Physical quantity and Measurement complete solution Class 8 icse

 

Chapter-2 – Quantity and Measurement

Objective Questions

Question 1e true or false for each statement:

(a) Equal volumes of two different substances have equal masses.

(b) The density of a piece of brass will change by changing its size or shape.

(c) The density of a liquid decreases with increase in its temperature.

(d) Relative density of water is 1.0.

(e) Relative density of a substance is expressed in g cm^-3.

(f) When a body is immersed in a liquid, the buoyant force experienced by the body is equal to the volume of the liquid displaced by it.

(g) A body experiences the same buoyant force while floating in water or alcohol.

(h) A body experiences the same buoyant force when it floats or sinks in water.

(i) A body floats in a liquid when its weight becomes equal to the weight of the liquid displaced by its submerged part.

(j) A body while floating, sinks deeper in a liquid of low density than in a liquid of high density.

Answer

(a) False
Correct Statement — Equal volumes of two different substances have different masses.

(b) False
Correct Statement — The density of a piece of brass will not change by changing its size or shape.

(c) True

(d) True

(e) False
Correct Statement — Relative density has no unit.

(f) False
Correct Statement — When a body is immersed in a liquid, the buoyant force experienced by the body is equal to the weight of the liquid displaced by the immersed part of the body.

(g) True

(h) False
Correct Statement — The buoyant force experience by the body is different when floating or sinking in water.

(i) True

(j) True

Question 2

Fill in the blanks:

(a) 1 kg is the mass of ............... mL of water at 4°C.

(b) Mass = density x ............... .

(c) The S.I. unit of density is ............... .

(d) Density of water is ............... kg m^-3.

(e) 1 g cm^-3 = ............... kg m^-3.

(f) The density of a body which sinks in water is ............... than 1000 kg m^-3.

(g) A body sinks in a liquid A, but floats in a liquid B. The density of liquid A is ............... than the density of liquid B.

(h) A body X sinks in water, but a body Y floats on water. The density of the body X is ............... than the density of body Y.

(i) The buoyant force experienced by a body when floating in salt-water is ............... that when floating in pure water.

(j) The weight of a body floating in a liquid is ............... .

Answer

(a) 1000

(b) volume

(c) kg m^-3

(d) 1000

(e) 1000

(f) more

(g) less

(h) more

(i) equal to

(j) zero

Question 3

Match the following:

Column A	Column B
(a) kg m-3	(i) relative density
(b) No unit	(ii) sinks in alcohol
(c) Relative density	(iii) floats on water
(d) Iron	(iv) density
(e) Wood	(v) density bottle

Answer

Column A	Column B
(a) kg m-3	(iv) density
(b) No unit	(i) relative density
(c) Relative density	(v) density bottle
(d) Iron	(ii) sinks in alcohol
(e) Wood	(iii) floats on water

Question 4a

Select the correct alternative:

The correct relation is:

1. Density = Mass x Volume
2. Mass = Density x Volume
3. Volume = Density x Mass
4. Density = Mass + Volume

Answer

Mass = Density x Volume

Reason — The density of a substance is its mass per unit volume.

Question 4b

Select the correct alternative:

The relative density of alcohol is 0.8. Its density is:

1. 0.8
2. 800 kg m^-3
3. 800 g cm^-3
4. 0.8 kg m^-3

Answer

800 kg m^-3

Reason —

Relative density of alcohol=Density of alcoholDensity of water0.8=Density of alcohol 1000 kg m⁻³ Density of alcohol=0.8×1000=800 kg m−3 Relative density of alcohol=Density of water Density of alcohol​0.8=1000 kg m−3 Density of alcohol ​ Density of alcohol=0.8×1000=800 kg m⁻³

Question 4c

Select the correct alternative:

A block of wood of density 0.8 g cm^-3 has a volume of 60 cm^-3. The mass of block is:

1. 60.8 g
2. 75 g
3. 48 g
4. 0.013 g

Answer

48 g

Reason —

Density=Mass Volume 0.8=Mass 60

Mass=0.8×60=48g

Density=Volume x Mass

​0.8 x 60

Mass=0.8×60=48g

Question 4d

Select the correct alternative:

The density of aluminium is 2.7 g cm^-3 and that of brass is 8.4 g cm^-3. The correct statement is:

1. Equal masses of aluminium and brass have equal volumes.
2. The mass of a certain volume of brass is more than the mass of equal volume of aluminium.
3. The volume of a certain mass of brass is more than the volume of equal mass of aluminium.
4. Equal volumes of aluminium and brass have equal masses.

Answer

The mass of a certain volume of brass is more than the mass of equal volume of aluminium.

Reason —

Density=MassVolumeDensity=VolumeMass​

So mass of certain volume of brass will be more than the mass of same volume of aluminium as density of brass is more than density of aluminium.

Question 4e

Select the correct alternative:

A density bottle has a marking 25 mL on it. It means that:

1. the mass of the density bottle is 25 g.
2. the density bottle will store 25 mL of any liquid in it.
3. the density bottle will store 25 mL of water, but more volume of liquid denser than water.
4. the density bottle will store 25 mL of water, but more volume of a liquid lighter than water.

Answer

The density bottle will store 25 mL of any liquid in it.

Reason — A density bottle can store specific volume of any density of liquid. Since this bottle is marked 25 mL so it can store 25 mL of any density liquid in it.

Question 4f

Select the correct alternative:

The correct statement is:

1. The buoyant force on a body is equal to the volume of the liquid displaced by it.
2. The buoyant force on a body is equal to the volume of the body.
3. The buoyant force on a body is equal to the weight of the liquid displaced by it.
4. The buoyant force on a body is always equal to the weight of the body.

Answer

The buoyant force on a body is equal to the weight of the liquid displaced by it.

Reason — According to Principle of Floatation, the buoyant force is equal to the weight of the liquid displaced by the immersed part of the body.

Question 4g

Select the correct alternative:

A piece of wood floats on water. The buoyant force on wood will be:

1. zero
2. more than the weight of the wood piece.
3. equal to the weight of the wood piece.
4. less than the weight of the wood piece.

Answer

equal to the weight of the wood piece

Reason — According to Law of Floatation when a body floats on water the weight of the body is equal to the buoyant force.

Question 4h

Select the correct alternative:

The weight of a body is more than the buoyant force experienced by it, due to a liquid. The body will:

1. sink
2. float with its some part outside the liquid.
3. float just below the surface of liquid.
4. float with whole of its volume above the surface of liquid.

Answer

sink

Reason — According to the Principle of Floatation, if the weight of the body is more than the buoyant force acting on the body the body will sink into the liquid.

Short/Long Answer Questions

Question 1

Define the term density of a substance.

Answer

The density of a substance is its mass per unit volume i.e.

Density=MassVolumeDensity=VolumeMass​

Question 2

Name the S.I. unit of density. How is it related to g cm^-3.

Answer

The S.I. unit of density is kg m^-3 (kilogram per cubic metre).
1 g cm^-3 = 1000 kg m^-3.

Question 3

The density of brass is 8.4 g cm^-3. What do you mean by this statement?

Answer

The statement means one cubic centimeter volume of brass has mass of 8.4 g.

Question 4

Arrange the following substances in order of their increasing density:

Iron, Cork, Brass, Water, Mercury.

Answer

Cork < Water < Iron < Brass < Mercury.

The density of cork is the least and that of mercury is the most.

Question 5

How does the density of a liquid (or gas) vary with temperature?

Answer

When temperature increases, volume of most of the liquid increases, so density decreases. Similarly when temperature decreases volume of liquids decreases so density increases. But exception is water which contracts on heating from 0°C to 4°C and expands on heating above 4°C. Water has maximum density at 4 degree celsius.

Question 6

A given quantity of a liquid is heated. Which of the following quantity will vary and how?

(a) mass
(b) volume
(c) density

Answer

(a) Mass of the liquid remains unaffected on heating.

(b) Volume of the liquid increases with increase in temperature.

(c) The density of liquid decreases on heating because its volume increases and density is inversely proportional to volume.

Question 7

Describe an experiment to determine the density of the material of a coin.

Answer

The density of the material of a coin is determined by the following experiment:

1. We know density is mass per unit volume. So, the first mass of the coin is measured by a common beam balance. Let it be M gram.
2. Volume of the coin is measured by a measuring cylinder as shown in the figure given below.
3. Take a measuring cylinder and fill it partly with water.
4. Note the level of water. Let it be V₁ mL.
5. Tie the coin with a thread and gently lower the coin inside the measuring cylinder. Take care that no water splashes out.
6. Note the level of water again. Let it be V₂ mL.
7. Find the volume of the coin (V) by subtracting V₁ from V₂.
   V = V₂ - V₁
   Density of coin is calculated by the formula: Density=MassVolumeDensity=VolumeMass​

D=MV g cm−3D=VM​ g cm−3

Question 8

Describe an experiment to determine the density of a liquid.

Answer

The density of a liquid is determined by first finding its mass and then its volume.

1. Take an empty beaker and measure its mass by using a common beam balance. Let the mass be M₁ g.
2. Take some amount of liquid and pour it into the empty beaker. Measure its mass again. Let it be M₂ g.
3. The difference between M₂ and M₁ gives the mass(M) of the liquid. M = M₂ - M₁.
4. Pour the liquid from beaker into a measuring cylinder. Note the volume of the liquid. Let it be V mL or V g cm^-3.
5. Calculate the density of the liquid using the following formula:

Density=MassVolumeDensity=VolumeMass​

D=MV g cm−3D=VM​ g cm−3

Question 9

What is a density bottle? How is it used to find the density of a liquid?

Answer

A density bottle is a specially designed bottle which is used to determine the density of a liquid. The figure of the density bottle is shown below:

Determination of Density of a Liquid using the Density Bottle

1. Measure the mass of empty density bottle using a beam balance. Let it be M₁ g.
2. Remove the stopper of the bottle and fill it with water. Replace the stopper, wipe the outside of bottle dry and measure its mass again. Let it be M₂ g.
3. Empty the bottle and dry it. Fill it with the given liquid, replace the stopper and wipe the outside of bottle dry. Measure its mass. Let it be M₃ g.
4. Calculate the mass of water (M₂ - M₁) and the mass of liquid (M₃ - M₁).
5. Since density of water is 1 g cm^-3, mass of water contained in bottle is the volume of bottle. Thus,
       Mass of liquid = M₃ - M₁
       Volume of liquid = M₂ - M₁

∴Density of liquid=Mass of liquidVolume of liquid=M3−M1M2−M1 g cm−3∴Density of liquid=Volume of liquidMass of liquid​=M2​−M1​M3​−M1​​ g cm−3

Question 10

Define the term relative density of a substance.

Answer

The relative density of a substance is defined as the ratio of the density of a substance to the density of water. The symbol used for relative density is R.D. Thus,

R.D.=Density of the substanceDensity of waterR.D.=Density of waterDensity of the substance​

Question 11

What is the unit of relative density?

Answer

Relative density has no unit because it is the ratio of two densities.

Question 12

Distinguish between density and relative density.

Answer

Sl. No.	Density	Relative Density
(i)	Density of a substance is defined as the mass of a unit volume of that substance.	The relative density of a substance is defined as the ratio of the density of a substance to the density of water.
(ii)	Units are kg m-3, g cm-3.	Relative Density has no units.

Question 13

Explain the meaning of the statement 'Relative density of aluminium is 2.7'.

Answer

Relative density of aluminium is 2.7 means a piece of aluminium of any volume has mass 2.7 times that of an equal volume of water.

Question 14

How does the density of a body and that of a liquid determine whether the body will float or sink into that liquid?

Answer

If the density of a body is less than the density of the liquid, the body will float on the surface of the liquid. But if the density of a body is more than the density of liquid the body will sink in the liquid.

Question 15

A cork piece floats on water's surface while an iron nail sinks in it. Explain the reason.

Answer

The density of cork is less than the density of water so it floats on water surface but the density of iron nail is more than the density of water so it sinks in water.

Question 16

Which of the following will sink or float on water? (Density of water = 1 g cm^-3)

1. Body A having density 500 kg m^-3
2. Body B having density 2520 kg m^-3
3. Body C having density 1100 kg m^-3
4. Body D having density 0.85 g cm^-3

Answer

1. Body A will float on water because its density (500 kg m^-3) is less than the density of water (1 g cm^-3 or 1000 kg m^-3).
2. Body B will sink in water because its density (2520 kg m^-3) is more than the density of water (1 g cm^-3 or 1000 kg m^-3).
3. Body C will sink in water because its density (1100 kg m^-3) is more than the density of water (1 g cm^-3 or 1000 kg m^-3).
4. Body D will float on water because its density (0.85 g cm^-3) is less than the density of water (1 g cm^-3).

Question 17

State the law of floatation?

Answer

The law of floatation states that — "When a body floats in a liquid, the weight of the liquid displaced by its immersed part is equal to the total weight of the body".
i.e., while floating
Weight of the floating body = Weight of the liquid displaced by its immersed part (i.e., buoyant force)

Question 18

The density of water is 1.0 g cm^-3. The density of iron is 7.8 g cm^-3. The density of mercury is 13.6 g cm^-3. Answer the following:

(a) Will a piece of iron float or sink in water?

(b) Will a piece of iron float or sink in mercury?

Answer

(a) Iron will sink in water as density of iron (7.8 g cm^-3) is more than the density of water (1 g cm^-3).

(b) Iron will float in mercury as the density of iron (7.8 g cm^-3) is less than the density of mercury (13.6 g cm^-3).

Question 19

The diagram given below shows a body floating in three different liquids A, B and C at different levels.

(a) In which liquid does the body experience the greatest buoyant force?

(b) Which liquid has the least density?

(c) Which liquid has the highest density?

Answer

(a) The buoyant force is same in each case as the weight of the body is same in each case and the buoyant force is equal to the weight of the liquid displaced by the immersed part of the body which balances the weight of the body.

(b) Liquid A has the least density as the maximum part of the body is immersed in liquid A.

(c) Liquid C has the highest density as the body immerses the least in liquid C.

Question 20

For a floating body, how is its weight related to the buoyant force?

Answer

When a body floats in a liquid, the weight of the liquid displaced by its immersed part (i.e. the buoyant force) is equal to the total weight of the body.

Question 21

Why does a piece of ice float on water?

Answer

Density of ice is 0.9 g cm^-3 and density of water is 1.0 g cm^-3. Since density of ice is less than density of water, so ice floats on water.

Question 22

Explain why an iron needle sinks in water, but a ship made of iron floats on water.

Answer

An iron needle is solid and density of iron is more than that of water. Weight of iron needle is more than the buoyant force of water on it so it sinks in water. Although ship is made of iron but it has hollow space that is filled with air which make its average density less than that of water, so ship floats on water.

Question 23

It is easier to swim in sea water than in river water. Explain the reason.

Answer

Sea water contains salts so its density is more than the density of river water. The weight of a person gets balanced by the less immersed part of his body in seawater as compared to that in river water. Thus, it is easier to swim in seawater than in river water.

Question 24

Icebergs floating on seawater are dangerous for ships. Explain the reason.

Answer

Density of ice (0.9 g cm^-3) is less than the density of seawater (1.02 g cm^-3). So, an iceberg floats in seawater with its large portion submerged inside the water and only a little portion of it is above the surface of water. Thus, a ship can collide with the invisible part of the iceberg under the surface of water. Hence, it is dangerous.

Question 25

Explain why it is easier to lift a stone underwater than in air.

Answer

When a stone is under water then its weight is less than its weight in air because it experiences an upward buoyant force. Due to apparent loss in weight, it is lighter in water which makes it easier to lift in water than in the air.

Question 26

What is a submarine? How can it be made to dive in the water and come to the surface of water?

Answer

A submarine is a water tight boat which can travel under water like a ship and it is provided with water tanks.

To make the submarine dive, the tanks are filled with water so that the average density of the submarine becomes greater than the density of sea water, so it sinks into the water.

To make the submarine rise to the surface of water, these tanks are emptied. This makes the average density of the submarine less than the density of sea water, so the submarine rises up to the surface of water.

Question 27

A balloon filled with hydrogen rises in air. Explain the reason.

Answer

The density of hydrogen gas is less than the density of air. So the buoyant force experienced by the balloon due to air is greater than the weight of the balloon. Due to this net upward force balloon rises up in air.

Numericals

Question 1

The density of air is 1.28 g litre^-1. Express it in:

(a) g cm^-3

(b) kg m^-3

Answer

(a) The density of air is 1.28 g litre^-1.

1 litre = 1000 cm³

∴ Density of air in g cm^-3 = 1.28100010001.28​

Density of air = 0.00128 g cm^-3

(b) Density of air in kg m^-3 = Density of air in g cm ^-3 x 1000 = 0.00128 x 1000 = 1.28 kg m^-3

∴ Density of air in kg m^-3 = 1.28 kg m^-3.

Question 2

The dimensions of a hall are 10 m x 7 m x 5 m. If the density of air is 1.11 kg m^-3, find the mass of air in the hall.

Answer

Density of air = 1.11 kg m^-3

Volume of hall = 10 m x 7 m x 5 m = 350 m³

Mass = ?

We know,

Density=MassVolumeMass=Density×Volume=1.11×350=388.5 kgDensity=VolumeMass​Mass=Density×Volume=1.11×350=388.5 kg

Hence, the mass of air in the hall = 388.5 kg.

Question 3

The density of aluminium is 2.7 g cm^-3. Express it in kg m^-3.

Answer

Density of Aluminium = 2.7 g cm^-3.

Density of aluminium in kg m^-3 = 2.7 x 1000 = 2700 kg m^-3

Hence, Density of aluminium = 2700 kg m^-3.

Question 4

The density of alcohol is 600 kg m^-3. Express it in g cm^-3.

Answer

Density of alcohol in g cm−3=Density of alcohol in kg m−31000=6001000=0.6 g cm−3Density of alcohol in g cm−3=1000Density of alcohol in kg m−3​=1000600​=0.6 g cm−3

Hence, Density of Alcohol in g cm^-3 = 0.6 g cm^-3.

Question 5

A piece of zinc of mass 438.6 g has a volume of 86 cm³. Calculate the density of zinc.

Answer

Mass of zinc = 438.6 g

Volume of zinc = 86 cm³

Density of zinc = ?

Density of Zinc=Mass of ZincVolume of Zinc=438.686=5.1 g cm−3Density of Zinc=Volume of ZincMass of Zinc​=86438.6​=5.1 g cm−3

So, Density of Zinc = 5.1 g cm^-3.

Question 6

A piece of wood of mass 150 g has a volume of 200 cm³. Find the density of wood in

(a) C.G.S. unit

(b) S.I. unit

Answer

Mass of wood = 150 g

Volume of wood = 200 cm³

Density of wood = ?

(a) C.G.S. unit of density is g cm^-3.

Density of Wood=Mass of WoodVolume of Wood=150200=0.75 g cm−3Density of Wood=Volume of WoodMass of Wood​=200150​=0.75 g cm−3

Density of wood = 0.75 g cm^-3.

(b) S.I. unit of density is kg m^-3.

Density of Wood in kg m^-3 = Density of Wood in g cm^-3 x 1000 = 0.75 x 1000 = 750 kg m^-3

So, Density of Wood = 750 kg m^-3.

Question 7

Calculate the volume of wood of mass 6000 kg if the density of wood is 0.8 g cm^-3.

Answer

Density of wood = 0.8 g cm^-3 or 800 kg m^-3

Mass of wood = 6000 kg

Volume of wood = ?

Density of Wood=Mass of WoodVolume of WoodVolume of Wood=Mass of woodDensity of wood=6000800=7.5 m3Density of Wood=Volume of WoodMass of Wood​Volume of Wood=Density of woodMass of wood​=8006000​=7.5 m3

So, Volume of wood = 7.5 m³.

Question 8

Calculate the density of a solid from the following data:

(a) Mass of solid = 72 g

(b) Initial volume of water in measuring cylinder = 24 mL

(c) Final volume of water when solid is completely immersed in water = 42 mL

Answer

Mass of solid = 72 g

Volume of solid = Final volume of water - Initial volume of water

Volume of solid = 42 - 24 = 18 mL

Density of Solid=Mass of SolidVolume of Solid=7218=4Density of Solid=Volume of SolidMass of Solid​=1872​=4

So, Density of solid = 4.0 g cm^-3.

Question 9

The mass of an empty density bottle is 21.8 g, when filled completely with water it is 41.8 g and when filled completely with liquid it is 40.6 g. Find:

(a) the volume of density bottle.

(b) the relative density of liquid.

Answer

Mass of empty density bottle (M₁) = 21.8 g

Mass of bottle with water (M₂) = 41.8 g

Mass of bottle with liquid (M₃)= 40.6 g

(a) Mass of water in bottle = M₂ - M₁ = 41.8 - 21.8 = 20 g

Density of water is 1 g cm^-3, so volume of density bottle = 20 mL

(b) Mass of liquid in bottle = M₃ - M₁ = 40.6 - 21.8 = 18.8 g

R.D. of liquid=Mass of LiquidMass of eq. vol. of water=18.820=0.94R.D. of liquid=Mass of eq. vol. of waterMass of Liquid​=2018.8​=0.94

Relative density of liquid = 0.94.

Question 10

From the following observations, calculate the density and relative density of a brine solution.

Mass of empty density bottle = 22 g

Mass of bottle + water = 50 g

Mass of bottle + brine solution = 54 g

Answer

Mass of empty density bottle (M₁) = 22 g

Mass of bottle + water (M₂) = 50 g

Mass of bottle + brine solution (M₃) = 54 g

Mass of water in bottle = M₂ - M₁ = 50 - 22 = 28 g

Density of water is 1 g cm^-3, so volume of density bottle = 28 cm³.

Mass of brine solution in bottle = M₃ - M₁ = 54 - 22 = 32 g

Density of brine soln.=Mass of brine soln. Vol. of brine soln.=3228=1.14 g cm−3Density of brine soln.= Vol. of brine soln.Mass of brine soln.​=2832​=1.14 g cm−3

So, density of brine solution = 1.14 g cm^-3.

R.D. of brine soln.=Mass of brine soln.Mass of eq. vol. of water=3228=1.14R.D. of brine soln.=Mass of eq. vol. of waterMass of brine soln.​=2832​=1.14

So, Relative density of brine solution is 1.14.

Question 11

The mass of an empty density bottle is 30 g, it is 75 g when filled completely with water and 65 g when filled completely with a liquid. Find:

(a) Volume of density bottle,

(b) Density of liquid and

(c) Relative density of liquid.

Answer

Mass of empty density bottle (M₁) = 30 g

Mass of bottle with water (M₂) = 75 g

Mass of bottle with liquid (M₃)= 65 g

(a) Mass of water in bottle = M₂ - M₁ = 75 - 30 = 45 g

Density of water is 1 g cm^-3, so volume of density bottle = 45 mL.

(b) Mass of liquid in bottle = M₃ - M₁ = 65 - 30 = 35 g

Density of liquid=Mass of liquid Vol. of liquid=3545=0.77 g cm−3Density of liquid= Vol. of liquidMass of liquid​=4535​=0.77 g cm−3

So, density of liquid = 0.77 g cm^-3.

(c) R.D. of liquid=Mass of liquidMass of eq. vol. of water=3545=0.77R.D. of liquid=Mass of eq. vol. of waterMass of liquid​=4535​=0.77

So, Relative density of liquid is 0.77.

==========================

Good Luck

 

 

 

Force and Pressure

Objective Questions

Question 1

Write true or false for each statement:

(a) The S.I. unit of force is kgf.

(b) A force always produces both the linear and turning motions.

(c) Moment of force = force x perpendicular distance of force from the pivoted point.

(d) Less force is needed when applied at a farther distance from the pivoted point.

(e) For a given thrust, pressure is more on a surface of large area.

(f) The pressure on a surface increases with an increase in the thrust on the surface.

(g) A man exerts same pressure on the ground whether he is standing or he is lying.

(h) It is easier to hammer a blunt nail into a piece of wood than a sharply pointed nail.

(i) The S.I. unit of pressure is pascal.

(j) Water in a lake exerts pressure only at its bottom.

(k) A liquid exerts pressure in all directions.

(l) Gases exert pressure in all directions.

(m) The atmospheric pressure is nearly 10⁵ Pa.

(n) Higher we go, greater is the air pressure.

Answer

(a) False
Correct Statement — The S.I. unit of force is newton.

(b) False
Correct Statement — A force can produce either linear motion or turning motion.

(c) True

(d) True

(e) False
Correct Statement — Pressure is defined as thrust per unit area. Hence, for a given thrust, pressure is less on a surface of large area.

(f) True

(g) False
Correct Statement — A man exerts more pressure while standing than lying on the ground as area of contact is less in case of standing as compared to lying on the ground.

(h) False
Correct Statement — Incase of blunt nail, the thrust exerted on hammering acts on a larger area so less pressure acts on the piece of wood making it difficult to hammer the nail.

(i) True

(j) False
Correct Statement — Water in a lake exerts pressure in all directions.

(k) True

(l) True

(m) True

(n) False
Correct Statement — The higher we go, lesser is the air pressure.

Question 2

Fill in the blanks:

(a) 1 kgf = ............... N (nearly).

(b) Moment of force = ............... x distance of force from the point of turning.

(c) In a door, handle is provided ............... from the hinges.

(d) The unit of thrust is ............... .

(e) Thrust is the ............... force acting on a surface.

(f) Pressure is the thrust acting on a surface of ............... area.

(g) The unit of pressure is ............... .

(h) Pressure is reduced if ............... increases.

(i) Pressure in a liquid ............... with the depth.

(j) The atmospheric pressure on earth surface is nearly ............... .

Answer

(a) 10

(b) Force

(c) Farthest

(d) newton

(e) normal

(f) unit

(g) pascal

(h) surface area

(i) increases

(j) 10⁵ Pa

Question 3

Match the following:

Column A	Column B
(a) Camel	(i) broad and deep foundation
(b) Truck	(ii) broad feet
(c) Knife	(iii) six or eight tyres
(d) High building	(iv) sharp cutting edge
(e) Thrust	(v) atm
(f) Moment of force	(vi) N
(g) Atmospheric pressure	(vii) N m

Answer

Column A	Column B
(a) Camel	(ii) broad feet
(b) Truck	(iii) six or eight tyres
(c) Knife	(iv) sharp cutting edge
(d) High building	(i) broad and deep foundation
(e) Thrust	(vi) N
(f) Moment of force	(vii) N m
(g) Atmospheric pressure	(v) atm

Question 4a

Select the correct alternative:

S.I. unit of moment of force is:

1. N
2. N cm
3. kgf m
4. N m

Answer

N m

Reason — S.I. Unit of moment of force = S.I. Unit of force x S.I. Unit of distance = N x m
So unit of moment of force = N m.

Question 4b

Select the correct alternative:

To obtain a given moment of force for turning a body, the force needed can be decreased by:

1. applying the force at the pivoted point.
2. applying the force very close to the pivoted point.
3. applying the force farthest from the pivoted point.
4. none of the above.

Answer

Applying the force farthest from the pivoted point.

Reason — Larger the perpendicular distance of the point at which the force is applied from the pivoted point, less force is needed to obtain a given moment of force for turning a body.

Question 4c

Select the correct alternative:

The unit of thrust is:

1. kgf
2. kg
3. g
4. m s^-1

Answer

kgf

Reason — The gravitational unit of thrust is kgf.

Question 4d

Select the correct alternative:

The unit of pressure is:

1. N x m
2. kgf
3. N m^-2
4. kgf m²

Answer

N m^-2

Reason —

Unit of Pressure = Unit of ForceUnit of AreaUnit of AreaUnit of Force​ = Nm2m2N​

So, unit of pressure is N m^-2

Question 4e

Select the correct alternative:

The pressure and thrust are related as:

1. Pressure = Thrust
2. Pressure = Thrust x Area
3. Pressure = ThrustAreaAreaThrust​
4. Pressure = AreaThrustThrustArea​

Answer

Pressure = ThrustAreaAreaThrust​

Reason — Pressure is defined as thrust acting per unit area.

Question 4f

Select the correct alternative:

A body weighing 5 kgf, placed on a surface of area 0.1 m², exerts a thrust on the surface equal to:

1. 50 kgf
2. 5 kgf
3. 50 kgf m^-2
4. 5 kgf m^-2

Answer

5 kgf

Reason — A body when placed on a surface exerts a thrust on the surface which is equal to its own weight. So thrust = weight of the body = 5 kgf.

Question 4g

Select the correct alternative:

The feet of lizards act like:

1. moving pads
2. drilling pads
3. suction pads
4. none of the above

Answer

suction pads

Reason — Lizards can stay and move on walls because their feet behave like suction pads so they remain pressed against the wall due to atmospheric pressure.

Question 4h

Select the correct alternative:

Pressure exerted by a liquid is due to its:

1. weight
2. mass
3. volume
4. area

Answer

Weight

Reason — Liquids have weight and pressure is weight acting per unit area.

Question 4i

Select the correct alternative:

Pressure inside a liquid increases with:

1. increase in depth.
2. decrease in depth.
3. decrease in density.
4. none of the above.

Answer

increase in depth

Reason — Pressure of liquid at a point increases with the height of the liquid column above it, so as depth increases pressure inside liquid increases.

Question 4j

Select the correct alternative:

The atmospheric pressure at sea level is nearly:

1. 10 Pa
2. 100,000 Pa
3. 100 Pa
4. 10,000 Pa

Answer

100,000 Pa

Reason — At sea level, the atmospheric pressure is 100,000 Pa which is 76 cm of mercury column.

Question 4k

Select the correct alternative:

Nose bleeding may occur at a high altitude because:

1. the atmospheric pressure decreases.
2. the oxygen content of atmosphere decreases.
3. the atmospheric pressure increases.
4. there are strong air currents at the high altitude.

Answer

the atmospheric pressure decreases

Reason — At higher altitudes, atmospheric pressure decreases but our body pressure remains the same. So, inside pressure becomes more than atmospheric pressure which causes nose bleeding.

Short/Long Answer Questions

Question 1

Define force. State its S.I. unit.

Answer

Force is that cause which changes the state of the body (either the state of rest or the state of motion) or changes the size or shape of the body. The S.I. unit of force is newton (N).

Question 2

State two effects of a force when applied on a body.

Answer

The two effects of a force when applied on a body are —

(i) It can change the shape or size of a body.

(ii) It can stop a moving body or it can move a stationary body.

Question 3

How does the effect of a force differ when it is applied on

(a) a rigid body

(b) a non-rigid body?

Answer

(a) When a force is applied on a rigid body, it produces only change in motion of the body. The force does not cause any change in the inter-spacing between its constituent particles.

(b) When a force is applied on a non-rigid body, it causes both change in its size or shape and motion in it.

Question 4

State the effect of force F in each of the following diagrams (a) and (b).

Answer

(i) In figure (a), when a force F is applied on the ball, the ball starts moving in a straight line in the direction of force F and continues to move in that direction. This is called linear motion.

(ii) In figure (b), when a force F is applied on the wheel pivoted at a point, the force turns the wheel in an anticlockwise direction about the axis of rotation producing turning motion.

Question 5

Define the term moment of force.

Answer

The moment of force is equal to the product of the magnitude of the force and the perpendicular distance of the force from the pivoted point.

Question 6

State the S.I. unit of moment of force.

Answer

The S.I. unit of moment of force is newton x metre (N m).

Question 7

State two factors which affect moment of force.

Answer

The two factors which affect moment of force are —

(i) Magnitude of force.

(ii) Perpendicular distance of force from the pivoted point.

Question 8

In figure below, a force F is applied in a direction passing through the pivoted point O of the body. Will the body rotate? Give reason to support your answer.

Answer

No, the body will not rotate because
Turning effect = Force x perpendicular distance of the force from the pivoted point.
= F x 0 = 0.
Perpendicular distance is 0 or force is parallel to the point of application of force, hence the body will not rotate.

Question 9

Write the expression for the moment of force about a given axis of rotation.

Answer

The expression for the moment of force about a given axis of rotation is —

Moment of force about a given axis = Force x perpendicular distance of force from the axis of rotation.

Question 10

State one way to decrease the moment of a given force about a given axis of rotation.

Answer

The moment of given force about a given axis of rotation is decreased by either decreasing the force or by decreasing the perpendicular distance of force from the axis of rotation.

Question 11

State one way to obtain greater moment of a given force about a given axis of rotation.

Answer

One way to obtain greater moment of a given force is to increase the distance from the axis of rotation where the force would act.

Question 12

What do you mean by the clockwise and anti-clockwise moment of force?

Answer

If the effect on the body is to turn it clockwise, the moment of force is called a clockwise moment. If the effect on the body is to turn it anti-clockwise, the moment of force is called an anti-clockwise moment.

Question 13(a)

Explain the following:

The spanner (or wrench) has a long handle.

Answer

A spanner has a long handle to produce a large turning effect by applying a small amount of force at the end of the handle as shown in the figure below:

Question 13(b)

Explain the following:

The steering wheel of a vehicle is of large diameter.

Answer

It is easier to turn the steering wheel of large diameter as larger the diameter, larger is the perpendicular distance from the axis of rotation so less force is applied to generate same amount of turning effect as compared to steering wheel of small diameter.

Question 13(c)

Explain the following:

The hand flour grinder is provided with a handle near the rim.

Answer

The hand flour grinder is provided with a handle near the rim which is at a maximum distance from the axis of rotation so that it can easily be rotated at its centre by applying a small force at the handle.

Question 13(d)

Explain the following:

It is easier to open the door by pushing it at its free end.

Answer

It is easier to open the door by pushing it at its free end because larger the perpendicular distance from the axis of rotation, less is the force required to open it.

Question 13(e)

Explain the following:

A potter turns his wheel by applying a force through the stick near the rim of wheel.

Answer

A potter's wheel is pivoted at the centre. When the potter applies a force through the stick near the rim of the wheel, the point of application of force is at a maximum distance from the axis of rotation, hence less force is required to rotate the wheel.

Question 14

What is thrust?

Answer

If a force is applied on a surface in a direction perpendicular to the surface, the force is called thrust. In other words, thrust is the force acting normally on the surface.

Question 15

State the unit of thrust.

Answer

The units of thrust are kilogram force(kgf), gram force(gf) and newton(N) (same as that of weight or force).

Question 16

On what factors does the effect of thrust on a surface depend?

Answer

The effect of thrust depends on the area of the surface on which it acts. Smaller the area of the surface on which a thrust acts larger is its effect and effect of thrust is less on a larger surface area.

Question 17

Define the term pressure and state its unit.

Answer

Pressure is defined as the thrust acting per unit area.

Pressure = ThrustAreaAreaThrust​

Its unit is pascal (Pa) or newtonmetre2metre2newton​ (N m^-2).

Question 18

How is thrust related to pressure?

Answer

Pressure is thrust acting per unit area. Thrust is directly proportional to pressure. Greater the thrust, greater is the pressure and smaller the thrust, smaller is the pressure.
Pressure = ThrustAreaAreaThrust​

Question 19

Name two factors on which the pressure on a surface depends.

Answer

The two factors on which the pressure on a surface depends are:

1. Area of the surface on which thrust acts.
2. Magnitude of thrust acting on the surface.

Question 20

When does a man exert more pressure on the floor: while standing or while walking?

Answer

A man exerts more pressure on the surface while walking than standing because while walking one foot is in contact with the floor so surface area is less but while standing both foot are in contact with the floor providing a large surface area. Since pressure is inversely proportional to surface area so pressure exerted while walking is more as compared to pressure exerted on standing.

Question 21

Why do camels or elephants have broad feet?

Answer

The broad feet of camels and elephants increases the surface of the feet in contact with the ground. This reduces the pressure exerted on the ground as pressure is inversely proportional to surface area.
Thus, this helps camels to walk easily on the desert sand without their feet sinking in it. The large body weight of elephants is supported by their broad feet helping them to move around comfortably.

Question 22

A sharp pin works better than a blunt pin. Explain the reason.

Answer

A sharp pin has less surface area as compared to a blunt pin, so for a given applied force sharp pin exerts more pressure than blunt pin and penetrates a surface like wood or wall easily. So, it works better than a blunt pin.

Question 23

Why is the bottom part of the foundation of a building made wider?

Answer

The bottom part of the foundation of a building is made wider so that the weight of the building will act on larger surface area exerting less pressure on the ground (as pressure is inversely proportional to surface area). This avoids sinking of buildings into the earth.

Question 24

It is easier to cut with a sharp knife than with a blunt one. Explain.

Answer

A sharp knife has sharp edges that provides less surface area of contact as compared to a blunt knife which has more surface area of contact. So pressure applied by the force is more in sharp knife which makes it easier to cut than a blunt one.

Question 25

A gum bottle rests on its base. If it is placed upside down,

(i) How does the thrust change?

(ii) How does the pressure change?

Answer

(i) A gum bottle has broader base and narrower neck and when it is placed upside down there will be no change in thrust as thrust is the weight of the body that is exerted on the surface.

(ii) When gum bottle is placed upside down, surface area is less so pressure will be more as pressure is inversely proportional to surface area.

Question 26

Explain the following:

(i) Sleepers are used below the rails.

(ii) A tall building has wide foundations.

Answer

(i) Wide wooden sleepers are placed below the railway tracks to increase the surface area of the tracks in contact with the ground. This reduces the pressure exerted by the rails on the ground.

(ii) The foundation of a tall building is made wider because the weight of the building will act on larger surface area exerting less pressure on the ground (as pressure is inversely proportional to surface area). This avoids sinking of building into the earth.

Question 27

Describe an experiment to show that a liquid exerts pressure at the bottom of the container in which it is kept.

Answer

The following experiment demonstrates that a liquid exerts pressure at the bottom of the container in which it is kept:

1. Take a glass tube and tie a balloon at its lower end.
2. Hold it vertically straight as shown in the above figure.
3. Pour some water in the tube.

Observations — After sometime we will notice that the balloon bulges out as shown in figure (b) above.

Conclusion — Balloon bulges out because water column exerts pressure at its bottom. The force on the balloon is equal to the weight of the water column known as thrust.
So, Pressure = ThrustAreaAreaThrust​ = WAAW​, where W is weight of the water column (thrust) and A is surface area.

Question 28

Describe a suitable experiment to demonstrate that a liquid exerts pressure sideways also.

Answer

The following experiment demonstrates that a liquid exerts pressure sideways also:

1. Take a glass tube closed at one end and having an opening in its side near the bottom.
2. Tie a deflated balloon at the side opening of the tube.
3. Hold the tube vertically straight as shown in the above figure.
4. Pour some water in the tube.

Observations — After sometime we will notice that the balloon bulges out as shown in figure (b) above.

Conclusion — From the above experiment it is concluded that a liquid exerts pressure sideways also.

Question 29

Describe a simple experiment to show that at a given depth, a liquid exerts same pressure in all directions.

Answer

The following experiment demonstrates that at a given depth, a liquid exerts same pressure in all directions:

1. Take a balloon and fill it with water.
2. Tie the mouth of the balloon.
3. Make holes in the balloon by inserting pins at several places in all directions.

Water comes out through each hole which shows that liquid exerts same pressure in all directions.

Question 30

State two factors on which the pressure at a point in a liquid depends.

Answer

The two factors on which the pressure at a point in a liquid depends are:

1. The height of the liquid column — Liquid pressure increases with the height of the liquid column above the point.
2. The density of the liquid — Liquid pressure increases with the increase in the density of the liquid.

Question 31

Describe an experiment to show that the liquid pressure at a point increases with the increase in height of the liquid column above that point.

Answer

The experiment is described below:

1. Take a glass tube open at both ends. Hold it vertically.
2. Tie a balloon at its lower end.
3. Pour some water in the tube.
4. Balloon starts bulging out as shown in figure (a) above.
5. Add more water in the tube.
6. Balloon will bulge more as shown in figure (b) above.

Conclusion — From the above experiment it is concluded that when the height of the water column increases, more pressure is exerted on the balloon and it bulges out more. This shows that the liquid pressure at a point increases with the increase in height of the liquid column above that point.

Question 32

Which fact about liquid pressure does the diagram in figure below illustrate?

Answer

The diagram demonstrates that liquid pressure at a point increases with the height of the liquid column above it as water flowing out from upper hole falls near the cylinder while water from lower hole falls far from the cylinder.

Question 33

Describe an experiment to show that liquid pressure depends on the density of liquid.

Answer

The following experiment demonstrates that liquid pressure depends on the density of liquid:

1. Take two identical glass tubes open at both the ends. Mark them A and B.
2. Hold both the tubes vertically and tie an inflated balloon to both the tubes at their lower ends.
3. In tube A pour some water.
4. In tube B pour some concentrated sugar solution such that its height is same as that of height of water in tube A.

Observation — It is noticed that the balloon attached to tube B bulges out more as compared to balloon attached to tube A as shown in the above figure.

Conclusion — Density of sugar solution is more than water so it exerts more pressure which shows that liquid pressure depends on the density of liquid.

Question 34

A dam has broader walls at the bottom than at the top. Give a reason.

Answer

The pressure at a point due to a liquid increases with the increase in height of the liquid column above it, so to withstand the increasing pressure of water a dam has broader walls at the bottom than at the top.

In the figure given below, the increasing length of arrows in water represents the increasing pressure on the wall of the dam towards the bottom.

Question 35

What do you mean by atmospheric pressure?

Answer

The thrust on unit area of the earth surface due to the column of air is called the atmospheric pressure.

Question 36

Write the numerical value of the atmospheric pressure on the earth surface in pascal.

Answer

The atmospheric pressure on the earth surface in pascal is 1.013 x 10⁵ Pa.

Question 37

We do not feel uneasy even under the enormous atmospheric pressure. Give a reason.

Answer

The blood in the veins of our body exerts a pressure which is slightly more than the atmospheric pressure which makes the effect of atmospheric pressure ineffective. So we do not feel uneasy even under the enormous atmospheric pressure.

Question 38

Describe a simple experiment to illustrate that air exerts pressure.

Answer

The following experiment demonstrates that air exerts pressure:

1. Take a glass filled with water up to its brim and place a post card on top of it.
2. Press the palm of your one hand on top of post card and then invert the filled glass upside down.
3. Now gently remove your hand from the post card to release it.

Conclusion — It is observed that post card does not fall from the glass because the atmospheric pressure is acting upwards on the post card from outside the glass which overcomes the pressure on post card due to water in the glass.

Question 39

Describe the crushing tin can experiment. What do you conclude from this experiment?

Answer

Crushing tin can experiment

1. Take a thin walled tin can with airtight stopper. Remove the stopper.
2. Fill the can partially with water. Heat the can over the flame of a burner till water boils.
3. Now the air pressure inside and outside the can is same.
4. When steam comes out of the opening put the stopper.
5. Remove the can from the burner.
6. Place the can in the tub and pour cold water on the can.

It is observed that the can collapses. The reason is that when steam comes out it takes away most of the air from the can. When cold water is poured on the can, steam condenses into water leaving a partial vacuum in the can. The air pressure outside is more than inside which exerts force on the can causing it to collapse.

From this experiment it is concluded that air exerts pressure.

Question 40(a)

Give reasons for the following:

A balloon collapses when air is removed from it.

Answer

When air is removed from the balloon, the pressure inside it becomes much less than the outside atmospheric pressure and hence the balloon collapses.

Question 40(b)

Give reasons for the following:

Water does not run out of a dropper unless its rubber bulb is pressed.

Answer

There are two forces acting on the water inside the dropper one is liquid pressure from inside and atmospheric pressure from outside. Atmospheric pressure acting from outside balances the liquid pressure from inside so water does not come out from the dropper. When bulb is pressed the liquid pressure increases than atmospheric pressure and water comes out from the dropper.

Question 40(c)

Give reasons for the following:

Two holes are made in a sealed oil tin to take out oil from it.

Answer

Two holes are made in a sealed oil tin to take out oil from it because through one hole atmospheric pressure acts due to air entering through it and through another hole the oil comes out easily.

Question 41

How does the atmospheric pressure change with altitude?

Answer

The atmospheric pressure decreases with increase in altitude. So, as we go higher the atmospheric pressure decreases.

Numericals

Question 1

Find the moment of force of 20 N about an axis of rotation at a distance of 0.5 m from the force.

Answer

Given:
Force f = 20 N
Distance d = 0.5 m
Moment of force = ?

Moment of force = force f x distance d
= 20 x 0.5
= 10 N m

So, the moment of force = 10 N m.

Question 2

The moment of a force of 25 N about a point is 2.5 N m. Find the perpendicular distance of force from that point.

Answer

Given:
Force f = 25 N
Moment of force = 2.5 N m
Perpendicular distance d = ?

Moment of force=force f×distance dDistance d=Moment of forceForce f=2.525=0.1 mMoment of force=force f×distance dDistance d=Force fMoment of force​=252.5​=0.1 m

So, perpendicular distance of force from the point = 0.1 m or 10 cm.

 

Question 3

A spanner of length 10 cm is used to unscrew a nut by applying a minimum force of 5.0 N. Calculate the moment of force required.

Answer

Given:
Force f = 5.0 N
Distance d = 10 cm = 0.1 m
Moment of force = ?

Moment of force = force f x distance d
= 5.0 x 0.1
= 0.5 N m

So, Moment of force = 0.5 N m.

Question 4

A wheel of diameter 2 m can be rotated about an axis passing through its centre by a moment of force equal to 2.0 N m. What minimum force must be applied on its rim?

Answer

Given:
Moment of force = 2.0 N m
Diameter = 2 m so radius = 1 m
Therefore perpendicular distance d = radius = 1 m
Force f = ?

Moment of force=force f×distance dForce f=Moment of forcedistance d=2.01=2 NMoment of force=force f×distance dForce f=distance dMoment of force​=12.0​=2 N

So, the minimum force required is 2 N.

Question 5

A normal force of 200 N acts on an area 0.02 m². Find the pressure in pascal.

Answer

Given:
Force (Thrust) f = 200 N
Area A = 0.02 m²
Pressure = ?

Pressure=ThrustArea=2000.02=10000 PaPressure=AreaThrust​=0.02200​=10000 Pa

So Pressure = 10000 Pa.

Question 6

Find the thrust required to exert a pressure of 50,000 Pa on an area of 0.05 m².

Answer

Given:
Pressure = 50000 Pa
Area = 0.05 m²
Thrust = ?

Pressure = ThrustAreaAreaThrust​

Thrust = Pressure x Area = 50000 x 0.05 = 2500 N

So, Thrust = 2500 N.

Question 7

Find the area of a body which experiences a pressure of 50,000 Pa by a thrust of 100 N.

Answer

Given:
Pressure = 50,000 Pa
Thrust = 100 N
Area = ?

Pressure=ThrustAreaArea=ThrustPressure=10050000=0.002 m2Pressure=AreaThrust​Area=PressureThrust​=50000100​=0.002 m2

So Area of body = 0.002 m² or 2 x 10^-3 m².

Question 8

Calculate the pressure in pascal exerted by a force of 300 N acting normally on an area of 30 cm².

Answer

Given:
Force (Thrust) = 300 N
Area = 30 cm² = 30100001000030​ = 0.003 m²
Pressure = ?

Pressure=ThrustArea=3000.003=1,00,000 PaPressure=AreaThrust​=0.003300​=1,00,000 Pa

So Pressure = 1,00,000 Pa or 10⁵ Pa.

Question 9

How much thrust will be required to exert a pressure of 20,000 Pa on an area of 1 cm²?

Answer

Given:
Pressure = 20000 Pa
Area = 1 cm² = 110000100001​ = 0.0001 m² or 10^-4 m²
Thrust = ?

Pressure = ThrustAreaAreaThrust​
Thrust = Pressure x Area
= 20000 x 10^-4
= 2 N

So Thrust = 2 N.

Question 10

The base of a container measures 15 cm x 20 cm. It is placed on a table top. If the weight of the container is 60 N, what is the pressure exerted by the container on the table top?

Answer

Given:
Weight = Thrust = 60 N
Area = 15 cm x 20 cm = 300 cm² = 3001000010000300​ = 0.03 m²
Pressure = ?

Pressure=ThrustArea=600.03=2000 PaPressure=AreaThrust​=0.0360​=2000 Pa

So, Pressure = 2000 Pa.

Question 11

Calculate the pressure exerted on a surface of 0.5 m² by a thrust of 100 kgf.

Answer

Given:
Thrust = 100 kgf
Area = 0.5 m²
Pressure = ?

Pressure=ThrustArea=1000.5=200 kgf m−2Pressure=AreaThrust​=0.5100​=200 kgf m−2

So Pressure = 200 kgf m^-2.

Question 12

A boy weighing 60 kgf stands on a platform of dimensions 2.5 cm x 0.5 cm. What pressure in Pascal does he exert?

Answer

Given:
Thrust = Weight = 60 kgf
1 kgf = 10 N
∴ 60 kgf = 60 x 10 = 600 N
So, Thrust = 600 N

Area = 2.5 cm x 0.5 cm = 1.25 cm² = 1.2510000100001.25​ = 0.000125 m²
Pressure =?

Pressure=ThrustArea=6000.000125=48,00,000 PaPressure=AreaThrust​=0.000125600​=48,00,000 Pa

So, Pressure = 48,00,000 or 4.8 x 10⁶ Pa.

Question 13

Figure below shows a brick of weight 2 kgf and dimensions 20 cm x 10 cm x 5 cm placed in three different positions on the ground. Find the pressure exerted by the brick in each case.

Answer

First case:

Thrust = Weight = 2 kgf
Area = 20 cm x 10 cm = 200 cm²
Pressure = ?

Pressure=ThrustArea=2200=0.01 kgf cm−2Pressure=AreaThrust​=2002​=0.01 kgf cm−2

So Pressure exerted by the brick in the first figure is 0.01 kgf cm^-2.

Second case:

Thrust = Weight = 2 kgf
Area = 5 cm x 10 cm = 50 cm²
Pressure = ?

Pressure=ThrustArea=250=0.04 kgf cm−2Pressure=AreaThrust​=502​=0.04 kgf cm−2

So Pressure exerted by brick in second figure is 0.04 kgf cm^-2.

Third case:

Thrust = Weight = 2 kgf
Area = 20 cm x 5 cm = 100 cm²
Pressure = ?

Pressure=ThrustArea=2100=0.02 kgf cm−2Pressure=AreaThrust​=1002​=0.02 kgf cm−2

So, Pressure exerted by brick in third figure is 0.02 kgf cm^-2.

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Energy (Chapter-3)

Objective Questions

Question 1

Write true or false for each statement:

(a) A coolie does no work against the force of gravity while carrying a luggage on a plane road.

(b) The energy stored in water of a dam is kinetic energy.

(c) The energy of a flying kite is kinetic energy.

(d) Work done by a boy depends on the time in which he does work.

(e) Power spent by a body depends on the time for which it does work.

Answer

(a) True

(b) False
Correct Statement — The energy stored in water of a dam is potential energy.

(c) True

(d) False
Correct Statement — Work done by a boy depends on the magnitude of force applied and distance moved by the body in the direction of force.

(e) True

Question 2

Fill in the blanks:

(a) Work is said to be done by a force only when ............... .

(b) Work done = Force x ............... .

(c) The energy of a body is its capacity to do ............... .

(d) The S.I. unit of energy is ............... .

(e) The potential energy of a body is due to its ............... and kinetic energy of a body is due to its ............... .

(f) Gravitational potential energy U = mass x force of gravity on unit mass x ............... .

(g) Kinetic energy = 1221​ x mass x ............... .

(h) Power P = ............... / time taken.

(i) The S.I. unit of power is ............... .

(j) 1 H.P. = ............... W.

Answer

(a) the body moves

(b) distance moved in the direction of force

(c) work

(d) joule

(e) state of rest or position, state of motion

(f) vertical height

(g) (speed)²

(h) Work done by the body

(i) watt (W)

(j) 746

Question 3

Match the following:

Column A	Column B
(a) A stone at a height	(i) power
(b) A moving ball	(ii) joule
(c) Energy	(iii) work done in 1 sec
(d) Power	(iv) potential energy
(e) watt	(v) kinetic energy

Answer

Column A	Column B
(a) A stone at a height	(iv) Potential energy
(b) A moving ball	(v) Kinetic energy
(c) Energy	(ii) joule
(d) Power	(iii) work done in 1 sec
(e) watt	(i) power

Question 4a

Select the correct alternative:

The S.I. unit of work is:

1. second
2. metre
3. joule
4. newton

Answer

joule

Reason — S.I. unit of work = S.I. unit of force x S.I. unit of distance = newton(N) x metre(m) or joule(J).

Question 4b

Select the correct alternative:

No work is done by a force if the body:

1. moves in the direction of force
2. does not move
3. moves in opposite direction
4. none of these

Answer

does not move

Reason — Work is said to be done if the force applied on the body makes the body move but if there is no motion in the body then work done is zero.

Question 4c

Select the correct alternative:

Two coolies A and B do some work in time 1 minute and 2 minute respectively. The power spent is:

1. same by both coolies
2. is more by coolie A than by B
3. is less by coolie A than by B
4. nothing can be said

Answer

is more by coolie A than by B

Reason — Power spent = Work donetime takentime takenWork done​.
Since coolie A takes less time than coolie B so power spent by coolie A is more than coolie B.

Question 4d

Select the correct alternative:

The expression of power P is:

1. P = mgh
2. P = 1221​ mv²
3. P = F x d
4. P = F x dttd​

Answer

P = F x dttd​

Reason — Power = Work donetime takentime takenWork done​ and Work done = F x d
So P = F x dttd​.

Question 4e

Select the correct alternative:

1 H.P. is equal to:

1. 1 W
2. 1 J
3. 764 J
4. 746 W

Answer

746 W

Reason — 1 H.P. = 746 W.

Question 4f

Select the correct alternative:

When a boy doubles his speed, his kinetic energy becomes:

1. half
2. double
3. four times
4. no change

Answer

four times

Reason — Kinetic energy = 1221​ mv²
If speed gets doubled then kinetic energy becomes four times.

Question 4g

Select the correct alternative:

A boy lifts a luggage from height 2 m to 4 m. The potential energy will become:

1. half
2. double
3. one-third
4. one-fourth

Answer

double

Reason — Potential energy (P.E.) is given by expression mgh.
In first case P.E. = mg2 or 2mg [h=2 m]
In second case P.E. = mg4 or 4mg [h=4 m]
So potential energy gets doubled.

Short/Long Answer Questions

Question 1

Define work.

Answer

The work done by a force on a body is equal to the product of the force applied and the distance moved by the body in the direction of force i.e.,

Work done = Force x distance moved in the direction of force

Question 2

When does a force perform work?

Answer

A force performs work when it changes the position of the body or it changes the size or shape of the body.

Question 3

State two conditions when no work is done by a force.

Answer

The two conditions where no work is done by a force are:

1. If the force applied on a body does not move the body i.e. displacement is zero.
2. If the displacement produced by applied force is normal to the direction of force.

Question 4

In which of the following cases is work being done:

(a) A boy pushing a heavy rock

(b) A boy climbing up the stairs

(c) A coolie standing with a box on his head

(d) A girl moving on the road.

Answer

Work is done in the following two cases:

(b) A boy climbing up the stairs: Work is done by the boy as the boy changes his position.

(d) A girl moving on the road: Work is done as displacement is produced by the body.

Question 5

A coolie is moving on a road with a luggage on his head. Does he perform work against the force of gravity? Give reason for your answer.

Answer

A coolie with a luggage on his head and moving on a road does no work against force of gravity as the direction of motion of the coolie is perpendicular to the direction of force of gravity.

Question 6

The moon is revolving around the earth in a circular path. How much work is done by the moon?

Answer

Work done by the moon is zero as the force of attraction on moon by earth is normal to the direction of motion of moon.

Question 7

Write the expression for work done by a force.

Answer

Work done = Force x distance moved in the direction of force
W = F x d

Question 8

State the S.I. unit of work and define it.

Answer

The S.I. unit of work is joule (J). One joule of work is said to be done if one newton force when acting on a body moves it by 1 metre in the direction of force.

Question 9

State two factors on which the work done on a body depends.

Answer

The two factors on which the work done on a body depends are:

1. The magnitude of the force applied.
2. The distance moved by the body in the direction of force.

Question 10

Define the term energy.

Answer

The energy of a body is its capacity to do work. The energy of a body in a state is equal to the work done on the body to bring it to that state.

Question 11

State the S.I. unit of energy.

Answer

The S.I. unit of energy is joule(J).

Question 12

Define 1 joule of energy.

Answer

A body is said to possess an energy of 1 joule if it can do one joule work or if one joule work is done on it.

Question 13

How is work related to energy?

Answer

There is a direct relationship between work and energy. To do more amount of work we need to spend more energy. Similarly, the work done on a body in changing its state is said to be the energy possessed by the body.

Question 14

What are the two kinds of mechanical energy?

Answer

The two kinds of mechanical energy are:

1. Potential energy
2. Kinetic energy

Question 15

What is potential energy? State its unit.

Answer

Potential energy of a body is the energy possessed by it due to its state of rest or position. Its unit is joule (J).

Question 16

Give one example of a body that has potential energy, in each of the following:

(a) due to its position at a height,

(b) due to its elongated stretched state.

Answer

(a) A stone placed at a height has potential energy because of its position. It is known as gravitational potential energy.

(b) A stretched rubber band has potential energy due to its elongated stretched state. It is known as elastic potential energy.

Question 17

State two factors on which the potential energy of a body at a certain height above the ground depends.

Answer

The potential energy of a body at a certain height above the ground depends on the following factors:

1. The mass of the body — Greater the mass of the body, greater is the potential energy of the body.
2. Its height above the ground — More the height of the body above the ground, more is the potential energy.

Question 18

Two bodies A and B of masses 10 kg and 20 kg respectively are at the same height above the ground. Which of the two has greater potential energy?

Answer

Body B has greater potential energy.

Reason — Both bodies, A and B are at the same height above the ground. Value of g is also same for both. As the mass of body B (20 kg) is greater than that of body A (10 kg) hence potential energy of body B is greater.

Question 19

A bucket full of water is on the first floor of your house and another identical bucket with same quantity of water is kept on the second floor. Which of the two has greater potential energy?

Answer

The bucket kept on second floor has greater potential energy.

Reason — The two buckets are identical containing the same amount of water so their mass is same. Value of g is also same for the two buckets. As the bucket on second floor is at a greater height hence its potential energy is also greater.

Question 20

Write the expression for the gravitational potential energy explaining the meaning of the symbols used?

Answer

Gravitational potential energy (U) = mgh

where,
m is mass of the body,
g is acceleration due to gravity and
h is height of the body above ground level.

Question 21

A body of mass m is moved from ground to a height h. If force of gravity on mass of 1 kg is g newton, find:

(a) the force needed to lift the body,

(b) the work done in lifting the body and

(c) the potential energy stored in the body.

Answer

(a) When a body of mass m is raised to height h above ground, a force is applied.

Force needed to lift the body (F) = weight of the body

If g is the force of gravity on mass of 1 kg, then the force of gravity on mass m kg will be mg N.

F = mg N

(b) Force = mg N; distance moved = h
Work done = force x distance moved = mg x h = mgh joule

(c) The work done against the force of gravity in lifting the body to a height h is stored in the body in form of its gravitational potential energy.

Potential energy (U) = mgh joule

Question 22

Define the term kinetic energy. Give one example of a body which possess kinetic energy.

Answer

Kinetic energy of a body is the energy possessed by it due to its state of motion. Example: A fast moving stone has kinetic energy which has the capacity of breaking a window pane.

Question 23

State two factors on which the kinetic energy of a moving body depends.

Answer

The two factors on which the kinetic energy of a moving body depends are:

1. The mass of the body — Greater the mass of the body, higher is its kinetic energy.
2. The speed of the body — More the speed of the body, higher is its kinetic energy.

Question 24

Two toy cars A and B of masses 200 g and 500 g respectively are moving with the same speed. Which of the two has greater kinetic energy?

Answer

The toy car B has greater kinetic energy than toy car A.

Reason — The two toy cars A and B are moving with the same speed. As the toy car B has greater mass (500 g) than toy car A (200 g) hence its kinetic energy is also greater.

Question 25

A cyclist doubles his speed. How will his kinetic energy change: increase, decrease or remain the same?

Answer

Kinetic energy of cyclist will increase as speed is doubled because more the speed of the body, more is its kinetic energy.

Question 26

Write the expression for the kinetic energy of a body explaining the meaning of the symbols used.

Answer

The expression for the kinetic energy of a body (K.E.) = 1221​ mv²
Where, m is mass of the body and
v is speed of the body.

Question 27

A ball of mass m is moving with a speed v. What is its kinetic energy?

Answer

Kinetic energy of ball (K.E.) = 1221​ mv²
Where, m is mass of the ball and
v is speed of the ball.

Question 28

Name the form of energy stored in a wound up spring of a watch.

Answer

Potential Energy.

Question 29

Can a body possess energy even when it is not in motion? Explain your answer with an example.

Answer

Yes, a body can possess energy even when it is not in motion.

Example — A stone at rest placed at a height above the ground has potential energy. Water stored in a dam has potential energy.

Question 30

Name the type of energy (kinetic or potential) possessed by the following:

(a) A moving cricket ball.

(b) A stone at rest on the top of a building.

(c) A compressed spring.

(d) A moving bus.

(e) A bullet fired from a gun.

(f) Water flowing in a river.

(g) A stretched rubber band.

Answer

(a) A moving cricket ball has kinetic energy due to its state of motion.

(b) A stone at rest on the top of a building has potential energy due to its raised position.

(c) A compressed spring has potential energy due to its compressed state.

(d) A moving bus has kinetic energy due to its state of motion.

(e) A bullet fired from a gun has kinetic energy due to its state of motion.

(f) Water flowing in a river has kinetic energy due to its state of motion.

(g) A stretched rubber band has potential energy because of its stretched position.

Question 31

Give an example to show the conversion of potential energy to kinetic energy when put in use.

Answer

Consider a ball placed at a height. It will have only potential energy and no kinetic energy.

If the ball is released from the height, it falls down and the vertical height of the ball from the ground decreases. Therefore, the potential energy decreases and it changes to kinetic energy due to which the speed of the ball increases. During the fall, the ball has both the potential energy and the kinetic energy. As the ball reaches the ground, the potential energy becomes zero and it changes entirely into kinetic energy.

The below figure shows the conversion of potential energy into kinetic energy during the vertical free fall of a ball at various positions A, B and C.

Question 32

State the energy changes that occur in a watch spring while it unwinds.

Answer

In a wound up watch spring the energy stored is potential energy. When the watch spring unwinds itself, the potential energy changes into kinetic energy and this kinetic energy is used to move the arms of the watch.

Question 33(a)

Give reasons for the following:

No work is done if a man is pushing against a wall.

Answer

Work is said to be done only when there is change in position or size and shape of the body. When a man pushes the wall it does not move so no work is done.

Question 33(b)

Give reasons for the following:

Hammer drives a nail into the wood only when it is lifted up and then struck.

Answer

When a hammer is lifted it has potential energy due to its raised position and then when it is struck it drives the nail into the wood due to its potential energy.

Question 33(c)

Give reasons for the following:

A horse and a dog are running with the same speed. Which one of them has more kinetic energy than the other?

Answer

A horse has more kinetic energy than a dog. Kinetic energy of a body depends on mass and speed of the body. Since both dog and horse have same speed but mass of horse is more than that of dog so horse has more kinetic energy than a dog.

Question 33(d)

Give reasons for the following:

A teacher moving around in the class is doing work but a child standing and reading a book is not doing any work.

Answer

As the teacher moves around in the class, he/she is in a state of motion, there is change in his/her position due to application of force. Hence, work is done by the teacher.
On the other hand, a child standing and reading a book is stationary. There is no change in his/her position. Hence, no work is done by the child.

Question 34

State the energy changes in the following while in use:

(a) An electric bulb

(b) An electric oven

(c) A loudspeaker

(d) A microphone

(e) An electric motor

Answer

(a) When an electric bulb glows, the electrical energy changes into heat and light energy.

(b) In electric oven, electrical energy changes into heat energy.

(c) In loudspeaker, electrical energy is converted into sound energy.

(d) A microphone converts sound energy into electrical energy.

(e) An electric motor converts electrical energy into mechanical energy.

Numericals

Question 1

A force of 30 N acts on a body and moves it through a distance of 5 m in the direction of force. Calculate the work done by the force.

Answer

Given:
Force (F) = 30 N
Distance (d) = 5 m
Work done (W) = ?

Work done = Force x distance = 30 N x 5 m = 150 J

So, the work done by the force = 150 J.

Question 2

A man lifts a mass of 20 kg to a height of 2.5 m. Assuming that the force of gravity on 1 kg mass is 10 N, find the work done by the man.

Answer

Given:
Mass (m) = 20 kg
Distance (d) = Height = 2.5 m
Force of gravity on mass of 1 kg = 10 N
Work done (W) = ?

Force (F) = mg = 20 x 10 = 200 N

Work done = Force x distance = 200 x 2.5 = 500 J

So, the work done by the man = 500 J.

Question 3

A body when acted upon by a force of 10 kgf moves to a distance 0.5 m in the direction of force. Find the work done by the force. Take 1 kgf = 10 N.

Answer

Given:
1 kgf = 10 N
10 kgf = 10 x 10 = 100 N
So, Force (F) = 100 N
Distance (d) = 0.5 m
Work done (W) = ?

Work done = Force x distance = 100 x 0.5 = 50 J

So, the work done by the force = 50 J.

Question 4

Two bodies of same masses are placed at height h and 2h. Compare their gravitational potential energy.

Answer

Given:
Mass of first body = Mass of second body = m
Height of first body = h
Height of second body = 2h
g is same for both the bodies.

We know Gravitational potential energy = mgh

Gravitation potential energy of first body (U₁) = mgh
Gravitation potential energy of second body (U₂)= mg2h

On comparing both gravitational potential energy:

U1U2=mghmg2h=12U2​U1​​=mg2hmgh​=21​

So, gravitational potential energy of first body : gravitational potential energy of second body = 1:2.

Question 5

Find the gravitational potential energy of 2.5 kg mass kept at a height of 15 m above the ground. The force of gravity on mass 1 kg is 10 N.

Answer

Given:
Mass (m) = 2.5 kg
Height (h) = 15 m
Force of gravity on mass 1 kg = 10 N

Gravitational potential energy (U) = mgh = 2.5 x 10 x 15 = 375 J

So, gravitational potential energy = 375 J.

Question 6

The gravitational potential energy stored in a box of weight 150 kgf is 1.5 x 10⁴ J. Find the height of the box. Take 1 kgf = 10 N.

Answer

Given:
Gravitational potential energy (U) = 1.5 x 10⁴ J = 15000 J
Weight = 150 kgf = 150 x 10 = 1500 N
height h = ?

U = mgh

15000 = 1500 x h

h = 150001500150015000​

h = 10 m

So the height of the box = 10 m.

Question 7

The potential energy of a body of mass 0.5 kg increases by 100 J when it is taken to the top of a tower from the ground. If the force of gravity on 1 kg = 10 N, what is the height of the tower?

Answer

Given:
Potential energy (U) = 100 J
Mass (m) = 0.5 kg
Force of gravity on 1 kg mass = 10 N
height (h) = ?

U = mgh

100 = 0.5 x 10 x h

h = 10055100​

h = 20 m

So height of the tower = 20 m.

Question 8

A body of mass 60 kg is moving with a speed 50 m s^-1. Find its kinetic energy.

Answer

Given:
Mass (m) = 60 kg
Speed (v) = 50 m s^-1
Kinetic energy = ?

Kinetic energy = 1221​ x mv²

= 1221​ x 60 x (50)²

= 30 x 2500

= 75000 J or 7.5 x 10⁴ J

So kinetic energy = 7.5 x 10⁴ J.

Question 9

A truck of mass 1000 kg increases its speed from 36 km h^-1 to 72 km h^-1. Find the increase in its kinetic energy.

Answer

Given:

Mass (m) = 1000 kg

1 km h^-1 = 518185​ m s^-1

36 km h^-1 = 518185​ x 36 = 10 m s^-1

So, initial speed (v₁) = 10 m s^-1

1 km h^-1 = 518185​ m s^-1

72 km h^-1 = 518185​ x 72 = 20 m s^-1

So, final speed (v₂) = 72 km h^-1 = 20 m s^-1

Increase in its kinetic energy = ?

Increase in kinetic energy = 1221​ m[(v₂)² - (v₁)²]

= 1221​ x 1000 x [(20)² - (10)²]

= 500 x [400 - 100]

= 500 x 300

= 150000 J or 1.5 x 10⁵ J

So increase in kinetic energy = 1.5 x 10⁵ J.

Question 10

A car is moving with a speed of 15 km h^-1 and another identical car is moving with a speed of 30 km h^-1. Compare their kinetic energy.

Answer

Speed of first car = 15 km h^-1
Speed of second car = 30 km h^-1
Mass of both cars = m

Kinetic energy of first car (K₁) = 1221​ x mv²

= 1221​ x m x (15)²

= 22522225​ x m

= 112.5 m J

Kinetic energy of second car (K₂) = 1221​ x mv²

= 1221​ x m x (30)²

= 90022900​ x m

= 450 m J

Comparing the kinetic energy we get:

K1K2=112.5 m450 m=14K2​K1​​=450 m112.5 m​=41​

So Kinetic energy of first car (K₁) : Kinetic energy of second car (K₂) = 1 : 4.

Question 11

A pump raises water by spending 4 x 10⁵ J of energy in 10 s. Find the power of pump.

Answer

Given:
Work done by pump = Energy spent = 4 x 10⁵ J or 400000 J
time = 10 s
Power = ?

Power = Work donetime takentime takenWork done​

= 4000001010400000​

= 40000 W or 4 x 10⁴ W

So power spent by the pump = 4 x 10⁴ W.

Question 12

It takes 20 s for a girl A to climb up the stairs while girl B takes 15 s for the same job. Compare:

(a) The work done and

(b) The power spent by them.

Answer

(a) Both the girls move the same distance and force is also equal on both. Hence, the work done by both the girls is the same.
∴ Work done by girl A : Work done by girl B = 1 : 1

(b) Power spent = Work donetime takentime takenWork done​

We know,

Work done by girl A = Work done by girl B = W

Power spent by girl A (P_A) = W2020W​

Power spent by girl B (P_B) = W1515W​

Comparing the power spent by girl A and girl B:

PAPB=W20W15=1520=34PB​PA​​=15W​20W​​=2015​=43​

∴ Power spent by girl A : Power spent by girl B = 3:4.

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