replacing one or more H-atom/atoms of hydrocarbon by some other hetero atom or groups of atoms containing hetero atoms. After replacement, a new compound is formed which has properties different from the parent hydrocarbon.

Examples: For methane, if one hydrogen atom is replaced by an – OH group, then a compound is methyl alcohol (CH₃OH). The -OH group is known as the alcoholic functional group.
Functional group is organic compound:
1. Alcohol: – OH (hydroxy group)

4. Carboxylic acid : -COOH

Question 4.
Homologous series.
Answer:
The length of the carbon chains in carbon compounds is different their chemical properties are very much similar due to the presence of the same functional group in them. The series of compounds formed by joining the same functional group in place of a particular hydrogen atom on the chains having sequentially increasing length is called homologous series. Two adjacent members of the series differ by only one -CH₂– (methylene) unit and their mass differ by 14 units.

The homologous series of straight chain alkanes can be represented by the general formula C_nH_{2n + 2}
The members of this series are as follows:

Methane – CH₄Ethane – C₂H₆	– These differ by – CH₂ units
Ethane – C₂H₆Propane – C₃H₈	– These differ by – CH₂ units
Butane – C₄H₁₀Pentane – C₅H₁₂	– These differ by – CH₂ units

Characteristics:
(1) In homologous series while going in an increasing order of the length of carbon chain
(a) one methylene unit (-CH₂-) gets added
(b) molecular mass increases by 14 u
(c) number of carbon atoms increases by one.
(2) Chemical properties of members of a homologous series show similarity due to the presence of the same functional group in them.
(3) Each member of the homologous series can be represented by the same general molecular formula.
(4) while going in an increasing order of the length there is gradation in the physical properties i.e., the boiling and melting points.

Question 5.
Polymerization.
Answer:
(1) The reaction by which monomer molecules are converted into a polymer is called polymerization. A macromolecule formed by regular repetition of a small unit is called polymer. The small unit that repeats regularly to form a polymer is called monomer. The important method of polymerization is to make a polymer by joining alkene type of monomers.

(2) When ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst, it polymerizes to form polyethylene or polythene (plastic).

(3) The polymer polystyrene is used to make thermocoal articles. The polymer polyvinyl chloride is used to make P.V.C. pipes, doormats, etc. The polymer teflon is used to make nonstick cookware. The polymer polypropylene is used to make injection syringe, furniture, etc.

Give scientific reasons:

Question 1.
Carbon atoms are capable of forming an unlimited number of compounds.
Answer:

Carbon has the property of catenation. Two or more carbon atoms can share some of their valence electrons to form (single, double and triple) bonds.
The straight chains or branched chains or rings may have different shapes and sizes. This results in formation of many compounds. Hence, carbon atoms are capable of forming an unlimited number of compounds.

Question 2.
Ethylene is an unsaturated hydrocarbon.
Answer:
(1) Ethylene (CH₂ = CH₂) contains a double bond between carbon atoms.
(2) Thus, the valencies of the two carbon atoms are not fully satisfied by single covalent bonds. Hence, ethylene is an unsaturated hydrocarbon.

Question 3.
Naphthalene burns with a yellow flame.
Answer:
(1) Naphthalene is an unsaturated compound. In unsaturated hydrocarbon the proportion of carbon is larger than that of saturated hydrocarbon. As a result, some unburnt carbon particles are also formed during combustion of unsaturated compounds.

(2) In the flame. these unburnt hot carbon particles emit yellow light and therefore the flame appears yellow. Hence, naphthalene burns with a yellow flame.

Question 4.
The colour of iodine disappears in the reaction between vegetable oil and iodine.
Answer:
(1) Vegetable oils (unsaturated compound) contains a multiple bond as their functional group. They undergo addition reaction to form a saturated compound as the product.
(2) The addition reaction of vegetable oil with iodine takes place instantaneously at room temperature. The colour of iodine disappears in this reaction. This iodine test indicates the presence of a multiple bond in vegetable oil.

Question 5.
The hydrogenation of vegetable oil in the presence of nickel catalyst forms vanaspati ghee.
Answer:
(1) The molecules of vegetable oil contain long and unsaturated carbon chains. These unsaturated hydrocarbons contain a multiple bond as their functional group. They undergo addition reaction to form a saturated compound as the product.

(2) When vegetable oil (unsaturated compound) is hydrogenated in the presence of nickel catalyst, the addition reaction takes place, vanaspati ghee (saturated compound) is formed.

Distinguish between the following:

Question 1.
Saturated hydrocarbons and Unsaturated hydrocarbons.
Answer:
Saturated hydrocarbons:

In saturated hydrocarbons, the carbon atoms are linked to each other only by single covalent bonds.
They contain only a single bond.
They are chemically less reactive.
Substitution reaction is a characteristic property of these hydrocarbons.
Their general formula is C_nH_{2n + 2}.

Unsaturated hydrocarbons:

In unsaturated hydrocarbons, the valencies of carbon atoms are not fully satisfied by single covalent bonds.
They contain carbon to carbon double or triple bonds.
They are chemically more reactive.
Addition reaction is a characteristic property of these hydrocarbons.
Their general formula is C_nH_2nor C_nH_{2n – 2}

Question 2.
Open chain hydrocarbons and closed chain hydrocarbons.
Answer:
Open chain hydrocarbons:

A hydrocarbon in which the chain of carbon atoms is not cyclic is called an open chain hydrocarbon.
All aliphatic hydrocarbons contain open chains.

Closed chain hydrocarbons:

A hydrocarbon in which the chain of carbon atoms is present in a cyclic form or ring form is called a closed chain hydrocarbon.
All aromatic hydrocarbons contain closed chains.

Question 3.
Alkane and Alkene.
Answer:
Alkane

Alkanes in which the carbon atoms are linked to each other only by single bonds.
The general formula of an alkane is C_nH_{2n + 2}
They are chemically less reactive.

Alkene:

Alkenes in which carbon atoms are linked to each other by double bonds.
The general formula of an alkene is C_nH_2n.
They are chemically more reactive.

Project:

Question 1.
Prepare a list of carbon compounds which occur in nature and discuss their uses in daily life.

10th Std Science Part 1 Questions And Answers:

Std 8 History Chapter 11 Questions And Answers Struggle for Equality Maharashtra Board

February 6, 2025January 5, 2025 by Bhagya

Balbharti Maharashtra State Board Class 8 History Solutions Chapter 11 Struggle for Equality Notes, Textbook Exercise Important Questions and Answers.

Class 8 History Chapter 11 Struggle for Equality Questions And Answers Maharashtra Board

Struggle for Equality Class 8 Questions And Answers Chapter 11 Maharashtra Board

Class 8 History Chapter 11 Struggle for Equality Textbook Questions and Answers

1. Rewrite the statements by- choosing the appropriate options:
(Lala Lajpat Rai, Sane Guruji, Dr. Rakhmabai Janardan Save, Deenbandhu Mitra, Babasaheb Bole)

Question 1.
…………….. founded the Red Cross Society at Rajkot.
Answer:
Dr. Rakhmabai Janardan Save

Question 2.
…………….. was the president of mill workers union at Ammalner.
Answer:
Sane Guruji

Question 3.
The president of the first session of AITUC was ……………..
Answer:
Lala Lajpat Rai

2. Write short notes:

Question 1.
Social Work of Maharshi Vitthal Ramji Shinde :
Answer:

The goal of Maharshi Vitthal Ramji Shinde was to make the Dalits self respectful, well educated and engaged in work.
He worked to destroy the delusive ideas regarding Dalits in the minds of the upper caste.
He founded the ‘Depressed Classes Mission’ for the progress of Dalits in 1906.
To achieve this, he started Marathi schools and work schools in parts of Parel, Deonar in Mumbai.
He actively took part in various movement which were related to the welfare of the Dalits like Satyagraha for entry in Parvati temple at Pune, Shetkari Parishad of Dalits and Federal Electorates.

Question 2.
Reforms of Rajarshi Shahu Maharaj in the state of Kolhapur :
Answer:

Rajarshi Shahu Maharaj worked for abolishing of caste discrimination.
He made revolutionary declaration for reservation in the state of Kolhapur.
He made law for free and compulsory primary education.
He worked to remove restrictions on inter dining, intermarriage and change of occupation in the caste system.
He passed the Act of inter-caste marriage and made it legal in state.
He abolished ‘Balutedari System’, by publishing a declaration in the Government Gazette of the state of Kolhapur on 22 February, 1918.
This allowed people to practice any occupation and freed them from type of social slavery.

3. Explain the following statements with reasons:

Question 1.
The Government decided to crush down the Communist movement.
Answer:

The Communist Party was formed in India in 1925.
The young communist leaders started building militant organisations of workers and peasants.
The British government started feeling the danger of Communist movement and hence decide to crush it.

Question 2.
Dr. Babasaheb Ambedkar started newspapers like ‘Muknayak’, ‘Bahishkrut Bharat’, etc.
Answer:

Dr. Babasaheb Ambedkar wanted to establish a society based on principles of Liberty, Equality and Fraternity.
He intended to carry out movement to establish self-respect and social equality for Dalits.
It was necessary to create awakening in society in this respect and to voice their grief.
The newspaper was an integral part of Dr. Babasaheb Ambedkar’s movement.
Hence, he started many newspapers so that his ideas and thoughts reach the society.

Question 3.
There aroused a need of a nationwide workers union.
Answer:

In the later half of the 19th century, textile mills, railway companies and such other industries were started in India.
The workers group were not aroused on large scale.
As there were no unions, the workers were not organised.
After the First World War, there was rise of worker class in India due to industrialisation.

Therefore, a necessity was felt for a nationwide worker union.

4. Answer the following questions in brief:

Question 1.
Why was the struggle for equality important in the making of Modern India?
Answer:

India achieved freedom because of freedom struggle.
The scope of this struggle was not limited to achieve political freedom but emancipation of man from all types of bondages.
Therefore in the course of this struggle, there was opposition to feudalism, social inequality and economic exploitation.
The principle of equality was given equal importance.
Various groups such as farmers, workers, women and Dalits came forward and started movements in order to get justice and bring about reforms.

This proves that the struggle for equality has great significance in the making of Modern India.

Question 2.
Write about the work of Sane Guruji in eastern Khandesh.
Answer:

In 1938, due to heavy rains crops were destroyed in eastern Khandesh.
He organised the farmers who were in miserable state.
Sane Guruji organised meetings and processions at many places in eastern Khandesh to have land revenue waived for the farmers.
He took out marches on the Collector office.
He tried to create a strong centre of workers union at Dhule-Ammalner.

Question 3.
How was the struggle built up by the workers decisive for the national movement?
Answer:

The beginning of the 19th century witnessed the emergence of working-class due to industrialisation. They made efforts to solve their problems.
During the anti-partition movement, workers carried out strikes from time to time to pressurise the British government.
The tea plantation workers in Assam launched an agitation against their wretched condition.
During the anti-partition movement, workers carried out strikes from time to time in support of Swadeshi.
In 1928, the Mill Workers Union went on strike for six months. Many such strikes were organised by the Railway workers, jute mill workers, etc.

Question 4.
Discuss the nature of reform movement related to women.
Answer:

Women had a secondary position in the Indian social system.
They were subjected to many injustices because of evil practices.
In the 20th century, many reforms were initiated for the betterment of women.
The reform movement was led by women and institutions formed by them.
They fought for issues such as the right to inheritance, right to vote through the medium of these institutions.
The involvement of women went on increasing. They played an active role in the national movement and in the revolutionary movement.
Due to the reform movement, women were included in the Provincial Ministries.

Do you Know?

Narayan Meghaji Lokhande:

Maharashtra Board Class 8 History Solutions Chapter 11 Struggle for Equality 1

The native place of Narayan Meghaji Lokhande was Kanhesar near Saswad in Pune district.

He formed the mill workers un ion known as ‘Bombay Mill Hands Association’ in 1890.
This workers union is believed to be the beginning of organised movement in India.
He was also the chairman of Mumbai branch of Satyashodhak Samaj founded by Mahatma Phule.
Due to his efforts, from 10 June 1890, the workers started getting weekly holiday on Sunday.

Dr. Anandibai Joshi:

Maharashtra Board Class 8 History Solutions Chapter 11 Struggle for Equality 2

She was the first Indian Female Doctor.
The death of her 10 days old son inspired her to study medicine and acquired M. D. degree in 1886.
She died of tuberculosis on February 1887 at Pune.

Project:

Question 1.
Read the biography of Dr. Anandibai Joshi.

Question 2.
Read the biography of Rajarshi Shahu Maharaj.

Class 8 History Chapter 11 Struggle for Equality Additional Important Questions and Answers

Rewrite the statements by- choosing the appropriate options :

Question 1.
……………. got a bill passed in the Bombay Provincial Assembly for the public water reservoirs to be opened to the untouchable.
Answer:
Babasaheb Bole

Question 2.
Neel Darpan written by …………….. brought to notice of the society wretched conditions of the peasants.
Answer:
Deenbandhu Mitra.

Name the following:

Question 1.
Through the magazine ‘Somavanshiya Mitra’ raised voice regarding issues of Muralis and Jogtins.
Answer:
Shivram Janbci Kamble

Question 2.
Chairman of Mumbai branch of Satyashodhak Samaj.
Answer:
Narayan Meghaji Lokhande

Question 3.
Movement of eradication of untouchability in Tamil Nadu
Answer:
Perriyar Ramaswamy

Question 4.
Active in international Communist Movement.
Answer:
Manavendranath Roy

Question 5.
Baba Ramachandra.
Answer:
Baba Ramchandra.

Rewrite the statements by choosing the appropriate options:

Question 1.
…………….. was the president of ‘Akhil Bharatiya Kisan Sabha’
(a) Sane Guruji
(b) Swami Sahajananda Saraswati
(c) Maharshi Vitthal Ramji Shinde
(d) Manavendranath Roy
Answer:
(b) Swami Sahajananda Saraswati

Question 2.
The session of the Congress was held in the rural part of …………… in Maharashtra.
(a) Nagpur
(b) Baramati
(c) Faizpur
(d) Miraj
Answer:
(c) Faizpur

Question 3.
……………… went on fast unto death to open the doors of Vitthal temple at Pandharpur for the Dalits,
(a) Sane Guruji
(b) Dr. Babasaheb Ambedkar
(c) Karmaveer Dadasaheb Gaikwad
(d) Babasaheb Bole
Answer:
(a) Sane Guruji

Question 4.
Due to efforts of …………….., the workers started getting weekly holiday on Sunday from 10 June, 1890.
(a) Shripad Amrut Dange
(b) Shashipad Banerjee
(c) Lala Lajpat Rai
(d) Narayan Meghaji Lokhande
Answer:
(d) Narayan Meghaji Lokhande

Question 5.
In 1881, …………….. wrote an article on Marx.
(a) Dr. Rammanohar Lohia
(b) Lokmanya Tilak
(c) Mahatma Gandhi
(d) Shripad Amrut Dange
Answer:
(b) Lokmanya Tilak

Question 6.
Condemned untouchability through his book ‘Vital Vidhwansan’
(a) Gopal Baba Walangkar
(b) Shivram Janba Kamble
(c) Thakkar Bappa
(d) Appasaheb Patwardhan
Answer:
(a) Gopal Baba Walangkar

Question 7.
Karmaveer Dadasaheb Gaikwad led Satyagraha for entry of Dalits in
(a) Kalaram temple
(b) Vitthal temple
(c) Parvati temple
(d) Bhavani temple
Answer:
(a) Kalaram temple

Question 8.
…………….. passed the Act of inter-caste marriage and gave it a legal acceptance in the state of Kolhapur.
(a) Mahatma Jyotirao Phule
(b) Rajarshi Shahu Maharaj
(c) Rajaram Maharaj
(d) Maharshi Vitthal Ramji Shinde
Answer:
(b) Rajarshi Shahu Maharaj

Answer the following questions in one sentence:

Question 1.
State the significance of ‘Neel Darpan’
Answer:
Deenbandhu Mitra in his play ‘Neel Darpan’ brought to the notice of the society the wretched conditions of the peasants who were forced to cultivate indigo.

Question 2.
What was the important work done by ‘Akhil Bharatiya Kisan Sabha’?
Answer:
The Akhil Bharatiya Kisan Sabha’ presented a declaration of peasants’ rights to the Indian National Congress.

Question 3.
What message did Lala Lajpat Rai give to the workers in the first session of AITUC?
Answer:
At the first session of AITUC Lala Lajpat Rai asked the workers to actively participate in the national movement.

Question 4.
Why did Dr. Ambedkar and his followers embrace Buddhism?
Answer:
Dr. Ambedkar and his followers embraced Buddhism as it advocated humanity and equality. ‘

Question 5.
Which institutions were founded by Pandita Ramabai?
Answer:
Pandita Ramabai had founded the ‘Arya Mahila Samaj’ and ‘Sharda Sadan’.

Question 6.
What message did Dr. Babasaheb Ambedkar give his followers?
Answer:
Dr. Babasaheb Ambedkar gave the inspirational message,’Be Educated, Be Organised and Be Agitated’ to his followers.

Do as Directed:

Complete the Concept Map:

Question 1.
Maharashtra Board Class 8 History Solutions Chapter 11 Struggle for Equality 3
Answer:

Question 2.
Maharashtra Board Class 8 History Solutions Chapter 11 Struggle for Equality 5
Answer:

Question 3.
Maharashtra Board Class 8 History Solutions Chapter 11 Struggle for Equality 7
Answer:

Complete the following table:

Question 1.

Founder	Organisation/ Institution
1. Maharshi Vitthal Ramji Shinde	…………………………………………
2. ……………………………….	Akhil Bharatiya Kisan Sabha
3. Ramabai Ranade	…………………………………………
4. ………………………………..	Independent Labour Party

Answer:

Founder	Organisation/ Institution
1. Maharshi Vitthal Ramji Shinde	Depressed Classes Mission
2. Prof. N.G. Ranga	Akhil Bharatiya Kisan Sabha
3. Ramabai Ranade	Seva Sadan institution
4. Dr. Babasaheb Ambedkar	Independent Labour Party

Write short notes:

Question 1.
The Meerut Conspiracy Case:
Answer:

The young communist leaders built militant organisations of workers and peasants.
The growing influence of Communism on Indian youth became a matter of concern for the British and they felt it was a danger.
The communist leaders Shripad Amrut Dange, Muzaffar Ahmed, Keshav Neelkanth Joglekar, etc. were arrested.
They were falsely charged with planning of a conspiracy to overthrow the British rule.
The trial took place at Meerut. So it was called as ‘Meerut Conspiracy Case’.
They were given different punishments.

Question 2.
Work of Rakhmabai Janardan Save:
Answer:

Dr. Rakhmabai Janardhan Save was the first practicing woman doctor in India.
She delivered lectures on health related issues of women.
She opened a branch of Red Cross Society at Rajkot.

Explain the following statements with reasons:

Question 1.
Narayan Meghaji Lokhande is described as ‘Father of Indian Workers Movement’.
Answer:

Industrialisation of India had started during the second half of the 19th century.
Shashipad Banerjee and Narayan Meghaji Lokhande organised workers at local level and made efforts to solve their problems.
Narayan Meghaji Lokhande formed the mill workers union known as ‘Bombay Mill Hands Association’ in 1890.
Due to his efforts, the workers started getting weekly holiday on Sunday from 10 June, 1890.
This was the beginning of organised movement in India.

Thus, his contribution to the working-class movement was so valuable that he is described as ‘Father of Indian Workers Movement’.

Question 2.
The Congress Socialist Party was formed in 1934.
Answer:

Many young activists of the Indian National Congress believed that the society should be built on the principles of economic and social equality.
Due to this, there was spread of socialist ideas and thoughts.
They believed that to protect interests of the people it was necessary to overthrow the British rule.
So, they decided to set up a Socialist Party within the Congress fold while they were in Nashik jail. Accordingly, the Congress Socialist Party was formed in 1934.

Answer the following questions in brief:

Question 1.
Write about the work of Rajarshi Shahu Maharaj for abolition of caste distinction.
Answer:
The following work was done by Rajarshi Shahu Maharaj in his state of Kolhapur for abolition of caste distinction:

He made revolutionary declaration for reservation in state of Kolhapur.
There were three restrictions in the caste system: inter dining, inter marriage and change of occupation.
Shahu Maharaj ate food from the hands of Dalit people and overthrew the restriction on inter dining.
He passed the Act of inter caste marriage and gave it a legal acceptance in his state.
He abolished ‘Balutedari System’ on 22 February, 1918 and granted permission to practice any occupation by anyone.

Question 2.
Explain the objectives of Dr. Babasaheb Ambedkar in launching the Dalit movement.
Answer:
The objectives of Dr. Babasaheb Ambedkar in launching the Dalit movement 5 is as follows:

Dr. Babasaheb Ambedkar aimed at building a society on the principles of Liberty, Equality and Fraternity.
He was of the view that unless the caste system is completely uprooted, the x injustice to the Dalits and inequality would x not end.
According to him, social equality is the right of Dalits.
He intended to carry out the movement based on self-respect.

Question 3.
Write about the work done by Mahatma Gandhi for eradication of untouchability.
Answer:

Mahatma Gandhi took up the issue of eradication of untouchability in his hands and put it up on the platform of Indian National Congress.
When in Yerwada jail, he debated with rigid Hindu Pandits stating that the religious texts of the Hindus do not support untouchability.
He gave inspiration to Harijan Sevak Sangh.
Many took inspiration from him and dedicated themselves to the work for 8 eradication of untouchability.

Answer the following in detail:

Question 1.
Give a brief account of the peasant’s movement during the pre-independence movement.
Answer:

The economic policies of the British protected the landlords and the moneylenders and adversely affected the farmers. The peasants revolted against the injustice in different parts of India.
The peasants in Bengal formed union and revolted against the compulsion to cultivate indigo.
The peasants in Maharashtra revolted in 1875 against the atrocities of the landlords and moneylenders.
In 1918, ‘Kisan Sabha’ was formed by the farmers in Uttar Pradesh.
The Mopla peasants in Kerala rose in great revolt.
With the initiative of Prof N.G. Ranga Akhil Bharatiya Kisan Sabha’ was established.
It presented a declaration of peasants rights to the Indian National Congress.
Thousands of peasants attended the session of Indian National Congress which was held in rural part of Faizpur.
In 1938, due to heavy rains crops were destroyed in eastern Khandesh. Sane Guruji organised the farmers who were in a miserable state and took out procession.
The peasants participated in large numbers in the revolutionary period of 1942.

Question 2.
Write information about the workers’ movement in the pre-independence movement.
Answer:

The beginning of the 19th century witnessed the emergence of the working-class due to industrialisation. They made efforts to solve their problems.
Shashipad Banerjee and Narayan Meghaji Lokhande organised workers at local level and made efforts to solve their problems.
Narayan Meghaji Lokhande described as ‘Father of Indian Workers Movement’ formed ‘Bombay Mill Hands Association’ in 1890. This was the beginning of organised movement in India.
The workers on the tea plantations in Assam launched an agitation against their wretched conditions.
In 1899, the workers of the Great Indian Peninsular Railway (GIP) called for a strike for their demands.
During the anti-partition movement, workers carried out strikes from time to time in support of Swadeshi.
In 1920, ‘All India Trade Union Congress’ (AITUC) was established.
In 1928, the Mumbai Mill workers went on strike for six months.
Many such strikes were carried out: by the railway workers, jute mill workers, etc.

Question 3.
Write about the development in the Dalit movement.
Answer:

The Dalit movement was launched to remove the injustice done to Dalits due to Indian social structure.
Mahatma Phule and Narayan Guru brought about awakening among the people regarding social inequality.
Gopal Baba Walangkar condemned untouchability through his book ‘Vital Vidhwansan’.
Periyar Ramaswamy started a movement for the eradication of untouchability in Tamil Nadu.
Maharshi Vitthal Ramji Shinde founded the ‘Depressed Classes Mission’ for the progress of the Dalits.
Rajarshi Shahu Maharaj worked to remove restrictions on inter dining, intermarriage and change of occupation.
Mahatma Gandhi presented the issue of untouchability before the Indian National Congress.
Dr. Babasaheb Ambedkar dedicated his entire life for the Dalit movement and gave it broader perspective.

Question 4.
What work was done for development of workers and Dalits by Dr. Babasaheb Ambedkar?
Answer:
The following work was done by Dr. Babasaheb Ambedkar for the development of workers and Dalits:

He founded the ‘Bahishkrut Hitkarni Sabha’ to establish social equality.
He led the Chavdar Lake Satyagraha in Mahad to make the water reservoir accessible to Dalits.
He burnt ‘Manusmriti’ that advocated social inequality.
In 1930, he started Satyagraha for the entry of Dalits in the Kalaram temple at Nashik.
He started periodicals such as ‘Muknayak’, ‘Bahishkrut Bharat’, Janata’, ‘Samata’ etc. to voice the grievances of the Dalits.
He established ‘All India Scheduled Caste Federation’ to put forth issues of the Dalits in effective manner.
He established ‘Independent Labour Party’ to oppose the laws that went against the interest of the workers.
Through the Constitution of India he made a significant contribution to the creation of a social structure based on equality in Modern India.

Question 5.
Identify the given incident depicted in the picture and write information on it.
image
Answer:

This picture is of the Satyagraha carried out by Dr. Babasaheb Ambedkar and his followers at Chavdar Lake in Mahad.
Though the bill regarding opening of public water reservoirs to the Dalits was passed they were denied access.
He believed that social equality was the right of Dalits.
Dr. Babasaheb Ambedkar and his 8 followers went to Chavdar lake and started a Satyagraha to make the lake water accessible to Dalits.

Question 6.
Why do you feel the structure of the society should be based on equality?
Answer:

I believe that it is the right of every human being to have society based on equality.
Due to inequality, rights of some groups are denied. They do not get opportunities for their development.
They face injustice in several forms. They do not get self-respect.
Farmers, workers, women and Dalits suffer due to inequality. So, it is important to have structure of the society based on equality.

8th Std History Questions And Answers:

Std 8 History Chapter 5 Questions And Answers Social and Religious Reforms Maharashtra Board

February 6, 2025January 5, 2025 by Bhagya

Balbharti Maharashtra State Board Class 8 History Solutions Chapter 5 Social and Religious Reforms Notes, Textbook Exercise Important Questions and Answers.

Class 8 History Chapter 5 Social and Religious Reforms Questions And Answers Maharashtra Board

Social and Religious Reforms Class 8 Questions And Answers Chapter 5 Maharashtra Board

Class 8 History Chapter 5 Social and Religious Reforms Textbook Questions and Answers

1. Rewrite the statements by choosing the appropriate options:
(Sir Sayyad Ahmad Khan, Maharshi Dhondo Karve, Abdul Latif, Swami Vivekananda, Maharshi Vitthal Ramji Shinde)

Question 1.
………….. established the Ramkrishna Mission.
Answer:
Swami Vivekananda

Question 2.
The Anglo-Mohammedan Oriental College was established by………….. .
Answer:
Sir Sayyad Ahmad Khan

Question 3.
The Depressed Classes Mission was founded by ……………. .
Answer:
Maharshi Vitthal Ramji Shinde

2. Complete the following table:

Question 1.

Answer:

3. Explain the following statements with reasons:

Question 1.
The social and religious reform movement began in India.
Answer:

With the spread of English education in India, there was spread of new ideas, new thoughts, new philosophy.
Indians got introduced to western thoughts and culture.
They wanted to create a society based on principles of Humanity, Equality and Fraternity.
They realised that the flaws like superstitions, casteism, old customs, class system and lack of critical outlook is responsible for the backwardness of India.
This association was responsible for social and religious reform in India.

Question 2.
Mahatma Phule conducted a strike of Barbers.
Answer:

There was a custom of Keshavapan, i.e. shaving head of widows in India.
In order to oppose this unjust custom, Mahatma Phule conducted a strike of Barbers.

4. Write short note:

Question 1.
Ramkrishna Mission :
Answer:

Swami Vivekananda, a close disciple of Ramkrishna Paramhansa, founded the Ramkrishna Mission in 1897.
The mission carried out social work like providing help to famine-stricken people, patients and gave medical help to the poor and worked for female education.
It taught people service to humanity is a true religion and worked towards spiritual progress of the people.

Question 2.
Reforms for women by Savitribai Phule:
Answer:

Savitribai Phule, wife of Mahatma Phule, advocated women’s education along with him.
She supported her husband in his efforts to start first school for girls at Bhide Wada in Pune.
She continued her work in the field of education though she faced severe criticism from the society.
She put great efforts in women reform movement which resulted in putting an end to many unjust practices.

Do you know?

Renaissance in other fields/areas :

Sr. No.	Field/ Area	Changes/Progress
1.	Literature	1) Stories and novels dealt with the themes related with social reforms. Writing by women authors. 2) Newspapers and magazines became the carriers of social reform and political awakening.
2.	Art	1) Music became people-oriented. 2) Traditional Indian style of painting was combined with western techniques.
3.	Science	1) Writing of books on science emphasized scientific outlook. 2) People realised the importance of experimentation and scientific outlook for progress.

Project:

Question 1.
Organise an essay competition on the topic ‘Education of women’.

Question 2.
Collect the paragraphs of social reformers.

Class 8 History Chapter 5 Social and Religious Reforms Additional Important Questions and Answers

Rewrite the statements by choosing the appropriate options:
(Sir Sayyad Ahmad Khan, Maharshi Dhondo Karve, Abdul Latif, Swami Vivekananda, Maharshi Vitthal Ramji Shinde)

Question 1.
Through the efforts of ……….. first women’s university was set up in the 20th century.
Answer:
Maharshi Dhondo Karve

Question 2.
………….. established The Mohammedan Literary Society in Bengal.
Answer:
Abdul Latif.

Name the following :

Question 1.
He founded Hindu College at Kolkata.
Answer:
Raja Rammohan Roy

Question 2.
First president of Prarthana Samaj.
Answer:
Dr. Atmaram Pandurang Tarkhadkar

Question 3.
‘Go Back to the Vedas’ was the slogan of this Institution.
Answer:
Arya Samaj

Question 4.
He represented Hinduism at the Parliament of Religions at Chicago in 1893.
Answer:
Swami Vivekananda.

Identify the wrong pair:

Answer:
Wrong pair: Dr. Keshav Baliram
Hedgewar – Founded Hindu Mahasabha
Corrected pair: Dr. Keshav Baliram
Hedgewar – founded Rashtriya
Swayam- Sevak Sangh.

Rewrite the statements by choosing the appropriate options:

Question 1.
Raja Rammohan Roy helped Governor General ……… to pass the Sati Prohibition Act.
(a) Lord Wellesley
(b) Lord Bentinck
(c) Robert Clive
(d) Lord Cornwallis
Answer:
Lord Bentinck

Question 2.
Gopal Ganesh Agarkar gave his staunch opinion about child marriage, law of consent in his newspaper ………… .
(a) Maratha
(b) Darpan
(c) Sudharak
(d) Dnyanoday
Answer:
Sudharak

Question 3.
……….. started the Nursing Course for Women through Seva Sadan Institute.
(a) Tarabai Shinde
(b) Ramabai Ranade
(c) Savitribai Phule
(d) Pandita Ramabai
Answer:
Ramabai Ranade

Question 4.
………… continued tradition of reformation in Sikh religion.
(a) Singh Sabha
(b) Akali movement
(c) Arya Samaj
(d) Prarthana Samaj
Answer:
Akali movement

Question 5.
Lokhitwadi advocated gender equality through his writings in ………… .
(a) Sudharak
(b) Kesari
(c) Shatpatre
(d) Darpan
Answer:
(c) Shatpatre

Do as Directed:

Complete the concept map:
Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

2. Complete the timeline:

Answer:

Answer the following in one sentence each :

Question 1.
What message was given by Swami Vivekanand to the Indian youth?
Answer:
‘Arise, Awake and stop not till the goal is achieved’ was the message given by Swami Vivekanand to the Indian youth.

Question 2.
Write about the work of Singh Sabha.
Answer:
The Singh Sabha worked to achieve reforms, to spread education among the Sikh community and bring in modernisation among them

Question 3.
What were the principles of Prarthana Samaj?
Answer:
The opposition to idol worship, monotheism and opposition to rituals were the principles of Prarthana Samaj.

Question 4.
Which social reformers worked for the cause of widow remarriage?
Answer:
Pandit Ishwarchandra Vidyasagar, Vishnushastri Pandit and Vireshlingam Pantalu worked for the cause of widow remarriage.

Question 5.
Who started ‘Anath Balikashram’?
Answer:
Maharshi Dhondo Keshav Karve started Anath Balikashram, an orphanage for girls, to give education to all women so that they become independent.

Question 6.
Who received the Nobel Prize and in which field?
Answer:
Rabindranath Tagore received Nobel in the field of literature and C. V. Raman for Science.

Question 7.
What was The Mohammedan Anglo Oriental College later known as?
Answer:
The Mohammedan Anglo Oriental College was later known as the Aligarh Muslim University.

Write short note:

Question 1.
Prarthana Samaj :
Answer:
(1) Paramhansa Sabha was dissolved and some of its members formed Prarthana Samaj.
(2) Dr. Atmaram Pandurang was its first President.
(3) They opposed idol worship, monotheism and advocated prayers and devotional songs instead of rituals in place of worship of God.
(4) The important contribution of Prarthana Samaj in reforming the society was that it started orphanages, women’s education institutes, night schools for workers and society for Dalits.
(5) The prestige of Prarthana Samaj rose immensely due to the enrollment of young graduates from Mumbai University.
(6) Justice Ranade, Dr. R. G. Bhandarkar carried the work of Prarthana Samaj forward.

Question 2.
Sir Sayyad Ahmed Khan :
Answer:

Sir Sayyad Ahmad Khan worked for the cause of Muslims.
He believed that the Muslims would not make progress without acquiring western education and science.
He founded ‘The Mohammedan Anglo Oriental College’ which later became Aligarh Muslim University.

Answer the following in 25 to 30 words:

Question 1.
Write about the contribution of Maharshi Vitthal Ramji Shinde.
Answer:

Maharshi Vitthal Ramji Shinde was member of Prarthana Samaj, contributed in reforming society.
He started the ‘Depressed Class Mission’.
He tried to solve problems in society through this mission.
He organised conference against the practice of Devdasi in Mumbai.

Question 2.
Write about the efforts taken to unite Hindu Society.
Answer:

Hindu Mahasabha was formed in 1915 to achieve respectful position of Hindu community and protect it.
Pandit Madan Mohan Malviya founded the ‘Banaras Hindu University’.
Dr. Keshav Baliram Hedgewar established Rashtriya Swayamsevak Sangh in 1925 at Nagpur to set up a disciplinary and virtuous organisation of Hindu youth.
Patit Pawan Temple built by V. D. Savarkar at Ratnagiri was open to all castes of Hindu religion. He also organized common dining programmes.

Question 3.
Give a brief account of the work of women social reformers for the emancipation of women.
Answer:
The women reformers contributed in the following way to improve the condition of women :

Savitribai Phule faced severe criticism of society but continued her work in the field of education.
Tarabai Shinde wrote the book ‘Stri Purush Tulana’ in which she fiercely put her views about the rights of women.
Pandita Ramabai founded the Sharada Sadan and took care of disabled women and children.
Ramabai Ranade founded the Seva Sadan Institute. She started the Nursing course for women as well as demanded the right to vote for them.

Question 4.
State the outcome of women reform movement.
Answer:

The women’s reforms movement resulted in putting an end to many unjust practices in the society.
They voiced their problems and made efforts to find solution to them.
The women got opportunities to prove their capabilities in different fields.
Women started expressing their ideas, thoughts through writing.
Their performance flourished in every sphere of life due to education.

Question 5.
What changes came about in the field of Science, Art and Literature during Indian Renaissance?
Answer:
The following changes were seen in the field of Science, Art and Literature during Indian Renaissance :
(A) Science :

C. V. Raman received the Nobel Prize in Science.
Many books were written on science which emphasized scientific outlook.
People realised the importance of experimentation and scientific outlook for progress of the country.

(B) Art :

Music became more popular and people-oriented.
A new school of painting combining traditional Indian style of painting with the western techniques emerged.

Rabindranath Tagore received the Nobel Prize in literature.
Stories and novels gave inspiration in gaining independence and expressed thoughts on social reforms.
Women took to writing.
New magazines and newspapers became sources of inspiration and political awakening.

Write short note :

Question 1.
The condition of women was miserable in the beginning of nineteenth century.
Answer:
The condition of women during the British period was very miserable in India, because :

They had no right to education.
There was no equality between men and women.
Women were victims of child marriage, dowry system, sati, Keshavapan, opposition to widow remarriage.

Answer the following in detail :

Question 1.
Write briefly about Indian Renaissance.
Answer:
1. The modern educated Indians realised that the unhealthy social conditions and customs like casteism, superstitions, old customs, class system and lack of critical outlook had arrested the progress of India.
2. Rise in the spread of new ideas, new thoughts, new philosophy marked the beginning of modern age.
3. It was necessary to eradicate the flaws and undesirable tendencies in order to create a new society based on principles of Humanity, Equality and Fraternity.
4. They started finding new ways for development of society and country. Educated thinkers started social awareness through writings.
5. This intellectual awakening in the contemporary society in India is called the Indian Renaissance.

Question 2.
Give a brief account of the work of social reformers for the betterment of women.
Answer:

Raja Rammohan Roy launched agitations against practice of Sati.
It led to the enactment of the Sati Prohibition Act in 1829.
He advocated widow remarriage and female education and opposed Purdah system.
Gopal Hari Deshmukh (Lokhitwadi) criticized the unjust social customs related to women and advocated equality of men and women through his writings in ‘Shatapatre’.
Mahatma Phule gave importance to 2 girl’s education. He started first school for i girls at Bhide Wada in Pune.
Through his writings Babasaheb Ambedkar exposed injustice inflicted on women.
Mahatma Gandhi advocated education for women.
Ishwar Chandra Vidyasagar, Vishnushastri Pandit and Vireshlingam Pantalu strove for the recognition of the right to remarriage for the widows.
Gopal Ganesh Agarkar gave his staunch opinion about child marriage and j law of consent in his newspaper ‘Sudharak’.
Maharshi Vitthal Ramji Shinde organised a conference to oppose practice of Devdasi.
Maharshi Dhondo Keshav Karve founded the Anath Balikashram for orphan girls and later the first Women’s University.

Question 3.
What would have happened if social reformers had not taken initiative for women education?
Answer:
We have seen many social reformers in the last 100-150 years. They not only insisted on women education but also took efforts to make it reality.
If they had not taken efforts towards women education then:

Women would have still remained illiterate and would have easily fallen prey to superstitions.
They would have to carry burden of age old customs and traditions.
Illiterate women could not contribute to the development of family, society and nation.
Today they work hand in hand with their male counterparts because they are educated.

Question 4.
What changes have been made in the life of women due to education?
Answer:
Education has brought lot of changes in the life of women.

Women started taking jobs, doing business which made them financially independent.
They are working and competing along with men in every field.
Educated women freed themselves from the clutches of superstitions.
Educated women have become strong enough to face the injustice of society.
The principle of equality is put into practice because of their education.
As woman got educated she contributed for development of her family and country.

Question 5.
Do you still feel there is need to make efforts for women’s education? If yes, then what efforts need to be made?
Answer:

I feel we still need to make efforts on girls’ education because among illiterates and less educated the number of women is more.
The number of illiterate girls in rural and tribal areas is more.
It is important to explain importance of girls’ education. Reforms are still required.
To make people understand the benefit of girls’ education, documentaries and advertisements should be made.
We need to take help of modern technology to achieve it.

8th Std History Questions And Answers:

Cell Biology and Biotechnology Class 10 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Bhagya

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 8 Cell Biology and Biotechnology Notes, Textbook Exercise Important Questions and Answers.

Std 10 Science Part 2 Chapter 8 Cell Biology and Biotechnology Question Answer Maharashtra Board

Class 10 Science Part 2 Chapter 8 Cell Biology and Biotechnology Question Answer Maharashtra Board

Question 1.
Fill in the blanks and complete the statements.
a. Methods like artificial insemination and embryo transplant are mainly used for ………..
(a) animal husbandry
(b) wild life
(c) pet animals
(d) for infertile women
Answer:
(a) animal husbandry

b. ……….. is the revolutionary event in biotechnology after cloning.
(a) Human genome project
(b) DNA discovery
(c) Stem cell research
(d) All the above
Answer:
(c) Stem cell research

c. The disease related with the synthesis of insulin is …………..
(a) cancer
(b) arthritis
(c) cardiac problems
(d) diabetes
Answer:
(d) diabetes

d. Government of India has encouraged the ……….. for improving the productivity by launching NKM-16.
(a) aquaculture
(b) poultry
(c) piggery
(d) apiculture
Answer:
(a) aquaculture

Question 2.
Match the pairs.

Column ‘A’	Column ‘B’
(1) Interferon	(a) Diabetes
(2) Factor VIII *	(b) Dwarfness
(3) Somatostatin	(c) Viral infection
(4) Interleukin	(d) Cancer
	(e) Haemophilia

[Note: In examination match the column question will have 2 components in Column A’ with 4 alternatives in Column B’.]
Answer:
(1) Interferon – Viral infection
(2) Factor VIII – Haemophilia
(3) Somatostatin – Dwarfness
(4) Interleukin – Cancer
[Note: Factor VIII* is an important protein factor and it should not be just factor as given in the textbook.]

Question 3.
Rewrite the following wrong statements after corrections:
a. Changes in genes of the cells are brought about in non-genetic technique.
Answer:
Non-genetic biotechnology involves use of either cell or tissue.

b. Gene from Bacillus thuringiensis is introduced into soyabean.
Answer:
Gene from Bacillus thuringiensis is introduced with gene of cotton.

Question 4.
Write short notes.
a. Biotechnology: Professional uses. (Commercial uses)
Answer:
(1) Biotechnology can be used in the following professional fields, viz. crop biotechnology, animal husbandry, human health, etc.

(2) In crop biotechnology, improvement in the yield and variety of agricultural field is done. The hybrid seeds, genetically modified crops, herbicide tolerant plants are some of the areas in which lot of biotechnological research is being done. By such research, high yielding and disease resistant varieties and varieties which can tolerate stresses such as alkalinity, weeds, cold and drought etc. are produced. BT cotton, BT Brinjal and golden rice are some GMO plants which have become popular in India.

Due to herbicide tolerant plants, the weeds are now selectively destroyed. By using biofertilizers, the use of chemical fertilizers is reduced. Use of bacteria such as Rhizobium, Azotobacter, Nostoc, Anaixiena and plants like Azolla the nitrogen fixation and phosphate solubilization abilities of the plants are improved.

(3) Animal husbandry is now using the methods of artificial insemination and embryo transfer by which the breeds of cattle are improved.

(4) To improve and to manage the human health, diagnosis ahd treatment of diseases have to be focussed. Diagnosis of diabetes, heart diseases and infectious diseases such as AIDS and dengue can be done rapidly due to biotechnology.

(5) The treatment and prevention of diseases need hormones, interferons, antibiotics and different vaccine which are now manufactured through biotechnology. Gene therapy is also used to treat hereditary disorders.

(6) Industrial products and clean technology to combat environmental pollution uses biotechnology practices.

(7) DNA fingerprinting has revolutionized the profession of forensic science.

b. Importance of medicinal plants.
Answer:

In Ayurveda practices, the natural remedies were used. Since India had great biodiversity and traditional knowledge of herbal medicinal uses, therefore, people depended on such medicinal plants.
In olden days, such herbs were collected by roaming in the jungles.
Such important medicinal herbs are now cultivated with care.
In entire world people have understood the importance of holy basil (tulsi), Adulsa, Jyesthmadh, etc.
In some of the allopathy medicines too, the plant extracts are used.
Medicines made from harmful chemicals have side effects and are not safe to be used unless there is medical supervision. Therefore, world-wide herbal remedies are gaining more popularity.

Question 5.
Answer the following questions in your own words.
a. Which products produced through biotechnology do you use in your daily life?
Answer:

The simplest use of biotechnology that we practice at home is making curd and buttermilk.
The primary type of biotechnology is used in the process of fermentation while making food stuffs, like bread, idli-dosa, dhokla, etc.
Nowadays, different types of cheese, paneer, yoghurt, energy drinks, etc. are produced with the help of biotechnology. We are consuming these in our daily life.
Seedless grapes, papaya, and watermelons are available in the market these days.
Violet cabbage, yellow capsicum and exotic vegetables used for salad are also biotechnology products.
The vaccines, antibiotics and the injections of human insulin are in regular use in many house-holds.

b. Which precautions will you take during spraying of pesticides?
Answer:

Pesticides are toxic chemicals. By using them indiscriminately, they contaminate the water, soil and also crops.
The D.D.T., chloropyriphos and malathion are very dangerous. They spread through the food chain causing biomagnification.
Therefore, we shall not use such insecticides and pesticides. We shall use organic pesticides. Excessive use will be avoided.
At the time of spraying, nose, eyes and skin will be covered and protected.
Care will be taken not to allow children or domestic animals to come in, contact with a pesticide.

c. Why some of the organs in human body are most valuable?
Answer:

The body can be in best health,if all the vital organs of the body are also in the best condition.
Brain, kidney, heart, liver, etc. are some such vital organs which are most essential for proper metabolism and functioning of the body. The sense organs of the body are also of utmost importance, especially eyes.
One cannot survive if any of these vital organs are not functioning properly. Some of the organs like brain will never regenerate too.
Some of the organs can be brought back to functionality by performing surgeries. However, any problem with these vital organs make life miserable, therefore, they are said to be valuable.

d. Explain the importance of fruit processing in human life?
Answer:
(1) Fruits are perishable food stuff. They are spoilt soon if not consumed immediately. Hence for storage and usage for a long term, their preservation is absolutely essential.
(2) For year-long use of the fruits they are dried, salted, packed in air tight containers, used for preparing jams and jellies or condensed into pulps or syrups. Beverages, pickles, sauce, and various other products made from the fruits are largely used by us.
(3) The preserved products also fetch financial benefits.
(4) In national and international markets, Indian fruits like mangoes are in great demand. We can get foreign currency through exports of fruits and fruit products. The local horticulturists get good benefit from their orchards.
(5) Processed fruit products also give vitamins and minerals that help in maintaining good health. Thus fruit processing is important for human life.

e. Explain the meaning of vaccination.
Answer:

Vaccination is the administering of vaccine. Vaccine is the ‘antigen’, given to a person or even to animals for acquiring immunity against particular pathogens or diseases.
In olden days, vaccipes were prepared with the help of completely or partially killed pathogens. But this method causes some inconvenience. Some persons were allergic to such raw vaccines or they contracted the same disease through such vaccines.
Hence in recent times the vaccines are produced by using biotechnology. These vaccines are artificial which are synthesised in the laboratories.
The antigen is produced with the help of gene of the pathogen. Such vaccine becomes safe for administering.
These antigenic proteins are injected to people to make their immune systems strong. This process of vaccination is absolutely safe. The vaccines are more thermostable and active for a long period of time.

Question 6.
Complete the following chart.

Answer:

Question 7.
Write the correct answer in blank boxes.

Answer:

Question 8.
Identify and complete the following correlations:
a. Insulin : Diabetes : : Interleukin : …………….
Answer:
Insulin : Diabetes : : Interleukin : Cancer

b. Interferon : ………. : : Erythropoietin : Anaemia.
Answer:
Interferon : Viral infection : : Erythropoietin : Anaemia.

c. ……….. : Dwarfness : : Factor VIII : Haemophilia.
Answer:
Somatostatin : Dwarfness : : Factor VIII : Haemophilia.

d. White revolution : Dairy : : Blue revolution : ………
Answer:
White revolution : Dairy : : Blue revolution : Fishery

Question 9.
Write a comparative note on usefulness and harmfulness of biotechnology.
(OR)
“Biotechnology is not only beneficial but it has some harmful effects too”. Express your opinion about this statement.
Answer:
(1) Biotechnology has proved to be useful in the field of agriculture, medicine, clean technology and industrial products.
(2) Due to various biotechnological experiments, the food production is increased substantially. The milk and milk products are now freely available. People no longer die of hunger due to abundant food supply.
(3) The sophisticated vaccines have stopped the spread of epidemics.
(4) The diseases like diabetes can be controlled due to human insulin injections that can be manufactured by biotechnology.
(5) The problems of pollution control, solid waste management and fuels are partially tackled by biotechnological alternatives.
(6) Though all such positive aspects are there, the biotechnology also poses some problems. The genetic changes are breaking the principles of nature. By inserting human genes in bacteria or virus, the products that are needed only for humans are produced.
(7) Human cloning is also a debatable issue. It will cause social and ethical problems. The new generations formed by cloning will have mothers but no fathers. If man tries to manipulate the genomes of other living organisms, it will cause disturbances in the natural balance. The long ternT effects of all such genetic manipulations can be disastrous. Thus, according to some views, biotechnology can be dangerous too.

Projects: (Do it your self)

Project 1.
Visit the organic manuring projects nearby your place and collect more information.

Project 2.
What will you do to increase public awareness about organ donation in your area?

Project 3.
Collect information about ‘green corridor’. Make a news-collection about it.

Can you recall? (Text Book Page No. 88)

Question 1.
What is cell?
Answer:
The structural and functional unit of the body is called a cell.

Question 2.
What is tissue? What are the functions of tissue?
Answer:
Tissue is a group of cells that performs a similar and definite function. E.g. The muscular tissues in the body perform contraction and extensions thereby helping in locomotion. The conducting tissues of the plants like xylem and phloem transport the water and food respectively.

Question 3.
Which technique in relation to tissues have you studied in earlier classes?
Answer:
The technique of tissue culture and genetic engineering has been studied last year. Tissue culture is ‘Ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium’. Genetic engineering and its use has also been studied under, ‘Introduction to biotechnology’.

Question 4.
Which are the various processes in tissue culture?
Answer:
Various step-wise processes are done while performing the-tissue culture. These processes are primary treatment, reproduction/cell division/multiplication, shooting or rooting, primary hardening, secondary hardening, etc.

Observe: (Text Book Page No. 88)

Question 1.
Assign names in the figure given below. Explain the various stages those are kept blank:

Answer:

Tissue Culture: Tissue culture is the technique in which ‘ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium’ is done. While performing experiments of tissue culture, a liquid, solid or gel-like ” medium prepared from agar, is used. Such medium supplies nutrients and energy necessary for tissue culture technique. Different processes are to be done while performing tissue culture, viz. primary treatment, reproduction or multiplication, shooting and rooting, primary hardening, secondary hardening, etc. From the source plant, required tissues are taken out and all the processes are carried in an aseptic medium in laboratory.

(Use your brain power. (Text Book Page No. 89)

Question 1.
Just like the grafting in plants, is the organ transplantation possible in humans?
Answer:
The grafting as done in case of plants, cannot be done in human beings. But the transplantation of certain organs can be done. Liver, kidney, heart, eyes, etc. can be transplanted. But for these transplantations the donor and the recipient should match with each other in respect of their bloodr groups, age, disease condition, etc. In future, the stem cell research can bring about certain changes in the field of transplantations.

(Text Book Page No. 94)

Question 1.
What will happen if the transgenic potatoes are cooked before consumption?
Answer:
Some types of transgenic potatotes that contain edible vaccine against Hepatitis can be cooked. The cooking does not destroy the antigen incorporated into these transgenic potatoes. But according to some scientists, transgenic potatoes with enterotoxin vaccine, if cooked shows denaturation of vaccine.

Choose the correct alternative and write its alphabet against the sub-question number:

Question 1.
The property of stem cells is called ………….
(a) diversity
(b) equality
(c) differentiation
(d) pluripotency
Answer:
(d) pluripotency

Question 2.
Cell ……….. starts from 14^th day of conception.
(a) development
(b) specialization
(c) growth
(d) differentiation
Answer:
(d) differentiation

Question 3.
Availability of ………… is an important requirement in organ transplantation.
(a) doctor
(b) clinic
(c) donor
(d) ambulance
Answer:
(c) donor

Question 4.
The toxin which is lethal for ……….. was produced in leaves and bolls of BT cotton.
(a) bollworm
(b) locust
(c) birds
(d) frogs
Answer:
(a) bollworm

Question 5.
Transgenic raw potatoes generate the immunity against ………… disease.
(a) plague
(b) cholera
(c) leprosy
(d) TB
Answer:
(b) cholera

Rewrite the following wrong statements after corrections:

Question 1.
High-class varieties of crops have been developed through the technique of transplantation.
Answer:
High-class varieties of crops have been developed through the technique of tissue-culture.

Question 2.
Earlier, insulin was being collected from, the pancreas of pigs.
Answer:
Earlier, insulin was being collected from the- pancreas of horses.

Question 3.
Malaria arises due to genetic changes in hepatocytes.
Answer:
Phenylketonuria (PKT) arises due to genetic changes in hepatocytes.

Question 4.
The E.coli bacteria are useful for cleaning the hydrocarbon and oil pollutants from soil and water.
Answer:
The Pseudomonas bacteria are useful for cleaning the hydrocarbon and oil pollutants from soil and water.

Question 5.
Various essential elements like N, P, K are removed and hence become unavailable to the crops due to earthworms and fungi.
Answer:
Various essential elements like N, P, K become available to crops due to earthworms and fungi.

Question 6.
We do not have any tradition that cures the diseases with the help of natural resources.
Answer:
We have a great tradition of ayurveda that cures the diseases with the help of natural resources.

Match the pairs:

Question 1.

Scientist	Contribution
(1) Dr. Anand Mohan Chakravarti	(a) Wheat production in America
(2) Dr. M. S. Swaminathan	(b) White revolution
	(c) Green revolution in India
	(d) Cleaning the oil spill

Answer:
(1) Dr. Anand Mohan Chakravarti – Cleaning the oil spill
(2) Dr. M.S. Swaminathan – Green revolution in India

Question 2.

Organism	Substance that is absorbed
(1) Pseudomonas	(a) Uranium and arsenic
(2) Pteris vitata	(b) Selenium
	(c) Arsenic
	(d) Hydrocarbons

Answer:
(1) Pseudomonas – Hydrocarbons
(2) Pteris vitata – Arsenic

Find the odd man out:

Question 1.
Green revolution, Industrial revolution, White revolution, Blue revolution
Answer:
Industrial revolution. (All others are concerned with food.)

Question 2.
DDT, malathion, chloropyriphos, Humus
Answer:
Humus. (All others are insecticides.)

Question 3.
Sodium, Aluminium, Potassium, Phosphorus
Answer:
Aluminium. (All others are essential elements for plant growth.)

Question 4.
Diabetes, Anaemia, Leukaemia, Thalassemia
Answer:
Diabetes. (All other diseases involve reduction in the number of blood cells.)

Question 5.
Drying, Salting, Cooking, Soaking with sugar
Answer:
Cooking. (All others are food preservative methods.)

Identify and complete the following correlations:

Question 1.
White revolution : Increase in dairy production : : Green revolution : ………. (March 2019)
Answer:
White revolution : Increase in dairy production : : Green revolution : Increase in agricultural production or crop yield

Question 2.
Nostoc, Anabaena : Biofertilizers : : Alfalfa : ………..
Answer:
Nostoc, Anabaena : Biofertilizers : : Alfalfa : Phytoremediation.

Give definition/Give meanings:

Question 1.
Stem cell or what are stem cells?
Answer:
The special cells having pluripotency and ability to divide and differentiate into new cells are called stem cells. They are present in multicellular living beings.

Question 2.
Biotechnology.
Answer:
Technology that brings about artificial genetic changes and hybridization in organisms for human welfare is called biotechnology.

Question 3.
Genetically modified crops.
Answer:
Crops having desired characters are developed by integrating foreign gene with their genome, such crops have modified genome and are known as genetically modified crops.

Question 4.
Golden rice.
Answer:
Biotechnologically developed variety of rice in which gene synthesizing the vitamin A (Beta carotene) has been incorporated and which contains 23 times more amount of beta carotene than that of the normal variety is called golden rice. It was developed in 2005.

Question 5.
Vaccine.
Answer:
The ‘antigen’ containing material given to a person or animal to acquire either permanent or temporary immunity against a specific pathogen or disease is called a vaccine.

Question 6.
Cloning.
Answer:
Production of replica of any cell or organ or entire organism through biotechnological process is called cloning.

Question 7.
DNA fingerprint.
Answer:
The nucleotide sequence present on the DNA of each person is unique just like the fingerprint, thus for establishing the identity of any person DNA can be analysed, this technique is known as DNA fingerprinting.

Question 8.
Green revolution.
Answer:
All the methods applied for harvesting maximum yield from minimum land are collectively referred to as green revolution.

Question 9.
White revolution.
Answer:
Achieving the self-sufficiency in dairy business, by performing various experiments for quality control, bringing about newer dairy products and their preservation and thus raising economic standards is called white revolution.

Question 10.
Blue revolution.
Answer:
The aquaculture practices to increase the yield of edible aquatic organisms is called blue revolution.

Name the following:

Question 1.
Research institutes involved with cell science.
Answer:

National Centre of Cell Science, Pune
Instem, Bengaluru.

Question 2.
Sources of stem cells.
Answer:

Umbilical cord
Embryonic cells
Redbone marrow
Adipose connective tissue and blood of adult human being.

Question 3.
Types of Stem cells.
Answer:

Embryonic stem cells
Adult stem cells.

Question 4.
Organs that can be donated.
Answer:
Eyes, heart, pancreas, liver, kidneys, skin, J bones, lungs.

Question 5.
Organisms used as biofertilizers.
Answer:
Rhizobium, Azotobacter, Nostoc, Anabaena, Azolla.

Question 6.
Two main methods used in animal husbandry.
Answer:

Artificial insemination
Embryo transfer.

Question 7.
Two important aspects of human health management.
Answer:

Diagnosis
Treatment of diseases.

Question 8.
Place where DNA fingerprinting research is done in India.
Answer:
Centre of DNA fingerprinting and Diagnostics, Hyderabad.

Question 9.
One benefit of biotechnology to the agriculture.
Answer:
Expenses on the pesticides are reduced.

scientific reasons:

Question 1.
Nowadays, safer vaccines are being produced.
Answer:

Before the advent of biotechnology, the vaccines were made from inactive or dead pathogens of that disease.
But now the vaccine is made artificially using biotechnological processes.
Such vaccines produced some disease symptoms in some cases.
The antigen of the disease is researched upon and its genetic code is found out.
A similar antigen is made in the laboratories which is used as a vaccine.
Such vaccines are more thermostable and remain active for longer duration. Therefore, the vaccines are now safer.

Question 2.
Awareness about organ donation after death is increasing.
Answer:

Due to accidents or illness, some of the vital organs may get damaged and may not work to fullest capacity.
In such cases, if organ transplantation is done, it will be very helpful for that needy patient.
The dead person’s organs can be used for organ transplantation and a life can be saved.
Many government and social organizations are spreading awareness about such donations. Therefore, gradually the awareness about organ transplantation is increasing.

Answer the following questions:

Question 1.
Write two uses of biotechnology related to human health. (Board’s Model Activity Sheet)
Answer:

Biotechnology is used to manufacture vaccines for controlling diseases.
Different hormones such as insulin, somatotropin and somatostatin can be prepared in laboratories by using new biotechnological processes. The clotting factors are also manufactured through such techniques.

Question 2.
Answer the following questions:
(a) What is biotechnology?
(b) Explain any two commercial applications of it. (March 2019)
Answer:
(a) Biotechnology: Technology that brings about artificial genetic changes and hybridization in organisms for human welfare is called biotechnology.

(b)

The treatment and prevention of diseases need hormones, interferons, antibiotics and different vaccine which are now manufactured through biotechnology. Gene therapy is also used to treat hereditary disorders.
Industrial products and clean technology to combat environmental pollution uses biotechnology practices.
DNA fingerprinting has revolutionized the profession of forensic science.

Question 3.
What is mainly included under biotechnology?
Answer:
Biotechnology includes the following main areas:

Abilities of microbes are used in producing yoghurt from milk and making alcohol from molasses.
Production of antibiotics and vaccines, etc. is carried out by with the help of specific cells using their productivity.
Bio-molecules like DNA and proteins are used for human welfare.
By performing gene manipulation, plants, animals and products of desired quality are produced. Genetically modified bacteria are used to produce human hormones such as Human Growth Hormone and insulin.
Tissue culture is a non-genetic technique which is used for production of new cells or tissues. Hybrid seeds are also produced in a similar way.

Question 4.
What are edible vaccines?
Answer:

Edible vaccines are those which are given as a food by incorporating them into the food-stuff.
Such edible vaccines are produced through biotechnology.
Transgenic potatoes are produced with the help of biotechnology which contain vaccine that act against bacteria like Vibrio cholera, Escherichiatoli.
If raw potatoes are consumed, then the immunity is generated in the body of a person. However, eating only raw potatoes generates the immunity against cholera and the disease caused due to E. coli.

Question 5.
What is DNA fingerprinting? Explain it in brief. Where is this technique used? Give any two examples. (Board’s Model Activity Sheet)
Answer:

As the fingerprints are unique for every individual, similarly the nucleotide sequence in the DNA molecule is also unique.
By knowing this sequence, one can find out the identity of any person. Such technique to establish the identity of a person by taking into consideration the nucleotide sequence is called DNA fingerprinting.
Its main use is in forensic sciences to confirm the identity of the criminal.
Similarly, identity of parents in case of disputed parentage for any child can be understood by taking DNA fingerprints of both the parents and a child.

Write short notes on:

Question 1.
Uses of stem cells.
Answer:
Stem cells are used for following purposes:

In regenerative therapy stem cells are used.
In case of diseased conditions like diabetes, myocardial infarction, Alzheimer’s disease, Parkinson’s disease, etc., stem cells can be used to replace the damaged or functionless cells.
In conditions such as anaemia, thalassaemia, leukaemia, etc. there is always the need of newer blood cells. Here, stem cells can be used to restore the number of blood cells.
In techniques of organ transplantation stem cells can be used and they can help in the transplantation of new organs such as kidney and liver The defective organs can be replaced by those that are produced with the help of stem cells and transplanted.

Question 2.
Cloning.
Answer:

Cloning is the modern technique in which there is production of replica of any cell or organ or entire organism is done.
There are two types of cloning, viz. (i) Reproductive cloning and (ii) Therapeutic cloning.
Reproductive cloning: In reproductive cloning, a clone is produced by fusion of a nucleus of diploid somatic cell with the enucleated ovum of anybody. In the process, the sperm or male gamete is not needed.
Therapeutic cloning: This technique is largely used for treatment purpose. Stem cells are derived from the cell formed in laboratory by the union of somatic cell nucleus with the enucleated egg cell.
This technique is used for therapy of various diseases.
Gene cloning can also be done to form millions of copies of same gene. Such genes are used for gene therapy and other purposes.
Due to cloning technique, the inheritance of hereditary diseases can be controlled, continuation of generations can be achieved and certain characteristic genes can be enhanced.
However, for human cloning, there is world-wide opposition due to ethical reasons.

Question 3.
Dolly.
Answer:

Dolly was the first mammalian cloned sheep.
Dolly was born on 5th July 1996 in Scotland by the process of cloning.
The Finn Dorset sheep was chosen and her diploid nucleus from the udder cell was introduced into the ovum whose haploid nucleus was removed. This enucleated ovum was of Scottish sheep.
The egg was then introduced into uterus of another Scottish sheep and it grew into Dolly.
Dolly resembled exactly like Finn Dorset sheep whose diploid nucleus was used. None of the characters of Scottish sheep were seen in Dolly.
In this way, Dolly had three mothers but no father.
Dolly gave birth to many young ones. She died on 14th February 2003 due to cancer of the lungs.

Question 4.
Green revolution.
Answer:

In agriculture, different methods used to harvest maximum yield from minimum land, these methods are collectively called green revolution.
Dr. M.S. Swaminathan is called father of Green Revolution in India while Dr. Norman Borlaug has done the similar efforts in the U.S.
Before the Green Revolution in India, there was always the dearth of the food grains. The overflowing Indian population was badly affected due to poor quality and quantity of food.
But due to the Green Revolution in India, attention was focussed on the agricultural research.
Improvised dwarf varieties of wheat and rice, proper use of fertilizers and pesticides and water management were the proper methods that increased production of food grains.
This created abundance of the grains for Indian population.

Question 5.
White revolution.
Answer:

Few years back, there was scarcity of milk in various parts of India. At some places, milk and milk products were abundant but they did not reach all the consumers.
Dr. Verghese Kurien ^ho was then the founder director of Anand Milk Union Limited (AMUL) started thecooperative movement in the direction to produce “operation flood”, i.e. abundance of milk everywhere.
The use of biotechnology was also done to increase the milk production.
Dr. Kurien’s efforts have reached all-time high status as India is now self-sufficient in dairy business.
This is popularly known as White Revolution. Different experiments were performed for quality control, newer dairy products were thought off and preservation methods were improved.
This created White Revolution. AMUL from Anand has now reached international standards.

Question 6.
Blue Revolution.
Answer:

Utilization of aquaculture practices for obtaining edible and commercial aquatic organisms is called blue revolution.
In East Asian countries where water bodies and fish population is abundant, the aquaculture was started.
On similar lines, in India, the aquaculture of different fresh water and marine organisms is being done with the help of fishery scientists.
Government of India has vowed to increase the aquaculture production by encouraging the people for aquaculture by launching the program ‘Nil- Kranti Mission-2016’ (NKM-16).
Pisciculutre is culturing of fish, mariculture is culture of marine organisms such as prawns/shrimps and lobsters. Sea weeds, oysters, clams are also cultured.
For carrying out aquaculture, 50% to 100% subsidies are offered by the Government.
Fresh water fishes like rohu, catla and other edible varieties like shrimp and lobsters are being cultured on a large scale which can bring about Blue Revolution.

Complete the paragraph by choosing the appropriate words given in the bracket:

Question 1.
(degenerated, red bone marrow, adipose connective tissue, blastocyst, umbilical cord, Differentiation)
………… of stem cells form can form various tissues, in the body. Stem cells are present in the ………….. by which the foetus is joined to the uterus of the mother. Stem cells are also present in the ……….. stage of embryonic development. Stem cells are present in ……….. and ………… of adult human beings. It has become possible to produce different types of tissues and the ……… part of any organ with the help of these stem cells.
Answer:
Differentiation of stem cells form can form various tissues in the body. Stem cells are present in the umbilical cord by which the foetus is joined to the uterus of the mother. Stem cells are also present in the blastocyst stage of embryonic development. Stem cells are present in red bone marrow and adipose connective tissue of adult human beings. It has become possible to produce different types of tissues and the degenerated part of any organ with the help of these stem cells.

Paragraph-based questions :

1. Green corridor refers to a special road route that enables harvested organs meant for transplants to reach the destined hospital. A 45-year-old woman, a victim of a railway accident, was declared brain dead, her husband and children agreed to donate her kidneys, liver and heart. One of her kidneys was transplanted to a patient in MGM Hospital and the second kidney helped a patient in Jaslok hospital. Her liver helped the transplant of a patient in Wockhardt Hospital. And her heart was sent to Fortis to the patient on a super urgent priority list, transported via a green corridor covering 18km in less than 16 minutes. This was possible due to Green corridor.
Questions and Answers :

Question 1.
What is Green corridor?
Answer:
Green corridor is a special road route that enables harvested organs meant for transplants to reach the destined hospital

Question 2.
Which organs of brain-dead lady were transplanted?
Answer:
Two kidneys, liver and heart of the brain- dead lady were transplanted.

Question 3.
How many lives were saved from organs of one lady?
Answer:
Four patients lives were saved due to organ donation of one lady.

Question 4.
How was distance of 18km covered in 16 minutes? Why?
Answer:
The distance was covered because the concept of Green corridor was applied. The heart was sent from one hospital to another, where the recipient was kept ready. The quick transportation is necessary to keep heart in living condition.

Question 5.
Who takes the decision to donate the organs?
Answer:
The close relatives of deceased person take the decision to donate the organs.

2. Read the following extract and answer the questions that follow: (March 2019)
A liberal view behind the concept of organ and body donation is that after death our body should be useful to other needful persons so that their miserable life would become comfortable. Awareness about these concepts is. increasing in our country and people are voluntarily donating their bodies.

Life of many people can be saved by organ and body donation. Blinds can regain their vision. Life of many people can be rendered comfortable by donation of organs like liver, kidneys, heart, heart valves, skin, etc. Similarly, body can be made available for research in medical studies. Many government and social organizations are working towards increasing the awareness about body donation.
Questions and Answers :

Question 1.
What is the liberal view behind the organ and body donation?
Answer:
By body donation, research in medical studies is possible. The needy persons can get vital organs which can save their lives.

Question 2.
Name any four organs that can be donated.
Answer:
Liver, Kidneys, heart, eyes, skin, etc. can be donated.

Complete the following table:

Question 1.

Plant/Microbes	Functions
(1) Pteris vitata	______________________________
(2) Pseudomonas	______________________________
(3) ______________________________	Absorption of uranium and arsenic
(4) ______________________________	Absorption of radiations of nuclear waste

Answer:

Plant/Microbes	Functions
(1) Pteris vitata	Absorbs arsenic from soil.
(2) Pseudomonas	Separates hydrocarbon and oil from water and soil
(3) Sunflower	Absorption of uranium and arsenic
(4) Deinococcus radiodurans	Absorption of radiations of nuclear waste

Diagram/chart based questions:

Question 1.

(A) Which process is shown in the above figure? *
Answer:
The figure shows process to make transgenic

(B) Describe in brief the steps I, II, III and IV.
Answer:

Question 2.
Draw well labelled diagram of Stem cell therapy.
Answer:

Question 3.
Label the following diagram :
(i) Stem cells and organ transplantation,
Answer:

(ii) Organs that can be donated:
Answer:

Question 4.
(i) Which therapy is shown in the Fig. 8.5?

(ii) Which will be possible benefits of this therapy in organ transplantation ?
Answer:
(i) The figure 8.5 shows the ‘regenerative therapy’ using stem cells. Also called stem cell therapy.
(ii) With the help of above therapy organs like liver, kidney from stem cells can be redeveloped to replace the failed ones.

Activity based questions:

Question 1.
Bring a packet of ‘Balghuti’ from ayurveda shop. Learn the information about each component in it. Collect information about various other medicines and prepare the chart as shown below. (Try this: Textbook page no. 99)
Answer:

Question 2.
Give five examples of each of the fruiting and flowering plants developed through tissue culture and mention their benefits. (Make a list and discuss: Textbook page no. 93)
Answer:
I. Fruiting trees: Banana, Chikoo (Sapota), Tomato, Fig, Pineapple.

II. Flowering trees: Orchids, Roses, Chrysanthemum, Gerbera, Begonia, Carnation, Lili. Benefits of such plants may be varied. Mostly fruits developed are made seedless and tastier.

III. Benefits of plants produced through technique of tissue culture:

Techniques of tissue culture can produce more copies of same plant with better characters. ’ The plant grower likes to have bigger and more fruits from fruit trees. On the flowering trees, colourful flowers with good fragrance are favoured.
Plants which do not depend on particular climate and local seasonal changes are produced by tissue culture methods. This helps to rise the yield in an area which otherwise may not produce a specific crop.
For tissue culture, saplings and seedlings are made available throughout the year through laboratory. The limitations of getting natural seeds are not there thus planting can be done throughout the year.
Tissue culture techniques create the plants of uniform size, shape and yield. Since they are exactly alike, it becomes beneficial.
In lesser time period, the crops reach maturity.
The crops are pest and disease resistant.
Tissue culture techniques are cost effective and easy to carry out.

Question 3.
Which new species of the rice have been developed in India? (Collect Information: Textbook page no. 97)
Answer:

Species in 2015-16: High zinc species (DRR Dhan 45), Pusa 1592, Punjab basmati 3, Pusa 1609, Telangana Sona.
Species in 2014: CR Dhan 205, CR Dhan 306, CRR, 451.

Question 4.
Discuss about stem cells and organ transplantation in the class with the help of figures given on textbook page no. 90. (Observe: Textbook page no. 90)
Answer:

Organ transplantation:
Various organs in the human body either become less efficient or completely functionless due to various reasons like aging, accidents, infections, disorders, etc. Life of such person becomes difficult or even fatality may occur under such conditions. However, if a person gets the necessary organ under such conditions, its life can be saved.

Availability of donor is an important requirement in organ transplantation. Each person has a pair of kidneys. As the process of excretion can occur with the help of single kidney, person can donate another one. Similarly, skin from certain parts of the body can also be donated.

Various factors like blood group, diseases, disorders, age, etc. of the donor and recipient need to be paid attention during transplantation.

However, other organs cannot be donated during life time. Organs like liver, heart, eyes can be donated after death only. This has lead to the emergence of concepts like posthumous (after death) donation of body and organs.

Organ and Body Donation: human bodies are disposed off after death as per traditional customs. However due to progress in science, it has been realized that many organs remain functional for certain period even after death occurs under specific conditions. Concepts like organ donation and body donation have emerged recently after realization that such organs can be used to save the life of other needful persons. A liberal view behind the concept of organ and body donation is that after death, our body should be useful to other needful persons so that their miserable life would become comfortable. Awareness about these concepts is increasing in our country and people are voluntarily donating their bodies.

Life of many people can be saved by organ and body donation. Blinds can regain the vision. Life of many people can be rendered comfortable by donation of organs like liver, kidneys, heart, heart valves, skin. etc. Similarly, body can be made available for research in medical studies. Many government and social organizations are working towards increasing the awareness about body donation.

Question 5.
Which fruits processing industries you observe in your surrounding? What is their effect? (Make a list and discuss: Textbook page no. 99)
Answer:

Fruit Processing:
we are daily using various products prepared from fruits. All are consuming the products like chocolates, juices, jams and jellies. All these products can be produced by processing on fruits. Fruits are perishable agro-produce. It needs the processing in such a way that it can be used throughout the year. Fruit processing includes various methods ranging from storage in cold storage to drying, salting, air tight pucking, preparing murabba, evaporating, etc.

Projects: (Do it your self)

Project 1.
Collect information about various hybrid varieties of animals. What are their benefits? Make a presentation of various pictures and videos. (Use of ICT: Textbook page no. 93)

Project 2.
Visit the websites: http://www.who.int/transplantation/organ/en/ and www.organindia.org / approaching-the- transplant/and collect more information about ‘brain dead’, organ donation and body donation (Internet is my friend: Textbook page no. 90)

Project 3.
Collect more information about the Human Genome Project, one of the important projects in the world.
(Internet is my friend: Textbook page no. 95)

Project 4.
Collect the information and make the chart about the work of various state and national-level institutes related with biotechnology. (Internet is my friend: Textbook page no. 97)

10th Std Science Part 2 Questions And Answers:

Effects of Electric Current Class 10 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Bhagya

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 4 Effects of Electric Current Notes, Textbook Exercise Important Questions and Answers.

Std 10 Science Part 1 Chapter 4 Effects of Electric Current Question Answer Maharashtra Board

Class 10 Science Part 1 Chapter 4 Effects of Electric Current Question Answer Maharashtra Board

Class 10 Science 1 Chapter 4 Effects Of Electric Current Exercise Question 1.
Tell the odd one out. Give proper explanation.
a. Fuse wire, bud conductor, rubber gloves, generator.
Answer:
Generator. It converts mechanical energy into electric energy, the remaining three do not.

b. Voltmeter, Ammeter, gulvanometer, thermometer.
Answer:
Thermometer. It measures temperature, the remaining three measure electrical quantities.

c. Loud speaker, microphone, electric motor, magnet.
Answer:
Magnet. It exerts a force on a magnetic material, the remaining three convert one form of energy into another.

4 Effects Of Electric Current Exercise Question 2.
Explain the construction and working of the following. Draw a neat diagram and label it.
a. Electric motor
Answer:
Figure shows the construction of an electric motor. Here, a rectangular loop ABCD of copper wire with resistive coating is placed between the north pole and south pole or a strong magnet, such as a horseshoe magnet, such that the branches AB and CD are perpendicular to the direction of the magnetic field. The ends of the loop are connected to the two halves, X and Y, of split rings X and Y have resistive coating on their inner surfaces and are tightly fitted on the axle. The outer conducting surfaces of X and Y are in contact with two stationary carbon brushes, E and F, respectively.

Working:
1. When the circuit is completed with a plug key or switch, the current flows in the direction E → A → B → C → D → F. As the magnetic field is directed from the north pole to the south pole, the force on AB is downward and that on CD is upward by Fleming’s left hand rule. Hence, AB moves downward and CD upward. These forces are equal in magnitude and opposite in direction. Therefore, as observed from the side AD, the loop ABCD and the axle start rotating in anticlockwise direction.

2. After half a rotation, X and Y come in contact with brushes F and E respectively and the current flows in the direction EDCBAF. Hence the force on CD is downward and that on AB is upward. Therefore, the loop and the axle continue to rotate in the anticlockwise direction.

3. After every half rotation, the current in the loop is reversed and the loop and the axle continue to rotate in anti clockwise direction. When the current is switched off, the loop stops rotating after some time.

b. Electric Generator (AC)
Answer:
Figure shows the construction of an AC electric generator. Here, a coil ABCD of copper wire is kept between the pole pieces (N and S) of a strong magnet. The ends of the coil are connected to the conducting rings R₁ and R₂ via carbon brushes B₁ and B₂. The rings are fixed to the axle and there is a resistive coating in between the rings and the axle. The stationary brushes are connected to a galvanometer used to show the direction of the current in the circuit.

Working:
When the axle is rotated with a machine from outside, the coil ABCD starts rotating. Suppose the coil rotates in clockwise direction, as observed from the side AD. Then as the branch AB moves upward, the branch CD moves downward. By Fleming’s right hand rule, the induced current flows in the direction A → B → C → D and in the external circuit, it flows from B2 to B, through the galvanometer. The induced current is proportional to the number of turns of the copper wire in the coil.

After half a rotation, AB and CD interchange their places. Hence, the induced current flows in the direction D → C → B → A. As AB is always in contact with B₁ and CD is in contact with B₂, the current in the external circuit flows from B₁ to B₂ through the galvanometer. Thus, the direction of the current is the external circuit is opposite to that in the previous half rotation. The process goes on repeating and alternating current is generated.

4 Effects Of Electric Current Question 3.
Electromagnetic induction means
a. Charging of an electric conductor.
b. Production of magnetic field due to a current flowing through a coil.
c. Generation of a current in a coil due to relative motion between the coil and the magnet.
d. Motion of the coil around the axle in an electric motor.
Answer:
c. Generation of a current in a coil due to relative motion between the coil and the magnet.

Electric Current Question 4.
4. Explain the difference: AC generator and DC generator.
Answer:
AC generator:

In an AC generator, the rings used are not split.
The direction of the current produced reverses after equal intervals of time.

DC generator:

In a DC generator, split rings are used.
The current produced flows in the same direction all the time.

Question 5.
Which device is used to produce electricity? Describe with a neat diagram.
(1) Electric motor
(2) Galvanometer
(3) Electric generator (DC)
(4) Voltmeter
Answer:
Electric generator (DC).

Figure shows the construction of a DC generator.
Working: The axle is rotated with a machine from outside. When the armature coil of the generator rotates in the magnetic field, electric potential difference is produced in the coil due to electromagnetic induction. This produces a current as shown by the glowing of the bulb or by a galvanometer. The direction of the current depends on the sense of rotation of the coil.

In a DC generator, one brush is always in contact with the arm of the coil moving up while the other brush is in contact with the arm of the coil moving down in the magnetic field. Hence, the flow of the current in the circuit is always in the same direction and the current flows so long as the coil continues to rotate in the magnetic field.

[Note In the case of a DC generator, the current is in the same direction during both the halves of the rotation of the coil. The magnitude of the current does vary periodically with time. In this respect, it differs from the current supplied by an electric cell.]

Question 6.
How does the short circuit form? What is its effect?
Answer:
If a bare live wire (phase wire) and a bare neutral wire touch each other (come in direct contact) or come very close to each other, the resistance of the circuit becomes very small and hence huge (very high) electric current flows through it. This condition is called a short circuit or short circuiting.

In this case, a large amount of heat is produced and the temperature of the components involved becomes very high. Hence, the circuit catches fire.

Question 7.
Give scientific reasons:
a. Tungsten is used to make a solenoid type coil in an electric bulb.
Answer:
1. The intensity of light emitted by the filament of a bulb depends on the temperature of the filament. It increases with the temperature.

2. The melting point of the material used to make the filament of a bulb should be very high so that the filament can be heated to a high temperature by passing a current through it, without melting it. This enables us to obtain more light. The melting point of tungsten is very high.

Hence, tungsten is used to make a solenoid type coil (filament) in an electric bulb.

b. In the electric equipment producing heat e.g. iron, electric heater, boiler, toaster, etc. an alloy such as Nichrome is used, not pure metals.
Answer:
1. The working of heating devices such as a toaster and an electric iron is based on the heating effect of electric current, i.e., conversion of electric energy into heat by passage of electric current through a metallic conductor.

2. An alloy, such as Nichrome, has high resistivity and it can be heated to a high temperature without oxidation, in contrast to pure metals. Therefore, the coils in heating devices such as a toaster and an electric iron are made of an alloy, such as Nichrome, rather than a pure metal.

c. For electric power transmission, copper or aluminium wire is used.
Answer:
1. Copper and aluminium are good conductors of electricity.

2. Copper, and aluminium have very low resistivity. Hence, when an electric current flows through a wire of copper or aluminium, heat produced is comparatively low. Therefore, for electric power transmission, copper or aluminium wire is used.

d. In practice the unit kWh is used for the measurement of electric energy, rather than the joule.
Answer:
(1) If an electric device rated 230 V, 5 A is operated for one hour, electric energy used
= VIt = 230 V × 5 A × 3600 s = 4140000 joules.

(2) If this energy is expressed in kW.h, it will be \(\frac{4140000}{3.6 \times 10^{6}}\) kW·h = 1.15 kW·h (more convenient). 3.6 × 10⁶
Hence, in practice the unit kW·h is used for the measurement of electric energy, rather than the joule.

Question 8.
Which of the statements given below correctly describes the magnetic field near a long, straight current-carrying conductor?
(1) The magnetic lines of force are in a plane, perpendicular to the conductor in the form of straight lines.
(2) The magnetic lines of force are parallel to the conductor on all the sides of conductor.
(3) The magnetic lines of force are perpendicular to the conductor going radially outward.
(4) The magnetic lines of force are in concentric circles with the wire as the center, in a plane perpendicular to the conductor.
Answer:
The magnetic lines of force are in concentric circles with the wire as the centre, in a plane perpendicular to the conductor.

Question 9.
What is a solenoid? Compare the magnetic field produced by a solenoid with the magnetic field of a bar magnet. Draw neat figures and name various components.
Answer:
When a copper wire with a resistive coating is wound in a chain of loops (like a spring), it is called a solenoid.

Magnetic lines of force (magnetic field lines) due to a current carrying solenoid.
B: Battery, K: Plug key, I: Current, N: North pole, S: South pole

The magnetic field lines (magnetic lines of force) due to a current-carrying solenoid are similar to those of a bar magnet. One face of the coil acts as the south pole and the other face as the north pole.

[Note: A current-carrying coil, like a magnet, can be used to magnetise the rod of a given material such as carbon steel or chromium steel. With a strong megnetic field, permanent magnetism can be produced in these materials.]

Question 10.
Name the following diagrams and explain the concept behind them.

Answer:
(a) Fleming’s right hand rule:
Answer:
Stretch the thumb, the index finger and the middle finger of the right hand in such a way that they are perpendicular to each other. In this position, the thumb indicates the direction of the motion of the conductor, the index finger the direction of the magnetic field, and the middle finger shows the direction of the induced current.

[Note The induced current is maximum when the direction of motion of the conductor is at right angles to the magnetic field. ]

(b) Fleming’s left hand rule: The left hand thumb, index finger, and the middle finger are stretched so as to be perpendicular to each other. If the index finger is in the direction of the magnetic field, and the middle finger points in the direction of the current, then the direction of the thumb in the direction of the force on the conductor.

[Note: A magnetic field exerts a force on a current-carrying conductor. Electric current is the time rate of flow of electric charge. Thus, a magnetic field exerts a force on a moving charge. This property is used to accelerate charged particles such as protons, deuterons and alpha particles, as well as electrons, to very high energies. A machine used for this purpose is called a charged particle accelerator. It may be linear or circular in design and very big in size. Such high energy particles are used to study the structure of matter. ]

Question 11.
Identify the figures and explain their use.

Answer:
(a) Fuse:
A fuse protects electrical circuits and appliances by stopping the flow of electric current when it exceeds a specified value. For this, it is connected in series with the appliance (or circuit) to be protected. A fuse is a piece of wire made of an alloy of low melting point (e.g. an alloy of lead and tin). If a current larger than the specified value flows through the fuse, its temperature increases enough to melt it. Hence, the circuit breaks and the appliance is protected from damage.

[Note : The fuse wire is usually enclosed in a cartridge of an insulator such as glass or porcelain provided with metal caps. The current rating (such as 1 A, 2 A) may be printed on the cartridge. ]

(b) Miniature circuit breaker:
These days miniature circuit breaker (MCB) switches are used in homes. When the current in the circuit suddenly increases this switch opens and current stops. Different types of MCBs are in use. For the entire house, however the usual fuse wire is used.

(c) Figure shows the construction of a DC generator.

Here, an ammeter is shown instead of a bulb.
Working: The axle is rotated with a machine from outside. When the armature coil of the generator rotates in the magnetic field, electric potential difference is produced in the coil due to electromagnetic induction. This produces a current as shown by the glowing of the bulb or by a galvanometer. The direction of the current depends on the sense of rotation of the coil.

Question 12.
Solve the following examples.
a. Heat energy is being produced in a resistance in a circuit at the rate of 100 W. The current of 3 A is flowing in the circuit. What must be the value of the resistance?
Solution:
Data: P = 100 W, I = 3 A, R = ?, P = I²R
∴ Resistance, R = \(\frac{P}{I^{2}}=\frac{100 \mathrm{W}}{(3 \mathrm{A})^{2}}=\frac{100}{9} \Omega\) = 11.11 Ω

b. Two tungsten bulbs of wattage 100 W and 60 W power work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor?
Solution:
Data : P₁ = 100 W, P₂ = 60 W, V = 220 V,
I = ?, ∴ I = \(\frac{P}{V}\)
P = VI
∴ I₁ = \(\frac{P_{1}}{V}\) and I₂ = \(\frac{P_{2}}{V}\)
Current in the main conductor, I = I₁ + I₂ (parallel connection)

c. Who will spend more electrical energy? 500 W TV set in 30 mins, or 600 W heater in 20 mins?
Solution:
Data : P₁ = 500 W, t₁ = 30 min = \(\frac{30}{60}\) h
= \(\frac{1}{2}\) h, P₂ = 600 W, t₂ = 20 min = \(\frac{20}{60}\) h = \(\frac{1}{3}\) h
Electrical energy used = Pt
TV set : P₁t₁ = 500 W × \(\frac{1}{2}\) h = 250 W·h
Heater : P₂t₂ = 600 W × \(\frac{1}{3}\) h = 200 W·h
Thus, the TV set will spend more electrical energy than the heater.

d. An electric iron of 1100 W is operated for 2 hours daily. What will be the electrical consumption expenses for that in the month of April? (The electric company charges ₹ 5 per unit of energy.)
Solution:
Data: P = 1100 W, t = 2 × 30 = 60 h,
₹ 5 per unit of energy, expenses = ?

∴ Electrical consumption expenses = 66 units × ₹ 5 per unit = ₹ 330.

Project:
Do it your self.
Project 1.
Under the guidance of your teachers, make a ‘free-energy generator’.

Can you recall? (Text Book Page No. 47)

Question 1.
How do we decide that a given material is a good conductor of electricity or is an insulator?
Answer:
A material which has very low electrical resistance is called a good conductor of electricity. Examples: silver, copper, aluminium.
A material which has extremely high electrical resistance is called an insulator of electricity. Examples: rubber, wood, glass.

Question 2.
Iron is a conductor of electricity, but when we pick up a piece of iron resting on the ground, why don’t we get electric shock?
Answer:
When we pick up a piece of iron resting on the ground, we don’t get electric shock because that piece does not carry any electric current at that time.

Use your brain power! (Text Book Page No. 48)

Question 1.
If in the circuit, the resistor is replaced by a motor, in which form will the energy given by the cell get transformed into?
Answer:
The energy given by the cell will get transformed into the kinetic energy of the copper coil in the motor.

Use your brain power! (Text Book Page No. 60)

Question 1.
Draw the diagram of a DC generator. Then explain as to how the DC current is obtained.
Answer:
Figure shows the construction of a DC generator.

Working:
The axle is rotated with a machine from outside. When the armature coil of the generator rotates in the magnetic field, electric potential difference is produced in the coil due to electromagnetic induction. This produces a current as shown by the glowing of the bulb or by a galvanometer. The direction of the current depends on the sense of rotation of the coil.

Fill in the blanks and rewrite the completed statements:

Question 1.
Electric power = V2/…….
Answer:
Electric power = \(\frac{V^{2}}{R}\)

Question 2.
……….= 1 joule/1 second.
Answer:
1 watt = 1 joule /1 second.

Question 3.
1 kW.h =………J.
Answer:
1 kW.h = 3.6 x 10⁶ J.

Question 4.
According to Joule’s law, quantity of heat (H) produced by an electric current =……….
Answer:
According to Joule’s law, quantity of heat (H) produced by an electric current = I²Rt or VIt or \(\frac{V^{2}}{R}\)t

Question 5.
Magnetic effect of electric current was dicovered by………..
Answer:
Magnetic effect of electric current was dicovered by Hans Christian Oersted.

Question 6.
………..is expressed in oersted.
Answer:
Intensity of magnetic field is expressed in oersted.

Question 7.
Electromagnetic induction was discovered by………..
Answer:
Electromagnetic induction was discovered by Michael Faraday and independently by Joseph Henry.

Question 8.
A galvanometer is used for………
Answer:
A galvanometer is used for detecting the presence of current in a circuit, as well as for some electrical measurements.

Question 9.
In India, the frequency of alternating current is……….
Answer:
In India, the frequency of alternating current is 50 Hz or 50 cycles per second.

Question 10.
Electric motor converts electric energy into………energy.
Answer:
Electric motor converts electric energy into mechanical energy.

Question 11.
Electric generator converts………..energy into electric energy.
Answer:
Electric generator converts mechanical energy into electric energy.

Rewrite the following statements by selecting the correct options:

Question 1.
The device used for producing a current is called……….
(a) a voltmeter
(b) an ammeter
(c) a galvanometer
(d) a generator
Answer:
(d) a generator

Question 2.
At the time of short circuit, the current in the circuit………
(a) increases
(b) decreases
(c) remains the same
(d) increases in steps
Answer:
(a) increases

Question 3.
The direction of the magnetic field around a straight conductor carrying current is given by……
(a) the right hand thumb rule
(b) Fleming’s left hand rule
(c) Fleming’s right hand rule
(d) none of these
Answer:
(a) the right hand thumb rule

Question 4.
The resistance of a wire is 100 Ω. If it carries a current of 1A for 10 seconds, the heat produced will be……….
(a) 1000 J
(b) 10 J
(c) 0.1 J
(d) 10000 J
Answer:
(a) 1000 J

Question 5.
If 220 V potential difference is applied across an electric bulb, a current of 0.45 A flows in the bulb. What must be the power of the bulb? (Practice Activity Sheet – 1)
(a) 99 W
(b) 70 W
(c) 45 W
(d) 22 W
Answer:
(a) 99 W

Question 6.
Electromagnetic induction means
(a) charging of an electric conductor.
(b) production of magnetic field due to a current flowing through a coil.
(c) generation of a current in a coil due to relative motion between the coil and the magnet.
(d) motion of the coil around the axle in an electric motor.
Answer:
(c) generation of a current in a coil due to relative motion between the coil and the magnet.

Question 7.
Write the correct option by observing the figures. (Practice Activity Sheet – 2)

(a) Magnetic field in A is stronger.
(b) Magnetic field in B is stronger.
(c) Magnetic fields in A and B are same.
(d) Magnetic fields in A and B are weaker.
Answer:
(b) Magnetic field in B is stronger.
[Explanation : The resistance in circuit B is less (parallel combination) than that in A. Hence, the current in B is more than that in A. Therefore, the magnetic field in B is stronger than that in A.]

Question 8.
Observe the following diagram and choose the correct alternative: (March 2019)

(a) The intensity of magnetic field in A is larger than in B.
(b) The intensity of magnetic field in B is less than in A.
(c) The intensity of magnetic field in A and B is same.
(d) The intensity of magnetic field in A is less than in B.
Answer:
(d) The intensity of magnetic field in A is less than in B.

State whether the following statements are true or false. (If a statement is false, correct it and rewrite it.) :

Question 1.
Electric power = I²R.
Answer:
True.

Question 2.
Magnetic poles exist in pairs.
Answer:
True.

Question 3.
Electromagnetism was discovered by Oersted.
Answer:
True.

Question 4.
Magnetic field increases as we go away from a magnet.
Answer:
False. (Magnetic field decreases as we go away from a magnet.)

Question 5.
Magnetic lines of force cross each other.
Answer:
False. (Magnetic lines of force do not cross each other.)

Question 6.
Electric generator is used to generate current.
Answer:
True.

Question 7.
An electric motor converts mechanical energy into electric energy.
Answer:
False. (An electric motor converts electric energy into mechanical energy.)

Question 8.
In India, the frequency of AC is 50 Hz.
Answer:
True.

Question 9.
The electricity meter in the domestic electric circuit measures electrical energy consumption in kilowatt·hours.
Answer:
True.

Question 10.
Electric generator converts mechanical energy into electric energy.
Answer:
True.

Question 11.
Split rings are used in a DC generator and in an electric motor.
Answer:
True.

Question 12.
Electromagnetic induction was discovered by Coulomb.
Answer:
False. (Electromagnetic induction was discovered by Faraday and independently by Henry.)

Question 13.
Faraday found that electricity could produce rotational motion.
Answer:
True.

Tell the odd one out. Give proper explanation:

Quesrtion 1.
Find the odd one out and justify it.
Fuse wire, M.C.B., rubber gloves, generator. (Practice Activity Sheet – 3)
Answer:
Generator. It converts mechanical energy into electric energy. All others are related to safety measures to avoid mishap due to electricity.

Match the columns:

Column I	Column II
1. The right hand thumb rule	a. The direction of the force on a current-carrying conductor placed in a magnetic field.
2. Fleming’s right hand rule	b. The direction of the magnetic field around a straight conductor carrying a current.
3. Fleming’s left hand rule	c. The direction of induced current in a conductor.

Answer:
(1) The right hand thumb rule – The direction of the magnetic field around a straight conductor carrying a current.
(2) Fleming’s right hand rule – The direction of induced current in a conductor.
(3) Fleming’s left hand rule – The direction of the force on a current-carrying conductor placed in a magnetic field.

Name the following:

Question 1.
The negatively charged particle considered as a free particle moving in a metallic conductor.
Answer:
Electron.

Question 2.
The quantity expressed in ampere.
Answer:
Electric current.

Question 3.
The quantity expressed in ohm.
Answer:
Electric resistance.

Question 4.
The quantity expressed in volt.
Answer:
Electric potential.

Question 5.
The quantity expressed in joule.
Answer:
Work (and energy).

Question 6.
The quantity expressed in watt.
Answer:
Power.

Question 7.
The quantity expressed in kilowatt-hour.
Answer:
Electric energy.

Question 8.
A component used to control the current.
Answer:
Resistor.

Question 9.
An instrument used to measure electric current.
Answer:
Ammeter

Question 10.
An instrument used to measure electric potential difference.
Answer:
Voltmeter.

Question 11.
The ratio of the work done to the quantity of charge transferred.
Answer:
Electric potential difference.

Question 12.
An alloy of Ni, Cr, Mn and Fe.
Answer:
Nichrome.

Question 13.
The SI unit of resistance.
Answer:
The ohm.

Question 14.
A metal used to make the filament of an electric bulb.
Answer:
Tungsten.

Question 15.
An alloy used to prepare a coil of high resistance for use in electric appliances such as an electric heater.
Answer:
Nichrome.

Question 16.
Constituents of the alloy used to make a fuse wire.
Answer:
Lead and tin.

Question 17.
The unit same as the watt·second.
Answer:
The joule.

Question 18.
A unit for intensity of magnetic field.
Answer:
The oersted.

Question 19.
The scientist in whose honour the SI unit of power is named.
Answer:
James Watt

Question 20.
A device that converts electric energy into mechanical energy.
Answer:
Electric motor.

Question 21.
A device that converts mechanical energy into electric energy.
Answer:
Electric generator.

Answer the following questions in one sentence each :

Question 1.
What is the production of magnetism by an electric current called?
Answer:
The production of magnetism by an electric current is called electromagnetism.

Question 2.
Is magnetic field a scalar or a vector?
Answer:
Magnetic field is a vector.

Question 3.
In India, what is the time interval in which AC changes direction?
Answer:
In India, AC changes direction every \(\frac{1}{100}\) s.

Question 4.
What is the periodic time of AC in India?
Answer:
In India, the periodic time of AC \(\frac{1}{50}\) is

Answer the following questions:

Question 1.
Define electric power.
Answer:
Electric power is the electric work done per unit time.
OR
Electric power is the rate at which electric energy is used.

Question 2.
State the formula for electric power. Hence, obtain its SI unit.
Answer:
Electric power (P) =

The SI unit of work is the joule and that of time is the second. Hence, the SI unit of power is the joule per second. It is given the special name: the watt (W). One watt equals one joule per second.
W = VIt = I²Rt = \(\frac{V^{2}}{R}\) t
∴ P = W/t = VI = I²R = V²/R.
Here, V is the potential difference applied across an electrical appliance, R is the resistance of the appliance and I is the current through the appliance.

[ Note The SI unit of power, the watt, is named in honour of James Watt (1736-1819), British instrument maker and engineer. ]

Question 3.
What is the commercial unit of electric energy? Obtain the relation between this unit and the SI unit of energy.
Answer:
The commercial unit of electric energy is the kilowatt·hour (kW·h) and the SI unit of energy is the joule (J).
1 kW·h = 10³ \(\frac{J}{s}\) × 3600 s s
= 3.6 × 10⁶ J
[Note: The kilowatt-hour is often called simply the unit. (See the energy bill, i.e., the electricity bill.)]

Question 4.
What is one kilowatt.hour?
Answer:
One kilowatt-hour is the electric energy used in one hour by an electrical appliance of power one kilowatt. It is equal to 3.6 × 10⁶ J.

Question 5.
What is heating effect of electric current? What is its origin?
Answer:
The production of heat in a resistance due to the electric current flowing through it when it is connected in an electrical circuit, is called the heating effect of electric current.

When a potential difference is applied across a metallic conductor, free electrons in the conductor move from the end at the lower potential to the end at the higher potential giving rise to electric current. These electrons collide with the atoms and positive ions and transfer some kinetic energy to them. This energy is converted into heat. Hence, the temperature of the conductor begins to rise i.e., the conductor becomes hot. This is the origin of the heating effect of electric current.

Question 6.
Statement 1: Electric current (flow of electrons) creates heat in a resistor.
Statement 2: Heat in the resistor is created according to the law of energy conservation.
Explain Statement 1 with the help of Statement 2. (Practice Activity Sheet – 2)
Answer:
(1) When electrons flow through a resistor (during flow of electric current) electrons possess kinetic energy.
(2) During the flow of electrons there is a decrease in the kinetic energy of the electrons due to collisions with atoms, ions and molecules.
(3) According to the law of conservation of energy, this decrease in the kinetic energy of the electrons gets converted into heat.

Question 7.
State Joule’s law about heating effect of electric current.
Answer:
Joule’s law about heating effect of electric current: The quantity of heat produced in a conductor when a current flows through it is directly proportional to (1) the square of the current (2) the resistance of the conductor (3) the time for which the current flows.

Question 8.
Obtain the mathematical expression for the heat generated in a metallic conductor by electric current (Joule’s law).
Answer:
If V is the potential difference applied across a metallic conductor of resistance R, the current through the conductor, given by Ohm’s law, is
I = V/R ……(1)
The charge passing through the conductor in time t when the current I flows in the conductor is
Q = It…….(2)
The work done in this process is W = VQ …..(3)
From Eqs. (1), (2) and (3), we have,
W = (IR) (It) = I²Rt = VIt
= \(V\left(\frac{V}{R}\right) t=\frac{V^{2}}{R} t\)
This work is converted into heat.
When I is expressed in ampere, R in ohm, t in second and V in volt, W is expressed in joule. In that case,
W = I²Rt = VIt = \(\frac{V^{2}}{R}\) t (in joule)
Usually heat energy (H) is expressed in calorie. Using the relation 4.18 J = 1 cal, we have

This is the required equation.

Question 9.
Two dissimilar bulbs are connected in series. Which bulb will be brighter? (Hint: Consider the resistance of each bulb.)
Answer:
The bulb of higher resistance will be brighter, assuming that the filaments of the two bulbs have the same length and the same area of cross section, but are made of metals with different resistivities.

[Explanation Heat produced (H) in time t = I²Rt , where I is the current through a conductor and R is the resistance of the conductor. In a series combination, the current through each conductor is the same. ∴ H ∝ R for a given t. Hence, the bulb with higher R will become more hot and hence emit more light energy per second. Here it is assumed that the filaments of the two bulbs have the same length and the same area of cross section, but are made of metals with different resistivities.]

Question 10.
Name any six domestic appliances whose working is based on the heating effect of electric current.
(OR)
State applications of heating effect of electric current.
Answer:
Domestic appliances whose working is based on the heating effect of electric current:

Electric heater
Electric iron
Electric oven
Electric toaster
Electric kettle
Electric geyser
Fuse.

Some other applications of heating effect of electric current:

Electric bulb
Electric furnace
In industry for soldering, welding, cutting, drilling
In surgery for cutting tissues with a finely heated platinum wire.

Question 11.
Explain the application of heating effect of electric current in an electric bulb.
Answer:
In an electric bulb, there is a filament of metal such as tungsten having high melting point. When an electric current is passed through the filament, it becomes hot and emits light. The bulbs are usually filled with chemically inactive gases such as nitrogen and argon to prevent oxidation of the filament and hence prolong their life.

Question 12.
Why is tungsten used to make solenoid type coil in an electric bulb? (Practice Activity Sheet – 3)
Answer:
Tungsten is used to make solenoid type coil in an electric bulb for the following reasons:

Tungsten has high resistance and high melting point (nearly 3422° C).
Using current, it can be heated to high temperature so that it emits more light.

Question 13.
Explain the application of heating effect of electric current in an electric iron.
Answer:
In an electric iron, a coil of high resistance is held between mica sheets and placed inside a heavy metal block provided with a handle made of an insulator such as plastic. When an electric current is passed through the coil, it becomes hot. Mica is a good conductor of heat. Hence, heat produced in the coil is transferred to the metal block which can then be used for ironing clothes.
Mica is a bad conductor of electricity. Hence, there is no electrical contact between the coil and the metal block. Therefore, the person using the iron does not get an electric shock even if he or she happens to touch it by chance.

Question 14.
Take any electricity bill of your home. In the bill there is one table which shows the units consumed by you for the last eleven months. Find the average consumption of electricity in your home for each season (i.e., summer, winter and rainy season). Are they the same? Why?
Answer:
The units consumed, on an average, in a home are different for each season.
The energy requirement depends very much on the temperature of the surroundings. For example, a refrigerator, electric fans, an air conditioner, etc. are used more in summer than in winter or rainy season. On the contrary, an electric heater, geyser, etc., are used more in winter than in summer. Hence, there is variation in the average consumption of electricity from season to season.

Question 15.
Name the types of wires or cables used in the electric power supply provided by the State Electricity Board for houses and factories.
Answer:
The wires or cables used in the electric power supply provided by the State Electricity Board are of three types:

phase wire (or live wire, the wire that carries an electric current)
Neutral wire
The earth wire.

Question 16.
In a domestic electric supply in India, what is the potential difference between the live wire and the neutral wire?
Answer:
In a domestic electric supply in India, the potential difference between the live wire and the neutral wire is 220 V.
[Note: AC is used in domestic electric supply.]

Question 17.
Name the type of wire to which the main fuse is connected.
Answer:
The main fuse is connected to the live wire (phase wire).

Question 18.
What does the electricity meter measure?
Answer:
The electricity meter measures electric energy consumption. It is expressed in ‘units’, where 1 unit means 1 kilowatt·hour ( = 3.6 × 10⁶ joules).

Question 19.
Is the electric potential difference across each appliance (in a domestic electric circuit) the same?
Answer:
Yes, the electric potential difference across each appliance (in a domestic electric circuit) is the same.

Question 20.
Name the types of wire across which an electric appliance is connected.
Answer:
An electric appliance is connected across the live wire (phase wire) and the neutral wire.

Question 21.
Electrical appliances are connected in parallel. What are the advantages of this arrangement?
Answer:
In the parallel arrangement of electric appliances, the applied potential difference is the same in each case. Further, even if one of the appliances does not work or is removed for repairing, the other appliances can still be used.

Question 22.
In a domestic electric supply, if two bulbs are connected in series instead of parallel, what will happen if the filament of one of the bulbs breaks?
Answer:
In a domestic electric supply, if two bulbs are connected in series instead of parallel, if the filament of one of the bulbs breaks, there will be no current through the other bulb as well even if the circuit is switched on. Hence the good bulb will also not glow.

Question 23.
What is overloading? When does it occur? What does it cause? How can overloading be avoided?
Answer:
A flow of large amount of current in a circuit, beyond the permissible value of current, is called overloading.

It occurs when many electrical appliances of high power rating, such as a geyser, a heater, an oven, a motor, etc., are switched on simultaneously. This causes fire.

Overloading can be avoided by not connecting many electrical appliances of high power rating in the same circuit.

Question 24.
Explain the application of heating effect of electric current in a fuse.
Answer:
A fuse protects electrical circuits and appliances by stopping the flow of electric current when it exceeds a specified value. For this, it is connected in series with the appliance (or circuit) to be protected. A fuse is a piece of wire made of an alloy of low melting point (e.g. an alloy of lead and tin). If a current larger than the specified value flows through the fuse, its temperature increases enough to melt it. Hence, the circuit breaks and the appliance is protected from damage.

[Note: The fuse wire is usually enclosed in a cartridge of an insulator such as glass or porcelain provided with metal caps. The current rating (such as 1 A, 2 A) may be printed on the cartridge. ]

Question 25.
State the conclusions that can be drawn from Oersted’s experiment. (For reference, see the experiment described on page 51 of the textbook.)
Answer:
Conclusions that can be drawn from Oersted’s experiment:
1. An electric current produces a magnetic field around it. The moving charge in the conducting wire is a source of magnetic field.

2. The direction of the magnetic field produced by the current is the direction in which the north pole of the magnetic needle is deflected. Hence, from the experimental observations we can conclude that at any point near the current-carrying conductor, the magnetic field is perpendicular to (i) the length of the conductor and (ii) the line joining the conductor and the given point.

Question 26.
What is the effect on the magnetic needle in Oersted’s experiment, when (1) a current is passed through the wire (2) the current through the wire is increased (3) the current through the wire is stopped (4) the current through the wire is reversed (5) the distance between the magnetic needle and the wire is increased, keeping the current through the wire constant?
Answer:
In Oersted’s experiment, when there is no current in the wire, the magnetic needle is at rest along the north-south direction.
(1) When a current is passed through the wire, the needle is deflected.
(2) When the current through the wire is increased, the deflection of the needle increases.
(3) When the current through the wire is stopped, the needle comes to rest in its original position along the north-south direction.
(4) When the current through the wire is reversed, the needle is deflected in the direction opposite to that in the first case.
(5) When the distance between the magnetic needle and the wire is increased, keeping the current through the wire constant, the deflection of the needle becomes less.

Question 27.
State the factors on which the magnitude of the magnetic field due to a current-carrying conductor depends and how it depends.
Answer:
The magnetic field at a point due to a current-carrying conductor depends on the current through the conductor and the distance of the point from the conductor.

The magnitude of the magnetic field produced at a given point is directly proportional to the magnitude of the current passing through the conductor.
The magnitude of the magnetic field produced by a given current in the conductor decreases as the distance from the conductor increases.

[Note If the direction of the current is reversed, the direction of the magnetic field is also reversed.]

Question 28.
State the right hand thumb rule.
Answer:
Imagine that you have held a current-carrying straight conductor in your right hand in such a way that your thumb points in the direction of the current. Then turn your fingers around the conductor. The direction of the fingers in the direction of the magnetic lines of force produced by the current.

Question 29.
With a neat labelled diagram, describe the pattern of magnetic lines of force due to a current through a circular loop. Also explain how the magnetic field depends on the number of turns (n) in the loop.
Answer:
The pattern of magnetic lines of force due to a current through a circular loop is shown in Figure

(I: Current, R: Resistance, A: Ammeter)
1. It is seen that every point of the loop forms a centre of a large number of concentric magnetic lines of force forming a series. The circles are small near the wire and become large as we move away from the wire. At the centre of the loop, the arcs of these circles appear as straight lines because of very large radius of the circle.

2. The magnetic field produced by a current-carrying wire at a given point is directly proportional to the current through the wire. If the loop has n turns, the field produced is n times that produced by a single turn (assuming that all the turns have practically the same radius and are in the same plane). The reason is the current in each turn has the same direction and the field due to each turn contributes equally to the total field.

Question 30.
Write Fleming’s left hand rule.
Answer:
Fleming’s left hand rule: The left hand thumb, index finger, and the middle finger are stretched so as to be perpendicular to each other. If the index finger is in the direction of the magnetic field, and the middle finger points in the direction of the current, then the direction of the thumb in the direction of the force on the conductor.

[Note: A magnetic field exerts a force on a current-carrying conductor. Electric current is the time rate of flow of electric charge. Thus, a magnetic field exerts a force on a moving charge. This property is used to accelerate charged particles such as protons, deuterons and alpha particles, as well as electrons, to very high energies. A machine used for this purpose is called a charged particle accelerator. It may be linear or circular in design and very big in size. Such high energy particles are used to study the structure of matter. ]

Question 31.
What is electric motor?
Answer:
A device that converts electric energy into mechanical energy is called an electric motor.

Question 32.
State the principle on which the working of an electric motor is based.
Answer:
An electric motor works on the principle that a current-carrying conductor placed in a magnetic field experiences a force. In this case, the forces acting on different parts of the coil of the motor produce the rotational motion of the coil.

Question 33.
State the uses/applications of an electric motor.
Answer:
Uses/applications of an electric motor:

In domestic appliances such as a mixer, a blender, a refrigerator and washing machine.
In an electric fan, a hair dryer, a record player, a tape recorder and a blower.
In an electric car, a rolling mill, an electric crane, an electric lift, a pump, a computer and an electric train.

Question 34.
(i) Which principle is explained in this figure?

(ii) Which rule is used to find out the direction of force in this principle?
(iii) In which machine is this principle used? Draw a diagram showing the working of that machine. (Practice Activity Sheet – 2)
Answer:
(i) A force is exerted on a current-carrying conductor in the presence of a magnetic field.
(ii) Fleming’s left hand rule is used.
(iii) Electric motor.
Scientifically and technically correct figure.

Question 35.
Observe the following diagram and answer the questions. (Practice Activity Sheet – 1)
(a) Construction of which equipment does the following diagram show?

(b) On which principle does this equipment work?
(c) According to which law does the coil ABCD rotate?
(d) Write the law in your own words.
(e) Where is this equipment used?
Answer:
(a) Given diagram shows the construction of an electric motor.

(b) An electric motor works on the principle that a current-carrying conductor placed in a magnetic field experiences a force. In this case, the forces acting on different parts of the coil of the motor produce the rotational motion of the coil.

(d) Fleming’s left hand rule: The left hand thumb, index finger, and the middle finger are stretched so as to be perpendicular to each other. If the index finger is in the direction of the magnetic field, and the middle finger points in the direction of the current, then the direction of the thumb in the direction of the force on the conductor.

(e) Uses / applications of an electric motor: (1) In domestic appliances such as a mixer, a blender, a refrigerator and washing machine. (2) In an electric fan, a hair dryer, a record player, a tape recorder and a blower. (3) In an electric car, a rolling mill, an electric crane, an electric lift, a pump, a computer and an electric train.

Question 36.
Study the following principle and answer the questions. (Practice Activity Sheet – 4)
A force is excreted on a current-carrying conductor placed in a magnetic field. The direction of this force depends on both the direction of the current and the direction of the magnetic field. This force is maximum when the direction of the current is perpendicular to the direction of the magnetic field.
(a) By which law can we determine the direction of the force excreted on the current-carrying conductor?
(b) In which electrical equipment is this principle used?
(c) Draw a diagram representing the construction of this equipment.
(d) Write the working of this equipment in brief.
Answer:
(a) Fleming’s left hand rule.
(b) Electric motor,
(c)

(d) Working:
1. When the circuit is completed with a plug key or switch, the current flows in the direction E → A → B → C → D → F. As the magnetic field is directed from the north pole to the south pole, the force on AB is downward and that on CD is upward by Fleming’s left hand rule. Hence, AB moves downward and CD upward. These forces are equal in magnitude and opposite in direction. Therefore, as observed from the side AD, the loop ABCD and the axle start rotating in anticlockwise direction.

3. After every half rotation, the current in the loop is reversed and the loop and the axle continue to rotate in anticlockwise direction.
When the current is switched off, the loop stops rotating after some time.

Question 37.
What is a galvanometer used for? Explain in brief the working of a galvanometer.
Answer:
Galvanometer is a sensitive device used to detect the presence of current in a circuit as well as to determine the direction of the current in the circuit.

With suitable modification, it can be used to measure charge, current and voltage. Its working is based on the same principle as that of an electric motor. Here, a coil is pivoted (or suspended) between the pole pieces of a magnet and a pointer is connected to the coil. As the coil rotates when a current is passed through it, the pointer also rotates. The rotation of the coil and hence the deflection of the coil is proportional to the current. The pointer deflects on both sides of the Central zero mark depending on’ the direction of the current.

Question 38.
Take a coil AB having 10-15 turns. Connect the two ends of the coil to the galvanometer as shown in Figure. Take a strong bar magnet. (1) Move the north pole of the magnet towards the end B of the coil.

Observe the deflection of the pointer in the galvanometer. Note the direction of the deflection (i.e. right or left). (2) Now repeat this with the south pole of the magnet towards the end B of the coil. Again observe the deflection. Note its direction. (3) What will happen if instead of the magnet, the coil is moved? (4) If both the coil and the magnet are kept stationary, do you observe any deflection? (5) Compare the direction of the deflection when the north pole of the magnet is moved towards the end B of the coil with that when the end B of the coil is moved away from the north pole of the magnet. (6) What conclusions do you draw from the observations?
Answer:
Observations:
The two deflections, in parts (1) and (2) of the experiment, are in the opposite directions.

(3) If instead of the magnet, the coil is moved towards the stationary magnet, the deflection of the pointer in the galvanometer is observed in one direction, while if the coil is moved away from the magnet, the deflection is observed in the opposite direction. The effect of moving the north pole of the magnet towards the coil and the effect of moving the coil towards the north pole of the magnet are the same.

(4) If both the coil and the magnet are kept stationary, no deflection is observed.

(5) The two deflections are in opposite directions.

(6) Whenever there is relative motion of the coil and the magnet, electric potential difference is induced in the circuit which gives rise to, i.e., induces, an electric current in the circuit causing the deflection of the pointer in the galvanometer. The direction of the current and hence that of the deflection of the pointer in the galvanometer depends on which pole of the magnet faces the coil as well as the direction of relative motion.

[Note: If the velocity of the magnet is increased, the induced current increases, and hence the deflection of the pointer in the galvanometer increases.]

Question 39.
Take two coils of about 50 turns. Insert them over a nonconducting cylindrical roll as shown in Figure.

(A thick paper roll can be used.) Connect coil 1 to a battery with a plug key K. Connect coil 2 to a galvanometer G. (1) Plug the key and observe the deflection in the galvanometer. (2) Unplug the key and again observe the deflection.
Note your observations. What conclusions do you draw from these observations?
Answer:
Observations :
1. When the key is plugged, the galvanometer shows a momentary deflection. When the current in coil 1 becomes steady, the galvanometer shows zero deflection, i.e., its pointer returns to the zero mark at the centre of the scale.

2. When the key is unplugged, the galvanometer shows a momentary deflection in the opposite direction relative to that in part (1) of the experiment. When the current in coil 1 becomes zero as the circuit is broken on unplugging the key, the galvanometer shows zero deflection, i.e., its pointer returns to the zero mark at the centre of the scale.

Conclusions:
As the current in coil 1 changes, the magnetic field associated with the current changes. This induces an electric potential difference in coil 2 which gives rise to an electric current and hence the deflection of the galvanometer. The direction of the induced current and hence that of the deflection of the pointer in the galvanometer depends on whether the current through coil 1 increases or decreases with time.

When there is a steady current in coil 1, there is no change in the associated magnetic field and hence no production of induced potential difference in coil 2. In that case there is no current in coil 2 and hence the galvanometer shows zero deflection.

[Note : Coil 1 is called the primary coil while coil 2 is called the secondary coil. This is because when the current through coil 1 is changed, induced current appears in coil 2.]

Question 40.
What is electromagnetic induction? Who discovered it?
Answer:
The process by which a changing magnetic field in a conductor induces a current in another conductor is called electromagnetic induction. A current can be induced in a conductor either by moving it in a magnetic field or by changing the magnetic field around the conductor. Electromagnetic induction was discovered by Michael Faraday in 1831 and independently by Joseph Henry in 1830.

[Note Michael Faraday (1792-1867), British chemist and physicist, discovered the laws of electrolysis, electromagnetic induction, and a magneto-optical effect now known as the Faraday effect. His discoveries also include benzene and the liquefaction of chlorine. Joseph Henry (1797-1878), US physicist, in addition to the dicovery of electromagnetic induction, invented and constructed the first practical electric motor.]

Question 41.
State Faraday’s law of induction.
Answer:
Whenever the number of magnetic lines of force passing through a coil changes, a current is induced in the coil.

Question 42.
State Fleming’s right hand rule.
Answer:
Stretch the thumb, the index finger and the middle finger of the right hand in such a way that they are perpendicular to each other. In this position, the thumb indicates the direction of the motion of the conductor, the index finger the direction of the magnetic field, and the middle finger shows the direction of the induced current.

[Note The induced current is maximum when the direction of motion of the conductor is at right angles to the magnetic field. ]

Question 43.
Observe the following figure. If the current in the coil A is changed, will some current be induced in the coil B? Explain.

Answer:
If the current in the coil A is changed, there will be some current induced in the coil B.

Explanation:
When the current in the coil A is changed, the magnetic field associated with the current changes. This induces potential difference in the coil B. This gives rise to (i.e., induces) a current in the coil B. The greater the rate at which the current in the coil A is changed with respect to time, the greater is the current induced in the coil B as can be seen from the deflection of the pointer in the galvanometer. This phenomenon is known as electromagnetic induction.

Question 44.
What is a direct current (DC)?
Answer:
A nonoscillatory current that flows only in one direction is called a direct current (DC). It can change in magnitude, but its direction remains the same. [Fig. 4.26 (a) and (b)]

[Note: A direct current is obtained with an electric cell or a DC generator.]

Question 45.
What is an alternating current (AC)?
Answer:
A current that changes in magnitude and direction after equal intervals of time is called an alternating current (AC) (Fig. 4.27).

Electric current changes sinusoidally with time. Electric current and potential difference are shown by the symbol ~.
[Note: An alternating current is obtained with an AC generator.]

Question 45.
What is the value of frequency of AC in India?
Answer:
In India, the value of frequency of AC is 50 hertz.

Question 46.
What is the periodic time of AC in India?
Answer:
In India, the periodic time of AC is 0.02 s (=\(\frac{1}{50}\)s)

Question 47.
State one advantage of AC over DC.
Answer:
One advantage of AC over DC is that electric power can be transmitted over long distances without much loss of energy.

Question 48.
Name two appliances/devices in which a direct current is used.
Answer:
A direct current is used in a portable electric torch and radio.
[Note A Direct current is also in an electric bell, a wall clock, to prepare an electromagnet, for electrolysis, etc. ]

Question 49.
Name two appliances/devices in which an alternating current is used.
(OR)
State any two uses of an AC generator.
Answer:
An alternating current is used in an electric heater and a refrigerator.
[Note Alternating current is also used in an electric iron, a washing machine, an electric mixer, a food processor, an air-conditioner, an electric fan, etc.]

Question 50.
What is (1) an electric generator (2) an AC generator (3) a DC generator?
Answer:
(1) A device which converts mechanical energy into electric energy is called an electric generator.
(2) A generator which converts mechanical energy into electric energy in the form of an alternating current (AC) is called an AC generator.
(3) A generator which converts mechanical energy into electric energy in the form of a direct current (DC) is called a DC generator.

Question 51.
State the principle on which the working of an electric generator is based.
Answer:
The working of an electric generator is based on the principle of electromagnetic induction. When the coil of an electric generator rotates in a magnetic field, a current is induced in the coil. This induced current then flows in the circuit connected to the coil.
[Note An external agency is needed to rotate the coil of an electric generator.]

Question 52.
Show graphically variation of AC with time. Explain the nature of the graph.
Answer:

In this case, the frequency of the alternating current (AC) produced is 50 Hz. The coil completes 50 rotations every second. The time for one rotation of the coil is \(\frac{1}{50}\) second. It is called the periodic time or simply the period of AC. Positive current means the current flows in one direction and negative current means the current flows in the opposite direction in the external circuit. Here, the maximum value of AC is 5 A.

Question 53.
Observe the figure and answer the following questions.

(a) Identify the machine shown in the figure.
(b) Write a use of this machine.
(c) How transformation of energy takes place in this machine. (Practice Activity Sheet – 3)
Answer:
(a) The instrument shown in the figure is generator.
(b) This machine is used to generate electricity.
(c) The generator generates electricity through following transformation:
Mechanical Energy → Electrical Energy

Question 54.
Observe the following figure. Which bulb will fuse? (Practice Activity Sheet – 4)

Answer:
Bulb A.

Give scientific reasons:

Question 1.
In an electric iron, the coil of high resistance is kept between mica sheets.
Answer:
(1) Mica is a bad conductor of electricity and good conductor of heat.
(2) In an electric iron, the coil of high resistance is kept between mica sheets so that there is no electrical contact between the coil and the heavy metal block of the iron though there is heat transfer. This protects the user from getting an electric shock.

Question 2.
The material used for fuse has low melting point.
OR
A fuse should be made of a material of low melting point.
Answer:
1. A fuse is used to protect a circuit and the appliances connected in the circuit by stopping the flow of an excessive electric current. For this, a fuse is connected in series in the circuit.

2. When the current in the circuit passes through the fuse, its temperature increases. When the current exceeds the specified value, the fuse must melt to break the circuit. For this, the material used for a fuse has low melting point.

Distinguish between the following:

Question 1.
Direct current and Alternating current.
Answer:
Direct current:

Direct current flows only in one direction.
It cannot be used for large scale of electricity for household purpose.

Alternating current:

Alternating current reverses its direction periodically with time.
It is used in household electrical appliances such as an electric heater, an electric iron, a refrigerator, etc.

Question 2.
Electric motor and Electric generator.
Answer:
Electric motor:

A battery is used in an electric motor to pass a current through the coil.
In this case, a current-carrying coil is set in rotation due to the magnetic field.
Split rings are used in an electric motor.
In this case, electric energy is converted into mechanical energy.

Electric generator:

A battery is not used in an electric generator.
In this case, a potential difference and hence a current is produced when the coil is set into rotation in the magnetic field by an external agent.
Rings used in an AC generator are not split.
In this case, mechanical energy is converted into electric energy.

Solve the following examples/numerical problems:

Question 1.
An electric bulb is connected to a source of 250 volts. The current passing through it is 0.27 A. What is the power of the bulb?
Solution:
Data: V = 250 V, I = 0.27 A, P = ?
P = VI
= 250 V × 0.27 A
= 67.5 W
The power of the bulb = 67.5 W.

Question 2.
If a bulb of 60 W is connected across a source of 220 V, find the current drawn by it.
Solution:
Data: P = 60 W, V = 220 V, I = ?
P = VI

The current drawn by the bulb
= \(\frac{3}{11}\) A = 0.2727 A

Question 3.
A bulb of 40 W is connected across a source of 220 V. Find the resistance of the bulb.
Solution:
Data: P = 40 W, V = 220 V, R = ?

= 40 × 110 Ω = 1210 Ω
The resistance of the bulb = 1210 Ω

Question 4.
If the current passing through a bulb is 0.2 A and the power of the bulb is 20 W, find the voltage applied across the bulb.
Solution:
Data: I = 0.2 A, P = 20 W, V = ?
P = VI
∴ V = \(\frac{P}{I}=\frac{20 \mathrm{W}}{0.2 \mathrm{A}}\)
= 100 V
The voltage across the bulb = 100 V.

Question 5.
Two tungsten bulbs of power 50 W and 60 W work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor? (March 2019)
Solution:
Data: P₁ = 50 W, P₂ = 60 W, V = 220 V, I = ?

Current in the main conductor,
I = I₁ + I₂ ……….(parallel combination)

= 0.5 A

Question 6.
An electric iron rated 750 W is operated for 2 hours/day. How much energy is consumed by the electric iron for 30 days?
Solution:
Data: P = 750 W, t = 2\(\frac{\text { hours }}{\text { day }}\) for 30 days
The energy consumed = Pt = 750 × 2 × 30
= 1500 × 30
= 45000 W·h
= 45 kW·h
The energy consumed by the electric iron for 30 days = 45 kW·h.

Question 7.
If a TV of rating 100 W operates for 6 hours per day, find the number of units consumed in a leap year.
Solution:
Data: P = 100 W, t = 6\(\frac{\text { hours }}{\text { day }}\) × 366 days
= 2196 hours
1 unit = 1 kW·h = 1000 W·h

219.6 units are consumed in a leap year.

Question 8.
An electric appliance of rating 300 W is used 5 hours per day in the month of March. Find the number of units consumed.
Solution:
Data: P = 300 W, t = 5\(\frac{\text { hour }}{\text { day }}\) × 31 days
= 155 hours, 1 unit = 1 kW·h = 1000 W.h

= 46.5 units
46.5 units are consumed in the month of March.

Question 9.
A washing machine rated 300 W operates one hour/day. If the cost of a unit is ₹ 3.00, find the cost of the energy to operate the washing machine for the month of March.
Solution:
Data: P = 300 W, ₹ 3.00 per unit,
t = 1 \(\frac{\text { hour }}{\text { day }}\) × 31 days = 31 hours, 1 unit = 1 kW·h
= 1000 W·h, cost of the energy = ?

Cost = 9.3 units × ₹ 3.00 per unit = ₹ 27.9.
The cost of the energy to operate the washing machine for the month of March = ₹ 27.9.

Question 10.
Find the heat produced in joule if a current of 0.1 A is passed through a coil of resistance 50 Ω for two minutes. Keeping other conditions the same if the length of the wire is reduced to the original length (by cutting 4 the wire), what will be the heat produced?
Solution:
Data: I = 0.1 A, R = 50 Ω, t = 2 minutes = 2 × 60 s = 120 s, H = ?
H = I²Rt = (0.1A)² × 50 Ω × 120 s
= 0.01 × 50 × 120 J = 60 J
Heat produced = 60 joules.
In the second case, the resistance of the wire will be \(\frac{50 \Omega}{4}\)
Hence, the heat produced = \(\frac{60 \mathrm{J}}{4}\) = 15 J.

Question 11.
Calculate the heat produced in calorie when a current of 0.1 A is passed through a wire of resistance 41.8 Ω for 10 minutes.
Solution:
Data: I = 0.1 A, R = 41.8 Ω, t = 10minutes
= 10 × 60 s = 600 s, H = ?

Heat produced = 60 calories.

Question 12.
A potential difference of 250 volts is applied across a resistance of 1000 Ω in an electric iron. Find (1) the current (2) the heat produced in joule in 12 seconds.
Keeping other conditions the same, if the length of the wire in the iron is reduced to half the original length (by cutting the wire), what will be the current and heat produced?
Solution:
Data: V = 250 V, R = 1000 Ω, t = 12 s,
I = ? H = ?
(1) V = IR
∴ I = \(\frac{V}{R}=\frac{250 \mathrm{V}}{1000 \Omega}\) = 0.25 A
The current through the resistance = 0.25 A.
(2) H = I²RT
= (0.25 A)² × 1000 Ω × 12 s
= (\(\frac{1}{4}\) × 1000) × (\(\frac{1}{4}\) × 12) J
= 250 × 3J = 750 J
H = VIt = 250 V × 0.25 A × 12s = 250 × 3J = 750 J
The heat energy produced in the resistance in 12 seconds = 750 joules.

On cutting the wire, the resistance of the wire will become half the initial resistance. Hence, the current will become double the initial current as I = V/R and V is the same in both the cases. Therefore, the current in the wire will be 0.25 A × 2 = 0.5 A. (Hence, the heat produced will be VIt = 250 V × 0.5 A × 12 s = 250 × 6 J = 1500 J.)

Question 13.
A potential difference of 100 V is applied across a resistor of resistance 50 Ω for 6 minutes and 58 seconds. Find the heat produced in (i) joule (ii) calorie.
Solution:
Data: V = 100 V, R = 50 Ω, t = 6 minutes and 58 seconds = (6 × 60 + 58) s = (360 + 58)
= 418 s, H = ?

Heat generated = 83600 joules.

Heat produced = 2 × 10⁴ calories.

Numerical Problems For Practice:

Question 1.
When the voltage applied across a bulb is 200 V, the current passing through the bulb is 0.1 A. Find the power of the bulb.
Answer:
20 W

Question 2.
A bulb of 100 W is connected across a source of 200 V. Find the current drawn by it.
Answer:
0.5 A

Question 3.
A bulb of 60 W is connected across a source of 240 V. Find the resistance of the bulb.
Answer:
960 Ω

Question 4.
If the current passing through a bulb is 0.15 A and the power of the bulb is 30 W, find the voltage applied across the bulb.
Answer:
200 V

Question 5.
An electric appliance of rating 800 W is used 4 hours per day in the month of December. Find the number of units consumed.
Answer:
99.2 units

Question 6.
An electric appliance rated 400 W is used 5 hours per day in the month of June. If the cost of a unit is ₹ 3.00, find the energy bill for June.
Answer:
₹ 180

Question 7.
An electric bulb rated 60 W is used 10 hours per day for 20 days. If the cost of a unit is ₹ 3.00, find the energy bill.
Answer:
₹ 36

Question 8.
Two electric bulbs rated 60 W and 40 W respectively are used 5 hours per day for 20 days. If the cost of a unit is ₹ 4.00, find the cost of the energy used.
Answer:
₹ 40

Question 9.
Find the heat produced in joule if a current of 0.1 A is passed through a coil of resistance 25 Ω for one minute.
Answer:
15 J

Question 10.
Calculate the heat produced in calorie when a current of 0.1 A is passed through a wire of resistance 41.8 Ω for 5 minutes.
Answer:
30 calories

Question 11.
Calculate the heat produced in calorie when a current of 0.2 A is passed through a wire of resistance 41.8 Ω for 10 minutes.
Answer:
240 calories

Question 12.
Find the heat produced in calorie when a current of 0.2 A is passed through a wire of resistance 20.9 Ω for 10 minutes.
Answer:
120 calories

Question 13.
A potential difference of 100 V is applied across a wire of resistance 50 Ω for one minute. Find the heat produced in joule.
Answer:
1.2 × 10⁴ joules

Question 14.
A potential difference of 100 V is applied across a wire for two minutes. If the current through the wire is 0.1 A, find the heat produced in joule.
Answer:
1200 joules

Question 15.
A potential difference of 100 V is applied across a wire for 6 minutes and 58 seconds.
If the current through the wire is 0.1 A, find the heat produced in calorie.
Answer:
1000 calories

10th Std Science Part 1 Questions And Answers:

Std 8 History Chapter 7 Questions And Answers Non-co-operation Movement Maharashtra Board

February 6, 2025January 5, 2025 by Bhagya

Balbharti Maharashtra State Board Class 8 History Solutions Chapter 7 Non-co-operation Movement Notes, Textbook Exercise Important Questions and Answers.

Class 8 History Chapter 7 Non-co-operation Movement Questions And Answers Maharashtra Board

Non-co-operation Movement Class 8 Questions And Answers Chapter 7 Maharashtra Board

Class 8 History Chapter 7 Non-co-operation Movement Textbook Questions and Answers

1. Rewrite the statements by choosing the appropriate options :

Question 1.
Gandhiji began his career from the country of …………. .
(a) India
(b) England
(c) South Africa
(d) Myanmar
Answer:
(c) South Africa

Question 2.
The farmers started the no-tax movement in the district of ………….
(a) Gorakhpur
(b) Kheda
(c) Solapur
(d) Amravati
Answer:
(b) Kheda

Question 3.
As a protest to Jallianwala Baug massacre, Rabindranath Tagore returned the title of ………… bestowed upon him by the British government.
(a) Lord
(b) Sir
(c) Raobahadur
(d) Raosaheb
Answer:
(b) Sir

2. Answer the following questions in one sentence :

Question 1.
Which restriction were put up on the Blacks in South Africa according to the declaration of 1906?
Answer:
In 1906, the government declared that the Blacks in South Africa should compulsorily carry an identity card. This was the restriction imposed on their freedom.

Question 2.
Where did Gandhiji launch the first Satyagraha in India?
Answer:
The first Satyagraha in India was launched by Gandhiji at Champaran at Bihar in 1917.

Question 3.
Which officer gave orders of firing in the Jallianwala Baug?
Answer:
General Dyer gave orders of firing in the Jallianwala Baug.

3. Answer the following questions in 25-30 words :

Question 1.
Explain the philosophy of Satyagraha.
Answer:

Satyagraha, a novel technique, means insistence of truth.
The aim of Satyagraha was to make the unjust person aware of truth and justice and also to transform his views.
A Satyagrahi should never use violence and untruth meAnswer:

Question 2.
Why was the Swaraj Party formed?
Answer:
1. The members of the Indian National Congress like Motilal Nehru and Chittaranjan Das put an idea to contest elections and enter the Legislative Assembly and Provincial Legislative Council to obstruct the working of the government and oppose the unjust policies.
2. Therefore, they formed the Swaraj Party in 1922.

4. Explain the following statements with reasons :

Question 1.
Rowlatt Act was opposed by the people of India.
Answer:

The Rowlatt Act was enacted to suppress the growing discontent among the Indians and suggest measures about it.
This Act authorised the government to arrest any Indian and imprison him without trial or warrant.
There was no provision for appeal against the punishment given under the Act.
Due to such unjust provisions in the Rowlatt Act it was opposed by the people of India and called ‘Black Act’.

Question 2.
Gandhiji suspended the Non-co-operation Movement.
Answer:

In February, 1922 the police opened fire on a peaceful procession at Chauri-Chaura in Gorakhpur district of Uttar Pradesh.
In retaliation to this, the enraged mob set fire the police station in Chauri- Chaura.
Twenty-two policemen including one officer were killed in this incident.
Gandhiji was hurt by this incident. So he decided to suspend the Non-co-operation Movement on 12 February, 1922.

Question 3.
The Indians boycotted the Simon Commission.
Answer:

The reforms introduced by Montague Chelmsford Act of 1919 was unsatisfactory.
It created discontent among IndiAnswer: The British Government appointed the Simon Commission under the chairmanship of Sir John Simon in 1927.’
There was not a single Indian member on the commission. In protest of this, the Congress boycotted the Simon Commission.

Question 4.
Khilafat movement was started in India.
Answer:

The ruler of Turkey was regarded as the Khalifa or religious head of the Muslims all over the world.
To get the support of Indian Muslims, in the First World War, the British gave false assurance of not harming the Khalifa’s empire after the war.
But they did not keep their word. It caused resentment among the Indian Muslims. So, they started the Khilafat Movement.

Do you Know?
Mulshi Satyagraha (1920-1924):
Muishi Pethcz was a small viilage,with hilly terrain, approximately 40-45 kms away from Pune. The Tata Company decided to set up hydro-electricity project in this part. It was given permission by the British government. The dam would have displaced 52 villages and 11000 people. There was no consent taken from the people nor was there any programme for their rehabilitation. The farmers started their struggle against a mighty power and influential capitalists. Senapa’ti Pandurcing Mahadev Bapat led the struggle. People started caffing him Senapati Bapat. In spite of resisting it for four years, they were unsuccessful. The Satyagraha became inspirational not only in India but all over the world wherever the issue of rehabilitation came up.

Project:

Trace out the pledge of independence that was taken on 26th January 1930 and read it aloud collectively in the classroom.

Class 8 History Chapter 7 Non-co-operation Movement Additional Important Questions and Answers

Rewrite the statements by choosing the appropriate options :

Question 1.
The period between 1920-1947 in the Indian national movement is known as ………….
(a) Moderate Phase
(b) Extremist Phase
(c) Revolutionary Period
(d) Gandhian Era
Answer:
(d) Gandhian Era

Question 2.
…………. suggested Gandhiji to tour the entire nation when he returned from South Africa.
(a) Dadabhai Nowrojee
(b) Surendranath Banerjee
(c) Gopal Krishna Gokhale
(d) Lokmanya Tilak
Answer:
(c) Gopal Krishna Gokhale

Question 3.
After the death of Lokmanya Tilak in 1920, the reins of freedom movement went into the hands of ………….
(a) Lala Lajpat Rai
(b) Subash Chandra Bose
(c) Mahatma Gandhi
(d) Pandit Jawaharlal Nehru
Answer:
(c) Mahatma Gandhi

Question 4.
…………. Commission was appointed to enquire about Jallianwala Baug massacre.
(a) Hunter
(b) Simon
(c) Minto
(d) Rowlatt
Answer:
(a) Hunter

Question 5.
The Resolution of Non-co-operation Movement was put forth by Chittaranjan Das in ………. session of Indian National Congress.
(a) Lahore
(b) Nagpur
(c) Mumbai
(d) Amritsar
Answer:
(b) Nagpur

Question 6.
An officer named attacked Lalaji with a lathi at Lahore.
(a) Dyer
(b) Hunter
(c) Odwire
(d) Saunders
Answer:
(d) Saunders

Question 7.
……………., Secretary of State, criticised that Indians were incapable of drafting a constitution.
(a) Morley
(b) Minto
(c) Berkenhead
(d) Montague
Answer:
(c) Berkenhead

Identify the wrong pair and correct it:

Work – Leader

(1) President of the Lahore session of
Congress – Motilal Nehru

(2) Demonstrated against Simon
commission at Lahore – Lala Lajpat Rai

(3) Held satyagraha
in South Africa – Mahatma Gandhi

(4) Led the Mulshi
Satyagraha – Senapati Bapat
Answer:
Wrong pair : President of the Lahore session of Congress
– Motilal Nehru.
Corrected pair : President of the Lahore session of Congress
– Jawaharlal Nehru.

Name the following:

Question 1.
Leaders in other countries who were influenced by Satyagraha technique of Mahatma Gandhi.
(1) …………………………………….
(2) …………………………………….
Answer:
(1) Martin Luther King in USA
(2) Nelson Mandela in South Africa

Question 2.
Young Congress leaders who demanded Poorna Swaraj.
(1) …………………………………….
(2) …………………………………….
Answer:
(1) Pandit Jawaharlal Nehru
(2) Subash Chandra Bose

Question 3.
Eminent lawyers who boycotted court by giving up their practice
(1) …………………………………….
(2) …………………………………….
Answer:
(1) Pandit Motilal Nehru
(2) Chittaranjan Das.

Complete the graphical presentation :

Question 1.
Maharashtra Board Class 8 History Solutions Chapter 7 Non-co-operation Movement 1
Answer:

Question 2.
Maharashtra Board Class 8 History Solutions Chapter 7 Non-co-operation Movement 3
Answer:

Question 3.
Maharashtra Board Class 8 History Solutions Chapter 7 Non-co-operation Movement 5
Answer:

Answer the following questions in one sentence :

Question 1.
Which principles gave a new direction to the freedom movement?
Answer:
The principles of truth and non-violence gave new direction to the freedom movement.

Question 2.
How was the constructive programme beneficial?
Answer:
Due to the constructive programme, the national movement became more comprehensive in rural areas.

Question 3.
How did Lala Lajpat Rai react after he was attacked by the British officer?
Answer:
Lala Lajpat Rai said that Every blow on my body will prove a nail in the coffin of the British Empire, when he was attacked by the British officer.

Write short note :

Question 1.
Gandhiji’s work in South Africa :
Answer:

In 1893, Gandhiji went to South Africa for some legal work where he began his political work.
Many Indians had settled in South Africa for the purpose of trade and business.
They were treated as criminals and humiliating treatment was given to them.
Many restrictions were imposed on their freedom like it was compulsory for them to carry an identity card.
Gandhiji adopted the path of Satyagraha and gained justice for them.

Question 2.
Champaran Satyagraha :
Answer:

The farmers in Champaran region in Bihar were forced to cultivate indigo.
The cultivators suffered as they received a fixed amount as price from the plantation owners.
In 1917, Gandhiji went to Champaran and organized the farmers.
He launched agitation by adopting the technique of Satyagraha.
The farmers were given justice and thus Gandhiji was successful in his first satyagraha in India.

Question 3.
Work of Swaraj Party :
Answer:

The leaders of the Swaraj Party put up an idea of contesting elections and obstructing the work of the government.
They severely opposed unjust policies of the government.
They demanded that a Responsible Government should be given to India.
They demanded for Round Table Conference.
They made a resolution in Assembly which demanded the release of political leaders taken as prisoners.
Many resolutions were passed in the Central Legislative Assembly but were rejected by the government.

Question 4.
Nehru Report :
Answer:
1. Berkenhead, the Secretary of State of India, criticised that Indian leaders were incapable of drafting a unanimously accepted constitution for India.
2. This challenge was taken up by all party committee under the chairmanship of Pandit Motilal Nehru.
3. The proposals of the Nehru Report were :

implement Adult Franchise system in India,
establish self government colonies of India, and
division of provinces on linguistic division.

Explain the following statements with reasons :

Question 1.
No-tax campaign in Kheda district was successful.
Answer:

Due to continuous famines the crops had failed in Kheda district of Gujarat.
But the farmers were forced to pay land tax by the government.
Gandhiji suggested to the farmers that they should refuse to pay tax.
The farmers started the campaign to scrap the tax in 1918.
The tax was suspended by the British Government which made the movement successful.

Question 2.
The movement started by Mill workers at Ahmedabad was successful.
Answer:

During the First World War the inflation was very high.
The mill workers demanded for rise in salary which was refused by the mill owners.
Gandhiji advised the workers to go on hunger strike.
In the end, the mill owners had to back off and the salary of workers was increased.

Question 3.
The British government passed the Rowlatt Act in 1919.
Answer:

Indians extended help to the British government in the First World War.
The Indians expected that after the war is over there will be a system of governance for taking decision for the well-being of the IndiAnswer:
Moreover, the rising prices, increased taxes led to growing unrest among the Indians.
To suppress this discontent and suggest measures about it Rowlatt Act was passed in 1919.

Answer the following questions in 25-30 words :

Question 1.
Why was the Hunter Commission appointed?
Answer:

General Dyer ordered an unrestrained firing on the unarmed people who gathered for a meeting on 13th April 1919, at Jallianwala Baug.
About 400 people were killed and injured. The injured did not receive any medical help.
There was nationwide protest against this act.
Rabindranath Tagore gave up the title of Sir given by the British government.
The Indians demanded an enquiry against this massacre. So the British government appointed Hunter commission.

Question 2.
Why did Gandhiji support the Khilafat movement?
Answer:

Indian Muslims started a movement to support Khalifa known as the Khilafat Movement.
Gandhiji felt that if Hindus and Muslims unitedly start a national movement then the government can be brought to its senses.
Therefore, Gandhiji supported the Khilafat movement.
The Khilafat committee accepted Gandhiji’s proposal and extended support to the Non-co-operation movement.

Question 3.
How was Simon Commission opposed in India?
Answer:

The Indian National Congress boycotted the Simon Commission because there was not a single Indian in it.
In 1928, when the commission arrived in India, people strongly protested against it shouting ‘Simon go back’.
The police lathi charged the demonstrators. Lala Lajpat Rai himself led the demonstrators in Lahore.
He was injured in lathi charge and succumbed to his injuries.

Question 4.
Why was Gandhiji sent to six years of imprisonment?
Answer:

Gandhiji was arrested in March 1922 for writing three anti-national articles in ‘Young India’.
His trial was conducted in a special court set up in Ahmedabad.
The charges were levied against him. He was imprisoned for six years.

Read the passage and answer the questions given below :

Question 1.
What was the pledge taken by Indians on 26th January 1930?
Answer:
On 26th January 1930, people all over the country took the pledge for carrying out of freedom movement with non-violent means for attainment of independence of India.

Question 2.
Why was the demand of Poorna Swaraj passed in Lahore session?
Answer:
1. The objective of Dominion Status was not acceptable to Pandit Nehru, Subash Chandra Bose and other young leaders.
2. Due to their influence, the resolution of demanding complete independence i.e. ‘Poorna Swaraj’ was passed in the Lahore session.

Question 3
Why was the Lahore session historic?
Answer:

The Indian National Congress gave up the objective of Dominion Status.
A resolution of complete independence was passed and it became the objective of the national movement.
A resolution was passed to observe 26th January as the Independence Day.

Answer the following in detail :

Question 1.
Give an account of the Jallianwala Baug Massacre.
Answer:
Points :
(a) Preceding events
(b) Massacre
(c) Reactions

(a) Preceding events :

The British government resorted to suppressive measures as the protest was more intense in Punjab province.
Amritsar became the centre of this movement.
Gandhiji was prohibited from entering the province of Punjab.
General Dyer had banned public meetings in Amritsar.
Dr. Satyapal and Dr. Kitchelu were arrested for their involvement in Amritsar Hartal case.

(b) Massacre :

Despite the ban, a public meeting was held at Jallianwala Baug in Amritsar on 13th April 1919 on the occasion of Baisakhi.
General Dyer deployed troops around the single gate that led to the ground. .
Without giving any warning, he ordered unrestrained firing on the unarmed people who had assembled for the meeting.
1660 rounds were fired. The firing continued until the ammunition was over.
About 400 men and women were killed.
Curfew was announced after the firing which made it impossible for medical help to reach.

Military law was imposed in entire Punjab. Many were imprisoned.
This led to discontent among the IndiAnswer:
Rabindranath Tagore gave away the title of ‘Sir’ given by the British.
As Indians demanded the enquiry, the British government appointed the Hunter Commission.

Question 2.
Give an account of the Non-co-operation.
Answer:

According to Gandhiji, the British rule lasted in India only because of the co-operation extended by the IndiAnswer:
If the Indians withdraw this co-operation and adopt complete non co-operation then the British government will collapse.
With this intention, Gandhiji devised and executed Non-co-operation Movement.
The Indian National Congress passed the resolution of Non-co-operation Movement at its Nagpur session in 1920.
The reins of the movement was given in the hands of Gandhiji.
According to this resolution there would be : (i) Boycott of government offices and courts, (ii) Boycott of government schools and colleges, (iii) Boycott of foreign goods and clothes.
Boycott and bonfire of foreign clothes was seen at several places and demonstration in front of shops selling such goods.
Motilal Nehru, Chittaranjan Das, etc. boycotted court by giving up their practice.
Schools and colleges imparting national education were started.
Elections were boycotted.
The movement spread from the planters in Assam to rail workers in Bengal.

Question 3.
Do you feel that the nation still needs the constructive programme started by Mahatma Gandhi? Why?
Answer:

Gandhiji implemented the constructive programme along with Non¬co-operation Movement.
It included Hindu-Muslim unity, prohibition of alcohol, removal of untouchability, etc.
In spite of completion of 70 years of independence, these problems still exist.
Communal and religious riots take place. It causes great loss of life and property.

Therefore it is necessary to implement constructive programme in present times.

8th Std History Questions And Answers:

Metallurgy Class 10 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Bhagya

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy Notes, Textbook Exercise Important Questions and Answers.

Std 10 Science Part 1 Chapter 8 Metallurgy Question Answer Maharashtra Board

Class 10 Science Part 1 Chapter 8 Metallurgy Question Answer Maharashtra Board

Question 1.
a. Alloy of sodium with mercury.
Answer:
Silver amalgam.

b.Molecular formula of common ore of aluminium.
Answer:
Al₂O₃.nH₂O

c. The oxide that forms salt and water by reacting with both acid and base.
Answer:
Aluminium oxide (Al₂O₃).

d. device used for grinding an ore.
Answer:
The device used for grinding an ore is grinding mill.

e. The nonmetal having electrical conductivity.
Answer:
Graphite having electrical conductivity.

f. The reagent that dissolves noble metals.
Answer:
Aqua regia is the reagent that dissolves noble metals like gold and platinum.

Question 2.
Make pairs of substances and their properties.

Column I	Column II
Substance	Property
(1) Potassium bromide	(a) Combustible
(2) Gold	(b) Soluble in water
(3) Sulphur	(c) No chemical reaction
(4) Neon	(d) High ductility
	(e) Magnetic ingredient

Answer:
(1) Potassium bromide – Soluble in water
(2) Gold – High ductility
(3) Sulphur – Combustible
(4) Neon – No chemical reaction

Question 3.
Identify the pairs of metals and their ores from the following.

Column I (ores)	Column II (metals)
(1) Bauxite	(a) Mercury
(2) Cassiterite	(b) Aluminium
(3) Cinnabar	(c) Tin
	(d) Copper

Answer:
(1) Bauxite – Aluminium
(2) Cassiterite – Tin
(3) Cinnabar – Mercury

Question 4.
Explain the terms.
a. Metallurgy
Answer:
Metallurgy: The process used for extraction of metals in their pure form from their ores, then metals are further purified by different methods of purification. All the process is called metallurgy.

b. Ores.
Answer:
Ores: The minerals from which metals are extracted profitably and conveniently are called ores.
Examples: Bauxite (Al₂O₃.H₂O), Cinnabar (HgS).

c. Minerals.
Answer:
Minerals: The naturally occurring compounds of metals along with other impurities are known as minerals.
Examples: Rocks are composed of mixtures of minerals. Talc and granite are minerals.

d. Gangue.
Answer:
Gangue: Ores contain metal compounds with some of the impurities like soil, sand, rocky material, etc. These impurities are called gangue.

Question 5.
Write scientific reasons.
a. Lemon or tamarind is used for cleaning copper vessels turned greenish.
Answer:

Copper undergoes oxidation in air to form black copper oxide. Copper oxide reacts slowly with carbon dioxide in air and gains a green coat. This green substance is copper carbonate.
Lemon and tamarind contain acid. The acid dissolves the green coating of basic copper carbonate present on the surface of a tarnished copper utensil and makes it shiny again.

b. Generally the ionic compounds have high melting points.
Answer:

The ionic compounds exist in solid state and are hard due to strong electrostatic force of attraction between oppositely charged ions.
The intermolecular force of attraction is high in ionic compounds and large energy is required to overcome it. Therefore, ionic compounds have high melting points.

c. Sodium is always kept in kerosene.
(OR)
Why is sodium stored in kerosene?
Answer:

Sodium reacts so vigorously with atmospheric oxygen that it catches fire if kept in the open.
It does not react with kerosene and sinks in it. Hence, to protect sodium and to prevent accidental fires it is always kept in kerosene.

d. Pine oil is used in the froth floatation process.
Answer:

In the concentration of an ore by froth floatation process, the ore is mixed with water and pine oil. When air is bubbled through the mixture a froth is formed.
The mineral particles in the ore are preferentially wetted by the oil and float on the top in the froth.
The gangue particles are wetted by water and settle down. Thus the mineral can be separated from the gangue and the ore is concentrated.

e. Anodes need to be replaced from time to time during the electrolysis of alumina.
Answer:

During electrolysis of alumina, the oxygen liberated at the carbon anode reacts with graphite rods (carbon anode) and forms carbon dioxide.
As the anodes get oxidised during electrolysis of alumina, they are continuously eroded. Hence, it is necessary to replace anodes from time to time.

Question 6.
When a copper coin is dipped in silver nitrate solution, a glitter appears on the coin after some time. Why does this happen? Write the chemical equation.
Answer:
When a copper coin is dipped in a silver nitrate solution, more reactive copper displaces silver from silver nitrate solution. The silver so liberated deposits on the copper coin. As a result, a shiny coat of silver is formed on the coin.
Cu_(s) + 2AgNO_3(aq) → Cu(NO₃)_2(aq) + 2Ag_(s)

Question 7.
The electronic configuration of metal ‘A’ is 2, 8, 1 and that of metal ‘B’ is 2, 8, 2. Which of the two metals is more reactive? Identify these metals. Write their reaction with dilute hydrochloric acid. (Practice Activity Sheet – 1)
Answer:
If the number of electrons in the outermost orbit is less, then the metal is more reactive. Metal A contains one electron in the outermost shell, while metal B contains two electrons. Hence, metal A is more reactive than metal B.

Metal A is sodium and metal B is magnesium. Reactions of Na and Mg with dil. HCl are,

Question 8.
Draw a neat labelled diagram.
a. Magnetic separation method.
Answer:

b. Forth floatation.
Answer:

c. Electrolytic reduction of alumina.
Answer:

d. Hydraulic separation method.
Answer:

Question 9.
Write chemical equation for the following events.
a. Aluminium came in contact with air.
Answer:
When aluminium is exposed to air, it develops a thin oxide layer of aluminium.

b. Iron filings are dropped in aqueous solution of copper sulphate.
Answer:
When iron filings are dropped in copper sulphate solution, more reactive iron displaces copper from copper sulphate solution. The iron filings get coated with reddish brown copper metal and the blue colour of copper sulphate fades gradually and ferrous sulphate is formed.

c. A reaction was brought about between ferric oxide and aluminium.
Answer:
The reaction between ferric oxide and iron produces aluminium oxide and iron. It is a thermite reaction and is highly exothermic.
It produces a large amount of heat, which is released to melt oxygen and aluminium. This reaction is used in welding of machineries. It is also used in warfare to make grenades.

The chemical reaction for the above is as follows:
3Fe₃O₂ + 4Al → 2Al₂O₃ + 6Fe

d. Electrolysis of alumina is done.
Answer:
During electrolysis of alumina, aluminium is deposited at the cathode. Molten aluminium being heavier than the electrolyte, is collected at the bottom of the tank. Oxygen gas is liberated at the anode.
Anode reaction: 2O^–^– → O₂ + 4e^– (Oxidation)
Cathode reaction: Al³⁺ + 3e^– → Al_(l) (Reduction)

e. Zinc oxide is dissolved in dilute hydrochloric acid.
Answer:
Zinc oxide is dissolved in dilute hydrochloric acid, zinc chloride and water are formed.

Question 10.
Complete the following statement using every given options.
During the extraction of aluminium
a. Ingredients and gangue in bauxite.
b. Use of leuching during the concentration of ore.
c. Chemical reaction of transformation of bauxite into alumina by Hall’s process.
d. Heating the aluminium ore with concentrated caustic soda.
Answer:
c. Chemical reaction of transformation of bauxite into alumina by Hall’s process.

Question 11.
Divide the metals Cu, Zn, Ca, Mg, Fe, Na, Li into three groups, namely, reactive metals, moderately reactive metals and less reactive metals.
Answer:
Reactive metals: Na, Li, Ca
Moderately reactive metals: Zn, Fe, Mg,
Less reactive metals: Cu

Project: (Do it your self)

Collect metal vessels and various metal articles. Write detailed informciton. write the steps in the procedure that can be done in the laboratory for giving glitter to these. Seek guidance from your teacher.

Can you recall? (Text Book Page No. 93)

Question 1.
what are the physical properties of metals and nonmetals?
Answer:
Properties of metals:

Solid state (Exception: Mercury and gallium)
Typical lustre
Malleability and ductility
Hardness (Exception: Lithium, sodium and potassium)
Good conductors of heat and electricity
High melting and boiling points (On the other hand, the melting and boiling points of the metals sodium, potassium, mercury and galium are very low.)
Sonorous and produce sound on striking a hard surface.

Properties or nonmetals:

Gaseous or solid state (Exception: Bromine in liquid state)
Lack of any typical lustre (Exception: Iodine and Diamond)
Brittleness in the solid state (Exception: Diamond is the hardest natural substance)
Bad conductors of heat and electricity (Exception: Graphite) (Diamond is good conductor of heat)
Low melting and boiling points.

Can you recall? (Text Book Page No. 102)

Question 1.
What is the electronic definition of oxidation and reduction?
Answer:
When a metal loses çlectrons the process is called an oxidation while when a nonmetal gains electrons, it is called a reduction,
Na → Na⁺ + e^– (oxidation)
Cl^– + e^– → Cl (reduction)

Can you recall? (Text Book Page No. 106)

Question 1.
What is meant by corrosion?
Answer:
Corrosion is degradation of a material due toreaction with its environment.

Question 2.
Have you observed the following things?
(1) Old iron bars in the builthngs.
Answer:
When old iron bars in the buildings are exposed to moist air for a long time, they acquire a coating of browm nlaky substance called rust. (Fe₂O₃.H₂O)

(2) Copper vessels not cleaned for a long time.
Answer:
If copper vessels are not cleaned for a long time, they react with moist carbon dioxide in the air, lose their shine and gain a green coat of copper carbonate. (CuCO3)

Question 3.
Silver ornaments or idols exposed to air for a long time.
Answer:
When silver ornaments or idols are kept exposed to air for a long time, silver reacts with sulphur in the air to form a coating of black silver sulphide. (Ag₂S)

Question 4.
Old vehicles fit to be thrown away.
Answer:
The metallic parts of the body of old cars are corroded, eaten up and sometimes become perforated. The old cars also lose the original colour due to formation of flakes of rust.

Use your brain power! (Text Book Page No. 98)

Question 1.
In the reaction between chlorine and HBr a transformation or HBr into Br₂ takes place. Can this transformation be called oxidation? What Is the oxidant that brings about this oxidation?
Answer:
The conversion of HBr to Br₂ is an oxidation process. In the above reaction, Cl₂ in the oxidant.

Can you tell? (Text Book Page No. 104)

Question 1.
what are the moderately reactive metals?
Answer:
In the middle of the reactivity series, metals such as iron. zinc, lead, copper are moderately reactive.

Question 2.
In which form to the moderately reactive metals occur in nature?
Answer:
The moderately reactive metals which occur in nature are in the form of their sulphide salts or carbonate salts.

Think about it (Text Book Page No. 106)

Question 1.
Why do silver articles turn blackish while copper vessels turn greenish on keeping in air for long time?
Answer:

Silver articles turn blackish on exposure to air for a long time. This is because of silver sulphide (Ag₂S) laver formed on the silver articles by the reaction of silver with hydrogen sulphide.
Carbon dioxide in moist air reacts with copper vessel. Copper loses its lustre due to formation of greenish layer of copper carbonate (CuCO₃) on its surface.

Question 2.
Why do pure gold and platinum always glitter?
Answer:
Gold and platinum are noble metals as they do not react with moisture, O₂ and CO₂ from air also acids and alkalis, therefore, pure gold and platinum always glitter.

Use your brain power! (Text Book Page No. 103)

Question 1.
Write the electrode reaction for electrolysis of molten magnesium chloride and calcium chloride.
Answer:
(1) Magnesium chloride (MgCl₂):
MgCl₂ → Mg²⁺ + 2Cl^–
At the cathode: Mg²⁺ + 2e^– → Mg
At the anode: 2Cl^– → Cl₂ + 2e^–

(2) Calcium chloride (CaCl2):
At the cathode: Ca²⁺ + 2e^– → Ca
At the anode: 2Cl^– → Cl₂ + 2e^–

Can you tell? (Text Book Page No. 106)

Question 1.
Which measures would you suggest to stop the corrosion of metallic articles or not allow the corrosion to start?
Answer:
Various types of methods are used to protect metals from corrosion. Almost in all the methods, special attention is paid so that iron does not rust. It is possible to lower the rate of the process of rusting of iron. Corrosion of metals can be stopped by detaching the air from metals.

Some methods are as follows :

To fix a layer of some substance on the metal surface so that the contact of the metal with moisture and oxygen in the air is prevented and no reaction would occur between them.
To prevent corrosion of metals by applying a layer of paint, oil, grease or varnish on their surface. For example, corrosion of iron can be prevented by this method.

Question 2.
What is done so to prevent rusting of iron windows and iron doors of your house?
Answer:
To prevent rusting of iron windows and iron doors in the house, they are painted so that they do not rust.

Question 3.
What is done so to prevent rusting or iron windows and iron doors of your house?
Answer:
To prevent rusting of iron windows and iron doors in the house, they are painted so that they do not rust.

Use your brain power! (Text Book Page No. 107)

Question 1.
Can we permanently prevent the rusting of an iron article by applying a layer of paint on its surface?
Answer:
The method of painting is alright for some time. We cannot protect the articles permanently from rusting by painting them.

Question 2.
Why do new iron sheets appear shiny?
Answer:
The new iron sheets appear shiny because a layer of non-corrosionable metal is fixed on the surface of corrosionable metal.

Collect information. (Text Book Page No. 108)

Question 1.
What are the various alloys used in daily life? Where are those used?
Answer:

Various alloys	Uses
1. Bronze	It is used to prepare: Coins, utensils, medals, statues
2. Brass	Pipes, condenser tubes, utensils worshipping God.
3. Stainless steel	Utensils, tools, dairy equipment, boilers.
4. Steel	Construction of bridges and buildings, cutting tools, blades.
5. Tungsten steel	High speed cutting tools
6. Amalgam	Silver amalgam used by dentists
7. Duralumin	Bodies of aircraft, buses, kitchenwares
8. Aluminium bronze	Pigment in ink and paint
9. German silver	Electrical heaters, resistors
10. Gun metal	Guns, boiler fittings
11. Magnelium	Beams of scientific balances, aircraft parts.
12. Gold with copper or nickel or silver or platinum	Jewellery

Question 2.
What are the properties that the alloy used for minting coins should have?
Answer:
The alloy used for minting coins should have excellent wear resistance and anti-corrosion properties.

Fill in the blanks:

Question 1.
……………has the highest melting point.
Answer:
Tungsten has the highest melting point.

Question 2.
Mercury and…………are two metals in the liquid state at room temperature.
Answer:
Mercury and galium are two metals in the liquid state at room temperature.

Question 3.
…………is the hardest natural substance.
Answer:
Diamond is the hardest natural substance.

Question 4.
The naturally occurring compounds of metals along with other impurities are known as………….
Answer:
The naturally occurring compounds of metals along with other impurities are known as minerals.

Question 5.
The minerals from which metals are extracted profitably and conveniently are called…………..
Answer:
The minerals from which metals are extracted profitably and conveniently are called ores.

Question 6.
An ore contains some of the impurities like soil, sand, etc. These impurities are called…………
Answer:
An ore contains some of the impurities like soil, sand, etc. These impurities are called gangue.

Question 7.
The process of extraction of a metal from its ore is called……….
Answer:
The process of extraction of a metal from its ore is called metallurgy.

Question 8.
Bauxite is a common ore of………..
Answer:
Bauxite is a common ore of aluminium.

Question 9.
…………. process is used for the purification of bauxite.
Answer:
Bayer’s process is used for the purification of bauxite.

Question 10.
During the electrolysis of alumina, ………..is liberated at the anode.
Answer:
During the electrolysis of alumina, oxygen is liberated at the anode.

Question 11.
The reaction of iron oxide with aluminium is known as…………..reaction.
Answer:
The reaction of iron oxide with aluminium is known as thermit reaction.

Question 12.
The process of coating a thin layer of zinc on iron is known as…………
Answer:
The process of coating a thin layer of zinc on iron is known as galvanising.

Question 13.
The metal that produces a sound on striking a hard surface is said to be………….
Answer:
The metal that produces a sound on striking a hard surface is said to be sonorous.

Question 14.
The process in which carbonate ores are changed into oxides by heating strongly in limited air is known as …………….
Answer:
The process in which carbonate ores are changed into oxides by heating strongly in limited air is known as calcination.

Question 15.
…………compounds are insoluble in solvents like kerosene and petrol.
Answer:
Ionic compounds are insoluble in solvents like kerosene and petrol.

Question 16.
…………… is used to obtain pure metals from impure metals.
Answer:
Electrolys is method is used to obtain pure metals from impure metals.

Question 17.
Corrosion can be prevented-by putting a layer of…………metal on corrosionable metal.
Answer:
Corrosion can be prevented by putting a layer of non-corrosionable metal on corrosionable metal.

Rewrite the following statements by selecting the correct options:

Question 1.
………… is a metal.
(a) Mg
(b) S
(c) P
(d) Br
Answer:
Mg is a metal.

Question 2.
………. is a nonmetal.
(a) Au
(b) Hg
(c) Br
(d) Cu
Answer:
Br is a nonmetal.

Question 3.
………… is a metalloid.
(a) Aluminium
(b) Antimony
(c) Zinc
(d) Mercury
Answer:
Antimony is a metalloid.

Question 4.
Metalloids have properties of ………..
(a) metals
(b) nonmetals
(c) both metals and nonmetals
(d) neither metals nor nonmetals
Answer:
Metalloids have properties of both metals and nonmetals.

Question 5.
…………. is a good conductor of electricity.
(a) Bromine
(b) Iodine
(c) Graphite
(d) Sulphur
Answer:
Graphite is a good conductor of electricity.

Question 6.
………. is a metal which is in liquid form at ordinary temperature and pressure.
(a) Magnesium
(b) Sodium
(c) Scandium
(d) Mercury
Answer:
Mercury is a metal which is in liquid form at ordinary temperature and pressure.

Question 7.
………. is an amphoteric oxide.
(a) Na₂O
(b) MgO
(c) ZnO
(d) SO₂
Answer:
ZnO is an amphoteric oxide.

Question 8.
……….. is an acidic oxide.
(a) Na₂O
(b) CO₂
(c) FeO₃
(d) H₂O
Answer:
CO₂ is an acidic oxide.

Question 9.
………. is a basic oxide.
(a) CO₂
(b) K₂O
(C) SO₂
(d) Al₂O₃
Answer:
K₂O is a basic oxide.

Question 10.
………… is an ore of aluminium.
(a) Cryolite
(b) Bauxite
(c) Haematite
(d) Aluminium carbonate
Answer:
Bauxite is an ore of aluminium.

Question 11.
Bronze is an alloy of ………..
(a) copper and tin
(b) copper and zinc
(c) copper and iron
(d) iron and nickel
Answer:
Bronze is an alloy of copper and tin.

Question 12.
An alloy prepared from iron, nickel and chromium is known as …………
(a) brass
(b) bronze
(c) stainless steel
(d) amalgam
Answer:
An alloy prepared from iron, nickel and chromium is known as stainless steel.

Question 13.
…………. is an allotropic form of a nonmetal which conducts electricity.
(a) Sulphur
(b) Graphite
(c) Chlorine
(d) Iodine
Answer:
Graphite is an allotropic form of a nonmetal which conducts electricity.

Question 14.
………….. has an oxide which is soluble in sodium hydroxide.
(a) Calcium
(b) Magnesium
(c) Iron
(d) Zinc.
Answer:
Zinc has an oxide which is soluble in sodium hydroxide.

Question 15.
………… prevents the rusting of iron.
(a) Copper
(b) Zinc
(c) Aluminium
(d) Silver
Answer:
Zinc prevents the rusting of iron.

Question 16.
………….. is obtained by the reduction of its oxide by carbon.
(a) Zinc
(b) Aluminium
(c) Sodium
(d) Potassium
Answer:
Zinc is obtained by the reduction of its oxide by carbon.

Question 17.
………….. is used as an anode during the electrolytic reduction of bauxite.
(a) Sulphur
(b) Graphite
(c) Platinum
(d) Aluminium
Answer:
Graphite is used as an anode during the electrolytic reduction of bauxite.

Question 18.
Silver gets corroded due to ………… in air.
(a) oxygen
(b) hydrogen sulphide
(c) carbon dioxide
(d) nitrogen
Answer:
Silver gets corroded due to hydrogen sulphide in air.

Question 19.
…………. is the hardest substance and has the highest melting and boiling points.
(a) Iodine
(b) Sulphur
(c) Diamond
(d) Phosphorus
Answer:
Diamond is the hardest substance and has the highest melting and boiling points.

Question 20.
Jewellery articles are gold plated ………….
(a) to prevent corrosion
(b) to prevent rusting of the base metal
(c) to make articles attractive
(d) all of these
Answer:
(d) all of these

Question 21.
To show that zinc is more reactive than copper, the correct procedure is to ………..
(a) prepare copper sulphate solution and dip a zinc strip in it
(b) prepare zinc sulphate solution and dip a copper strip in it
(c) heat together zinc and copper strips
(d) add dil. nitric acid to both the strips
Answer:
To show that zinc is more reactive than copper, the correct procedure is to prepare copper sulphate solution and dip a zinc strip in it.

Question 22.
Iron is ………
(a) more reactive than zinc
(b) more reactive than aluminuium
(c) less reactive than copper
(d) less reactive than aluminium
Answer:
Iron is less reactive than aluminium.

Question 23.
A solution of Al₂(SO₄)₃ in water is …………
(a) blue
(b) pink
(c) green
(d) colourless
Answer:
A solution of Al₂(SO₄)₃ in water is colourless.

Question 24.
A solution of ………… in water is blue in colour.
(a) CuSO₄
(b) FeSO₄
(c) ZnSO₄
(d) Al₂(SO₄)₃
Answer:
A solution of CuSO₄ in water is blue in colour.

Question 25.
A solution of …………. n water is green in colour.
(a) CuSO₄
(b) FeSO₄
(c) ZnSO₄
(d) Al₂(SO₄)₃
Answer:
A solution of FeSO₄ in water is green in colour.

Question 26.
What would be the correct order if Zn, Fe, Al and Cu are arranged in increasing order of reactivity?
(a) Cu, Fe, Zn, Al
(b) Al, Cu, Fe, Zn
(c) Zn, Al, Cu, Fe
(d) Fe, Zn, Al, Cu (Practice Activity Sheet – 2)
Answer:
(a) Cu, Fe, Zn, Al

Question 27.
During the extraction of aluminium ……….
(a) Ingredients and gangue in bauxite
(b) Use of leaching during the concentration of ore
(c) Chemical reaction of transformation of bauxite into alumina by Hall’s process.
(d) Heating the aluminium ore with concentrated caustic soda.
Answer:
During the extraction of aluminium Chemical reaction of transformation of bauxite into alumina by Hall’s process.

Question 28.
A solution of CuSO₄ in water is ………… in colour.
(a) pink
(b) blue
(c) colourless
(d) green
Answer:
A solution of CuSO₄ in water is blue in colour.

Question 29.
Which of the following process is to be carried out to avoid the formation of greenish layer on brass vessels due to corrosion?
(a) Plating
(b) Anodization
(c) Tinning
(d) Alloying (Practice Activity Sheet – 3)
Answer:
(c) Tinning

State whether the following statements are True or False (If a statement is false, correct it and rewrite it.):

Question 1.
Metals are known as sonar metals.
Answer:
True.

Question 2.
Diamond is the softest natural substance.
Answer:
False. (Diamond is the hardest natural substance.)

Question 3.
Electrolysis method is used to obtain pure metals from impure metals.
Answer:
True.

Question 4.
Iodine and diamond are lustrous substances.
Answer:
True.

Question 5.
Aqua Regia is a mixture of conc. HCl and conc. HNO₃ in the ratio of 1:3.
Answer:
False. (Aqua Regia is a mixture of conc. HCl and conc. HNO₃ in the ratio of 3:1.)

Question 6.
Corrosion of metals can be stopped by detaching the air from metals.
Answer:
True.

Question 7.
Due to corrosion a greenish layer forms on the surface of copper or brass vessel.
Answer:
True.

Question 8.
Ionic compounds are soluble in kerosene.
Answer:
False. (Ionic compounds are soluble in water and insoluble in kerosene.)

Question 9.
Ionic compounds in the solid state conduct electricity.
Answer:
False. (Ionic compounds in the solid state do not conduct electricity.)

Question 10.
Mercury, silver and gold are very reactive metals.
Answer:
False. (Mercury, silver and gold are least reactive metals.)

Question 11.
In electroplating a metal is coated with another metal using electrolysis.
Answer:
True.

Question 12.
In anodising method. the copper or aluminium article is used as anode.
Answer:
True.

Question 13.
Silver plated spoon, gold plated ornaments are the examples of alloying.
Answer:
False. (Silver plated spoon, gold plated ornaments are the examples of electroplating)

Question 14.
silver amalgam is mainly used by dentists.
Answer:
True.

Question 15.
Aluminium oxide is an acidic oxide.
Answer:
False. (Aluminium oxide is an amphoteric oxide.)

Question 16.
copper reacts with moist carbon form copper carbonate.
Answer:
True.

Question 17.
Corrosion is degradation of a reaction with its environment.
Answer:
True.

Find the correlation in the given pair and rewrite the answer:

Question 1.
Brass : Copper and Zinc :: Bronze :………….
Answer:
Brass : Copper and Zinc :: Bronze : Copper and tin

Question 2.
Tinning : Tin :: Galvanizing :…………….
Answer:
Tinning : Tin :: Galvanizing : Zinc

Question 3.
Pressure cooker : Anodizing :: Silver plated spoons :…………..
Answer:
Pressure cooker : Anodizing :: Silver plated spoons : Electro-plating

Question 4.
The sulphides ores are strongly heated in air : Roasting :: The carbonates ores are strongly heated in a limited supply of air :………….
Answer:
The sulphides ores are strongly heated in air : Roasting :: The carbonates ores are strongly heated in a limited supply of air : Calcination.

Question 5.
Sulphide ores : Froth floatation method : Cassiterite ore :………..
Answer:
Sulphide ores : Froth floatation method : Cassiterite ore : Magnetic separation method.

Find the odd one out:

Question 1.
Sodium, Potassium, Silver, Sulphur
Answer:
Sulphur. (All except sulphur, others are metals.)

Question 2.
Boron, Chlorine, Bromine, Fluorine
Answer:
Boron. (All except boron, others are nonmetals.)

Question 3.
Copper, Iron, Mercury, Brass
Answer:
Brass. (All except brass, others are metals.)

Question 4.
Brass, Bronze, Phosphorus, Stainless steel
Answer:
Phosphorus. (All except phosphorus, others are alloys.)

Question 5.
Magnesium chloride, Sodium chloride, Water, Zinc chloride
Answer:
Water. (All except water, others are ionic compounds.)

Question 6.
Tinning, Anodization, Alloying, Froth floatation (March 2019)
Answer:
Froth floatation. (All except froth floatation, others are processes of coating a thin layer of metal on the surface of other metals.)

Match the following:

Question 1.

Column I	Column II
(1) ZnS	(a) Cuprous sulphide
(2) HgS	(b) Bauxite
(3) Cu₂S	(c) Zinc blend
(4) Al₂O₃.H₂O	(d) Cinnabar
	(e) Cryolite

Answer:
(1) ZnS – Zinc blend
(2) HgS – Cinnabar
(3) Cu₄S – Cuprous sulphide
(4) Al₂O₃.H₂O – Bauxite.

Question 2.

Column I	Column II
(1) Copper and zinc	(a) Stainless steel
(2) Copper and tin	(b) Zinc amalgam
(3) Iron, nickel and chromium	(c) Bronze
(4) Mercury and zinc	(d) Brass
	(e) Steel

Answer:
(1) Copper and zinc – Brass
(2) Copper and tin – Bronze
(3) Iron, nickel and chromium – Stainless steel
(4) Mercury and zinc – Zinc amalgam.

Question 3.

Column I	Column II
(1) Galvanising	(a) Pressure cooker
(2) Tinning	(b) Silver plated spoons
(3) Electroplating	(c) Coating of tin on copper
(4) Anodizing	(d) Coating of Zn on iron

Answer:
(1) Galvanising – Coating of Zn on iron
(2) Tinning – Coating of tin on copper
(3) Electroplating – Silver plated spoons
(4) Anodizing – Pressure cooker.

Translate the following statements into chemical equations and then balance them:

Question 1.
steam is passed over aluminium.
Answer:

Question 2.
Extraction of copper from its sulphide ore.
Answer:

Question 3.
Thermit reaction.
Answer:

Question 4.
Magnesium reacts with hot water.
Answer:

Question 5.
what happens when aluminium oxide dissolves in aqueous sodium hydroxide?
Answer:

Question 6.
Zinc reacts with sulphuric acid.
Answer:

Question 7.
Iron reacts with sulphuric acid.
Answer:

Name the following:

Question 1.
A metal which forms an amphoteric oxide.
Answer:
Aluminium forms an amphoteric oxide.

Question 2.
An alloy of copper and zinc.
Answer:
An alloy of copper and zinc is termed as brass.

Question 3.
A compound which is added to lower the fusion temperature.
Answer:
Cryolite (AlF₃, 3NaF) and fluorspar (CaF₂) are added to lower the fusion temperature.

Question 4.
A metal which does not react with cold water but reacts with steam.
Answer:
Aluminium does not react With cold water but reacts with Steam.

Question 5.
A common ore of aluminium.
Answer:
Bauxite (Al₂O₃.H₂O) is a common ore of aluminium.

Question 6.
A metal which is in liquid state at ordinary temperature.
Answer:
Mercury is in liquid state at ordinary temperature.

Question 7.
Two metals which are malleable.
Answer:
Iron and aluminium are malleable metals.

Question 8.
Two metals which are ductile.
Answer:
Gold and silver are ductile metals.

Question 9.
Two metals which are good conductors of heat.
Answer:
Silver and copper are good conductors of heat.

Question 10.
Two metals which are good conductors of electricity.
Answer:
Copper and aluminium are good conductors of electricity.

Question 11.
Two metals which are used for making cooking vessels.
Answer:
Copper and aluminium are used in making cooking vessels.

Question 12.
Two metals having low melting points.
Answer:
Sodium and potassium have low melting points.

Question 13.
Two highly reactive metals.
Answer:
Sodium and potassium are highly reactive metals.

Question 14.
A nonmetal which is in liquid state at room temperature.
Answer:
Bromine is in liquid state at room temperature.

Question 15.
Two ionic compounds.
Answer:
Sodium chloride (NaCl) and magnesium chloride (MgCl₂) are ionic compounds.

Question 16.
The process of heating the sulphide ore to a high temperature in the excess of air.
Answer:
In roasting, sulphide ore is heated to a high temperature in the excess of air.

Question 17.
The process of heating the carbonate ore to a high temperature in limited air.
Answer:
In calcination, carbonated ore is heated to a high temperature in limited air.

Question 18.
The compound formed by the reaction between aluminium oxide and sodium hydroxide.
Answer:
Sodium aluminate is formed by the reaction between aluminium oxide and sodium hydroxide.

Question 19.
Two metals which are found in the free state in nature.
Answer:
Gold (Au) and silver (Ag) are found in the free state in nature.

Question 20.
A metal which has the highest melting point.
Answer:
Tungsten has the highest melting point.

Question 21.
Two nonmetals which are lustrous.
Answer:
Iodine and diamond are lustrous in nature.

Answer the following questions in one sentence each:

Question 1.
State the property of the metals due to which they can be drawn into wires.
Answer:
The property of the metal due to which they can be drawn into wires is called ductility.

Question 2.
State the property of the metals due to which they can be beaten into thin sheets.
Answer:
The property of the metals due to which they can be beaten into thin sheets is called malleability.

Question 3.
Which is the hardest substance?
Answer:
Diamond which is a form of carbon is the hardest substance.

Question 4.
What material is used to coat electrical wires?
Answer:
PVC (Polyvinyl chloride) is used to coat electrical wires.

Question 5.
State two metals which can be cut easily with a knife.
Answer:
Sodium and potassium are soft metals and can be cut easily with a knife.

Question 6.
Which of the following metals react with cold water?
Sodium, iron, copper, potassium.
Answer:
Sodium and potassium metals react with cold water.

Question 7.
Which of the following metals do not react with cold water or hot water?
Sodium, potassium, aluminium, iron.
Answer:
Aluminium and iron do not react with cold water or hot water.

Question 8.
State two metals which displace hydrogen from dilute acids and two metals which do not do so.
Answer:
Metals which displace hydrogen from dilute acids are: Magnesium and zinc.
Metals which do not displace hydrogen from dilute acids are: Copper and silver.

Question 9.
Arrange the following metals in the increasing order of their activity:
Copper, Silver, Aluminium, Iron. (Practice Activity Sheet – 1)
Answer:
The arrangement of metals in the increasing order of their activity:
Silver < Copper < Iron < Aluminium

Question 10.
Write the chemical equation for the reaction of hot iron with steam.
Answer:

Question 11.
Complete the following reactions:
(1) Zn_(s) + H₂O_(g) → _____________

Answer:
(1) Zn_(s) + H₂O_(g) → ZnO_(s) + H₂_(g)

Question 12.
Complete the following reactions:

Answer:

Question 13.
3MnO₂ + 4Al → 3Mn + 2Al₂O₃ + heat.
Identify the substances undergone oxidation and reduction reactions.
Answer:
MnO₂ is reduced to Mn.
Al is oxidised to Al₂O₃

Question 14.
State the impurities present in the bauxite ore.
Answer:
The main impurities present in the bauxite ore are silica (SiO₂) and iron oxide (Fe₂O₃).

Question 15.
Write the formula of (i) bauxite (ii) cryolite.
Answer:
The formula of bauxite is Al₂O_3·H₂O and that of cryolite is (Na₃AlF₆).

Question 16.
What is galvanization?
Answer:
The process of coating a thin layer of zinc on iron or steel is called galvanization.

Question 17.
Name the reaction in which aluminium is used as a reducing agent.
Answer:
The thermite reaction in which iron oxide is reduced by aluminium. Aluminium is used as a reducing agent in the thermit reaction.

Question 18.
What are the constituents of bronze?
Answer:
Copper and tin are the constituents of bronze.

Question 19.
State the term used to express the purity of gold.
Answer:
The purity of gold is expressed in carat.

Question 20.
What is meant by amalgam?
Answer:
The amalgam is an alloy in which one of the metals is mercury.

Question 21.
What is meant by electroplating?
Answer:
A process in which a less reactive metal is coated on a more reactive metal by electrolysis is called electroplating.

Question 22.
Why are metals called electropositive elements?
Answer:
Metals are reactive. They lose electrons and become positively charged ions. Therefore, metals are called electropositive elements.

Answer the following questions:

Question 1.
Distinguish between the physical properties of metals and nonmetals with respect to the following points:
(1) Physical state (2) Lustre (3) Ductility and malleability (4) Conduction of heat and electricity (5) Hardness (6) Melting and boiling points.
Answer:
(1) Physical state: Under ordinary conditions, metals are generally solids. Exceptions: mercury and gallium are liquids. Under ordinary conditions, nonmetals may be solids or gases. Exception: bromine is in liquid state.

(2) Lustre: Metals usually have a high lustre (called metallic lustre). They can be polished to give a highly reflective surface. With the exceptions of gold and copper, metals usually have silvery grey colour. Nonmetals lack lustre, exceptions: graphite and iodine. Some nonmetals are colourless and others possess a variety of colours.

(3) Ductility and malleability: Metals are ductile and malleable. Nonmetals are not ductile and mallfeable.

(4) Conduction of heat and electricity: Metals are good conductors of heat and electricity. Nonmetals are bad conductors of heat and electricity. Exception: Graphite is a good conductor of electricity.

(5) Hardness: Metals are usually hard, but not brittle, exceptions: sodium, potassium, lead, zinc. Nonmetals are brittle in the solid state, exception: diamond.

(6) Melting and boiling points: The melting and boiling points of metals are high, exceptions: sodium, potassium, mercury, gallium. The melting and boiling points of nonmetals are low, exceptions: carbon, silicon.

Question 2.
Write any three physical properties of nonmetals.
Answer:

Nonmetals may be solid or gaseous.
Nonmetals lack lustre. They are not ductile and malleable.
The melting and boiling points of nonmetals are low.
Nonmetals are bad conductors of heat and electricity.

Question 3.
Metals are good conductors of heat. Explain why.
Answer:
(1) The electrons in the outermost shells of atoms of a metal are free to move throughout the metal.
(2) When a metal is heated, these electrons start moving with higher velocity and conduct heat. Hence, metals are good conductors of heat.

Question 4.
Metals are good conductors of electricity. Explain why.
Answer:
(1) The electrons in the outermost shells of atoms of a metal are free to move throughout the metal.
(2) When a potential difference is applied between the ends of a metal wire, the net movement of the electrons in a particular direction, from a point at lower potential to a point at higher potential, constitutes an electric current. Hence, metals are good conductors of electricity.

Question 5.
A metal can be drawn into a wire. Explain why.
Answer:

The property due to which a substance can be drawn into a thin wire without cracking or breaking is called ductility.
Metals are ductile. Thus, a metal can be drawn into a wire.

Question 6.
A metal can be hammered into a thin sheet. Explain why.
Answer:

The property due to which a substance can be hammered (or rolled) into a thin sheet without cracking is called malleability.
Metals are malleable. Thus, a metal can be hammered to form a thin sheet.

Question 7.
How do metals react with oxygen?
Answer:
Metals combine with oxygen on heating in air and metal oxides are formed.
Metal + Oxygen → Metal oxide

Question 8.
How does a metal react with water?
Answer:
Sodium and potassium react vigorously with water to evolve hydrogen. Calcium reacts with water slowly and less vigorously to evolve hydrogen and the metal floats on water. Magnesium reacts with hot water to evolve hydrogen. Aluminium, iron and zinc do not react with cold or hot water but they react with steam to evolve their oxides and hydrogen.
2Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g) + heat energy
2K_(s) + 2H₂O_(l) → 2KOH_(aq) + H_2(g) + heat energy
2Ca_(s) + 2H₂O_(l) → 2Ca(OH)_2(aq) + H_2(g)
Mg_(s) + 2H₂O_(hot) → Mg(OH)_2(aq) + H_2(g)
2Al_(s) + 3H₂O _steam → Al₂O₃_(s) + 3H_2(g)
3Fe_(s) + 4H₂O _steam → Fe₃O₄ + 4H_2(g)
Zn_(s) + H₂O _steam → ZnO_(s) + H_2(g)

Question 9.
(9) How does a metal react with an acid?
Answer:
Reaction of metals with acids: Metals react with dilute hydrochloric acid or dilute sulphuric acid to form metal chloride or metal sulphate and hydrogen gas. The rate of evolution of H2 is maximum in case of magnesium. The reactivity decreases in the order
Mg > Al > Zn > Fe.
Mg_(s) + 2HCl_(aq) → MgCl₂_(s) + H_2(g)
2Al_(s) + 6HCl_(aq) → 2AlCl_3(aq) + 3H_2(g)
Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)
Fe_(s) + 2HCl_(aq) → FeCl_2(aq) + H_2(g)
Fe_(s) + H₂SO_4(aq) → FeSO₄ + H_2(g)
Zn_(s) + H₂SO_4(aq) → ZnSO₄ + H_2(g)
Mg_(s) + H₂SO_4(aq) → MgSO_4(aq) + H_2(g)

Question 10.
How does a metal react with nitric acid?
Answer:
Metals react with nitric acid to form nitrate salts. Depending on the concentration of nitric acid, various oxides of nitrogen (NO, NO₂) are formed.

Question 11.
Arrange the following metals in the decreasing order of chemical reactivity:
Cu, Mg, Fe, Ca, Zn, Na.
Answer:
The reactivity of metal decreases in the following order:
Na > Mg > Ca > Zn > Fe > Cu.

Question 12.
What is meant by aqua regia?
Answer:
Aqua regia is a highly corrosive and fuming liquid. It is a freshly prepared mixture of conc. HCl and conc. HNO₃ in the ratio of 3:1. Most of the substances dissolve in it. Aqua regia is a reagent which dissolves gold and platinum.

Question 13.
How does a metal react with salts of other metals?
Answer:
The reaction of metals with solutions of salts of other metals is the displacement reaction. If a metal A displaces other metal B from the solution of its salt, it means that the metal A is more reactive than the metal B.
Metal A + Salt solution of metal B → Salt solution of metal A + metal B

In this reaction Fe has displaced Cu from CuSO₄. It means Fe is more reactive than Cu.

Question 14.
Explain the reactivity series of the metals.
Answer:
The arrangement of metals in decreasing order of their reactivity in the form of a series is called the reactivity or activity series of the metals.
The most reactive metal is placed at the top of the list and least reactive metal is placed at the bottom of the list.

On the basis of reactivity, we can classify metals into the following categories:

High reactivity metals
Moderately reactive metals
Less reactive metals.

1. Extraction of High reactivity metals: The metals which are placed at the top of the reactivity series are very reactive. They are never found in nature as free elements, e.g., sodium, potassium, calcium and aluminium. These metals are obtained by electrolytic reduction.

2. Extraction of Moderately reactive metals: The metals in the middle of reactivity series such as iron, zinc, lead, copper are moderately reactive. These elements are present as sulphides or carbonates in nature. Generally metals are obtained from their oxide as compared to their sulphides and carbonates.

3. Extraction of Less reactive metals: The metals which are placed at the bottom of the reactivity series are least reactive. They occur in free state, e.g. gold, silver and copper. Copper and silver are also found in the combined state as sulphide and oxide ores. These metals are obtained from their ores by just heating the ores in air.

Question 15.
Atomic number of metal “A” is 11, while atomic number of metal “B” is 20. Which of them will be more reactive? Write the chemical reaction of dilute HCl with metal “A”. (Practice Activity Sheet – 2)
Answer:
Metal ‘A’ is more reactive than metal ‘B’.
Atomic number of metal ‘A’ is 11, hence it is Na.
2Na + 2HCl → 2NaCl_(aq) + H_2(g)

Question 16.
How does a metal react with a nonmetal?
Answer:
By oxidation of a metal, cations are formed, on the other hand by reduction of a nonmetal, anions are formed. The ionic compound is formed due to the metal losing electrons while the nonmetal accepts the electrons. The ionic compound of sodium chloride is formed as sodium loses one electron while chlorine accepts one electron.

Similarly Mg and K form ionic compounds MgCl₂ and KCl.

Question 17.
How do nonmetals react with oxygen?
Answer:
Nonmetals combine with oxygen to form acidic oxides. In some cases, neutral oxides are formed.

Question 18.
How do nonmetals react with water?
Answer:
Nonmetals do not react with water, (exception : halogen). Chlorine dissolves in water giving hypochlorous acid.

Question 19.
How do nonmetals react with dilute acids?
Answer:
Nonmetals do not react with dilute acid, (exception: halogen). Chlorine reacts with dil. hydrobromic acid to form bromine and HCl.

Question 20.
How do nonmetals react with hydrogen?
Answer:
Nonmetals react with hydrogen under certain conditions (such as proper temperature, pressure, catalyst, etc.)

Question 21.
What is meant by an ionic compound?
Answer:
The compound formed from two units, namely cation and anion is called an ionic compound.

Question 22.
What is meant by an ionic bond?
Answer:
The cation and anion being oppositely charged, there is an electrostatic force of attraction between them, this force of attraction between cation and anion is called the ionic bond.

Question 23.
State the general properties of ionic compounds.
Answer:

Ionic compounds are solids and hard due to strong electrostatic force of attraction between oppositely charged ions.
They are generally brittle. When pressure is applied they break into pieces.
They have high melting and boiling points, due to intermolecular force of attraction is high in ionic compounds.
They are soluble in water and insoluble in solvents such as kerosene and petrol.
Ionic compounds cannot conduct electricity when in solid state, they are electrically neutral. They conduct electricity in the molten state and also in an aqueous solution.

Question 24.
Explain the following terms:
1. Concentration of ores
2. Roasting
3. Calcination
4. Refining
Answer:
1. Concentration of ores: The process of separating gangue from the other ores is called concentration of ores.
2. Roasting: The process of heating an ore to a high temperature in excess of air and converting it into its oxide is called roasting.
Examples: ZnS (zinc blend), PbS (Galena)

3. Calcination: The process of heating an ore in a limited supply of air and converting it into its oxide is called calcination.
Example: Zinc carbonate (ZnCO₃)
ZnCO₃ → ZnO + CO₂↑
4. Refining: The metal obtained by chemical reduction contains impurities. The process of electrolysis method is used to obtain pure metals from impure metals is known as refining.

Question 25.
State two methods of concentration of ores in which the heavy particles of ores can be separated from the light gangue particles by the gravitational method.
Answer:

Wilfley table method
Hydraulic separation method are two methods of concentration of ores in which the heavy particles of ores can be separated from the light, gangue particles by the gravitational method.

Question 26.
What are the different methods used for removing gangue from ores?
(OR)
Write the five methods of concentration of ores.
Answer:

Wilfley table method
Hydraulic separation method
Magnetic separation method
Froth floatation method
Leaching method.

Question 27.
Write short notes on: (1) Wilfley table method (2) Hydraulic separation method (3) Magnetic separation method (4) Froth floatation method (5) Leaching method.
Answer:
(1) Wilfley table method : (Separation based on gravitation) This method of separation uses the Wilfley table, it is made by fixing narrow and thin wooden wedges/blocks on inclined surface with low slope. The table is kept continuously vibrating.

Lumps of the ore is made powdered ore by using ball mill. This powdered ore is poured on the table and a stream of water is simultaneously released from the upper side. This result in the lighter gangue particles getting carried away along with the flowing water, while the heavier particles in which proportion of minerals is more and proportion of gangue particles is less, are blocked by the wooden wedges and is collected through the slits between them.

(2) Hydraulic separation method: The hydraulic separation method is based on the working of a mill. This is a tapering vessel similar to that used in a grinding mill. It opens in a tank like a container that is tapering on the lower side. The tank has an outlet for water on the upper side and a water inlet on the lower side.

Finely ground ore is added to the tank. A fast stream of water is released in the tank from the lower side. The lighter gangue particles flow out along with the water stream from the outlet on the upper side of the tank and are collected separately, simultaneously the heavy particles of the ore are collected at the bottom from the lower side of the tank. This method is based on the law of gravitation, wherein particles of the same size are separated by their weight with the help of water.

(3) Magnetic separation method: Electro-magnetic machine is used in this method. The main parts of this machine are two types of iron rollers and the conveyor belt continuously moving around them. One of the rollers is nonmagnetic while the other is electromagnetic. The conveyor belt moving around the rollers is made up of leather or brass (nonmagnetic). The powdered ore is poured at that end of the conveyor belt which is on the side of the nonmagnetic roller. Two collector vessels are placed below the magnetic roller.

The particles of the nonmagnetic part in the ore are not attracted towards the magnetic roller. Therefore, they are carried out further along the belt and fall in the collector vessel which is away from the magnetic roller. Simultaneously the particles of the magnetic ingredients of the ore stick to the magnetic roller and therefore fall in the collector vessel near the belt.

In this way the magnetic and nonmagnetic particles in the ore are separated because of their magnetic nature. For example, cassiterite is a tin ore. It contains mainly the nonmagnetic ingredient stannic oxide (SnO₂) and the magnetic ingredient ferrous tungstate (FeWO₄). These are separated by the electromagnetic method.

(4) Froth floatation method: The froth floatation method is based on the two opposite properties, hydrophilic and hydrophobic, of the particles. The metal sulphides particles get wet mainly with oil due to their hydrophobic property. The gangue particles get wet with water due to the hydrophilic property.

In this method the finely ground mineral is put into a big tank containing a lot of water. The finely powdered ore and vegetable oil such as pine oil, eucalyptus oil are mixed with water for formation of froth. The pressurised air is blown through the mixture. There is an agitator rotating around its axis in the centre of the floatation tank. The agitator is used as per the requirement. Bubbles are formed due to the blown air.

A foam is formed from oil, water and air bubbles together, due to the agitating. This foam rises to the surface of the water and floats. Hence this method is called froth floatation. Sulphide minerals float with the foam on water as they get and can be removed. The gangue particles are wetted by water, settles down at the bottom. This method is used for concentration of zinc blend (ZnS) and copper pyrite (CuFeS₂).

(5) Leaching: Leaching is the first step in the extraction of the metals like aluminium, gold and silver from their ores. In this method the ore is soaked in a particular solution for long time. The ore dissolves in that solution due to specific chemical reaction. The gangue, however, does not react and therefore does not dissolve in that solution. It can be separated easily.

For example, concentration of bauxite, the aluminium ore, is done by leaching method. Bauxite is soaked in aqueous NaOH or aqueous Na₂CO₃ which dissolves the main ingredient alumina in it. This means that bauxite is leached by sodium hydroxide.

Question 28.
Draw a neat labelled diagram of the arrangement of the equipment used in (1) Wilfley table method (2) Hydraulic separation method.
Answer:
1. Wilfley table method:

2. Hydraulic separation method:

Question 29.
Complete the following flow chart and answer the questions below:

(i) In which method pine oil is used?
(ii) Explain that method of concentration in brief. (Practice Activity Sheet – 2)
Answer:

(i) Pine oil is used in froth floatation method.

(ii) The finely powdered ore and vegetable oil such as pine oil, eucalyptus oil are mixed with water for formation of froth. The pressurised air is blown through the mixture. The agitator is used as per the requirement. Bubbles are formed due to the blown air. A foam is formed from oil, water and air bubbles together, due to the agitating. This foam rises to the surface of the water and floats. Hence this method is called froth floatation. Sulphide minerals float with the foam on water as they get and can be removed. The gangue particles are wetted by water, settles down at the bottom. This method is used for concentration of zinc blend (ZnS) and copper pyrite (CuFeS₂).

Question 30.
A tapping vessel opens in a tank like container that is tapering on the lower side. The tank has an outlet for water on the upper side and a water inlet on the lower side. Finely ground ore is released in the tank. A forceful jet of water is introduced in the tank from lower side and gangue particles and pure ore are separated by this method.
(i) The above description is of which gravitation separation method?
(ii) Draw labelled diagram of this method. (March 2019)
Answer:
(i) Hydraulic separation method.
(ii)

Question 31.
How are sodium, magnesium and potassium obtained from their molten chloride salts?
Answer:
The metals sodium, calcium and magnesium are obtained by electrolysis of their molten chloride salts. In this process metal is deposited on the cathode while chlorine gas is liberated at the anode.

Question 32.
Name the main ore of aluminium.
Answer:
Bauxite (Al₂O₃·H₂O) is the main ore of aluminium.

Question 33.
What is bauxite? What are the main impurities found in this ore?
Answer:
Bauxite (Al₂O₃·H₂O) is hydrated aluminium oxide. It contains 30% to 70% Al₂O₃. The main impurities present in it are iron oxide (Fe₂O₃) and sand (SiO₂).

Question 34.
From which ore is aluminium extracted? What are the stages in its extraction (give only names)?
Answer:
Aluminium is extracted from bauxite (Al₂O₃·nH₂O). Stages id the extraction: (i) Concentration of ore, i.e., conversion of bauxite into alumina, (ii) Electrolytic reduction of alumina.

Question 35.
Describe Bayer’s process for concentration of bauxite.
Answer:
(1) Bayer’s process is used to obtain pure aluminium oxide from bauxite.
(2) Bauxite is then concentrated by chemical separation. Bauxite contains impurities like iron oxide (Fe₂O₃) and silica (SiO₂).
(3) Bauxite ore is powdered and heated with sodium hydroxide under high pressure for 2 to 8 hours at 140 °C in the digester. The aluminium oxide being amphoteric in nature present in bauxite reacts with sodium hydroxide to form water soluble sodium aluminate. This means that bauxite leached by sodium hydroxide. Silica reacts with sodium hydroxide to form soluble sodium silicate. The basic iron oxide (Fe₂O₃) in the gangue remains unaffected. It is separated by filtration.

(4) The filtrate containing sodium aluminate and sodium silicate is stirred with water and then cooling to 50° C. It is hydrolysed to give precipitate of aluminium hydroxide.

(5) Aluminium hydroxide is then filtered, washed with water, dried and then calcinated by heating at 1000 °C to get pure aluminium oxide.

Question 36.
Describe Hall’s process for concentration of bauxite.
Answer:
In Hall’s process the ore is powdered and then it is leached by heating with aqueous sodium carbonate in the digester to form water soluble sodium aluminate. Then the insoluble impurities are filtered opt. The filtrate is warmed and neutralised by passing carbon dioxide gas through it. This result in precipitation of aluminium hydroxide.

The precipitate of Al(OH)₃ obtained in this processes is filtered, washed, dried and then calcinated by heating at 1000 °C to obtain alumina.

Question 37.
Describe the process of preparation of aluminium by the electrolysis of alumina.
(OR)
Draw and label the diagram of electrolysis of alumina and explain the electrolytic reduction of alumina.
Answer:
Electrolytic reduction of alumina:
(1) The electrolytic cell consists of a rectangular steel tank lined from inside with graphite.

(2) The carbon lining (graphite) acts as a cathode. The anode consists of graphite rods suspended in the molten electrolyte.
(3) Alumina has very high melting point ( > 2000 °C). The electrolysis of alumina is carried out at a low temperature by dissolving it in molten cryolite (Na₃AlF₆). The solution of alumina in cryolite and small amount of fluorspar (CaF₂) is added in the mixture to lower its melting point up to 1000 °C.
(4) On passing an electric current, alumina is electrolysed.
(5) Molten aluminium is collected at the cathode, while oxygen gas is evolved at the anode.
The electrode reactions are shown below:
Al₂O₃ → 2Al³⁺ + 3O^2-
Anode reaction: 2O^2- → O_2(g) + 4e^–
Cathode reaction: Al³⁺ + 3e^– → Al
The molten aluminium is heavier than the electrolyte. Therefore, it sinks to the bottom of the electrolyte and is removed from time to time. About 99% pure aluminium is obtained by this process. The oxygen gas liberated reacts with carbon anode and forms carbon dioxide. As the anode gets oxidised during the electrolysis of alumina, it has to be replaced from time to time.

Question 38.
In the extraction of aluminium:
(i) Name the process of concentration of bauxite.
Answer:
The process of concentration of bauxite is known as Bayer’s process.

(ii) Write the cathode reaction in electrolytic reduction of alumina.
Answer:
At the cathode: Al³⁺ + 3e^– → Al.

(iii) Write the function and formula of cryolite in the extraction of aluminium.
Answer:
Cryolite is added to the molten mixture of alumina to reduce the melting point to about 1000 °C.
The formula of cryolite is (Na₃AlF₆) or AlF₃, 3NaF.

(iv) write an equation for the action of heat on aluminium hydroxide.
Answer:

(v) Draw the diagram of extraction of aluminium.
Answer:

(vi) Write the anode reaction in electrolytic reduction of alumina.
Answer:
Al₂O₃ → 2Al³⁺ + 3O^2-
At Anode: 2O^2- → O_2(g) + 4e^–

(vii) Write the cathode reaction in electrolytic reduction of alumina.
Answer:
Al₂O₃ → 2Al³⁺ + 3O^2-
Cathode: Al³⁺ + 3e^– → Al(l)

Question 39.
What happens when aluminium ore is heated with caustic soda? Write the balanced chemical equation for the same.
Answer:
When aluminium ore is heated with caustic soda solution under high pressure for 2 to 8 hours and at 140 °C to 150 °C, aluminium oxide from aluminium ore, being amphoteric in nature, dissolves in caustic soda solution to form sodium aluminate.

Question 40.
How is zinc extracted from its ore zinc sulphide or zinc carbonate?
Answer:
The crude zinc sulphide ore is heated strongly in excess of air. Zinc sulphide is converted into zinc oxide. This process is known as roasting.

(OR)
The crude zinc carbonate ore is heated strongly in limited supply of air. Zinc carbonate is converted into zinc oxide. This process is known as calcination.

The zinc oxide is reduced to zinc by using a reducing agent such as carbon.

Question 41.
How is copper extracted from its sulphide ore?
Answer:
Copper is found as cuprous sulphide (Cu₂S) in nature. When Cu₂S is heated in air, copper is obtained.

Question 42.
How is mercury extracted from cinnabar?
(OR)
Extraction of mercury from its ore cinnabar and write the corresponding chemical reaction.
Answer:
Cinnabar is an ore of mercury. When cinnabar is heated (roasted), it is converted into mercuric oxide (HgO). Mercuric oxide is then reduced to mercury on further heating.

Question 43.
Show the steps involved in the extraction of moderately reactive metals from their sulphide ores.
Answer:
Moderately reactive elements are present as sulphides or carbonates in nature.

Question 44.
In the reactivity series of metals, some metals are misplaced. Rearrange these metals in the decreasing order of their reactivity.

Answer:

Question 45.
Complete the table, if a metal reacts with the reagent then mark ✓ and if not then ✗.

Metal	Ferrous sulphate	Silver nitrate	Copper sulphate	Zinc sulphate
Cu
Al

Answer:

Metal	Ferrous sulphate	Silver nitrate	Copper sulphate	Zinc sulphate
Cu	✗	✓	✗	✗
Al	✓	✓	✓	✓

Question 46.
Explain the term corrosion with a suitable example.
(OR)
What is corrosion ?
Answer:
The process in which a metal is destroyed gradually by the action of air, moisture or a chemical (like an acid) on its surface is called corrosion.
(OR)
Corrosion is degradation of a material due to reaction with its environment.
The major problem of corrosion occurs with iron, as it is used as a structural material in construction, bridges, shipbuilding.
Iron gets covered by reddish brown flakes when exposed to atmosphere. This is an example of corrosion.

Question 47.
Explain the different methods to prevent corrosion of metals.
Answer:
(1) Corrosion of a metal can be prevented if the contact between metal and air is cut off.
(2) Corrosion of a metal is prevented by coating with something which does not allow moisture and oxygen to react with it.
(3) A layer of oil or paint or grease is applied on the surface of a metal to prevent corrosion. The rusting or corrosion of iron can be prevented by this method.
(4) Corrosion is also prevented by coating a corrosive metal with a noncorrosive metal. Galvanising, tinning, electroplating, anodising and alloying are the different methods in which a metal is coated with a noncorrosive metal to prevent corrosion.

Question 48.
Write three methods of preventing rusting of iron.
Answer:

The rusting of iron can be prevented by painting, oiling, greasing or varnishing its surface.
Galvanisation is another method of protecting iron from rusting by coating iron with a thin layer of zinc.
Corrosion of iron is prevented by coating iron with noncorrosive substance like carbon. This process is termed as alloying.

Question 49.
What is meant by an alloy? Give two examples with chemical composition.
Answer:
The homogeneous mixture formed by mixing a metal with other metals or nonmetals in certain proportion is called an alloy.
Examples:
1. Bronze: Bronze is an alloy formed from 90% copper and 10% tin. Bronze statues stay well in sun and rain.
2. Stainless steel: Stainless steel alloy is made from 74% iron, 18% chromium and 8% carbon. This alloy does not get stained with air or water and does not rust.

Write short notes on the following:

Question 1.
Galvanizing.
Answer:
(1) The process of coating a thin layer of zinc on iron or steel is called galvanization.
(2) In this method corrosion of zinc occurs first because zinc is more electropositive than iron. After a few years zinc layer goes away and the iron layer gets exposed and starts rusting.
(3) In galvanization an iron object is dipped into molten zinc. A thin layer of zinc is formed all over the iron object. Examples: Shiny iron nails, pin, iron pipes.

Question 2.
Tinning.
Answer:
The process of coating a thin layer of tin (molten tin) on copper or brass is called tinning. Cooking vessels made of copper and brass get a greenish coating due to corrosion. The greenish substance is copper carbonate and it is poisonous. If buttermilk or curry is placed in such a vessel it gets spoiled. Therefore, these vessels are coated with tin to prevent corrosion.

Question 3.
Electroplating.
Answer:

The process in which a less reactive metal is coated on a more reactive metal by electrolysis is called electroplating.
Examples: Silver-plated spoon, gold-plated jewellery.
(1) Which process will you study with the help of above material and solutions.
Answer:
With the help of above material and solutions, electroplating process is studied.

(2) Define the process.
Answer:
The process in which less reactive metal is coated on a more reactive metals by electrolysis is called electroplating.

(3) Write the anode and cathode reactions.
Answer:
At anode: Ag → Ag⁺ + e^–
At cathode: Ag⁺ + e^– → Ag

Question 4.
Anodizing.
(OR)
Identify the process shown in the above diagram and explain it in brief. (Practice Activity Sheet – 3)

Answer:
The anodizing technique is an application of electrolysis. In this method copper or aluminium article is used as anode and it is coated with a strong film of their oxides by means of electrolysis. This oxide layer is strong and uniform all over the surface. This thin film protects the metals from corrosion. The protection can be further increased by making the oxide layer thicker during the anodization.
Examples: Kitchen articles such as ; anodized pressure cooker and anodized pan.

Question 5.
Alloying.
Answer:
A homogenous mixture of two or more metals or a metal and a nonmetal in a definite proportion is called an alloy. The physical properties of an alloy are different from those of its constituents. Alloys are corrosion resistant. Alloy decreases the intensity of corrosion of metals.

Examples: Brass is made from copper and zinc, 90 % Copper and 10 % tin are used to make an alloy called bronze, Stainless steel is made from 74% iron, 8 % carbon and 18 % chromium.

Distinguish between: (Two points of distinction)

Question 1.
Metals and Nonmetals.
Answer:
Metals:

Metals have a lustre.
They are generally good conductors of heat and electricity.
They are generally solids at room temperature.
Exception: Mercury and gallium are liquids.
Metals form basic oxides.

Nonmetals:

Nonmetals have no lustre.
Exception: Iodine and Diamond.
They are bad conductors of heat and electricity.
Exception: Graphite.
They are generally gases and solids at room temperature.
Exception: Bromine is a liquid.
Nonmetals form acidic or neutral oxides.

Question 2.
Roasting and Calcination.
Answer:
Roasting:

In this process, the ore is heated strongly in the presence of air.
In this process, sulphide ore is converted into metal oxide.
During this process SO₂ is given out.

Calcination:

In this process, the ore is heated strongly in the limited supply of air.
In this process, carbonate ore is converted into metal oxide.
During this process CO₂ is given out.

Give scientific reasons for the following:

Question 1.
Calcium floats on water during the reaction with water.
Answer:

Calcium reacts with water less vigorously hence the heat evolved is not sufficient for hydrogen to catch fire.
Instead, calcium floats on water because the bubbles of hydrogen gas formed stick to the surface of the metal.

Question 2.
Common salt has high melting and boiling points.
Answer:

Common salt is an ionic compound. Common salt is solid and hard due to strong electrostatic attraction between oppositely charged Na+ and Cl- ions.
A large amount of energy is required to break the strong intermolecular attraction (strong ionic bond). Hence, common salt has high melting and boiling points.

Question 3.
Metals are good conductors, while non-metals are poor conductors of electricity.
Answer:

The electrons in the outermost orbit of atoms of a metal are free to move throughout the metal.
When a potential difference is applied between the ends of a metal wire, the movement of the electrons constitutes an electric current. Hence, metals are good conductors of electricity.
Nonmetals involve covalent bonding and do not have free electrons like metals. Hence, nonmetals are poor conductors of electricity.

Question 4.
Sodium is more reactive than aluminium.
Answer:

If the number of electrons in the outermost orbit of an atom of a metal is less, the metal is more reactive.
Sodium has electronic configuration (2, 8, 1) and aluminium has electronic configuration (2, 8, 3). The number of electrons in the outermost orbit of sodium and aluminium atoms are 1 and 3, respectively. Hence, sodium is more reactive than aluminium.

Question 5.
When zinc granules are added to copper sulphate solution, the blue coloured solution turns colourless.
Answer:

Zinc is more reactive than copper.
When zinc granules are added to copper sulphate solution, they displace copper from the copper sulphate solution to form zinc sulphate solution. As zinc sulphate is colourless, the blue coloured solution of copper sulphate disappears.

Question 6.
When an iron nail is dipped into a copper solution, a shiny coat of copper is formed on the nail.
Answer:

Iron is more reactive than copper.
When an iron nail is dipped into copper sulphate solution, iron displaces copper from the copper sulphate solution. The copper so liberated deposits on the iron nail. As a result, a shiny coat of copper is formed on the nail.

Question 7.
Cryolite (Na3AlF6) and fluorspar (CaF₂) are added to the electrolytic mixture containing pure alumina.
Answer:
(1) Alumina has very high melting point ( > 2000 °C). Cryolite (Na₃AlF₆) and fluorspar (CaF₂) lower the fusion temperature of the mixture containing alumina from 2000 ° C to 1000 ° C, thereby saving electrical energy.
(2) They increase the conductivity and the mobility of the fused mixture. Hence, cryolite and fluorspar are added to the electrolytic mixture containing pure alumina.

Question 8.
Air is bubbled through the mixture in Froth floatation process.
Answer:
(1) In the froth floatation process, in a tank water, ore and an oil are mixed. When air is bubbled through the mixture the oil forms froth.
(2) The mineral particles are wetted by the oil and float on the surface.
(3) The gangue particles are wetted by water and settle down. Hence, the ore can be concentrated. Hence, air is bubbled through the mixture in froth floatation process.

Question 9.
Silver amalgam is used for filling dental cavities.
Answer:
(1) Silver is a soft metal and wears off on constant usage particularly due to abrasion. Silver amalgam is an alloy of silver with mercury.
(2) It is a hard substance. It is nontoxic. Besides these properties it is a lustrous shining substance. It melts at a comparatively low temperature and can therefore conveniently fill in the cavities. Hence, silver amalgam is used for filling dental cavities.

Explain the following reactions with the help of balanced equations:

Question 1.
Out of sodium and sulphur which is a metal? Explain its reaction with oxygen. (March 2019)
Answer:
Sodium is a metal. Sodium reacts with oxygen in air at room temperature to form sodium oxide.

Question 2.
Magnesium burns in air.
Answer:
When magnesium bums in air, it combines with oxygen, emitting intense light and heat to form magnesium oxide.

Question 3.
Copper reacts with air.
Answer:
Copper tarnishes in moist air and forms black coloured oxide when strongly heated.

Question 4.
Sodium reacts with water.
Answer:
When sodium reacts with water, it evolves hydrogen which immediately catches fire producing a lot of heat.

Question 5.
Calcium reacts with water.
Answer:
Calcium reacts with water less vigorously to form hydrogen gas and calcium hydroxide. In this reaction, the heat evolved is not sufficient for hydrogen to catch fire.

Question 6.
Steam is passed over aluminium.
Answer:
When steam is passed over aluminium, hydrogen gas and aluminium oxide are formed.

Question 7.
Steam is passed over iron.
Answer:
When steam is passed over iron, iron (III) oxide and hydrogen gas are formed.

Question 8.
Magnesium reacts with dil. hydrochloric acid.
Answer:
When magnesium reacts with dil. hydrochloric acid, magnesium chloride is formed and hydrogen gas is evolved.

Question 9.
Aluminium is treated with dil. hydrochloric acid.
Answer:
When aluminium is treated with dil. hydrochloric acid, aluminium chloride and hydrogen gas are formed.

Question 10.
Zinc reacts with dil. hydrochloric acid.
Answer:
When zinc reacts with dil. hydrochloric acid, zinc chloride and hydrogen gas are formed.

Question 11.
Iron is treated with dil. hydrochloric acid.
Answer:
When iron reacts with dil. hydrochloric acid, ferrous chloride and hydrogen gas is formed.

Question 12.
Copper is reacted with cone, nitric acid.
Answer:
When copper is reacted with cone, nitric acid, copper nitrate and reddish brown nitrogen dioxide are formed.

Question 13.
Copper is reacted with dil. nitric acid.
Answer:
When copper is reacted with dil. nitric acid, copper nitrate and nitric oxide are formed.

Question 14.
Sodium metal is reacted chlorine gas.
Answer:
When sodium metal is reacted with chlorine, sodium chloride an ionic compound is formed.

Question 15.
Sulphur burns in air.
Answer:
When sulphur burns in air, it combines with oxygen to form acidic sulphur dioxide.

Question 16.
Chlorine dissolves in water.
Answer:
When chlorine dissolves in water, hypochlorous acid is formed.

Question 17.
Chlorine is treated with hydrobromic acid.
Answer:
When chlorine is treated with hydrobromic acid, chlorine displaces bromine from hydrobromic acid.
Cl_2(g) + 2HBr_(aq) → 2HCl_(aq) + Br_2(aq)

Question 18.
Hydrogen gas is passed over boiling sulphur.
Answer:
When hydrogen gas is passed over boiling sulphur, sulphur combines with hydrogen to form hydrogen sulphide which has rotten egg smell.

Question 19.
Sodium aluminate is treated with water.
Answer:
When sodium aluminate is treated with water, it is hydrolysed to give a precipitate of aluminium hydroxide.

Question 20.
Dry aluminium hydroxide is ignited at 1000 °C.
Answer:
When dry aluminium hydroxide is ignited at 1000 °C, alumina (Al₂O₃) formed.

Question 21.
Zinc sulphide is heated strongly in excess of air.
Answer:
When zinc sulphide is heated strongly in excess of air, it forms zinc oxide and sulphur dioxide gas.

Question 22.
Zinc carbonate is heated strongly in a limited supply of air.
Answer:
When zinc carbonate is heated strongly in a limited supply of air, it gives zinc oxide and carbon dioxide.

Question 23.
Zinc oxide is treated with carbon.
Answer:
When zinc oxide is treated with carbon, it is reduced to zinc. In this reaction, carbon acts as reducing agent.

Question 24.
Manganese dioxide is heated with aluminium.
Answer:
When manganese dioxide is heated with aluminium, manganese dioxide is reduced to manganese and large amount of heat is evolved.

Question 25.
Cinnabar is heated in air.
Answer:
When cinnabar is heated in air, it forms mercuric oxide and sulphur dioxide.

Question 26.
Cuprous sulphide is heated in air.
Answer:
When cuprous sulphide is heated in air, cuprous oxide is formed. Cuprous oxide is reduced to copper in the presence of ore.

Project: (Do it your self)

Project 1.
Which metals are used in day to day life? What are its uses?

Project 2.
Which nonmetals are used in day to day life? What are its uses?

10th Std Science Part 1 Questions And Answers:

Life Processes in Living Organisms Part – 2 Class 10 Questions And Answers Maharashtra Board

January 6, 2025January 5, 2025 by Bhagya

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part – 2 Notes, Textbook Exercise Important Questions and Answers.

Std 10 Science Part 2 Chapter 3 Life Processes in Living Organisms Part – 2 Question Answer Maharashtra Board

Class 10 Science Part 2 Chapter 3 Life Processes in Living Organisms Part – 2 Question Answer Maharashtra Board

Question 1.
Complete the following chart.

Asexual reproduction	Sexual reproduction
1. Reproduction that occurs with the help of somatic cells is called asexual reproduction.	1. __________________________________________ ____________________________________________
2. __________________________________________ ____________________________________________	2. Male and female parent are necessary for sexual reproduction.
3. This reproduction occurs with the help of mitosis only.	3. __________________________________________ ____________________________________________
4. __________________________________________ ____________________________________________	4. New individual formed by this method is genetically different from parents.
5. Asexual reproduction occurs in different individuals by various methods like binary fission, multiple fission, budding, fragmentation, regeneration, vegetative propagation, spore production, etc.	5. __________________________________________ _____________________________________________ _____________________________________________ _____________________________________________

Answer:

Asexual reproduction	Sexual reproduction
1. Reproduction that occurs with the help of somatic cells is called asexual reproduction.	1. Reproduction that occurs due to fertilization of gametes is called sexual reproduction.
2. For asexual reproduction only one parent is necessary.	2. Male and female parents are necessary for sexual reproduction.
3. This reproduction occurs with the help of mitosis only.	3. This reproduction occurs with the help of both mitosis and meiosis.
4. New individual formed by this method is genetically identical with parents.	4. New individual formed by this method is genetically different from parents.
5. Asexual reproduction occurs in different individuals by various methods like binary fission, multiple fission, budding, fragmentation, regeneration, vegetative propagation, spore production, etc.	5. Sexual reproduction occurs in two steps: First formation of haploid gametes by meiosis and then fertilization of these haploid gametes to form diploid zygote. There are no subtypes in the sexual reproduction.

Question 2.
Fill in the blanks.
a. In humans, sperm production occurs in the organ …………
(a) prostate gland
(b) testis
(c) ovaries
(d) Cowper’s gland
Answer:
(b) testis

b. In humans, …………. chromosome is responsible for maleness.
(a) X
(b) Y
(c) Z
(d) O
Answer:
(b) Y

c. In male and female reproductive system of human, …………. gland is same.
Answer:
There is no similar gland in male and female reproductive system. There may be some homologies but there is no similarity.

d. Implantation of embryo occurs in …………
(a) ovaries
(b) fallopian duct
(c) uterus
(d) vagina
Answer:
(c) uterus

e. …………type of reproduction occurs without fusion of gametes.
(a) Asexual
(b) sexual
(c) Fertilization
(d) Gamete formation
Answer:
(a) Asexual

f. Body breaks up into several fragments and each fragment begins to live as a new individual.
This is ………. type of reproduction.
(a) regeneration
(b) fragmentation
(c) binary fission
(d) budding
Answer:
(b) fragmentation

g. Pollen grains are formed by division in locules of anthers.
(a) meiosis
(b) mitosis
(c) amitosis
(d) binary
Answer:
(a) meiosis

Question 3.
Complete the paragraph with the help of words given in the bracket:
(Luteinizing hormone, endometrium of uterus, follicle stimulating hormone, estrogen, progesterone, corpus luteum)
Growth of follicles present in the ovary occurs under the effect of ………….. This follicle secretes estrogen. Ovarian follicle along with oocyte grows/regenerates under the effect of estrogen. Under the effect of ………….., fully grown up follicle bursts, ovulation occurs and …………….. is formed from remaining part of follicle. It secretes ……………. and ………. Under the effect of these hormones, glands of ……….. are activated and it becomes ready for implantation.
Answer:
Growth of follicles present in the ovary occurs under the effect of follicle stimulating hormone. This follicle secretes estrogen. Ovarian follicle along with oocyte grows/regenerates under the effect of estrogen. Under the effect of Luteinizing hormone, fully grown up follicle bursts, ovulation occurs and corpus luteum is formed from remaining part of follicle. It secretes estrogen and progesterone. Under the effect of these hormones, glands of endometrium of uterus are activated and it becomes ready for implantation.

Question 4.
Answer the following questions short.
a. Explain with examples types of asexual reproduction in unicellular organism.
Answer:

There are different methods of asexual reproduction in different unicellular animals.
(1) Binary fission: The process in which the parent cell divides to form two similar daughter cells
is binary fission. It takes place either by mitosis or amitosis. When there are favourable conditions and abundant food supply then the organisms undergo binary fission. Prokaryotes, Protists and eukaryotic 5 cell-organelle like mitochondria and chloroplasts perform binary fission.

Based on axis of fission there are three subtypes of binary fission.
(a) Simple binary fission: The plane of division is not definite, it can be in any direction due to lack of specific shape as in Amoeba.
(b) Transverse binary fission: The plane of J division is transverse, as in Paramoecium.
(c) Longitudinal binary fission: The plane of division is in length-wise direction as in Euglena.

(2) Multiple fission:

During unfavourable conditions when there is lack of food, multiple fission is shown by amoeba. Amoeba forms protective covering and becomes encysted. Inside the cyst, amoeba undergoes repeated nuclear division. This is followed by cytoplasmic divisions. Many amoebulae are formed which remain dormant inside the cyst. When favourable conditions reappear, they come out by breaking the cyst.

(3) Budding in yeast:

Yeast is unicellular fungus that performs budding. The parent cell produces two daughter nuclei by mitotic division. This results in a small bulgingbud on the surface of parent cell. One daughter nucleus enters the bud. It then grows and upon becoming big it separates from the parent cell to have independent life as new yeast cell.

b. Explain the concept of IVF.
Answer:
(1) IVF means In Vitro Fertilization (IVF)
(2) This is the technique in the modern medical field where childless couples can be blessed by their own child.
(3) IVF technique is used for childless couples who are faced with problems such as less sperm count, obstacles in oviduct, etc.
(4) The IVF technique is done by removing the oocyte from the mother and artificially fertilizing by the sperms collected from father. This fertilization is done in a test-tube. Thus it is also called test tube baby. The embryo formed is implanted in uterus of real mother or a surrogate mother at appropriate time.

c. Which precautions will you follow to maintain the reproductive health?
Answer:
About reproductive health one should have scientific and authentic information. The cleanliness of body is very essential but keeping the mind clean is also important to maintain good reproductive health. One should be careful about sexual relationships. These things should not be experimented in young age. Mistakes committed like these can change the sexual health forever. The cleanliness and hygiene during menstruation, the cleanliness of genitals and other private parts are the aspects of personal hygiene. When living in a society, one should always be away from cross-infections of venereal type.

d. What is menstrual cycle? Describe it in brief.
Answer:

Menstrual cycle is the events of cyclic changes that takes place with the interval of 28 to 30 days in mature woman.
Hormones from pituitary, FSH (Follicle Stimulating Hormone) and LH (Luteinizing Hormone) and hormones from ovary, estrogen and progesterone control the menstrual cycle.
Due to influence of FSH, the ovarian follicle grows along with the oocyte that is present in it.
This growing follicle produces estrogen.
Under the influence of estrogen, the uterine inner layer called endometrium grows or regenerates. In the meantime the development of follicle is completed.
LH from pituitary stimulates the bursting of ovarian follicle and releases the mature oocyte out of the follicle and the ovarian wall. This process is called ovulation.
The empty ovarian follicle after the ovulation becomes corpus luteum. Corpus luteum produces hormone progesterone.
Under the influence of progesterone, the glands from uterine endometrium start secreting. The oocyte if fertilized is implanted over this endometrium.
If oocyte is not fertilized, the corpus luteum becomes a degenerate body called corpus albicans. The corpus albicans cannot secrete estrogen and progesterone.
Due to lack of these hormones, the endometrial layer of the uterus collapses. The tissue debris, along with unfertilized egg is given out through the vagina as menstrual flow. This results in bleeding for about 5 days.
If woman is not pregnant, then this menstrual cycle keeps on repeating with regularity.

Question 5.
In case of sexual reproduction, newborn show similarities about characters. Explain this statement with suitable examples.
Answer:
(1) Sexual reproduction occurs due to two different gametes. One male gamete is from father while the other female gamete is from mother.
(2) Both the gametes are produced by meiosis.
(3) When the gametes unite it is called process of fertilization which produces diploid zygote.
(4) Due to the chromosomes of parents, their DNA pass to the next generation through such fertilization. Therefore, the characters of newborn show similarities with parents.

Question 6.
Sketch the labelled diagrams.
a. Human male reproductive system.
Answer:

b. Human female reproductive system.
Answer:

c. Flower with its reproductive organs.
Answer:

d. Menstrual cycle.
Answer:

Question 7.
Give the names.
a. Hormones related with male reproductive system.
Answer:
Follicle stimulating hormone and ICSH or Luteinizing hormone secreted by pituitary gland, testosterone secreted by testis.

b. Hormones secreted by ovary of female reproductive system.
Answer:
Estrogen and progesterone.

c. Types of twins.
Answer:
Monozygotic twins, Siamese twins and Dizygotic twins.

d. Any two sexual diseases.
Answer:
Gonorrhea and Syphilis.

e. Methods of family planning.
Answer:
Copper T, condoms, oral contraceptive pills.

Question 8.
Gender of child is determined by the male? partner of couple. Explain with reasons whether this statement is true or false.
(OR)
“A couple shall have a male child or female child totally depends upon husband”. Prove truthfulness of this statement with scientific reason.
Answer:

Sex determination in Human being
(1) The statement Gender of child is determined by the male partner of couple is true.
(2) It is clearly seen from the diagram that there are two types of sperms produced by males. One sperm has a X chromosome while the other has a Y chromosome, apart from autosomes. The mother on the other hand has all X bearing oocytes. Thus the sperm that fertilizes the oocyte decides the sex of the child.
(3) If X bearing sperm fertilizes the oocyte, daughter is born and when Y bearing sperm fertilizes the oocyte, son is born.
(4) Thus father or male partner is responsible for the determination of the sex.

Question 9.
Explain asexual reproduction in plants.
Answer:

Vegetative propagation is the method of asexual reproduction in plants.
It takes place with the help of vegetative parts like root, stem, leaf and bud.
Potato, suran (Amorphophallus) and other tubes propagate with the help of ‘eyes’ which are buds. These eyes are present on the stem tubers.
In case of plants like sugarcane and grasses, buds present on nodes perform vegetative propagation.
Plants like Bryophyllum performs vegetative propagation with the help of buds present on leaf margin.

Question 10.
Modern techniques like surrogate mother, sperm bank and IVF technique will help the human beings. Justify this statement.
OR
Despite various diagnostic tests, a couple could not have a child. In this situation, which remedies will you suggest? (July 2019)
Answer:
(1) Some couples want a child but they are not able to bear one due to various problems either in mother or in father. In such cases modern techniques such as IVF, surrogacy and sperm bank are useful in conceiving a child.

(2) These methods are as follows:
(i) Surrogacy: In woman if there is problem regarding the implantation of embryo in uterus, then help of another women is taken. This women is called surrogate mother.

Oocyte from real mother is taken out and fertilized with sperms collected from her husband. These gametes are fertilized outside in a test-tube and then the fertilized zygote is implanted in the surrogate mother.

(ii) In Vitro Fertilization (IVF) is done when there are problems like less sperm count or obstacles in oviduct. In IVF, fertilization is done in the test-tube. The embryo formed is implanted in uterus of woman for further growth.

(iii) Sperm bank: If man has problems with the sperm production, then the sperms are collected from the sperm bank. Sperm bank is the place where the donor’s donate the sperms and such sperms are kept stored. The donor’s identity is kept secret and he should also be physically and medically fit person.

Question 11.
Explain sexual reproduction in plants.
(OR)
Explain double fertilization in plants.
Answer:

Plants reproduce sexually with the help of flowers.
Androecium and gynoecium are male and female parts of the flowers respectively.
In the carpel, the ovule undergoes meiosis and forms embryo sac.
A haploid egg cell and two haploid polar nuclei are present in each embryo sac.
The pollen grains from the anther reach the stigma of flower by the process of pollination. They germinate here on the stigma.
As a result of germination, long pollen tube and two male gametes are formed.
The pollen tube travels through the style of flower and the male gametes present in the pollen tube are transferred till the embryo sac in ovary. Upon reaching there, tip of the pollen tube bursts releasing two male gametes in embryo sac.
One male gamete unites with the egg cell and forms zygote. While other male gamete unites with two polar nuclei forming the endosperm.
Because there are two nuclei participating in this process, therefore it is called double fertilization.
After fertilization ovule develops into seed and ovary forms a fruit. When the seed again gets favourable conditions, it can produce a new plant.

Activities: (Do it your self)

Question 1.
Collect the official data about present and a decade old population of various Asian countries and plot a graph of that data. With the help of it, draw your conclusions about demographic changes.

Question 2.
With the help of your teacher, compose and present a road show to increase the awareness about prenatal gender detection and gender bias.

Can you recall? (Text Book Page No. 22)

Question 1.
Which are the important life processes in living organisms?
Answer:
The important life processes in living organisms are respiration, circulation, nutrition, excretion, sensation and response through nervous system.

Question 2.
Which life processes are essential for production of energy required by body?
Answer:
The oxidation of nutrients that are absorbed in body is done because of oxygen supplied to cells by respiratory and circulatory system. This helps in liberation of energy. Thus respiration, circulation and nutrition are the life processes that are essential for production of energy required by body.

Question 3.
Which are main types of cell division? What are the differences?
Answer:
The main types of cell division are mitosis and meiosis. In mitosis, the chromosome number remains the same. 2 daughter cells are obtained from one cell. In meiosis, the chromosome number is reduced to half. From one cell, four daughter cells are obtained.

Question 4.
What is the role of chromosomes in cell division?
Answer:
Due to chromosomes, the DNA from parental cells enter into daughter cells. The hereditary, characters are transmitted to next generation by cell division.

Can you recall? (Text Book Page No. 22)

Question 1.
What do we mean by maintenance of species?
Answer:
Maintenance of species means a species undertakes successful reproduction and produces individuals of its own kind. This keeps the species existing on the earth.

Question 2.
Whether the new organism is genetically exactly similar to earlier one that has produced it?
Answer:
No. The new organism produced from the old one is not genetically exactly similar to the parents. In meiotic cell division there is crossing over in the homologous chromosomes. This produces genetic recombination. Thus the new organism is different from the earlier one. However, if the reproduction is of asexual type, then the young one is exactly similar to the parents.

Question 3.
Who determines whether the two organisms of a species will be exactly similar or not?
Answer:
The type of reproduction, whether it is asexual or sexual, the type of crossing over, the extent of genetic recombination, etc. determine the similarity among the parent organisms and their, offspring. Based on this genetic recombination the two organisms of a species do not show exact similarity. However, in case of monozygotic twins there is exact similarity. In asexual reproduction to there is similarity.

Question 4.
What is the relationship between the cell division and formation of new organism of same species by earlier existing organism?
Answer:
In the process of reproduction, there is division of chromosomes. Due to cell division, the gametes are formed. The union of gametes produce new offspring. In sexual reproduction, all these processes take place due to cell division. In asexual reproduction too there is cell division. Growth of new organism also occurs due to cell division.

Let’s Think: (Text Book Page No. 26)

Question 1.
What would have been happened if the male and female gametes had been diploid?
Answer:
Diploid (2n) gametes if united, they will form 4n, i.e. tetraploid variety. Such zygote will show severe abnormality. The chromosome number will not be maintained.

Question 2.
What would have been happened if any of the cells in nature had not been divided by meiosis?
Answer:
If meiosis does not happen the gametes produced will be diploid. This will create abnormality.

Can you recall? (Text Book Page No. 28)

Question 1.
Which different hormones control the functions of human reproductive system through chemical coordination?
Answer:
Pituitary gland secretes FSH and LH. LH is known as ICSH in males, as its function in the male body is different. From the gonads of male and female, hormones are secreted which are essential for male and female reproductive functions respectively. These hormones are testosterone secreted from testis in males and estrogen and progesterone secreted from the ovaries in females. Testosterone is essential for masculinity as well as for sperm production while female hormones are essential for changes in the female body leading to motherhood.

Question 2.
Which hormones are responsible for changes in human body occurring during onset of sexual maturity?
Answer:
Testosterone in male body and estrogen in female body are responsible for maturity onset changes in human body.

Question 3.
Why has the Government of India enacted the law to fix the minimum age of marriage as 18 in girls and 21 in boys?
Answer:
The full growth of female body is not completed till the age of 18. Till 18 years of age the physical and emotional maturity is not attained. Therefore, she is not suitable for marriage, sexual relationship and pregnancy. Similarly, boy attains complete growth only the age of 21. Therefore, to keep individuals and their progeny safe and healthy the Government of India enacted the law to fix the minimum age of marriage as 18 in girls and 21 in boys.

Can you recall? (Text Book Page No. 31)

Question 1.
Which hormone is released from pituitary of mother once the foetal development is completed?
Answer:
The hormone oxytocin is released from the posterior pituitary of mother once the foetal development is completed.

Question 2.
Under the effect of that hormone, which organ of the female reproductive system starts to contract and thereby birth process (parturition) is facilitated?
Answer:
Due to oxytocin, uterus contracts involuntarily and the baby is expelled out. Thus initiation of birth process is possible due to contractions of uterus.

Use your brain power. (Text Book Page No. 24)

Question 1.
Does the parent cell exist after asexual reproduction-fission?
Answer:
In fission, the parent cell divides into two. This nucleus and cytoplasm, both are divided. Thus, parent cell does not exist any longer, it is converted into new cells.

Choose the correct alternative and write that alphabet against the sub-question number:

Question 1.
Pranav and Pritee are twins in your class. They belong to ……….. twins type.
(a) monozygotic
(b) dizyotic
(c) siamese
(d) none of the above
Answer:
(b) dizyotic

Question 2.
Longitudinal binary fission is seen in …………..
(a) Paramoecium
(b) Euglena
(c) Amoeba
(d) Spirogyra
Answer:
(b) Euglena

Question 3.
Yeast cell performs asexual reproduction by ……………..
(a) fragmentation
(b) budding
(c) binary fission
(d) regeneration
Answer:
(b) budding

Question 4.
Carrot and raddish undergoes …………. with the help of their roots.
(a) vegetative propagation
(b) fragmentation
(c) budding
(d) regeneration
Answer:
(a) vegetative propagation

Question 5.
Androecium and gynoecium are ……….. whorl of the flower.
(a) accessory
(b) essential
(c) external
(d) internal
Answer:
(b) essential

Question 6.
Flowers without stalk are called ……….. flowers.
(a) stalkless
(b) sessile
(c) incomplete
(d) complete
Answer:
(b) sessile

Question 7.
………….. on the inner surface of fallopian ducts (oviducts) push the egg towards uterus.
(a) Cilia
(b) Tentacles
(c) Flagella
(d) Fibres
Answer:
(a) Cilia

Question 8.
Pregnant mother supplies nourishment to her foetus through …………..
(a) breasts
(b) uterus
(c) placenta
(d) ovaries
Answer:
(c) placenta

Question 9.
The length of a sperm is about …………. micrometers.
(a) 400
(b) 5
(c) 60
(d) 600 (July ’19)
Answer:
(c) 60

Question 10.
Vegetative propagation is performed with the help of ……….. in sweet potato.
(a) root
(b) stem
(c) leaf
(d) flower
Answer:
(a) root

Question 11.
Which of the following is not a unisexual flower?
(a) Coconut
(b) Papaya
(c) Gulmohar
(d) Maize
Answer:
(c) Gulmohar

Write whether the following statements are true or false, with the suitable reason:

Question 1.
Absence of genetic recombination is an advantage whereas fast process is drawback of asexual reproductive method.
Answer:
False. (Absence of genetic recombination is a drawback whereas fast process is advantage of asexual reproductive method.)

Question 2.
Prokaryotes show fission which occurs either by mitosis or amitosis.
Answer:
True. (Prokaryotes show fission by both the methods, i.e. mitosis and amitosis.)

Question 3.
During favourable conditions multiple fission is performed by amoeba.
Answer:
False. (During unfavourable conditions multiple fission is performed by amoeba.)

Question 4.
Any encysted Amoeba or any other protist is called ‘Cyst’.
Answer:
True. (Cyst is the tough capsule like structure which keeps the protists dormant inside it. This helps the organisms to tide over unfavourable conditions.)

Question 5.
If the body of Sycon breaks up accidentally into only large and few fragments, then only each fragment develops into new Sycon.
Answer:
False. (If the body of Sycon breaks up accidentally into many fragments, each fragment develops into new Sycon. Because the capacity to regenerate is very strong in poriferan Sycon, even a small piece of parent Sycon can give rise to entire new individual.)

Question 6.
Pollen tube reaches the zygote via style.
Answer:
False. (Pollen tube reaches the embryo sac via style. Later, double fertilization takes place and the zygote and endosperm are formed.)

Question 7.
There is glucose sugar in the semen.
Answer:
False. (There is fructose sugar in the semen. Glucose is not present in semen.)

Question 8.
Out of 2 – 4 million ova, approximately only 400 oocytes are released up to the age of menopause.
Answer:
True. (During the reproductive span of the woman, from menarche to menopause only one oocyte per one month is released in the span of 30 to 35 years.)

Question 9.
If the oocyte is fertilized, secretion of estrogen and progesterone stops completely.
Answer:
False. (If the oocyte is not fertilized, there is no need of corpus luteum which secretes progesterone. In absence of conception, the progesterone is not needed, thus corpus luteum degenerates and forms corpus albicans.)

Question 10.
During menstruation there is need of rest along with special personal hygiene.
Answer:
True. (During phase of menstruation there is pain and bleeding in woman. Her body is also susceptible for infections. There is weakness and hence she needs rest along with special personal hygiene.)

Find the odd one out:

Question 1.
Circulation, Excretion, Sensation, Reproduction.
Answer:
Reproduction. (All others are processes necessary for survival of the individual.)

Question 2.
Budding in hydra, Regeneration, Binary fission, Fragmentation
Answer:
Binary fission. (All others are processes of asexual reproduction in multicellular organisms.)

Question 3.
Carrot, Radish, Potato, Sweet potato.
Answer:
Potato. (All others are edible roots.)

Question 4.
Vas eferens, vas deferens, prostate gland, epididymis.
Answer:
Prostate gland. (All others are duct systems in male reproductive system.)

Question 5.
Prostate gland, Bartholin’s gland, Cowper’s gland, Epididymis.
Answer:
Bartholin’s glands. (All others are parts of male reproductive system.)

Question 6.
Stigma, style, pollen, ovary.
Answer:
Pollen. (All others are parts of gynoecium.)

Identify the correlation between the first two words and suggest the suitable words in the fourth place:

Question 1.
Amoeba : Fission : : Hydra : ………….
Answer:
Amoeba : Fission : : Hydra : Budding

Question 2.
Transverse binary fission : Paramoecium : : Longitudinal binary fission : ………… (July ‘19)
Answer:
Transverse binary fission : Paramoecium : : Longitudinal binary fission : Euglena

Question 3.
Calyx : Sepals : : Corolla : ………….
Answer:
Calyx : Sepals : : Corolla : Petals

Question 4.
Accessory whorls : Calyx and corolla : : Essential whorls : ………..
Answer:
Accessory whorls : Calyx and corolla : : Essential whorls : Androecium and gynoeciuin

Question 5.
Bisexual flower : Hibiscus : : Unisexual flower : ………….
Answer:
Bisexual flower : Hibiscus : : Unisexual flower : Papaya

Question 6.
FSH : Development of qocyte : : LH : ………….
Answer:
FSH : Development of qocyte : : LH : Ovulation

Define the following/Give meanings of the following:

Question 1.
Budding in yeast.
Answer:
Budding in yeast: Budding is the asexual reproductive process in which a! small bulge or bud appears on the surface of parent cell as seen in unicellular yeast.

Question 2.
Budding in hydra.
Answer:
Budding in hydra: Budding in hydra is asexual reproductive process in which an outgrowth is formed by repeated divisions of regenerative cells of body wall called bud.

Question 3.
Regeneration.
Answer:
Regeneration: Regeneration is the asexual reproduction in Planaria in which the body is broken up into two parts and resulting each part regenerates remaining part of the body.

Question 4.
Fragmentation.
Answer:
Fragmentation: Fragmentation is the asexual type of reproduction in which the body of parent organism breaks up into many fragments. Each fragment can start living independently.

Question 5.
Vegetative propagation.
Answer:
Vegetative propagation: Vegetative propagation is a type of asexual reproduction in plants that takes place with the help of vegetative parts like root, stem, leaf and bud.

Question 6.
Fertilization.
Answer:
Fertilization: The process by which two haploid gametes unite to form a diploid zygote is called fertilization.

Question 7.
Pedicel.
Answer:
Pedicel: The stalk of the flower which is for the support is called pedicel.

Question 8.
Pollination.
Answer:
Pollination: Transfer of pollen grains from anther to the stigma is called pollination.

Question 9.
Self-Pollination.
Answer:
Self-Pollination: Pollination involving only one flower or two flowers borne on same plant is called self-pollination.

Question 10.
Cross-Pollination.
Answer:
Cross-Pollination: Pollination involving two flowers borne on two plants of same species is cross-pollination.

Question 11.
Endosperm.
Answer:
Endosperm: Endosperm is the nourishing substance formed by the union of second male gamete with two polar nuclei at the time of fertilization in plants.

Question 12.
Embryo sac.
Answer:
Embryo sac: There are many ovules in the ovary, the structure formed in each of the ovule by meiosis is called embryo sac.

Question 13.
Menopause.
Answer:
Menopause: Stoppage of functioning of female reproductive system due to lack of synthesis of hormones due to advancing age is called menopause.

Question 14.
Placenta.
Answer:
Placenta: An organ developed in the uterus of the pregnant mother, through which the embryo is given nourishment is called placenta.

Question 15.
Menstrual cycle.
Answer:
Menstrual cycle: The repetitive changes at the interval of every 28-30 days in female reproductive system that take place after puberty, form menstrual cycle.

Question 16.
Corpus luteum.
Answer:
Corpus luteum: Corpus luteum is the secondary structure that is formed from empty ovarian follicle after ovulation. This corpus luteum produces progesterone and thereby maintains pregnancy.

Question 17.
Corpus albicans.
Answer:
Corpus albicans: Corpus albicans is the degenerate body which is formed from corpus luteum, if the ovum is not fertilized.

Question 18.
Ovulation.
Answer:
Ovulation: Bursting of mature ovarian follicle under the influence of hormones to release the oocyte is called ovulation.

Question 19.
IVF.
Answer:
IVF: In Vitro Fertilization is the technique in which fertilization is brought about outside the female body but in the test-tube and the embryo is implanted in uterus of woman.

Question 20.
Sperm bank.
Answer:
Sperm bank: Sperm bank is the place where semen donated by the desired men is collected after their thorough physical and medical check-up and stored at sub-zero temperatures in sterile conditions.

Name the following/Give the names:

Question 1.
Different glands associated with male reproductive system.
Answer:
Seminal vesicles, Prostate gland, Cowper’s or bulbourethral glands.

Question 2.
Agents of pollination.
Answer:
Biotic: Insects, birds, few animals.
Abiotic: Water and wind.

Question 3.
Components of semen.
Answer:
Secretion of prostate gland seminal vesicles and Cowper’s glands along with sperms.

Question 4.
Two accessory whorls in flower.
Answer:
Calyx and corolla.

Question 5.
Two essential whorls in flower.
Answer:
Androecium and gynoecium.

Question 6.
The modern techniques in reproduction.
Answer:
In Vitro Fertilization, Surrogate mother, Sperm bank.

Question 7.
Symptoms of gonorrhea.
Answer:
Painful and burning sensation during urination, oozing of pus through penis and vagina, inflammation of urinary tract, anus, throat, eyes, etc.

Question 8.
Symptoms of syphilis.
Answer:
Occurrence of chancre (patches) on various parts of body including genitals, rash, fever, inflammation of joints, alopecia, etc.

Write the functions of the following organs:

Question 1.
Sporangium.
Answer:
Storing the spores and releasing them by bursting.

Question 2.
Calyx.
Answer:
Protection of inner whorls of the flower.

Question 3.
Corolla.
Answer:
Attracting insects for pollination. Protecting inner whorls.

Question 4.
Androecium.
Answer:
Production of pollen grains, the male gametes of flower.

Question 5.
Gynoecium.
Answer:
Production of female gametes of flower. Participating in production of fruits.

Question 6.
Endosperm.
Answer:
Nourishment of the growing embryo.

Question 7.
Testis.
Answer:
Production of sperms and male hormone-testosterone.

Question 8.
Scrotum.
Answer:
Protection and temperature control of testis.

Question 9.
Seminal vesicles.
Answer:
Secretion of seminal fluid which forms major portion of semen. Nourishment of sperms.

Question 10.
Penis.
Answer:
Transferring of sperms to vagina at the time of intercourse. Release of urine at the time of urination.

Question 11.
Ovary.
Answer:
Production of oocytes and female hormones – estrogen and progesterone.

Question 12.
Uterus.
Answer:
Growth and development of foetus during pregnancy. Helping in parturition (childbirth) by contractions.

Question 13.
Fallopian tubes/ducts.
Answer:
Transporting the released oocyte after ovulation to the uterus. Providing space for fertilization of oocyte by sperm. Conception is possible only when sperm and oocyte meet in the fallopian tube.

Question 14.
Vagina.
Answer:
Passage for copulation/intercourse. Birth canal. Passage for menstrual flow.

Question 15.
Placenta.
Answer:
Supplying nourishment to the growing foetus.

Distinguish between:

Question 1.
Binary fission and Multiple fission.
Answer:
Binary fission:

Two new individuals are formed from one old individual at one time.
The division of nucleus and cytoplasm takes place initially.
The axis of division can be transverse, longitudinal or any one axis as it is in simple binary fission.
Formation of protective cyst does not take place.
Binary fission can be done during favourable period.

Multiple fission:

Many new individuals are formed from one old individual at one time.
Only nucleus divides initially followed by division of cytoplasm.
There is no exact axis for the fission.
Protective covering is formed around dividing amoebulae which is called cyst.
Multiple fission takes place only at the time of unfavourable period.

Question 2.
Human male and Human female reproduction system.
Answer:
Human male reproductive system:

Testis are essential organs which are located outside the abdomen in the scrotal sacs.
There is common urethra through which urine and semen, are passed out.
Reproductive system of male continues to work even in old age.
Sperms or male gametes are produced by meiosis in the testis.
Sperms are produced in millions at one time.
Three accessory glands are associated with the male reproductive system.
Testis secrete testosterone which is essential male hormone.

Human female reproductive system:

Ovaries are essential organs which are located along with all other organs inside the lower abdomen.
Urethra and vagina are two separate openings that open to outside.
Reproductive system works only till menopause.
Oocytes or ova are produced by meiosis in the ovaries.
Only single oocyte is produced per month.
Only one gland is associated with female reproductive system.
Ovaries produce estrogen and progesterone which are essential female hormones.

Question 3.
Monozygotic twins and Dizygotic twins.
Answer:
Monozygotic twins:

Two children developing from only one zygote are called monozygotic twins.
Monozygotic twins develop from same oocyte.
Gender of both the twins is same.
The monozygotic twins are genetically exactly alike.

Dizygotic twins:

Two children developing from two different zygotes are called dizygotic twins.
Dizygotic twins develop from two different oocytes.
Gender of both the twins can be same or can be different.
The dizygotic twins are genetically not exactly alike.

Give scientific reasons:

Question 1.
Individual developed by sexual reproduction always carry recombined genes of both the parents.
Answer:

In sexual reproduction, the haploid male and female gametes are united to form diploid zygote.
The zygote thus carries chromosomes of both parents which are transferred via male and female gametes.
While producing gametes, there is meiosis in which genetic recombination takes place.
Therefore, the individual developed by sexual reproduction always carry recombined genes of both the parents.

Question 2.
Flower is the structural unit of sexual reproduction in plant.
Answer:

Flower produces male and female gametes.
For this purpose there are essential whorls of androecium and gynoecium.
The double fertilization also takes place in flower.
Therefore, flower is called the structural unit of sexual reproduction in plants.

Question 3.
Fertilization in plants is called double fertilization.
Answer:

After pollination the pollen grains drop on the sticky stigma of the flower.
They germinate here producing two male gametes and a long pollen tube.
The male gametes travel through the pollen tube till they reach the embryo sac.
Here the male gametes are released by bursting the pollen tube. One male gamete unites with the egg cell to form zygote while the second male gamete unites with two polar nuclei forming endosperm.
In this way because two nuclei participate in the fertilization process, therefore it is called double fertilization.

Question 4.
By the age of 45 – 50 women gets menopause.
Answer:

By the age of 45-50, the secretion of hormones which control the functioning of the reproductive system is reduced gradually and then it stops.
This causes end of menstrual cycle. This results into menopause.

Question 5.
Older mothers have greater chance of conceiving abnormal children.
Answer:

In older women the menopausal age approaches.
The oocytes, released from ovaries during this phase are not normal.
Their meiotic cell division is abnormal and thus oocyte becomes abnormal too.
If such abnormal oocytes are fertilized, the baby is born with many genetic problems, e.g. Down’s syndrome or Turner’s syndrome.

Question 6.
Indians should follow family planning for controlling the population.
Answer:

There is severe population explosion in India. It has almost reached to 121 crores.
This results into unemployment, decreasing per capita income and increasing loan, stress on natural resources, etc.
Only by controlling population, the quality of life can be restored.
Therefore, Indians should follow family planning for controlling the population.

Answer the following questions in short:

Question 1.
How does reproduction take place in fungus Mucor?
Answer:

Mucor reproduces asexually by spore formation.
It has filamentous body that possess sporangia.
When the spores are formed, the sporangia burst. The spores are released which settle down at suitable Places.
They germinate in moist and warm place forming a new fungal colony.

Question 2.
What is the type of asexual reproduction shown in the diagram below? (Board’s Model Activity Sheet)
Answer:

Type of asexual reproduction shown in the diagram above is fragmentation in Spirogyra.

Question 3.
Describe the structure of a flower.

Answer:
(1) The structural unit of sexual reproduction in plants is flower. There are total four floral whorls. Of these, two are accessory floral whorls while two are essential floral whorls.
(2) Calyx and corolla are accessory whorls. They are protective in nature.
(3) Members of calyx are known as sepals. They are usually green in colour. They protect the inner whorls.
(4) The members of corolla are called petals. They can be of different colours.
(5) Androecium and gynoecium are essential whorls as they participate in sexual reproduction.
(6) The male whorl androecium is made up of stamens. Each stamen has a filament with anther located at the upper end. In the anther there are four locules. Inside the locules the meiosis takes place forming pollen grains. During suitable time, the pollen grains are released from anther lobes.
(7) Gynoecium is made up of carpels, either in separate form or are united. Each carpel is formed of ovary at the basal end hollow ‘style’ and the stigma at the tip of style. There are one or many ovules inside the ovary.
(8) In bisexual flowers both androecium and gynoecium are located in the same flower, e.g. Hibiscus.
(9) In unisexual flowers, androecium is present in male flowers and gynoecium is present in the female flowers, e.g. Papaya.

Question 4.
Describe the human male reproductive system.
Answer:

In human male reproductive system, the reproductive organs are as follows:

Testes, different types of duct systems and glands.
Testes are in pair. Each testis lies in the scrotum which lies outside to abdominal cavity.
Testes -consist of numerous seminiferous tubules. The germinal epithelium of seminiferous tubules form sperms by undergoing meiosis.
These sperm cells are immature.
They are pushed gradually through various duct systems till the penis.
This path is as follows:
Rete testis → vas efferentia → epididymis → vasa deferentia → Ejaculatory duct → urethra
As the sperms are travelling, they gradually become mature. They are made capable to perform process of fertilization.
Seminal vesicles (in pairs), Single prostate gland and a pair of Cowper’s glands secrete their secretions. These secretions and the sperms together form semen.
This semen is deposited in the vagina with help of penis.

Question 5.
Describe the human female reproductive system.
Answer:

All the organs of the human female reproductive system are located inside the lower abdomen.
There are pair of ovaries, pair of fallopian ducts and a single median uterus.
The uterus opens out by vagina. In vaginal walls there are Bartholin’s glands.
The urethra in female body is separate and not a common passage as in male body.
The free end of fallopian duct is funnel-like having an opening in the centre. The oocyte released from the ovary due to ovulation is picked up by this funnel.
The other end of fallopian duct opens into uterus. There are cilia on inner surface of oviduct. With the help of the cilia the oocyte is pushed to the uterus through the fallopian duct.
The fertilization of oocyte can take place only in the middle’part of the fallopian duct.
The lower end of uterus opens into vagina. The contractions of uterus help in the process of parturition.
Vagina is the birth canal as well as copulatory passage. It is also a passage for menstrual flow.

Question 6.
What problems cause infertility in couple?
Answer:

In woman if there are problems like irregularity in menstrual cycle, difficulties in oocyte production or implantation in uterus, obstacles in the oviduct, etc.
In man if there are no sperms in the semen, slow movement of sperms, or anomalies in the sperms then he becomes sterile.
But now with the help of advanced medical techniques these problems can be overcome and a childless couple can be parents.

Question 7.
Answer the following questions: (July 2019)
(a) In our country, there seems to be lack of awareness regarding reproductive health. Why?
(b) Write the symptoms of disease gonorrhea.
(c) What precautions will you take to maintain reproductive health?
Answer:
(a) There is lack of awareness about reproductive health among majority of people of our country. This is due to social customs, traditions, illiteracy, social taboo and shyness.

(b) Symptoms of gonorrhea are as follows:

Painful burning during urination.
Oozing of pus through penis or vagina.
Inflammation of urinary tract, anus, throat, eyes, etc.

(c) Precautions to maintain reproductive health are cleanliness and personal hygiene. Guarding against any sexual infections.

Question 8.
If a piece of bread is kept in a container in moist place for 2-3 days, (1) What will you see? (2) Write scientific name and a character of the organism you may observe.
Answer:
(1) If a piece of bread is kept in moist container we can see growth of fungus on it.
(2) Fungi belonging to species Mucor is seen. It has filamentous body and sporangia. Sporangia burst open to spread spores. It has saprophytic mode of nutrition as it devoid of chlorophyll.

Write short notes on:

Question 1.
Multiple fission.
Answer:

During unfavourable conditions when there is lack of food. multiple fission is shown by amoeba. Amoeba forms protective covering and becomes encysted. Inside the cyst, amoeba undergoes repeated nuclear division. This is followed by cytoplasmic divisions. Many amoebulae are formed which remain dormant inside the cyst. When favourable conditions reappear, they come out by breaking the cyst.

Question 2.
Regeneration.
Answer:

In developed animals like wall lizard the process of regeneration is used to restore the lost parts like tail or limbs. As the reproductive system is one of the fullfledged system in the body, the process of regeneration cannot be called type of reproduction.

But some primitive organisms such as Plarfaria use this method for procreation. Planaria breaks up its body into two parts. Each part has the ability to develop the lost part by process of regeneration. This forms two new Planaria.

Question 3.
Seed germination.
Answer:

Seed germination is the process in which the seed develops into a new plantlet. In the plants, after fertilization the ovule develops into seed and ovary turns into fruit. Seeds fallen on the ground due to bursting of the fruits start germinating. Only under favourable conditions in the soil, this germination takes place. The zygote present inside the seed uses food stored in endosperm of seed and hence develops further to produce a new plantlet.

Question 4.
Budding in hydra.
Answer:

In multicellular organisms asexual reproduction by budding is shown by hydra. In fully grown Hydra, at specific part of its body there is development of bud.

This development is only during favourable period. The bud is an outgrowth developed due to repeated divisions of regenerative cells of body wall. It grows up gradually to form a small hydra. Parent hydra’s dermal layers and digestive cavity are in continuity with those of the budding hydra. It receives all the nutrition from parent hydra.When the budding hydra grows sufficiently, it detaches from parent hydra. Then it leads an independent life.

Question 5.
Fragmantation.
Answer:

Fragmentation is one of the type of asexual reproduction in multicellular organisms. During fragmentation, the body of parent organism breaks up into many fragments. All the resulting fragments start to develop as an independent new organism. In alga Spirogyra, and sponge like Sycon asexual reproduction takes place by fragmentation. Spirogyra grow up very fast and break up into many small fragments when there are favourable conditions. Each newly formed fragment lives independently as a new Spirogyra. Similarly the body of Sycon if accidentally broken into many fragments, develops into new Sycon from each old fragment.

Question 6.
Monozygotic twins.
Answer:
The twins developed from a single embryo are called monozygotic twins. If within 8 days of zygote formation i.e. in early embryonic development cells of that embryo divide into two groups. Each one develop as two separate embryos forming two monozygotic twins. Monozygotic twins are genetically exactly similar to each other. The gender of the twins is also same.

The Siamese twins develop from monozygotic twins, if the embryonic cells are divided into two groups 8 days after the zygote formation. These are conjoined twins where some parts of body are joined to each other. Also some organs are common in Siamese twins.

Read the paragraph and answer the questions given below:

Reproduction is the process by which the living species continues its existence. Lower organisms carry out asexual reproduction while higher plants and animals always show sexual reproduction. Plants reproduce asexually by methods such as fragmentation, vegetative propagation, budding, spore formation. For sexual reproduction they form gametes. In animal kingdom, budding, fission of different types and parthenogenesis are some of the methods that do not require both the sexes. Though regeneration also forms new individual, it is not considered to be a reproductive process because, basically it is a repair process. The ability to regenerate is lost in higher phyla. In human beings | it is restricted only to wound healing. Sexual reproduction is also undergoing lots of experimentation such as cloning which may make females capable of producing their own baby without intervention of any male.

Questions and Answers:

Question 1.
How do living species continue their existence?
Answer:
Through the process of reproduction, living species continue their existence.

Question 2.
Which are asexual methods of reproduction in kingdom Animalia?
Answer:
Fission, budding and parthenogenesis are the asexual methods of reproduction in Kingdom Animalia.

Question 3.
Why is regeneration not true method of reproduction?
Answer:
Regeneration is the repair process than a reproductive process. It is not done with the intention of producing offspring, but is for healing or repairing the lost part.

Question 4.
What are methods of reproduction in plants?
Answer:
Plants reproduce by asexual as well as sexual methods. Asexual reproduction is by fragmentation, vegetative propagation, budding, spore formation, while by formation of gametes, sexual reproduction is done.

Question 5.
What is the modern method of reproduction aimed at in higher organisms?
Answer:
Cloning is the modern method of reproduction by which production of young one can be aimed at.

Diagram-based Questions:

Question 1.
Observe the figure 3.18 and answer the questions below: (Board’s Model Activity Sheet)

(a) What does the figure 3.18 indicate?
Answer:
The figure indicates the menstrual cycle in human female.

(b) Which human organs are involved in this process?
Answer:
The ovary and uterus are primarily involved in this process. But the pituitary gland also controls this cycle.

(c) Which hormones take part in this process?
Answer:
Following hormones regulate this menstrual cycle.
Pituitary hormones: Follicle Stimulating Hormone (FSH) and Luteinizing Hormone (LH).
Ovarian hormones: Estrogen and progesterone.

(d) What is the periodicity for these changes?
Answer:
The menstrual cycle shows repetitive changes every 28 to 30 days.

(e) The body of woman undergoing this process is impure, she should remain away from other people. What is your opinion about this statement? Give justification for your opinion.
Answer:
A menstruating woman is not at all with impure body. It is a natural process in which the endometrium of the uterus is sloughed off and repaired.

She should get enough rest and nutrition during this period. It is painful period in which there is a possibility of infections. Therefore, she should take ! hygienic care and rest till the bleeding persists. But blind faith and superstition to keep her away from others should not be followed.

Question 2.
Observe the diagram (Fig. 3.19) of menstrual cycle and answer the following questions:

(1) What is the period of menstruation?
Answer:
1 to 5 days is the period of menstruation.

(2) On which day does ovulation occur during menstrual cycle?
Answer:
Ovulation occurs on 14th or 15th day.

(3) During which period is corpus luteum active during menstrual cycle? Which hormones are secreted by corpus luteum ?
Answer:
Corpus luteum is active till the 28th day of menstrual cycle. During this time if there is no union of sperm and ovum, then corpus luteum degenerates. Corpus luteum secretes estrogen and progesterone.

(4) In menstrual cycle which reproductive organs undergo changes?
Answer:
Ovary and uterus undergo changes during menstrual cycle.

(5) Which period is said to be period of regeneration of endometrium?
Answer:
In menstrual cycle, days 5 to 14 are period of regeneration of endometrium.

(6) Which period is said to be period of secretions of glands in endometrium?
Answer:
Period of secretions of glands in endometrium is 15 to 28 days.

Question 3.
Observe the following picture and describe the type of reproduction shown in.

Answer:
In developed animals like wall lizard the process of regeneration is used to restore the lost parts like tail or limbs. As the reproductive system is one of the fullfledged system in the body. the process of regeneration cannot be called type of reproduction.

But some primitive organisms such as Planaria use this method for procreation. Planaria breaks up its body into two parts. Each part has the ability to develop the lost part by process of regeneration. This forms two new Planaria.

Question 4.
Answer the following questions: (March 2019)
(a) “Gender of child is determined by the male partner of couple.” Draw a diagram explaining the above statement.
Answer:

(b) Prepare a slogan for campaign against female foeticide.
Answer:

Save the girl child.
Daughters give lot of joy, it is not only the boy.

(c) In the following figure, explain how new fungal colonies of mucor are formed:

Answer:
Mucor is a fungus having filamentous body. The filaments bear sporangia. Mature sporangia burst and release spores. Spores germinate to form new hyphae upon getting favourable moist and warm place.

(d) Identify and state the type of reproduction represented in the above figure.
Ans. The spore formation is asexual type of reproduction seen in Mucor.

Question 5.
Write the type of asexual reproduction shown in the figure.

Answer:
The figure shows budding in yeast. Budding is the type of asexual reproduction.

Experiments:
(Try this: Text Book Pages 23 and 24)

(1) Observation of Paramoecia.
(2) Observation of yeast.
(3) Study of Hibiscus.
[For detailed information on practicals, refer to Vikas Science and Technology Experiment Book: Standard X]

Projects:

Project 1.
Use of ICT. (Textbook page no. 27)
Make an video album of pollination and show it in the class.

Project 2.
Internet is my friend. (Textbook page no. 33)
You may have read that sometimes a woman may deliver more than two offspring at a time. Collect more information from internet about reasons for such incidences.

Project 3.
Get information. (Textbook page no. 34)
Visit a public health centre nearby your place and collect the information through an interview of health officer about meaning and various methods of family planning.

10th Std Science Part 2 Questions And Answers:

Std 8 History Chapter 3 Questions And Answers Effects of British Rule Maharashtra Board

February 6, 2025January 5, 2025 by Bhagya

Balbharti Maharashtra State Board Class 8 History Solutions Chapter 3 Effects of British Rule Notes, Textbook Exercise Important Questions and Answers.

Class 8 History Chapter 3 Effects of British Rule Questions And Answers Maharashtra Board

Effects of British Rule Class 8 Questions And Answers Chapter 3 Maharashtra Board

Class 8 History Chapter 3 Effects of British Rule Textbook Questions and Answers

1. Rewrite the statements by choosing the appropriate options:

Question 1.
Portuguese, ………….., French, British participated in the competition of capturing Indian market.
(a) Austrian
(b) Dutch
(c) German
(d) Swedish
Answer:
(b) Dutch

Question 2.
In 1802, Peshwa ………….. signed the Subsidiary Alliance with the British.
(a) Bajirao I
(b) Sawai Madhavrao
(c) Peshwa Nanasaheb
(d) Bajirao II
Answer:
(d) Bajirao II

Question 3.
Jamshedjee Tata started the manufacturing of steel at Tata Iron and Steel Industry established in ………….. .
(a) Mumbai
(b) Kolkata
(c) Jamshedpur
(d) Delhi
Answer:
(c) Jamshedpur

2. Explain the following concept:

Question 1.
Civil Services :
Answer:

There was a need of bureaucrats to strengthen the British rule in India.
Lord Cornwallis introduced Civil Services which became an important part of the British government.
The territories occupied by the British were divided into districts for administrative convenience. The district administration was headed by Collector.
The officers appointed through the Civil Services (ICS) were taken into administrative services.

Question 2.
Commercialisation of Agriculture :
Answer:

In the pre-British period, farmers used to cultivate food grains to fulfill domestic need as well as need of the village.
The British government gave encouragements to the cultivation of cash crops like indigo, cotton, tobacco, tea, etc.
The shift from cultivation of foodgrains to profit-yielding cash crops is known as Commercialisation of Agriculture.

Question 3.
Economic Policy of British :
Answer:

Capitalist economy prevailed in England due to the Industrial Revolution.
This system was brought to India to nurture the British economy.
It resulted in the economic gains for England but exploitation and impoverishment of India.

3. Explain the following statements with reasons:

Question 1.
Farmers in India became bankrupt.
Answer:
The British made number of changes in the existing system to increase the revenue.

The payment of revenue was made compulsory in cash and within the prescribed time limit.
Land was confiscated if the revenue was not paid in time.
The land revenue collection differed from place to place which resulted in the exploitation of the farmers.
Farmers were forced to sell their produce to the merchants and middlemen at a low price in order to pay revenue.
Farmers mortgaged land to pay tax and became indebted to moneylenders.

In this way, the farmers became bankrupt in India.

Question 2.
There was decline of traditional industries in India.
Answer:

The British government levied heavy duty on the goods exported from India to England.
The goods imported from England were produced in factories on a large scale and at minimum cost.
The duty levied on them by British was far too less.
So, these goods were cheap as compared to traditional goods.
The Indian artisans found it difficult to compete with low priced British goods.
Eventually, this led to closing down of traditional industries in India.

4. Complete the following table:

Question.
Maharashtra Board Class 8 History Solutions Chapter 3 Effects of British Rule 1
Answer:

Do You Know?

Work done by Chhatrapati Pratapsingh :
Maharashtra Board Class 8 History Solutions Chapter 3 Effects of British Rule 3

Water tank was built on back side of Yevteshwar temple and Mahadara which supplied water to Satara city.
He built roads and planted trees on both the sides.
Sanskrit, Marathi and English was taught to girls and boys in schools.
Printing press was set up and many books were published.
A book titled ‘Sabhaniti’ was printed on polity in 1827.
A road connecting Satara to Mahabaleshwar to Pratapgad was built by him which was further extended to Mahad.aqs

Project:

Prepare detailed information with pictures about the development by British in administration, education, transport and communication in India.

Class 8 History Chapter 3 Effects of British Rule Additional Important Questions and Answers

Rewrite the statements by choosing the appropriate options:

Question 1.
………….. was the main centre of British in western India.
(a) Surat
(b) Cochin
(c) Goa
(d) Mumbai
Answer:
(d) Mumbai

Question 2.
With the treaty of ……………. in 1782 the first Anglo-Maratha war came to an end.
(a) Wadgaon
(b) Vasai
(c) Salbai
(d) Ahmednagar
Answer:
(c) Salbai

Question 3.
In 1848, Lord Dalhousie rejected the adoption policy and annexed the state of ………… .
(a) Pune
(b) Kolhapur
(c) Sangli
(d) Satara
Answer:
(d) Satara

Question 4.
A committee was set up to create a Code of Law, under the leadership of ………… .
(a) Lord Cornwallis
(b) Lord Macaulay
(c) Lord Warren Hastings
(d) Lord Robert Clive
Answer:
(b) Lord Macaulay

Question 5.
……………. started the first textile mill in 1853 at Mumbai.
(a) Jamshedji Tata
(b) Ratan Tata
(c) Kawasjee Nanabhoy Davar
(d) Jamshedji Jeejibhoy
Answer:
(c) Kawasjee Nanabhoy Davar

Question 6.
The first jute mill was set up at in Bengal.
(a) Kolkata
(b) Rishra
(c) Hooghli
(d) Dhakka
Answer:
(b) Rishra

Question 7.
The territory under the control of the English was divided into ………… for the convenience of administration.
(a) Subhas
(b) districts
(c) Paraganas
(d) mahajanpadas
Answer:
(b) districts

Question 8.
The process of giving stress on cultivation of cash crops instead of food grains is known as …………. of agriculture.
(a) Commercialisation
(b) Liberalisation
(c) Globalisation
(d) Rotation
Answer:
(a) Commercialisation

Question 9.
…………… wrote the book ‘Sabhaniti’ in 1827.
(a) Bajirao II
(b) Rango Bapuji Gupte
(c) Chhatrapati Shivaji Maharaj
(d) Chhatrapati Pratapsingh
Answer:
(d) Chhatrapati Pratapsingh

Question 10.
A German thinker …………… was a devout scholar of Indian religion, language and history.
(a) Max Mueller
(b) Lord Macaulay
(c) John Stuart Elphistone
(d) William Jones
Answer:
(a) Max Mueller

Name the following:

Question 1.
He started Subsidiary Alliance.
Answer:
Lord Wellesley

Question 2.
First Governor General according to the Regulating Act.
Answer:
Lord Warren Hastings

Question 3.
He started Dual Government system in Bengal.
Answer:
Robert Clive

Question 4.
Recommended English education in India.
Answer:
Lord Macaulay

Question 5.
Loyal officer of Chhatrapati Pratapsingh.
Answer:
Rango Bapuji Gupte.

Answer the following questions in one sentence each:

Question 1.
How were the officers in administrative services appointed?
Answer:
The officers in the administrative services were appointed through competitive examination known as Indian Civil Services (ICS).

Question 2.
Which principle was introduced by the British in the administration of Justice?
Answer:
The British introduced the principle of ‘equality before law’ in the administration of justice all over British India.

Question 3.
Which factors hindered the growth of new industries in India?
Answer:
The growth of new industries in India was hindered by the lack of British support, capital and experience of management.

Question 4.
Which cash crops were encouraged by the British Government?
Answer:
The British Government gave more encouragement to cash crops like cotton, indigo, tobacco, tea, etc.

Question 5.
Where were the universities established in 1857?
Answer:
The universities were established in Mumbai, Kolkata and Madras (Chennai) in 1857.

Question 6.
What was the effect of development of modern means of transport and communication ?
Answer:
The development of modern means of transport and communication helped to improve communication between the people and strengthened their sense of unity between Indians.

Question 7.
Which values shaped the new era in the 19th century Europe?
Answer:
The new era in the 19th century Europe was shaped on the values humanitarianism, democracy, nationalism and liberalism.

Do as Directed:

1. Complete the concept map:

Question 1.
Maharashtra Board Class 8 History Solutions Chapter 3 Effects of British Rule 4
Answer:

Question 2.
Maharashtra Board Class 8 History Solutions Chapter 3 Effects of British Rule 6
Answer:

Question 3.
Maharashtra Board Class 8 History Solutions Chapter 3 Effects of British Rule 8
Answer:

Question 4.
Maharashtra Board Class 8 History Solutions Chapter 3 Effects of British Rule 10
Answer:

2. Complete the following table:

Question 1.
Maharashtra Board Class 8 History Solutions Chapter 3 Effects of British Rule 12
Answer:

3. Complete the timeline:

Question 1.
Maharashtra Board Class 8 History Solutions Chapter 3 Effects of British Rule 14
Answer:

Explain the following concept:

Question 1.
Dual Government :
Answer:
1. The British East India Company in order to consolidate rule introduced new system of governance in India known as Dual Government.
2. It was introduced by Robert Clive in Bengal in 1765 wherein the East India Company took over the revenue collection. The maintenance of law and order was the responsibility of the Nawab of Bengal.

Answer the following in 25-30 words:

Question 1.
What were the conditions laid in Subsidiary Alliance?
Answer:
Lord Wellesley signed Subsidiary Alliance with many Indian rulers in 1798.
Its conditions were :

Indian rulers should keep the British army in their Court.
They have to pay the company towards the maintenance of these forces in cash or a part of their territory of equivalent amount of revenue.
Without the intervention of the British, they would not have any alliance or declare war with any power.
They should keep the British resident in their court.

Question 2.
What were the provisions in Regulating Act of 1773?
Answer:
According to Regulating Act of 1773:

The Governor of Bengal was designated as the Governor General. Lord Warren Hastings became the first Governor General of India.
It gave the Governor General controlling powers over the Bombay and Madras Presidencies.
It provided a committee of four members to assist the Governor General.

Question 3.
Write about the Pitt’s India Act of 1784.
Answer:

1. The Pitt’s India Act of 1784 established a permanent Board of Control to regulate and manage administration of Company in India.
2. The Board was authorized to issue directives to the Company regarding the governance of India.

Question 4.
Write about the Judicial system introduced by the British in India.
Answer:

The new Judicial system was introduced in India on the basis of the Judicial system in England.
Accordingly, each district had a Civil and Criminal Court for the respective cases.
High Courts were established to reconsider the judgements delivered by the District courts.

Question 5.
What were the functions of the military?
Answer:

The functions of the military were to defend the Indian territories under the control of the British.
It was expected to acquire new territories and quell any uprising/revolt against the British India.

Question 6.
What were the defects in the British Judicial System?
Answer:
The defects in the British Judicial System were as follows :

There were different laws and separate courts for the Europeans.
It was difficult for the Common people to understand the new laws.
It was very expensive to fight a legal case.
The cases got delayed and remained pending for years.

Question 7.
What reforms were introduced in the Judicial System by the British in India?
Answer:

In the pre-British period, laws were different from place to place.
There was difference in judgement on the basis of casteism.
A committee was set up to create a uniform code of law under the leadership of Lord Macaulay.
The Indian Penal Code was enforced all over British India with British principle of Equality before Law.

Question 8.
Write about introduction of the English education system?
Answer:

The British were in need of Indians who had received English education in order to run the administration.
English education was imparted in India according to the recommendations of Lord Macaulay in 1835.
Indians were introduced to western thoughts, modem reforms, science and technology.
They realised the need to study their history, culture, religions and also realised the drawbacks.
The western educated Indian middle class was responsible for initiating social reforms in later period.

Explain the following statements with reasons:

Question 1.
Britishers entered into Maratha politics.
Answer:
1. Marathas had a strong hold on the areas in and around Mumbai which was prime centre of British in western India.
2. British tried to acquire nearby territories but the Marathas checked their expansion.
3. It was only after the death of Peshwa Madhavrao, that they got entry in Maratha politics. Raghunathrao in his greed for Peshwaship sought their help.
This led to the entry of the British in Maratha politics.

Question 2.
The British Parliament introduced some laws to keep control over the affairs of the company.
Answer:

The ‘Dual Government’ system was introduced by Robert Clive in Bengal in 1765.
But, sometimes, the officers of the company pocketed money.
As monopoly of trade was given to East India Company, many trading companies in England envied them.
The working system of the company was criticised in the British Parliament. So, to keep control on the affairs of the company, the British Parliament introduced some laws.

Question 3.
Salary of company officers were increased.
Answer:

Lord Cornwallis introduced bureaucracy in India to strengthen the British rule in India.
He restricted the private trade carried out by the company officers. For this reason, he increased the salary of the company officers.

Answer the following in detail:

Question 1.
Give an account of the Anglo- Maratha war.
Answer:

Raghunathrao in his greed for the seat of Peshwa sought help of the British which facilitated their entry into Maratha politics.
Three wars were fought between Marathas and the British between 1774 to 1818.
As the Marathas fought unitedly, the Britishers were defeated in the first Anglo- Maratha War in 1782.
In 1802, Bajirao II entered into Subsidiary Alliance with the British known as the Treaty of Vasai.
The second Anglo-Maratha War took place as some Maratha Sardars opposed this treaty.
Marathas were defeated in the second Anglo-Maratha War.
The defeat of the Marathas led to the increase in interference of the British in the Maratha state.
Bajirao II declared war against the British as he could not tolerate their interference.
He lost the war and surrendered to the British in 1818.

Question 2.
Write about the public work done by Chhatrapati Pratapsingh.
Answer:
Chhatrapati Pratapsingh did the following public work :

Water tank was built by him on the back side of Yevteshwar temple and Mahadara which supplied water to Satara city.
He built roads and planted trees on both the sides.
Sanskrit, Marathi and English was taught to girls and boys in schools built by him.
Printing press was set up and many books were published.
A book titled ‘Sabhaniti’ was printed on polity in 1827.
A road connecting Satara – Mahabaleshwar – Pratapgad was built by him. This road was further extended to Mahad.

Question 3.
Compare the Land Revenue Policy in the pre-British period and during the British period.
Answer:
Maharashtra Board Class 8 History Solutions Chapter 3 Effects of British Rule 16

Question 4.
How did the British administrative system benefit India?
Answer:
There were many benefits of British administrative system in India.

The British introduced railways in India.
Telegraph system connected major cities and military stations got connected.
Postal system was also introduced.
These improved communication generated a sense of unity among the people of India.
Coal, metal, sugar, cement and chemical industries also developed gradually.
The western educated Indians learnt the values of humanism, rationalism, democracy, nationalism and liberalism.
Indians felt the need to study Indian history, religion and traditions.
Universities were established at Kolkata, Mumbai and Madras (Chennai).
The newly educated Indian led the social reform movement in later period.

Question 5.
According to you, what were the two positive and two negative effects of British rule on India?
Answer:
The positive effects were :

English education laid the foundation of future progress in India.
Indian adopted the principles of freedom, equality and humanity.

The negative effects were :

The British policy of ‘Divide and Rule’ deepened its roots.
Regional languages got neglected as English got importance.

8th Std History Questions And Answers: