NEET Foundation

Chapter 4 Work And Energy Multiple Choice Question And Answers

Direction: Choose the correct option for each questions. There is only one correct response for each question.

Question 1. We need ______ for other activities like playing, singing, reading, writing, thinking, jumping, cycling and running.

1. Energy
2. Work
3. Motion
4. None of the above

Answer. 1. Energy

Question 2. What is the unit of work?

1. Newton metre (Nm)
2. Coulomb
3. Ampere
4. None of the above

Answer. 1. Newton metre (Nm)

Question 3. Which is the biggest natural source of energy to us?

1. Moon
2. Stars
3. Sea
4. Sun

Answer. 1. Moon

Question 4. On which factor the work done on an object does not depend?

1. Displacement
2. Force applied
3. Angle between force and displacement
4. Initial velocity of the object

Answer. 4. Initial velocity of the object

Question 5. Water stored in a dam possesses

1. No energy
2. Electrical energy
3. Potential energy
4. Kinetic energy

Answer. 4. Kinetic energy

Question 6. A body is falling from a height h. After it has fallen a height h/2 , it will possess

1. Only potential energy
2. Only kinetic energy
3. Half potential and half kinetic energy
4. More kinetic and less potential energy

Answer. 3. Half potential and half kinetic energy

Question 7. How are Joule (J) and ergs (erg) related?

1. J = 10⁷ erg
2. 1 erg = 10^-7 J
3. 1 J = 10^-7 erg
4. None of the above

Answer. 3. 1 J = 10^-7 erg

Question 8. Work done = Force × ______.

1. Displacement
2. Acceleration
3. Velocity
4. Speed

Answer. 2. Acceleration

Question 9. 1 Joule = 1 ______.

1. Nm²
2. kg m/s²
3. Nm
4. N²m²

Answer. 1. Nm²

Question 10. Which form of energy does the flowing water possess?

1. Gravitational
2. Potential
3. Kinetic
4. electricity

Answer. 3. Kinetic

Question 11. 3730 watts = ______ h.p.

1. 5
2. 2
3. 746
4. 6

Answer. 3. 746

Question 12. The P.E. of a body at a certain height is 200  J. The kinetic energy possessed by it when it just touches the surface of the earth is

1. > P.E.
2. < P.E.
3. = P.E.
4. None of the above

Answer. 1. > P.E.

Question 13. Power is a measure of the

1. Rate of change of momentum
2. Force which produces motion
3. Change of energy
4. Rate of change of energy

Answer. 4. Rate of change of energy

Question 14. 1.5 kW = ______ watts.

1. 150
2. 15000
3. 1500
4. 15

Answer. 2. 15000

Question 15. What is the energy of the simple pendulum when it is at its mean position?

1. Potential energy
2. Kinetic energy
3. Both (a) and (b)
4. Sound energy

Answer. 3. Both (a) and (b)

Question 16. Name the physical quantity which is equal to the product of force and velocity.

1. Work
2. Power
3. Energy
4. Current

Answer. 2. Power

Question 17. What is the example of kinetic energy in the following options?

1. A moving bus
2. A moving particle in electric field
3. A stretched rubber band just released
4. All the above

Answer. 2. A moving particle in electric field

Question 18. A light and the heavy body have equal momenta, which one has greater kinetic energy?

1. A light body
2. A heavy body
3. Both have same K.E
4. None of the above

Answer. 4. Both have same K.E

Question 19. Sum of kinetic and potential energy is called

1. Mechanical energy
2. Chemical energy
3. Electrical energy
4. Magnetic energy

Answer. 1. Mechanical energy

Question 20. Kinetic energy is given by

1. F × S
2. \(\frac{1}{2} m V^2\)
3. Mgh
4. None of the above

Answer. 1. F × S

Question 21. 1 kJ equals to

1. 100 J
2. 10 J
3. 1000 J
4. None of the above

Answer. 2. 10 J

Question 22. Work done in raising an object from the ground to that point against gravity is called

1. gravitational potential energy
2. gravitational kinetic energy
3. gravitational energy
4. none of the above

Answer. 3. gravitational energy

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Question 23. The energy used in households, industries and commercial establishments are usually expressed in ______.

1. kilowatt hour
2. Joules
3. kilo joule
4. None of the above

Answer. 1. kilowatt hour

Question 24. Which is the formula of potential energy?

1. Mgh
2. \(\frac{1}{2} m V^2\)
3. F × S
4. None of the above

Answer. 1. Mgh

Question 25. What are the conditions needed for work to be done?

1. a force should act on an object and
2. the object must be displaced
3. there should be a chemical energy applied

1. 1 and 2
2. 2 and 3
3. 1 and 3
4. All the above

Answer. 1. 1 and 2

Question 26. Which of the following statements are true?

1. Work done by force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force.
2. Work has only magnitude and no direction.
3. Here the unit of work is newton metre (N m) or joule (J)
4. There are two conditions need to be satisfied for work to be done

1. 1 and 2
2. 2 and 4
3. 1, 2 and 3
4. All the above

Answer. 1. 1 and 2

Question 27. What is the unit of work?

1. Newton metre
2. Joule
3. Ampere
4. Coulomb

1. 1 and 3
2. 3 and 4
3. 1 and 2
4. None of the above

Answer. 4. None of the above

Question 28. Which statement is correct regarding energy?

1. The energy possessed by an object is measured in terms of its capacity of doing work.

2. The unit of energy is Joule.

3. 1 kJ equals 100 J.

1. 1 and 2
2. 2 and 3
3. 1 and 3
4. All the above

Answer. 3. 1 and 3

Question 29. Which of the following are the examples of kinetic energy?

1. A falling coconut
2. a speeding car
3. a rolling stone
4. a flying aircraft

1. 1 and 2
2. 2 and 3
3. 3 and 4
4. All the above

Answer. 1. 1 and 2

Question 30. Which of the following are examples of potential energy?

1. Flowing water
2. Blowing wind
3. A coiled spring
4. Wheels on roller skates before someone skates

1. 1 and 2
2. 3 and 2
3. 4 and 1
4. 3 and 4

Answer. 1. 1 and 2

Question 31. Which statement is correct?

1. Power is defined as the rate of doing work or the rate of transfer of energy.
2. Power = work/time
3. The unit of power is watt
4. 1 W = 1 J s^-1

1. 1 and 2
2. 2 and 4
3. 3 and 1
4. All the above

Answer. 4. All the above

Question 32. Work done in raising a box on a platform depends on

1. how fast it is raised
2. strength of the man
3.  negative work
4. zero work

Answer. 4. zero work

Question 33. Work done upon a body is

1. a vector quantity
2. a scalar quantity
3. always positive
4. always negative

Answer. 3. always positive

Question 34. Kilowatt hour (kWh) represents the unit of

1. power
2. impulse
3. momentum
4. none of these

Answer. 2. impulse

Question 35. When two unequal masses possess the same momentum, then kinetic energy of the heavier mass is _______ kinetic energy of the lighter mass.

1. same as
2. greater than
3. smaller than
4. much greater than

Answer. 4. much greater than

Question 36. The number of joules contained in 1 kWh is

1. 36 × 10²
2. 36 × 10³
3. 36 × 10⁴
4. 3.6 × 10⁶

Answer. 3. 36 × 10⁴

Question 37. A completely inelastic collision is one in which the two colliding particles

1. are separated after the collision
2. remain together after the collision
3. split into small fragments flying in all directions
4. none of the above

Answer. 4. none of the above

Question 38. A body moves through a distance of 3 m in the following different ways. In which case is the maximum work done?

1. When pushed over an inclined plane
2. When lifted vertically upward
3. When pushed over smooth rollers
4. When pushed on a plane horizontal surface

Answer. 2. When lifted vertically upward

Question 39. A truck and a car are moving on a smooth, level road such that the K.E. associated with them is same. Breakes are applied to both of them simultaneously. Which one will cover a greater distance before it stops?

1. Car
2. Truck
3. Both will cover the same distance
4. kinetic and potential energies

Answer. 2. Truck

Question 40. Two bullets P and Q, masses 10 and 20 g, are moving in the same direction towards a target with velocities of 20 and 10 m/s respectively. Which one of the bullets will pierce a greater distance through the target?

1. P
2. Q
3. Both will cover the same distance
4. Nothing can be decided

Answer. 3. Both will cover the same distance

Question 41. When the force applied and the displacement of the body are inclined at 90° with each other, then work done is

1. infinite
2. maximum
3. zero
4. unity

Answer. 1. infinite

Question 42. kg m² s^-2 represents the unit of

1. kinetic energy
2. work done
3. potential energy
4. all of these

Answer. 3. potential energy

Question 43. If sand drops vertically at the rate of 2 kg/sec on a conveyor belt moving horizontally with the velocity of 0.2 m/sec, then the extra force required to keep the belt moving is

1. 0.04 N
2. 0.08 N
3. 0.4 N
4. 0.2 N

Answer. 4. 0.2 N

Question 44. A body is dropped from a certain height to the ground. When it is halfway down, it possesses,

1. only K.E.
2. both K.E. and P.E.
3. only P.E.
4. zero energy

Answer. 3. only P.E.

Question 45. The energy required to raise a given volume of water from a well can be

1. mega watts
2. mega newton
3. mega joules
4. kilo watts

Answer. 2. mega newton

Question 46. If a force F is applied on a body and it moves with velocity v, then power will be

1. F ×v
2. \(\frac{F}{v^2}\)
3. \(\frac{F}{v}\)
4. F × v²

Answer. 3. \(\frac{F}{v}\)

Question 47. Which of the following graphs closely represents the P.E. (U) of a freely falling body and its height (h) above the ground?

Answer: 1

Question 48. The displacement x of a particle moving in one dimension, under the action of a constant force, is related to the time t by the equation, t x = + 3 where x is in metres and t in seconds. The displacement of the particle when its velocity is zero, is

1. 0
2. 6 m
3. 12 m
4. 18 m

Answer. 1. 0

Question 49. Asha lifts a doll from the floor and places it on a table. If the weight of the doll is known, what else does one need to know in order to calculate the work Asha has done on the doll?

1. Time required
2. Height of the table
3. Mass of the ball
4. Cost of the doll or the table

Answer. 1. Time required

Question 50. The work done in lifting a mass of 1 kg to a height of 9.8 m is

1. 1 J
2. (9.8)² J
3. 9.8 J
4. None of these

Answer. 2. (9.8)² J

Question 51. In which of the following cases, will the work done be maximum? The body is moved through a distances on the ground

Answer: 2

Question 52. Work done by a centripetal force

1. increases by decreasing the radius of the circle
2. decreases by increasing the radius of the circle
3. increases by increasing the mass of the body
4. is always zero

Answer. 2. decreases by increasing the radius of the circle

Question 53. Certain weight is attached with a spring. It is pulled down and then released. It oscillates up and down. Its K.E. will be

1. maximum in the middle of the movement
2. maximum at the bottom
3. maximum just before it is released
4. constant

Answer. 4. constant

Question 54. A 1  kg mass falls from a height of 10 m into a sand box. What is the speed of the mass just before hitting the sand box? If it travels a distance of 2 cm into the sand before coming to rest, what is the average retarding force?

1. 12 m/sec and 3600 N
2. 14 m/sec and 4900 N
3. 16 m/sec and 6400 N
4. 18 m/sec and 8100 N

Answer. 1. 12 m/sec and 3600 N

Question 55. If L, M denote the angular momentum and mass of a particle and p its linear momentum, which of the following can represent the kinetic energy of the particle moving in a circle of radius R?

1. \(\frac{L^2}{2 M}\)
2. \(\frac{p^2}{M}\)
3. \(\frac{L^2}{2 M R^2}\)
4. \(\frac{1}{2} M p\)

Answer. 2. \(\frac{p^2}{M}\)

Question 56. A particle of mass 4 m which is at rest and explodes into three fragments. Two of the fragments each of mass mare found to move with a speed v in mutually perpendicular directions. The total energy released in the explosion is

1. 2mv²
2. \(\frac{1}{2} m v^2\)
3. mv²
4. \(\frac{3}{2} m v^2\)

Answer. 3. mv²

Question 57. kWh represents the unit for

1. force
2.  power
3. time
4. energy

Answer. 4. energy

Question 58. Energy cannot be measured in

1. Js
2. Ws
3. kWh
4. erg

Answer. 4. erg

Question 59. A steam engine converts

1. heat energy into sound energy
2. heat energy into mechanical energy
3. mechanical energy into heat energy
4. electrical energy into sound energy

Answer. 1. heat energy into sound energy

Question 60. A man throws bricks to the height of 12 m where they reach with a speed of 12 m/s. If he throws the bricks such that they just reach this height, then what percentage of energy will he save?

1. 19%
2. 38%
3. 57%
4. 76%

Answer. 2. 38%

Question 61. Two bodies with masses MA and MB are moving with equal kinetic energy. Their linear momenta are numerically in a ratio |PA| : |PB|, the ratio of their masses will be

1. MB : MA
2. MA : MB
3. \(\sqrt{M_A}: \sqrt{M_B}\)
4. \(M_{\mathrm{A}}^2: M_{\mathrm{B}}^2\)

Answer. 2. MA : MB

Question 62. Mechanical work done is equal to (symbols have their usual meanings)

1. W = F/S
2. W = FS
3. W = F + S
4. W = F – S

Answer. 3. W = F + S

Question 63. Which of the following graphs best represents the graphical relation between momentum (P) and kinetic energy (K) for a body in motion?

Answer: 2

Question 64. An elevator is designed to lift a load of 1000 kg through 6 floors of a building on an average of 3.5 m per floor in 6 sec. Power of the elevator, neglecting other losses, will be

1. 3.43 × 10⁴ watt
2. 4.33 × 10⁴ watt
3. 2.21 × 10⁴ watt
4. 5.65 × 10⁴ watt

Answer. 4. 5.65 × 10⁴ watt

Question 65. When the momentum of a body increases by 100 %, then its K.E. increases by

1. 20 %
2. 40 %
3. 100 %
4. 300 %

Answer. 1. 20 %

Question 66. No work is said to have been done when an object moves at an angle of _____ with the direction of the force.

1. 0°
2. 90°
3. 180°
4. between 90° and 180°

Answer. 4. between 90° and 180°

Question 67. When a body is whirled in a circle, then work done on it is

1. positive
2. negative
3. zero
4. infinite

Answer. 2. negative

Question 68. A crane is used to lift 1000 kg of coal from a mine 100 m deep. If the time taken by the crane is 1 hr, then find the power of the crane, assuming the efficiency of the crane to be 80 %. (g = 9.8 m/s²)

1. 2567 watts
2. 2403 watts
3. 3403 watts
4. 3761 watts

Answer. 3. 3403 watts

Question 69. The flowing water of a river possesses

1. gravitational energy
2. potential energy
3. electrical energy
4. kinetic energy

Answer. 3. electrical energy

Question 70. The mass of an object P is double the mass of object Q. If both move with the same velocity, then the ratio of K.E. of P to Q is

1. 1 : 2
2. 2 : 1
3. 1 : 4
4. 4 : 1

Answer. 4. 4 : 1

Question 71. A truck can move up on a road having the gradient of 1 m rise for every 50 m with a speed of 15 km/hr. The resisting force is equal to 1/25 th the weight of the truck. How fast will the same truck move down the hill with the same horse power?

1. 30 km/hr
2. 45 km/hr
3. 60 km/hr
4. 75 km/hr

Answer. 2. 45 km/hr

Question 72. A body of mass 1 kg strikes elastically with another body at rest and continues to move in the same direction with one fourth the initial velocity. The mass of the other body is

1. 3 kg
2. 0.6 kg
3. 2.4 kg
4. 4 kg

Answer. 2. 0.6 kg

Question 73. A body rolling down a hill has

1. K.E. only
2. P.E. only
3. Neither K.E. nor P.E.
4. Both (a) and (b) above

Answer. 2. P.E. only

Question 74. A total of 4900 joules was expended in lifting a 50 kg mass. The mass was raised to the height of

1. 98 m
2. 960 m
3. 245 m
4. 10 m

Answer. 4. 10 m

Question 75. A ball of mass 200 g falls from a height of 5 m. What is its K.E. when it just reaches the ground?

1. 9.8 J
2. 98
3. 980 J
4. None of these

Answer. 4. None of these

Question 76. A stretched spring possesses

1. kinetic energy
2. elastic potential energy
3. electric energy
4. magnetic energy

Answer. 1. kinetic energy

Question 77. When a person climbs a hill, he possesses

1. only K.E
2. only P.E.
3. both K.E. and P.E.
4. none of these

Answer. 2. only P.E.

Question 78. An iron sphere of mass 30 kg has the same diameter as an aluminium sphere whose mass is 10.5 kg. The spheres are dropped simultaneously from a cliff. When they are 10m from the ground, they have the same

1. acceleration
2. momentum
3. potential energy
4. K.E.

Answer. 3. potential energy

Question 79. A stone of mass m kg is whirled in a vertical circle of radius 20 cm. The difference in the kinetic energies at the lowest and the topmost positions is

1. 4 mg joules
2. 0.4 mg joules
3. 40 mg joules
4. None of these

Answer. 1. 4 mg joules

Chapter 4 Work And Energy Long Answer Type Question And Answers

Question 1. A bullet of mass 50 g travelling horizontally with a speed of 200 ms^-1 strikes a glass pave 4 mm thick and falls dead, on emerging from it. How much work is done by the bullet against the opposing force of the glass pave?
Answer.

Given:

A bullet of mass 50 g travelling horizontally with a speed of 200 ms^-1 strikes a glass pave 4 mm thick and falls dead, on emerging from it.

Initial K.E. of the bullet = \(\frac{1}{2} m u^2\)

Final K.E. of the bullet = \(\frac{1}{2} m v^2\)

Loss in K.E. of the bullet = Initial K.E. – Final K.E.

= \(\frac{1}{2} m u^2-\frac{1}{2} m v^2\)

= \(\frac{1}{2} m\left(u^2-v^2\right)\)

By Work – Energy theorem,

Work done by a body = Loss in its K.E.

W = \(\frac{1}{2} m\left(u^2-v^2\right)\)

= \(\frac{1}{2} \times \frac{50}{1000}\left(200^2-0^2\right)\)

= \(\frac{1}{2} \times \frac{5}{100} \times 200 \times 200\)

= 1000J

= 1 x 10³ J

Work done by a body = 1 x 10³ J

Question 2. If a heavy truck (mass = M) and a light car (mass m which is less than M) possess same amount of momentum, then which one possess greater kinetic energy? Explain.
Answer.

Given

If a heavy truck (mass = M) and a light car (mass m which is less than M) possess same amount of momentum

We know that momentum ‘p’ of a body of mass ‘m’ moving with velocity v is given by

p = mv

Squaring on both sides,

p² = (mv)² = m²v²

p² = m⋅mv

= \(2 m\left(\frac{1}{2} m v^2\right)\)

p² = 2mE_k

So, K.E., \(E_\kappa=\frac{p^2}{2 m}\)

If p is same, i.e., constant, then

\(E_K=\frac{C}{m}\)

i.e., E_k is inversely proportional to mass.

It  means, a body with smaller mass possess more K.E.

∴ Car will possess more K.E. than the truck.

Question 3. A body fall from height H. If t₁ is time taken for covering the first half height and t₂ be the time taken for second half. Which of these relations is true for t₁ and t₂.

1. t₁ > t₂
2. t₁ < t₂
3. t₁= t₂
4. Depends on the mass of the body

Answer.

Given:

A body fall from height H. If t₁ is time taken for covering the first half height and t₂ be the time taken for second half.

Let H be the height, then

First Half

\(\frac{H}{2}=\left(\frac{1}{2}\right) g t_1^2\)   (1)

Or  \(\left(\frac{1}{2}\right) g t_1^2=\frac{H}{2}\)

Also v = gt₁

Second Half

\(\frac{H}{2}=v t_2+\left(\frac{1}{2}\right) g t_2^2\)

or \(\left(\frac{H}{2}\right)=g t_1 t_2+\left(\frac{1}{2}\right) g t_2^2\)

or \(\left(\frac{1}{2}\right) g t_2^2=\left(\frac{H}{2}\right)-g t_1 t_2\)   (2)

t₂² + 2t₁t₂ − t₁² = 0

or \(t_2=\frac{\left[-2 t_1+\sqrt{\left(4 t_1^2+4 t_1^2\right)}\right]}{2}\)

or \(t_2=\frac{\left[-2 t_1+2 t_1 \sqrt{2}\right]}{2}\)

t₂ = 0.4 t₁

so, t₁ > t₂

Hence a is correct.

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Question 4. A delivery boy wishes to launch a 2.0 kg package up an inclined plane with sufficient speed to reach the top of the incline. The plane is 3 m long and is inclined at 20. the coefficient of friction between the package and the inclined plane is .40.What minimum initial K.E. must the boy supply to the package given as sin20 = .342 cos20 = .940?
Answer.

Given:

A delivery boy wishes to launch a 2.0 kg package up an inclined plane with sufficient speed to reach the top of the incline.

The plane is 3 m long and is inclined at 20. the coefficient of friction between the package and the inclined plane is .40.

If the package travels the entire length S of the incline, the frictional force will perform work − μNS where μ is the coefficient of friction and N is normal reaction.

Let ‘h’ be the height of the incline plane then the gravitational potential energy of the package will increase by mgh.

Now let us assume ‘v’ be the speed given to the package so as to reach the top

Then kinetic energy at the initial point = \(\frac{1}{2} m v^2\)

Now applying work energy theorem

K.E_f – K.E_i = Work done by the gravitational force + Work done by the frictional force

Now since K.E_f = 0

Also Work done by the gravitational force

= –(change in gravitational potential energy)

= −mgh

Therefore

\(\frac{1}{2} m v^2\) = mgh – μNS

or \(\frac{1}{2} m v^2\) = mgh + μN

Now S = 3

N = mgcosθ

h = S sinθ

Substituting all the values

\(\frac{1}{2} m v^2\) = 42.2 J

Question 5. Deduce a formula for K.E of a body.
Answer.

A formula for K.E of a body:

Consider a body of mass ‘m’ starts moving from rest, with uniform velocity ‘u’. After a time interval ‘t’ its velocity becomes v.

If initial velocity of the body is u or vi = 0, final velocity vf = v and the displacement of body is ‘S’. Then

First of all we will find the acceleration of the body.

Using equation of motion

2aS = v_f² − v_i²

Substituting the above mentioned values

2aS = v² − 0

a = \(\frac{v^2}{2 S}\)

Now force is given by

F = ma

Substituting the value of acceleration

F = \(m\left(\frac{v^2}{2 S}\right)\)

As we know that,

Work done = F.S

Substituting the value of F

Work done = \(\left(\frac{m v^2}{2 S}\right)(S)\)

Work done = \(\frac{m v^2}{2}\)

or Work done = \(\frac{1}{2} m v^2\)

Since the work done in motion is called ‘Kinetic Energy’

i.e., K.E. = Work done

or \(\frac{1}{2} m v^2\)

Question 6. A body of mass 1.0 kg initially at rest is moved by a horizontal force of 0.5 N on a smooth frictionless table. Calculate the work done by the force in 10 s and show that this is equal to the change in K.E. of the body.
Answer.

Given:

A body of mass 1.0 kg initially at rest is moved by a horizontal force of 0.5 N on a smooth frictionless table.

Here, m = 1.0 kg, u = 0, F = 0.5 N, t = 10s

a = \(\frac{F}{m}=\frac{0.5}{1.0}=0.5 \mathrm{~ms}^{-2}\)

From S = \(u t+\frac{1}{2} a t^2\)

S = \(0+\frac{1}{2} \times 0.5(10)^2=2.5 \mathrm{~m}\)

Work done = F × S

= 0.5 × 25

= 12.5 J

From v = u + at = 0 + 0.5 × 10

= 5 ms–1

Change in K.E. = \(\frac{1}{2} m\left(v^2-u^2\right)\)

= \(\frac{1}{2} \times 1.0\left(5^2-0\right)\)

= 12.5 J

The work done by the force in 10 s and show that this is equal to the change in K.E. of the body = 12.5 J

Chapter 4 Work And Energy Short Answer Type Questions

Question 1. A force of 20 N acts on a body to displace it through 10 m on a level road. Calculate the work done if force and displacement make an angle 60 degree with each other.
Answer. 

Given:

A force of 20 N acts on a body to displace it through 10 m on a level road.

Here force F = 20 N, Displacement, S = 10 m,

θ = 60 degrees

Now Work done, W = FS cos θ

= 20 × 10 × cos 60

= 20 × 10 × 0.5 = 100 J

The work done if force and displacement make an angle 60 degree with each other = 100 J

Read And Learn More NEET Foundation Short Answer Questions

Question 2. Define work. Write the formula of work when force and displacement do not act in same direction.
Answer.

Work:

Work is the product of force and displacement.

When force and displacement do not act in same direction, the formula for work is given by,

W = FS cos θ

Where θ is the angle between force and displacement.

Question 3. Mention the difference between potential energy and kinetic energy. A sliding box of mass 1 kg slows down from 10 m/s to 4 m/s. Calculate the change in kinetic energy.
Answer.

Given:

The difference between potential energy and kinetic energy. A sliding box of mass 1 kg slows down from 10 m/s to 4 m/s.

Potential energy of a body is the energy possessed by a body due to its position while kinetic energy of a body is the energy possessed due to velocity of the body.

Change in kinetic energy,

ΔK = 1/2m(v² – u²)

ΔK = (42 – 10²)/2

= -84/2 = -42 J

The change in kinetic energy = -42 J

Question 4. Which physical quantity has Nm as its SI unit? Two different bodies of masses in the ratio of 1 : 2 are moving with same speed. What is the ratio of their kinetic energy?
Answer.

Given:

Two different bodies of masses in the ratio of 1 : 2 are moving with same speed

Work has the SI unit Nm.

Ratio of Kinetic energies

\(\frac{K_1}{K_2}=\frac{\left(\frac{1}{2} m_1 v_1^2\right)}{\left(\frac{1}{2} m_2 v_2^2\right)}\) \(\frac{K_1}{K_2}=\left(\frac{m_1}{m_2}\right)\left(\frac{v_1^2}{v_2^2}\right)\) \(\frac{K_1}{K_2}=\frac{1}{2} \times 1 \quad\left(\text { since } v_1=v_2 \& \frac{m_1}{m_2}=\frac{1}{2}\right)\)

So \(\frac{K_1}{K_2}\) = 0.5

Work And Energy Questions

Question 5. In lifting a body to 10 m, 800 J of work was done. Calculate its weight.
Answer.

Given:

In lifting a body to 10 m, 800 J of work was done.

Work done W = mgh

800 = mg × 10

mg = 800/10 = 80 N

Thus, weight of the body is 80 N.

Question 6. How much work is done in lifting a book weighing 800 g through a height of 1.5 m? (Take g = 10 ms^-2)
Answer.

We know that,

W = Force × displacement

⇒ W = mg × S

⇒ W = mgh (∵ S = h)

⇒ W = 0.8 kg × 10 ms^-2 × 1.5 m

⇒ W = 12 J

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Question 7. When can work done by a force be zero? Give examples.
Answer.

We know that, in general work done is given by

W = FS cos θ

Where θ is the angle between direction of force and displacement.

W = 0 = FS cos θ

⇒ either S = 0

or cos θ = 0

i.e., θ = 90°

So work done by a force is zero.

When either displacement is zero or angle between directions of force and displacement is zero.

Examples

1. When a force is applied to a body and it doesn’t move

Example: a wall.

2. Displacement is in horizontal direction and force in vertical direction. So, work done by gravity on a bucket, which a man is carrying along horizontal ground.

Question 8. How much work is done in pulling a trolley through 6 m, as shown in the below figure?

Answer.

We know that,

W = FS cos θ

= 100 N × 6 m × cos(60°)

= 100 × 6 × 0.5 Nm

= 300 J

Work Done = 300 J

Question 9. You are standing on roof top of a tower of height 25 m. What is the potential energy of a ball of mass 150 g kept on the ground? (Take g = 10 ms^-2)
Answer.

In this case, potential energy will be negative since the body is below the reference level which is on the roof top.

P.E. = mgh

= 0.150 kg × 10 ms^-2 × (−25 m)

= −37.5 J

Note: This numerical tells us that potential energy of a body can be negative.

Example: Potential energy of coal in a coalmine is negative with respect to the observer on ground.

Question 10. If the velocity of a body is doubled, how will its kinetic energy change? Compare new kinetic energy with the old one.
Answer.

Consider a body of mass ‘m’ moving with a velocity ‘v1’.

Then, its

\(\text { K.E. }=\frac{1}{2} m v_1^2\)

∴ \(E_1=\frac{1}{2} m v_1^2\)    (1)

Now, its velocity is doubled.

So, v₂ = 2v₁

∴ Its new kinetic energy

\(E_2=\frac{1}{2} m v_2^2\)

= \(\frac{1}{2} m\left(2 v_1^2\right)\)

= \(4 \cdot \frac{1}{2} m v_1^2\)

= 4 E₁ form (1)

Thus, its kinetic energy becomes four times.

Question 11. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed (in m/sec) of a particle is zero, than what is the speed (in m/sec) after 5 s?
Answer.

Given:

A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle.

We know that

\(\text { Power }=\frac{d W}{d t}\)

W = 0.5 × 5 = 2.5 = K.E._f − K.E._i

Since the initial speed of the particle is zero, the initial kinetic energy will also be zero.

⇒ \(2.5=\frac{M}{2}\left(v_f^2-v_i^2\right)\)

⇒ \(2.5=\frac{M}{2}\left(v_f^2-0\right)\)

⇒ \(v_f^2=2.5 \times \frac{2}{0.2}=25\)

⇒ v_f = 5

Question 12. Deduce the formula of P.E. of a body.
Answer.

P.E. of a body:

Consider an object of mass, m. Let it be raised through a height h from the ground. A force is required to do this. The minimum force required to raise the object is equal to the weight of the object mg.

The object gains energy equal to the work done on it. Let the work done on the object against gravity be W. That is, work done,

W = Force × Displacement

= mg × h

= mgh

Since work done on the object is equal to mgh, the energy equal to mgh units is gained by the object. This is the potential energy (EP) of the object.

E_p = mgh

Question 13. When a rubber is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx², where a and b are constants. What is the work done in stretching the un-stretched rubber b and by L?
Answer.

Given:

When a rubber is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx², where a and b are constants

We have

F = ax + bx²

W = Fdx

The work done by stretching the un-stretched rubber b and by L is

\(W=\int_0^L\left(a x+b x^2\right) d x\)

= \(\left[a\left(\frac{x^2}{2}\right)+b\left(\frac{x^3}{3}\right)\right]_0^L\)

= \(a\left(\frac{L^2}{2}\right)+b\left(\frac{L^3}{3}\right)\)

Chapter 3 Gravitation Long Answer Type Question And Answers

Question 1. Interconnected vessel as shown in figure is filled with an ideal liquid P₁ and P₂ are airtight pistons having areas A₁ = 2 cm² and A₂ = 5 cm² respectively. A weight of 3 kg is kept on piston P₁. Calculate

1. pressure acting on piston P₁
2. pressure on piston P₂
3. force with which P₂ moves up.

Answer.

Given:

Interconnected vessel as shown in figure is filled with an ideal liquid P₁ and P₂ are airtight pistons having areas A₁ = 2 cm² and A₂ = 5 cm² respectively.

A weight of 3 kg is kept on piston P₁.

Pressure P₁ exerted by the piston P₁ on the confined liquid is given by

P₁ = \(\frac{\text { Force }}{\text { Area }}=\frac{\text { Weight }}{\text { Area }}=\frac{M_g}{A_1}\)

= \(\frac{3 \times 10 \mathrm{~N}}{2 \times 10^{-1} \mathrm{~m}^2}\)

P₁ = 15 × 10⁴ Pa, (downwards)

Pressure acting on piston P₁  = 15 × 10⁴ Pa, (downwards)

By Pascal’s law, this pressure is communicated to the piston P₂.

∴ Upward pressure P₂ acting on piston P₂

= P₁

= 15 × 10⁴ Pa, (upwards)

We know that,

P = \(\frac{F}{A}\)

∴ F = PA

= P₂A₂

= \(15 \times 10^4 \frac{\mathrm{N}}{\mathrm{m}^2} \times 5 \times 10^{-4} \mathrm{~m}^2\)

= 75 N

Read and Learn More NEET Foundation Long Answer Questions

Question 2. Nine tenth of an iceberg having a total volume of 500 m3 is submerged in ocean. Calculate the buoyant force acting on the iceberg.

1. Take ρoceanwater = 1.02 × 10³ kgm^-3
2. g = 10 ms^-2

Answer.

Given:

Nine tenth of an iceberg having a total volume of 500 m3 is submerged in ocean

Buoyant force or upthrust is given by

U = Vρ g

where

V = Volume of immersed part of solid.

= \(\left(\frac{9}{10} \times 500\right) \mathrm{m}^3\)

= 450 m³

So U = V ρ g

= \(450 \mathrm{~m}^3 \times 1.02 \times 10^3 \frac{\mathrm{kg}}{\mathrm{m}^3} \times 10 \mathrm{~m}\)

= 45 × 1.02 × 10⁵ N

= 4.59 × 10⁶ N

The buoyant force acting on the iceberg = 4.59 × 10⁶ N

Question 3. A piece of metal weighs 200 gf in air and 180 gf in water. Finds its relative density.
Answer.

Given:

A piece of metal weighs 200 gf in air and 180 gf in water.

\(\text { R. D. of a solid }=\frac{\text { Weight in air }}{\text { Loss of weight in water }}\)

= \(\frac{W_1}{W_1-W_2}\)

= \(\frac{200 \mathrm{gf}}{(200-180) g f}\)

= \(\frac{200}{20}\)

∴ RD of metal = 10 (No unit).

Question 4. Imagine a planet whose both diameter and mass are one half of the Earth. The day’s temperature of this planet’s surface reaches up to 800 K. Find whether oxygen molecules are possible in the atmosphere of this planet.
Answer.

Given:

Imagine a planet whose both diameter and mass are one half of the Earth. The day’s temperature of this planet’s surface reaches up to 800 K.

Escape velocity,

v_e = \(\sqrt{2} G M / R\)

Let v_p = escape velocity on the planet

v_e = escape velocity on the earth

\(\frac{v_p}{v_e}=\sqrt{\left(\frac{M_p}{R_p} \times \frac{R_e}{M_r}\right)}=\sqrt{\left(\frac{1}{2} \times \frac{2}{1}\right)}=1\)

v_p = v_e = 11.2 km/s

From kinetic theory of gases

v_rms = \(\sqrt{3 R T / M})=\sqrt{3 N K T / M}=\sqrt{3 N K T / N m}\)

where N = Avogadro’s number

m = mass of oxygen molecule

K = Boltzmann constant

v_rms = \(\sqrt{3 R T / m}\)

= \(\sqrt{\left(\frac{\left(3 \times 1.38 \times 10^{-23} \times 800\right)}{5.3 \times 10^{-2 h}}\right)}\)

(m = 5.3 × 10−26 kg)

v_rms = 0.79 km/s

As v_rms is very small compared to the escape velocity on the planet, molecules cannot escape from the surface of the planet’s atmosphere.

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Question 5. Deduce the relationship between g and G.
Answer.

Relationship between g and G:

Let g = acceleration due to gravity at a planet

M = mass

R = radius

m = mass of object

By Newton’s law of motion, force on a body due to gravity on its surface

F = mass × acceleration due to gravity

= m g

By Newton’s gravitational law, force is

F = GMm/R²

Therefore,

GMm/R² = mg

Acceleration due to gravity g = GM/R²

Question 6. How much below the surface of earth does the acceleration due to gravity

1. reduce to 36%
2. reduce by 36%, of its value on the surface of earth, (radius of earth = 6400 km).

Answer. Case (1)

We have,

g_h = g (1 – d/R)

Or d = (g – g_h) R/g (1)

Here

g_h = 36/100 g (2)

Using (1) and (2)

g_h = 36/100 g

d = (1 – 36/100) R

d = (100 – 36)/100 × 6400

d = 4096 km

Case (2)

Here

g_h = g – 36/100 g

i.e., g_h = 64/100 g (3)

Using equation (1) and (3)

d = (g – 64/100 g) 6400/g

d = 100 – 64/100 × 6400

d = 2304 km

Question 7. Deduce gravitational force between

1. gravitational force between earth and the  sun and
2. gravitational force between the moon and  the earth.

Answer. (1) Gravitational force between earth and the sun

Mass of earth, m₁ = 6 × 10²⁴ kg

Mass of sun, m₂ = 2 × 10³⁰kg

Distance between sun and earth, R = 1.5 × 10¹¹m

Gravitational force between the sun and the earth,

F = \(G \frac{m_1 m_2}{r^2}\)

F = (6.67 × 10^-11 Nm² kg^-2 × 6 × 10²⁴ kg × 2 × 10³⁰kg)/(1.5 × 10¹¹m) ²

F = 3.6 × 10²² N

(2) Gravitational force between the moon and the earth

Mass of earth, m₁ = 6 × 10²⁴ kg

Mass of moon, m₂ = 7.4 × 10²² kg

Gravitational force between earth and the moon is

F = \(G \frac{m_1 m_2}{r^2}\)

F = (6.67 × 10^-11Nm²kg^-2× 6 × 10²⁴kg × 7.4 × 10²² kg)/(3.8 × 10⁸m)²

F = 2.05 × 10²⁰ N

Question 8. Explain why a sheet of paper falls slower than one that is crumpled into a ball?
Answer.

The sheet of paper falls slower than the one that is crumpled into a ball because the air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the sheet of paper is more than the resistance offered by air to the paper ball because the sheet has larger area.

Question 9. The escape velocity of a body on the surface of the earth is 11.2 km/s. A body is projected away with twice this speed. What is the speed of the body at infinity? Ignore the presence of other heavenly bodies.
Answer.

If v is the velocity of projection and v’ is the velocity at infinity, then we have by energy conservation principle.

\(\frac{1}{2} m v^2-\frac{G M m}{R}=\frac{1}{2} m v^{\prime 2}+0\)

Here v = 2v_e

Thus,

\(\left(\frac{1}{2}\right) \cdot 4 v_e^2-\frac{G M}{R}=\frac{1}{2} v^{\prime 2}\) \(2 v_e^2-\frac{G M}{R}=\frac{1}{2} v^{\prime 2}\)

Now, v_e = \(\frac{\sqrt{2 G M}}{R}\)

\(2 v_e^2-\frac{v_e^2}{2}=\frac{1}{2} v^{\prime 2}\)

Or, v’² = 3 v_e²

Or, v’ = \(\sqrt{3} v_e=\sqrt{3} \times 11.2 \mathrm{~km} / \mathrm{s}=19.4 \mathrm{~km} / \mathrm{s}\)

Question 10. Deduce the equation showing the variation in the value of g with depth below the surface of the earth.
Answer.

Let us consider a body of mass m at a depth h below the surface of earth. Then, radius of the inner solid sphere of the earth = R – h.

Volume of the inner solid sphere of the earth = \(\frac{4}{3} \pi(R-h)^3 d\)

If d is the average density of the earth, then

Mass of inner solid sphere of earth = \(\frac{4}{3} \pi(R-h)^3 d\).

According to the law of gravitation,

mg_d = \(G \times \frac{4}{3} \pi(R-h)^3 d \times m /(R-h)^2\)

This gives

g_d = \(G \times \frac{4}{3} \pi(R-h) d (1)\)

On the surface of earth,

g = \(\frac{G M}{R^2}\)

= \(G \times \frac{4}{3} \pi R^3 \frac{d}{R^2}\)

= \(G \times \frac{4}{3} \pi R d (2)\)

From equation (1) and (2)

g_d/g = \(G \times \frac{4}{3} \pi(R-h) \times d / G \times \frac{4}{3} \pi R d\)

= (R – h)/R

Or g_d = g (1 – h/R) or (R – h)/R < 1

So, g_d < g

Thus, the value of g at a depth inside the earth is less than that on the surface of the earth.

The value (R – h)/R decreases with the value of h, i.e. depth below the surface of the earth. So the value of g decreases as we go down below the surface of earth.

Question 11. A stone is dropped from the edge of the roof.

1. How long does it take to fall 4.9 m?
2. How fast does it move at the end of the fall?
3. How fast does it move at the end of 7.9 m?
4. What is its acceleration after 1 s and after 2 s?

Answer. (1) As the stone is dropped, its initial velocity,

u = 0, h = 4.9 m

a = g = 9.8 ms^-2, time, t =?

From

h = \(u t+\frac{1}{2} g \mathrm{t}^2\)

4.9 = \(0+\frac{1}{2} \times 9.8 t^2=4.9 t^2\)

Or \(f^2=\frac{4.9}{4.9}=1, t=\sqrt{1}=1 \mathrm{~s}\)

(2) Final velocity, v = ? at t = 1 s

From

v = u + gt

v = 0 + 9.8 × 1 = 9.8 m s^-1

(3) Let v be the final velocity when h = 7.9 m

From

v² – u² = 2ah

v² – 0 = 2 (9.8) 7.9

Or v = \(\sqrt{2 \times 9.8 \times 7.9}=\sqrt{154.84}\)

v = 12.4 m s^–1

(4) The acceleration of a freely falling body remains the same at all times, i.e., a = g = 9.8 ms^-1 after 1 s and 2 s.

NEET Physics Gravitation Questions

Chapter 3 Gravitation Short Answer Type Question And Answers

Question 1. Calculate the gravitational force of attraction between two small gold balls weighing 500 g and 2 kg kept on a horizontal table, with their centers 40 cm apart. Take G = 6.7 × 10^-11 Nm² kg^2-2
Answer.

Given:

Two small gold balls weighing 500 g and 2 kg kept on a horizontal table, with their centres 40 cm apart.

Given:  m₁ = 500g = 0.5 kg

m₂ = 2 kg

r = 40 cm = 0.40 m

and G = 6.7 × 10^-11 Nm² kg^-2

By Newton’s law of gravitation,

F = \(\frac{G m_1 m_2}{r^2}\)

= \(6.7 \times 10^{-11} \frac{\mathrm{Nm}^2}{\mathrm{~kg}^2} \times \frac{0.5 \mathrm{~kg} \times 2 \mathrm{~kg}}{(0.4 \mathrm{~m})^2}\)

= \(6.7 \times 10^{-11} \times \frac{0.5 \times 2}{0.16} \mathrm{~N}\)

= \(6.7 \times 10^{-11} \times \frac{100}{16} \mathrm{~N}\)

∴ F = 4.1875 × 10^-10 N

The gravitational force of attraction between two small gold balls = 4.1875 × 10^-10 N

Question 2. With what force earth attracts moon? With what force moon attracts earth?
Answer.

Take M_E = 6 × 10²⁴ kg

M_m = 7.5 × 10²² kg

r = 3,80,000 km = 3.8 × 10⁸ m

According to Newton’s law of gravitation,

F = \(\frac{G M_E M_m}{r^2}\)

= \(6.7 \times 10^{-12} \frac{\mathrm{Nm}^2}{\mathrm{~kg}^2} \times \frac{6 \times 10^{24} \mathrm{~kg} \times 7.5 \times 10^{22} \mathrm{~kg}}{\left(3.8 \times 10^5 \mathrm{~m}\right)^2}\)

= \(\frac{6.7 \times 6 \times 7.5 \times 10^{35}}{3.8 \times 3.8 \times 10^{16}} \mathrm{~N}\)

Therefore, F = 20.9 × 10¹⁹ N

According to Newton’s III law, moon will also attract earth towards itself with the same force i.e., 20.9 × 10¹⁹ N.

Read And Learn More NEET Foundation Short Answer Questions

Question 3. A body is thrown vertically upwards with a velocity of 49 ms^-1. Find

1. how long it takes to reach the highest point
2. maximum height attained by it.

Take g = 9.8 ms^-1 and neglect air-friction.

Answer. Given: u = 49 ms^-1

v = 0 (at the highest point)

a = −9.8 ms^-2

t = ?

h = S = ?

(1) We know that

v = u + at

0 = 49 + (−9.8) × t

9.8 t = 49

∴ t = 49

9 8. = 5 s

5 s it will takes to reach the highest point

(2) ∵ v² = u² + 2ah

0 = (49)² + 2 (−9.8) h

19.6 h = 49 × 49

∴ h = \(\frac{49 \times 49}{19.6}\)

∴ h = 122.5 m

Maximum height attained by it is 122.5 m

Question 4. A body is dropped from the top of a tower 200  m tall. Simultaneously, another body is thrown vertically upwards from its foot with a velocity of 100 ms^-1. Find when and where they meet or cross each other.
Take g = 10 ms^-2 and ignore air friction.
Answer. 

Given:

A body is dropped from the top of a tower 200  m tall. Simultaneously, another body is thrown vertically upwards from its foot with a velocity of 100 ms^-1.

Suppose the two bodies meet at point C after ‘x’ seconds.

Consider downward motion AC of first body.

∵ S = \(u t+\frac{1}{2} g t^2\)

∴ AC = \(0+\frac{1}{2} \times 10 \times x^2\)

∴ AC = 5x² (1)

Consider upward motion BC of second body.

∴ S = \(u t+\frac{1}{2} g t^2\)

∴ BC = \(100 x+\frac{1}{2}(-10) x^2\)

∴ BC = 100x − 5x² (2)

Adding equations (1) and (2), we get

AC + BC = 5x² + 100x − 5x²

∴ AB = 100x

i.e., 200 = 100x

∴ x = \(\frac{200}{100}=2 s\)

Substituting x = 25 in equation (i) we get,

AC = 5x² = 5 × 2² = 20 m

Therefore, the two bodies meet at a point 20 m below the top of the tower or (200 − 20) m = 180 m above the foot of the tower, 2 s after they embark on their journey.

Gravitation NEET Questions 

Question 5. A girl weighing 40 kg is wearing pointed sandals having a total area of 2 × 10^-4 m². How much pressure she will exert on ground? In which case, the ground is more likely to be damaged? Explain. (Take g = 10 ms^-2)
Answer.

Given:

A girl weighing 40 kg is wearing pointed sandals having a total area of 2 × 10^-4 m².

A boy of same weight wears shoe having area of 100 × 10^-4 m.

Pressure P_g exerted by girl on the ground due to her weight is given by

P_g = \(\frac{\text { Force }}{\text { Area }}=\frac{\text { Weight }}{\text { Area }}=\frac{M g}{A}\)

= \(\frac{40 \times 10 \mathrm{~N}}{2 \times 10^{-4} \mathrm{~m}^2}\)

P_g = 2 × 10⁶ Nm^-2or P_a Similarly pressure exerted by the boy on the ground due to his weight

P_b = \(\frac{\mathrm{Mg}_g}{A}=\frac{40 \times 10 \mathrm{~N}}{100 \times 10^{-4} \mathrm{~m}^2}\)

P_b = 4 × 10⁴ Nm^-2 or Pa

Since P_g > P_b, the girl is more likely to damage the ground than the boy

NEET Foundation Physics Chapter 3 

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Question 6. A tall cylinder contains water column of height 3 m. Calculate the pressure it exerts on the bottom of the cylinder. Take g = 10 ms^-2. Density of water = 1000 kg.
Answer.

Given:

A tall cylinder contains water column of height 3 m.

We know that, pressure exerted by a liquid column is given by,

P = h ρ g

= \(3 \mathrm{~m} \times 1000 \frac{\mathrm{kg}}{\mathrm{m}^3} \times 10 \frac{\mathrm{m}}{\mathrm{s}^2}\)

∴ P = 3 × 10⁴ Nm² or Pa

The pressure it exerts on the bottom of the cylinder = 3 × 10⁴ Nm² or Pa

Gravitation Short Answer Questions 

Question 7. It is often said that atmospheric pressure is 76cm of mercury column. What does it mean? Express this pressure in Pascal.
Take ρmercury = 13.6 × 10³ kgm^-3 and g = 9.8 ms^-2.
Answer.

The given statement means that the earth’s atmosphere exerts some pressure as exerted by a mercury column of height 76 cm.

∴ Atmospheric pressure = Pressure of 76 cm of mercury

= h ρ g

= 0.76 m × 13.6 × 10³ kgm^-3 × 9.8 ms^-2

= 1.013 × 10⁵ Pa

∴ Pressure in Pascal = 1.013 × 10⁵ Pa

Physics Gravitation Practice Problems 

Question 8. A ship having a total weight of 20,000 tonnes is floating on water. How much upthrust acts on the ship?
Take g = 10 ms^-2
Answer.

Ship is in equilibrium (i.e., at rest)

So, Upward force acting on the ship = Downward force acting on the ship.

U = W

= Mg

= \(20,000 \text { tonnes } \times 10^s \frac{\mathrm{kg}}{\text { tonne }} \times 10 \mathrm{~ms}^{-2}\)

= 2 × 10⁸ N

∴ Upthrust acts on the ship is 2 × 10⁸ N

Chapter 3 Gravitation Multiple Choice Question And Answers

Direction: Choose the correct option for each questions. There is only one correct response for each question.

Question 1. Choose the unit of relative density.

1. kg
2. G
3. Newton
4. None of the above

Answer. 4. None of the above

Question 2. The S.I. unit of thrust is ______.

1. Newton
2. m/s
3. Dyne
4. pascal

Answer. 1. Newton

Question 3. ______ is the quantity of matter contained in the body.

1. Weight
2. Mass
3. Acceleration
4. Force

Answer. 2. Mass

Question 4. The value of gravitational constant is ______

1. 6.6734 × 10^-11 N m²/kg²
2. 7.6734 × 10^-10 N m²/kg²
3. 8.6734 × 10^-11 N m/kg²
4. 9.6734 × 10^-11 N m²/kg

Answer. 1. 6.6734 × 10^-11 N m²/kg²

Question 5. When an object is thrown up, the force of  gravity ______.

1. Increases as it rises up
2. Becomes zero at the highest point
3. Is opposite to the direction of motion
4. Is in the same direction as the direction of motion

Answer. 3. Is opposite to the direction of motion

Read and Learn More NEET Foundation Multiple Choice Questions

Question 6. The acceleration due to gravity is zero at ______

1. Poles
2. Centre of earth
3. Equator
4. Sea level

Answer. 2. Centre of earth

Question 7. Mass of a body is ______ throughout the universe.

1. Constant
2. Variable
3. Zero
4. Negative

Answer. 1. Constant

Question 8. ______ is the thrust per unit area of surface.

1. Speed
2. Velocity
3. Pressure
4. Acceleration

Answer. 3. Pressure

Question 9. The gravitational force between two ­bodies is ______.

1. Always attractive
2. Always repulsive
3. Attractive only at large distance
4. Repulsive only at large distance

Answer. 1. Always attractive

Question 10. There is no atmosphere on the moon because

1. It revolves round the earth.
2. It is closer to the earth.
3. The escape velocity of moon is less than root mean square velocity of gas molecule here.
4. It is far away from the sun.

Answer. 3. The escape velocity of moon is less than root mean square velocity of gas molecule here.

Question 11. The upward force acting on an object submerged in a liquid is called ______.

1. Pressure
2. Force of friction
3. Thrust
4. Buoyant force

Answer. 4. Buoyant force

Question 12. Gravitational force is responsible for ______.

1. Keeping plants in their rad2
2. Keeping animals in their rad2.
3. Keeping plants on theft axes
4. For the motion of planets around the sun

Answer. 4. For the motion of planets around the sun

Question 13. The value of g decreases with ______.

1. Speed
2. Altitude
3. Acceleration
4. Velocity

Answer. 2. Altitude

Question 14. The distance from the centre of the earth to the centre of the moon is called as ______.

1. Orbital radius of the moon
2. Orbital length of the earth
3. Orbital radius of the earth
4. Orbital length of the moon

Answer. 1. Orbital radius of the moon

Question 15. The weight of an object in a satellite orbiting around the earth is ______.

1. Less than the actual weight
2. Greater than the actual weight
3. Zero
4. Actual weight

Answer. 3. Zero

Question 16. The acceleration of the moon is because of the ______

1. Gravitational force exerted by the planets.
2. Gravitational force exerted on the moon by the earth.
3. Gravitational force exerted on the earth by the moon.
4. Gravitational force exerted by the sun.

Answer. 2. Gravitational force exerted on the moon by the earth.

Question 17. Larger the volume of an object submerged in a fluid, greater is the ______.

1. Gravity
2. Upthrust.
3. Mass
4. Pressure

Answer. 2. Upthrust.

Question 18. Any substance that can flow is called ______.

1. Solid
2. Liquid
3. Fluid
4. Gas

Answer. 3. Fluid

Question 19. Buoyancy is denoted by the symbol ______.

1. F_A
2. F_B
3. F_C
4. F_D

Answer. 2. F_B

Question 20. Acceleration due to Gravity (g) on the ­surface of earth at poles is ______.

1. Maximum
2. Minimum
3. Zero
4. Same as that on equator

Answer. 1. Maximum

Question 21. The unit of (g) – acceleration due to Gravity, is ______.

1. N m/s²
2. N m/s
3. m/s²
4. kg m/s

Answer. 3. N m/s²

Question 22. In order to derive the law of gravitation, Newton assumed that the moon’s orbit is ______.

1. Straight
2. Parabolic
3. Uniform
4. Circular

Answer. 4. Circular

Question 23. The equation for the mass of the earth (ME) is ______.

1. M_E = R E² g/G
2. M_E = R E⁴ g/G
3. M_E = RG E² /g
4. M_E = R g E² /G

Answer. 1. M_E = R E² g/G

Question 24. If the density of an object is more than the density of the liquid, then it ______ in the liquid.

1. Floats
2. Sinks
3. Oscillates
4. Jumps

Answer. 2. Sinks

Question 25. A fluid contained in a vessel exerts pressure in ______.

1. All directions
2. Downwards
3. Upwards
4. Sideways

Answer. 1. All directions

Question 26. The S.I. unit of pressure is ______.

1. newton
2. dyne
3. pascal
4. metre

Answer. 3. pascal

Question 27. As per Newton, the force of attraction (F) between two particles is ______

1. Directly proportional to the product of their masses
2. Inversely proportional to the square of distance between the particles
3. Both are correct
4. None of the Above

Answer. 3. Both are correct

Question 28. The ______ is the measure of its inertia.

1. Mass
2. Velocity
3. Height
4. Distance

Answer. 1. Mass

Question 29. Choose the incorrect statement related to laws of liquid pressure

1. Pressure at a point inside the liquid increases with the depth from its free surface.
2. In a stationary liquid, pressure is same at all the points on a horizontal plane.
3. Pressure is different in all the directions about a point in liquid.
4. Pressure at same depth is in different liquids.

Answer. 3. Pressure is different in all the directions about a point in liquid.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 30. Pressure is a ______ quantity.

1. Scalar
2. Vector
3. Both
4. None of the above

Answer. 1. Scalar

Question 31. If mass is more, inertia will be ______.

1. Less
2. More
3. Equal
4. Independent

Answer. 2. More

Question 32. The force (F) on a body due to earth’s gravity is ______.

1. mg
2. mg/2
3. m/g
4. m/2g

Answer. 1. mg

Question 33. The value of g on moon’s surface is nearly ______.

1. one-third of the value of g on earth’s surface
2. one-fourth of the value of g on earth’s surface
3. one-sixth of the value of g on earth’s surface
4. one-eight of the value of g on earth’s surface

Answer. 3. one-sixth of the value of g on earth’s surface

Question 34. Force due to gravity acts ______.

1. Vertically downwards at the centre of gravity of the object
2. Vertically upwards at the centre of gravity of the object
3. in all the directions
4. None of the Above

Answer. 1. Vertically downwards at the centre of gravity of the object

Question 35. Archimedes’ principle is applicable for ______.

1. Solid
2. Liquid
3. Gas
4. All of the above

Answer. 

2. Liquid

3. Gas

Question 36. The magnitude of upthrust on an object due to liquid depends on the ______.

1. Volume of object submerged in the liquid
2. Density of the liquid in which the object is submerged
3. Volume of object above the liquid
4. Density of the solid

Answer.

1. Volume of object submerged in the liquid
2. Density of the liquid in which the object is submerged

Question 37. The pressure at a point inside a liquid depends on ______.

1. Depth of the point below the free surface
2. Density of liquid
3. Acceleration due to gravity
4. Surface area of liquid

Answer.

1. Depth of the point below the free surface
2. Density of liquid
3. Acceleration due to gravity

Question 38. The universal law of gravitation is

1. The force that binds us to the earth.
2. The motion of the moon around the earth.
3. The motion of planets around the sun.
4. The tides due to the moon and the sun.

Answer.  

1. The force that binds us to the earth.
2. The motion of the moon around the earth.
3. The motion of planets around the sun.
4. The tides due to the moon and the sun.

Question 39. The force of attraction (F) between two particles is ______.

1. Directly proportional to the product of their masses
2. Inversely proportional to the product of their masses
3. Directly proportional to the square of distance between them
4. Inversely proportional to the square of distance between them

Answer.

1. Directly proportional to the product of their masses

4. Inversely proportional to the square of distance between them

Question 40. Choose the correct statement

1. The weight of an object is the force with which the earth attracts the object.
2. The weight of an object is the force with which the earth repels the object.
3. Weight is a vector quantity.
4. Weight is a scalar quantity.

Answer.

1. The weight of an object is the force with which the earth attracts the object.

4. Weight is a scalar quantity.

Question 41. Choose the incorrect statement

1. Relative density of a substance is defined as the ratio of mass of the substance to the mass of an equal volume of water at 4°C.
2. Relative density (R.D.) of a substance is the ratio of the density of that substance to the density of water at 4°C.
3. Unit of relative density is Newton.
4. Unit of relative density is Dyne.

Answer.

3. Unit of relative density is Newton.

4. Unit of relative density is Dyne.

Question 42. Characteristic properties of upthrust are _____.

1. Larger the volume of object submerged in fluid, smaller is the upthrust.
2. Larger the volume of object submerged in fluid, greater is the upthrust.
3. For same volume inside the fluid more the density of fluid, greater is the upthrust.
4. The upthrust acts on an object in upward direction at the centre of buoyancy.

Answer.

2. Larger the volume of object submerged in fluid, greater is the upthrust.

3. For same volume inside the fluid more the density of fluid, greater is the upthrust.

4. The upthrust acts on an object in upward direction at the centre of buoyancy.

Question 43. Choose the incorrect statement

1. The normal force per unit area is called pressure.
2. The normal force per unit area is called thrust.
3. The normal force per unit area is called density.
4. The normal force per unit area is called upthrust.

Answer.

2. The normal force per unit area is called thrust.

3. The normal force per unit area is called density.

4. The normal force per unit area is called upthrust.

Question 44. Choose the correct statement

1. Pressure = Thrust/Area
2. Pressure is a scalar quantity
3. Pressure is a vector quantity
4. Pressure = Distance/Time

Answer. 

1. Pressure = Thrust/Area
2. Pressure is a scalar quantity

Question 45. The mass of earth is 6 × 10²⁴ kg and radius of earth is 6.4 × 10⁶ m. The magnitude of force between the mass of 1 kg and the earth is:

1. 9.770 N
2. 9.810 N
3. 9.830 N
4. 9.790 N

Answer. 1. 9.770 N

Question 46. With the increase in the density of a fluid, the upthrust experienced by a body immersed in it is

1. decreases
2. increases
3. remains same
4. none of these

Answer. 2. increases

Question 47. The acceleration due to gravity on earth is 9.81  ms^-2. If the acceleration due to gravity on the surface of moon is 1/6 of the earth, the weight of 10 kg mass on moon is:

1. 1.67 N
2. 16.35 N
3. 17.50 N
4. None of these

Answer. 2. 16.35 N

Question 48. The apparent weight of a body in a fluid is:

1. equal to weight of fluid displaced
2. volume of fluid displaced
3. difference between its weight in air and weight of fluid displaced
4. none of the above

Answer. 3. difference between its weight in air and weight of fluid displaced

Question 49. The average distance between the earth and moon is 3.84 × 10⁵ km. If the mass of moon is 7.4 × 10²² kg and that of earth is 6 × 10²⁴ kg, the gravitational force of the moon on the earth is:

1. 2.01 × 10²⁰ N
2. 20.1 × 10²¹ N
3. 20.1 × 10²⁰ N
4. none of these

Answer. 1. 2.01 × 10²⁰ N

Question 50. The phenomenon due to which a solid expereinces upward force when immersed in water is called:

1. floatation
2. buoyancy
3. density
4. none of these

Answer. 2. buoyancy

Question 51. The mass of earth is 6 × 10²⁴ kg, radius 6.4 × 10⁶ m and G = 6.7 × 10^-11 Nm² kg^-2. The acceleration due to gravity on the surface of earth is:

1. 9.70 ms^-2
2. 9.78 ms^-2
3. 9.89 ms^-2
4. 9.80 ms^-2

Answer. 4. 9.80 ms^-2

Question 52. The radius of earth is 6400 km and value of ‘g’ at its surface is 9.8 ms^-2. The value of ‘g’ at a height of 12,800 km will be:

1. 5.07 ms^-2
2. 1.09 ms^-2
3. 3.09 ms^-2
4. 4.08 ms^-2

Answer. 2. 1.09 ms^-2

Question 53. If ‘g’ = 9.8 ms^-2, the weight of body of mass 10 kg is:

1. 138 N
2. 158 N
3. 98 N
4. 78 N

Answer. 3. 98 N

Question 54. Imagine a heavenly body which has a mass twice of the earth and radius thrice of the earth. If a stone on earth weighs 900 N, its weight on the heavenly is:

1. 200 N
2. 600 N
3. 400 N
4. 500 N

Answer. 1. 200 N

Question 55. Two neutrons of mass 1.67 × 10^-27 kg are at distance of 10^-15 m from each other. The gravitational force of attraction between them is:

1. 4.86 × 10^-34 N
2. 5.86 × 10^-34 N
3. 1.86 × 10^-34 N
4. 2.86 × 10^-34 N

Answer. 3. 1.86 × 10^-34 N

Question 56. A spaceship is orbiting around the earth at height twice the radius of earth. If ‘g’ on earth is 9.8 ms^-2, the ‘g’ at the height is:

1. 2.1 ms^-2
2. 5.1 ms^-2
3. 3.1 ms^-2
4. 1.1 ms^-2

Answer. 4. 1.1 ms^-2

Question 57. The mass and radius of earth is 6 × 10²⁴ kg and 6400 km. If another heavenly body has half the mass and the radius of earth, the value of g on it will be:

1. 4.96 ms^-2
2. 19.6 ms^-2
3. 29.6 ms^-2
4. 39.6 ms^-2

Answer. 2. 19.6 ms^-2

Question 58. The radius of moon is 0.27 times the radius of earth, whereas of g_moon is 1.27 ms^-2 and g_earth is 9.8 ms^-2. The ratio of mass of earth to the mass of moon is:

1. 46 times
2. 58 times
3. 79 times
4. 67 times

Answer. 3. 79 times

Question 59. The weight of body on the surface of earth is 90  kgf. Its weight on the surface of another planet of mass 1/9 and radius half on the earth will be:

1. 40 kgf
2. 30 kgf
3. 35 kgf
4. 50 kgf

Answer. 1. 40 kgf

Question 60. When an object sinks in a liquid its:

1. buoyant force is more than the weight of object
2. buoyant force is less than the weight of object
3. buoyant force is equal to the weight of object
4. none of the above

Answer. 2. buoyant force is less than the weight of object

Question 61. If the diameter of earth becomes 3 times its present value, but mass does not change, the weight of an object will decrease to:

1. \(\frac{1}{3} \text { th }\)
2. \(\frac{1}{6} \text { th }\)
3. \(\frac{1}{8} \text { th }\)
4. \(\frac{1}{9} \text { th }\)

Answer. 4. \(\frac{1}{9} \text { th }\)

Question 62. A glass stopper suspended form the hook of a spring balance and immersed completely in water reads Wgf. If the same stopper is immersed in alcohol (density 0.80 gcm^-3), the reading on the spring balance will be:

1. less than Wgf
2. more than Wgf
3. Wgf
4. none of these

Answer. 2. more than Wgf

Question 63. The value of ‘g’ on the top of Mount Everest (height 8848 m) will be:

1. 9.77 ms^-2
2. 6.77 ms^-2
3. 6.67 ms^-2
4. 5.77 ms^-2

Answer. 1. 9.77 ms^-2

Question 64. The radius of earth is 6400 km. The value of ‘g’ will be 60% at the height of:

1. 5.76 × 10³ km
2. 1.86 × 10³ km
3. 1.47 × 10³ km
4. 3.86 × 10³ km

Answer. 2. 1.86 × 10³ km

Question 65. If radius of earth is R, at what height from the surface of earth the weight becomes half:

1. 0.414
2. 0.518 R
3. 0.514 R
4. 0.314 R

Answer. 1. 0.414

Question 66. Thrust acting perpendicularly on the unit surface area is called:

1. pressure
2. moment of force
3. down thrust
4. none of these

Answer. 1. pressure

Question 67. The pressure exerted on a given surface is:

1. directly proportional to the force acting on the surface
2. directly proportional to area of crosssection of surface
3. inversely proportional to area of crosssection of surface
4. both (a) and (c)

Answer. 4. both (a) and (c)

Question 68. Your weight recorded by an accurate weighing machine is 42 kgf. In actual practice your weight is:

1. more than 42 kgf
2. less than 42 kgf
3. 42 kgf
4. none of these

Answer. 1. more than 42 kgf

Question 69. One pascal is equal to:

1. N cm^-2
2. Nm^-2
3. Nm2
4. Nm^-1

Answer. 2. Nm^-2

Question 70. Tank trailors are provided with 16 wheels so as to:

1. increase pressure on the road
2. decrease pressure on the road
3. support the weight of the tank
4. none of the above

Answer. 2. decrease pressure on the road

Question 71. Pressure applied on liquids is transmitted with undiminished force:

1. in downward direction only
2. upward direction only
3. sides of containing vessel
4. in all directions

Answer. 4. in all directions

Question 72. The knives are often sharpened:

1. to decrease the area of cutting edge and hence increasing pressure
2. to decrease the area of cutting edge and hence decreasing pressure
3. to give it a shine
4. none of the above

Answer. 1. to decrease the area of cutting edge and hence increasing pressure

Question 73. The strap of a school bag is not made of thin string, because:

1. string is likely to break
2. string has less area of cross-section and hence exerts less pressure on shoulder
3. string has less area of cross-section and hence exerts more pressure on shoulder
4. none of the above

Answer. 3. string has less area of cross-section and hence exerts more pressure on shoulder

Question 74. When a body is wholly or partially immersed in a liquid it experiences a buoyant force which is equal to the:

1. volume of liquid displaced by it
2. weight of liquid displaced by it
3. both (a) and (b)
4. none of the (a) or (b)

Answer. 2. weight of liquid displaced by it

Question 75. The relative density of zinc metal is 4.2. Its density in SI system is:

1. 4.2 kg m^-3
2. 42 kg m^-3
3. 420 kg m^-3
4. 4200 kg m^-3

Answer. 4. 4200 kg m^-3

Question 76. Which is incorrect statement?

1. tides are formed in the sea due to gravitational pull of the sun and moon.
2. earth holds atmosphere on account of gravitational pull of earth.
3. hydrogen filled balloons rise up because, no gravitational pull acts on it
4. gravitational pull of the earth keeps us and other objects firmly on ground

Answer. 3. hydrogen filled balloons rise up because, no gravitational pull acts on it

Question 77. Two similar sheets of a paper Aand B are taken. The sheet B is crumpeld to form a ball. Both A and B are allowed to fall from a height of 10 m.

1. both A and B will reach ground at the same time.
2. A will reach earlier than B.
3. B will reach earlier than A
4. none of the above

Answer. 3. B will reach earlier than A

Question 78. A stone is released from the top of tower 19.6m high. If ‘g’ is 9.8 ms^-2, the final velocity of the stone on reaching the ground is:

1. 9.8 ms^-1
2. 28.4 ms^-1
3. 19.6 ms^-1
4. none of these

Answer. 3. 19.6 ms^-1

Question 79. A stone is thrown vertically upward with an initial velocity of 40 ms^-1. Taking g = 10 ms^-2, the maximum displacement and distance covered by the stone on reaching ground is:

1. displacement: zero; distance = 80 m
2. displacement: zero; distance covered = 160 m
3. displacement = 80 m; distance = 160
4. distance and displacement = 180 m.

Answer. 2. displacement: zero; distance covered = 160 m

Question 80. A cricket ball is vertically thrown upward, which rises to the height of 10 m. If g = 10 ms^-2, the time in which it rises up to the maximum height is:

1. 1.53 s
2. 1.63 s
3. 1.55 s
4. 1.43 s

Answer. 1. 1.53 s

Question 81. A plastic ball released under water comes up. It is because the weight of the plastic ball is:

1. equal to the weight of water displaced by it.
2. less than the weight of water displaced by it.
3. more than the weight of water displaced by it.
4. none of the above

Answer. 2. less than the weight of water displaced by it.

Question 82. A body has density 9.6 gcm^-3. Its density in SI system is:

1. 96 kgm^-3
2. 960 kgm^-3
3. 9600 kgm^-3
4. 96,000 kgm^-3

Answer. 3. 9600 kgm^-3

Question 83. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upward from the ground with a velocity of 25 ms^-1. If g = 10 ms^-2, the stones will meet at:

1. 25 m from ground after 5 s (g = 10 ms–2)
2. 20 m from ground after 4 s
3. 20 m from ground after 4.5 s
4. 20 m from ground after 3.8 s

Answer. 2. 20 m from ground after 4 s

Question 84. The SI unit of relative density is:

1. kgm^-3
2. gcm^-3
3. kgcm^-3
4. no unit

Answer. 4. no unit

Question 85. A stone thrown vertically upward returns back to the thrower in 6 sec. The maximum height attained by the stone is:

1. 25 m
2. 35 m
3. 40 m
4. 45 m

Answer. 4. 45 m

Question 86. The pressure exerted by 50 kg (g = 10 ms^-2) on an area of cross-section of 2 m² is:

1. 50 Pa
2. 200 Pa
3. 250 Pa
4. 1000 Pa

Answer. 3. 250 Pa

Question 87. A truck is of mass 50,000 kg (g = 10 ms^-2), its tyres exert a pressure of 2,500,000 Pa. The surface area of tyres in contact with ground is:

1. 2 m²
2. 0.2 m²
3. 2.5 m²
4. 2.75 m²

Answer. 2. 0.2 m²

Question 88. The force experienced by a body when partially or fully immersed in water is called:

1. apparent weight
2. upthrust
3. down thrust
4. none of these

Answer. 2. upthrust

Question 89. When a body is floating in a liquid:

1. the weight of body is less than the upthrust due to immersed part of body
2. the weight of body is more than the upthrust due to immersed part of body
3. the weight of body is equal to the upthrust due to immersed part of body
4. none of the above

Answer. 3. the weight of body is equal to the upthrust due to immersed part of body

Chapter 3 Gravitation Free Fall

Question 1. ______ is a natural phenomenon in which two objects having masses attract towards each other.

1. Gravitation
2. Speed
3. Velocity
4. Acceleration

Answer. 1. Gravitation

Question 2. The S.I. unit of G is ______.

1. m/s
2. Nm²/kg²
3. N
4. Dyne

Answer. 2. Nm²/kg²

Question 3. An object weighs 10 N when measured on the surface of earth. What should be its weight on moon?

1. 1.67 N
2. 4.18 N
3. 2.20 N
4. 16.7 N

Answer. 1. 1.67 N

Question 4. According to Newton, the force of attraction (F) between two particles is:

1. Inversely proportional to the product of their masses
2. Directly proportional to the their masses
3. Inversely proportional to the their masses
4. Directly proportional to the product of their masses

Answer. 4. Directly proportional to the product of their masses

Question 5. ______ between two bodies forms action reaction pair.

1. Kinetic energy
2. Acceleration
3. Gravitational force
4. Potential energy

Answer. 3. Gravitational force

Chapter 3 Gravitation Mass

Question 1. The force with which the earth ______ an object is called force due to gravity on the object.

1. attracts
2. repel
3. forms
4. moves

Answer. 1. attracts

Question 2. The value of g is maximum

1. at equator of earth
2. at poles of earth
3. in a mine
4. at a high hill

Answer. 2. at poles of earth

Question 3. Acceleration due to gravity ______ with depth below the surface of the earth.

1. increases
2. decreases
3. doubles
4. becomes half

Answer. 2. decreases

Question 4. A ball is thrown vertically upwards and attains a maximum height of 100  m. It was thrown with a speed.

1. 9.8 m/s
2. 44.3 m/s
3. 19.6 m/s
4. 98 m/s

Answer. 2. 44.3 m/s

Question 5. The value of g is _______ on different planets and satellites

1. different
2. same
3. equivalent
4. None of the above

Answer. 1. different

Chapter 3 Gravitation Thrust and Pressure

Question 1. Relative density has _________ unit.
Answer. No

Question 2. The relative density of water is

1. Zero
2. One
3. Two
4. Can’t say

Answer. 2. One

Question 3. If the relative density of an object is less than that of water, then it will

1. Rise
2. Float
3. Sink
4. Can’t say

Answer. 2. Float

Question 4. Archimedes’ principle is applicable to the objects in fluids. (True/False)
Answer. True

Chapter 3 Gravitation Practice Exercises

Question 1. Two objects of different masses falling freely near the surface of moon would

1. have same velocities at any instant
2. have different accelerations
3. experience forces of same magnitude
4. undergo a change in their inertia

Answer. 1. have same velocities at any instant

Question 2. The value of acceleration due to gravity

1. is same on equator and poles
2. is least on poles
3. is least on equator
4. increases from pole to equator

Answer. 3. is least on equator

Question 3. The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become

1. F/4
2. F/2
3. F
4. 2 F

Answer. 1. F/4

Question 4. A boy is whirling a stone tied with a string in an horizontal circular path. If the string breaks, the stone

1. will continue to move in the circular path
2. will move along a straight line towards the centre of the circular path
3. will move along a straight line tangential to the circular path
4. will move along a straight line perpendicular to the circular path away from the boy

Answer. 3. will move along a straight line tangential to the circular path

Question 5. An object is put one by one in three liquids having different densities. The object floats with 1/9 2/11 and 3/7 , and parts of their volumes outside the liquid surface in liquids of densities d₁, d₂ and d₃ respectively. Which of the following statement is correct?

1. d₁ > d₂ > d₃
2. d₁ > d₂ < d₃
3. d₁ < d₂ > d₃
4. d₁ < d₂ < d₃

Answer. 4. d₁ < d₂ < d₃

Question 6. In the relation F = GMm/d², the quantity G

1. depends on the value of g at the place of observation
2. is used only when the earth is one of the two masses
3. is greatest at the surface of the earth
4. is universal constant of nature

Answer. 4. is universal constant of nature

Question 7. Law of gravitation gives the gravitational force between

1. the earth and a point mass only
2. the earth and Sun only
3. any two bodies having some mass
4. two charged bodies only

Answer. 3. any two bodies having some mass

Question 8. The value of quantity G in the law of gravitation

1. depends on mass of earth only
2. depends on radius of earth only
3. depends on both mass and radius of earth
4. is independent of mass and radius of the earth

Answer. 4. is independent of mass and radius of the earth

Question 9. Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be

1. \(\frac{1}{4} \text { times }\)
2. 4 times
3. \(\frac{1}{2} \text { times }\)
4. unchanged

Answer. 2. 4 times

Question 10. The atmosphere is held to the earth by

1. gravity
2. wind
3. clouds
4. earth’s magnetic field

Answer. 1. gravity

Question 11. The force of attraction between two unit point masses separated by a unit distance is called

1. gravitational potential
2. acceleration due to gravity
3. gravitational field
4. universal gravitational constant

Answer. 4. universal gravitational constant

Question 12. The weight of an object at the centre of the earth of radius R is

1. zero
2. infinite
3. R times the weight at the surface of the earth
4. 1/R2 times the weight at surface of the earth

Answer. 1. zero

Question 13. An object weighs 10 N in air. When immersed fully in water, it weighs only 8 N. The weight of the liquid displaced by the object will be

1. 2 N
2. 8 N
3. 10 N
4. 12 N

Answer. 1. 2 N

Question 14. A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. In which of the following cases, pressure exerted by the brick will be

1. maximum when length and breadth form the base
2. maximum when breadth and width form the base
3. maximum when width and length form the base
4. the same in all the above three cases

Answer. 2. maximum when breadth and width form the base

Question 15. An apple falls from a tree because of gravitational attraction between the earth and apple. If F₁ is the magnitude of force exerted by the earth on the apple and F₂ is the magnitude of force exerted by apple on earth, then

1. F₁ is very much greater than F₂
2. F₂ is very much greater than F₁
3. F₁ is only a little greater than F₂
4. F₁ and F₂ are equal

Answer. 4. F₁ and F₂ are equal

Chapter 3 Gravitation

Gravitation

In our day to day life, we see that objects get attracted towards earth. When we throw a ball up in the air, it comes back towards earth. It means there is some force that compels objects to return back towards earth. This force is known as gravitational force. Gravitational force is a natural phenomenon where all objects having energy attract towards each other.

For example, stars, planet, sun, etc. Gravity is considered as the weakest among the four fundamental interactions of nature. The four fundamental interactions (fundamental forces) are gravitational, electromagnetic, strong nuclear and weak nuclear.

Due to gravitational force, earth and other planets are revolving around the sun in different orbit. In this chapter, we will understand about the gravitational force and its details.

Chapter 3 Gravitation

Gravitation is a natural phenomenon in which two objects having masses attract each other. We have studied that the moon revolves around the earth and the earth revolves around the sun. There are also other planets other than earth that rotate around the sun. The reason why planets revolve around the sun and why does not it go straight towards the sun is due to gravitational force.

Read and Learn More: NEET Foundation Notes

In 628, Brahmagupta, an Indian astronomer recognised gravity as a force of attraction. In 1600’s, Galileo Galilei, Robert Hooke and Johannes Kepler formulated the laws of gravity near the earth.

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While studying about the orbits of planets around the sun, Kepler determined that gravitational attraction varied with the separation and thus formulated his laws of planetary motion.

In 1687, Isaac Newton worked on planetary motion and its empirical measurements. His study established the law of universal gravitation.

Universal Law of Gravitation

As we know, each particle of universe attracts the other particle. The force of attraction between two particles due to their masses is called gravitational force of attraction. Sir Issac Newton gave the law ‘Newton’s universal law of gravitation’ for its magnitude.

According to Newton, gravitational force of attraction (F) between two particles is:

1. Directly proportional to the product of their masses
   i.e., F ∝ m₁m₂
2. Inversely proportional to square of the distance between them
   i.e.,  \(F \propto \frac{1}{r^2}\)

Combining both the equations,

\(F \propto \frac{m_1 m_2}{r^2}\)

\(F=G \frac{m_1 m_2}{r^2}\)

where G = constant of proportionality known as gravitational constant.

The value of G is same at all places and is not dependent on the nature of particles, temperature, medium and so on. Therefore, it is a universal constant and is therefore known as “universal gravitational constant”.

Unit and Value of Gravitational Constant

G = \(\frac{F \times r^2}{m_1 \times m_2}\)

\(\text { S.I. unit of } G=\frac{\text { Newton } \times \text { metre }^2}{\text { kilogram } \times \text { kilogram }}\)

= Nm2/kg2

Value of G = 6.67 × 10-11 N m²/kg²

If m₁ = 1 kg, m₂ = 1 kg, r = 1 m, then G = F

Therefore, G can be defined as:

Gravitational constant ‘G’ is numerically equal to the gravitational force of attraction ­between two masses, each of mass 1 kg placed at a distance of 1 m.

Gravitational force between two masses is:

1. Always attractive.
2. Directly proportional to the product of the masses.
3. Inversely proportional to the square of separation between them.
4. Significant between heavenly bodies but is insignificant between ordinary bodies because of small magnitude of G.

Kepler’s Law of Planetary Motion

Newton’s law of motion which state’s that the gravitational force between two masses varies inversely proportional to the square of the distance between them was proved by him on the basis of Kepler’s law of planetary motion.

There are three laws which explain the motion of planets. These laws are called Kepler’s law. They are:

1. Law of orbit: This is the 1^st law of planetary motion which states that each planet moves around sun in an elliptical orbit with the sun at one of the foci of the orbit. This can be shown by the below image.

2. Law of areas: This is the 2^nd law of Kepler which states that the line joining the sun and planet sweep out equal areas in equal interval of time. It means velocity of the planet around the sun is not constant. In the figure below the sun is at S, which is one of the foci of the elliptical orbit. If the time of travel of planet from P₁ to P₂ is same as that of P₃ to P₄, than according to this law, Area of OP₁P₂ = Area of OP₃P₄

Therefore arc P₁P₂ is smaller than arc P₃P₄. This means that the speed of the planet is greater when it is closer to the sun, than its speed when it is farther away from the sun.

3. Law of periods: The square of the time taken by a planet to complete a revolution around the sun is directly proportional to the cube of semi major axis of the elliptical orbit. This is the third law of Kepler’s planetary motion.

i.e., T² ∝ r³

Or T² = constant × r³

i.e., T²/r³ = constant

Importance of Universal Law of Gravitation

The universal law of gravitation is important because it is:

1. The force that binds us to the earth
2. The motion of the moon around the earth
3. The motion of planets around the Sun
4. The tides due to the moon and the Sun

Characteristics of Gravitational Force

1. Gravitational force acts at distance. It means it does not need any contact between the two objects or bodies.
2. Gravitational force between the two bodies varies inversely proportional to the square of the distance between them
3. Gravitational force between two bodies forms action reaction pair.

Chapter 3 Gravitation Free Fall

Any object that falls towards the earth is said to be in free fall. This happens due to gravitational force. On falling there is no change in the direction of motion of the objects but because of the earth’s attraction, there will be a change in magnitude of the velocity. Change in velocity causes acceleration.

Motion of Objects Under the Influence of Gravitational Force of the Earth

The acceleration of a vertically falling object (a) = +g

The acceleration of an object vertically thrown up (a) = –g

If an object falls from rest freely under gravity from a height h, u = 0 and acceleration a is replaced by g then

v = gt

s = \(\frac{1}{2} g t^2\)

v² = 2gh

If a body is thrown vertically downwards with an initial velocity u, then

v = u + gt

s = \(u t+\frac{1}{2} g t^2\)

v² = u² + 2gh

If an object is thrown vertically up with an initial velocity u, there will be retardation i.e. a = –g. Then, equations will be

v = u – gt

s = \(u t-\frac{1}{2} g t^2\)

v² = u² – 2gh

When the thrown object reaches its highest point, its final velocity v = 0, therefore

Maximum height hmax = \(\frac{u^2}{2 g}\)

Time taken to reach the highest point t = \(\frac{u}{g}\)

Object comes back to the earth in the same time t, therefore total journey time t’ = 2t = \(\frac{2 u}{g}\)

Total distance travelled by the object, h’ = 2 hmax = \(\frac{u^2}{g}\)

Force Due to Gravity

The force with which the earth attracts an object is called force due to gravity on the object. It always acts vertically downwards at the centre of gravity of the object. In vertical motion near the earth surface, the force of gravity of an object is assumed to be same throughout, assuming that the change in height of the object is much smaller than the radius of earth.

The force due to gravity on an object (with mass m) kept on the surface of the earth (mass is M and radius is R) is equal to the force of attraction between the earth and the object.

Equation

F = \(\frac{G M m}{R^2}\)

Let us assume M = 5.96 × 10²⁴ kg

R = 6.37 × 10⁶ m

m = 1 kg

F = \(\frac{\left(6.67 \times 10^{-11}\right) \times\left(5.96 \times 10^{24}\right) \times 1}{\left(6.37 \times 10^6\right)^2}\)

= 9.8 N

This shows that earth attracts an object of mass 1 kg by a force of 9.8 N towards its centre.

Gravitational Force between Earth and the Sun

Mass of earth, m₁ = 6 × 10²⁴ kg

Mass of sun, m₂ = 2 × 10³⁰ kg

Distance between sun and earth, R = 1.5 × 10¹¹ m

Gravitational force between the sun and the earth,

F = \(G \frac{m_1 m_2}{r^2}\)

F = (6.67 × 10^-11 Nm² kg^-2 × 6 × 10²⁴ kg × 2 × 10³⁰ kg)/(1.5 × 10¹¹ m)²

F = 3.6 × 10²² N

Gravitational Force between the Moon and the Earth

Mass of earth, m₁ = 6 × 10²⁴ kg

Mass of moon, m₂ = 7.4 × 10²² kg

Gravitational force between earth and the moon is

F = \(G \frac{m_1 m_2}{r^2}\)

F = (6.67 × 10^-11 Nm² kg^-2 × 6 × 10²⁴ kg × 7.4 × 10²² kg)/(3.8 × 10⁸ m)²

F = 2.05 × 10²⁰ N

As an object gets attracted towards the earth, similarly the earth too gets attracted ­towards the object. An object is free to move so it starts moving towards the earth but the earth has higher inertia therefore it does not move towards the object.

For example, when a ball is thrown from a top of a building, due to gravitational force it falls on the ground. Earth has higher inertia than the ball therefore the earth doesn’t move from its location.

Centre of Mass and Centre of Gravity

Centre of Mass is the point in body where its whole mass is assumed to be concentrated. The centre of mass of a homogeneous sphere or cube must lie at its geometrical centre. Each particle or portion of the body experiences the force of gravity. The net effect of all these forces is equivalent to the effect of a single force, mg acting through a point called centre of gravity of a body.

On the surface of the earth, or near it, where the force of gravity is constant, the centre of mass also becomes the centre of gravity.

Application of Newton’s Law of Gravitation

1. Determination of masses of planets and stars: Knowing the values of g, R and G, it is possible to determine accurately the mass of any planet or star by using relationship,

\(M=\frac{g R^2}{G}\)

2. Estimating the masses of double stars: A double star is a system consisting of two stars orbiting around their common centre of masses. From the extent of irregularity in the motion of star due to gravitational pull by some other star bound to it is used for estimating the masses of star. Such irregularity in motion is called wobble.

Acceleration due to Gravity

The scientist Galileo did analysis on the motion of different objects under the force of attraction of the earth. From his experiments, he concluded that if objects of different sizes, masses fall at the same time and from the same height in vacuum, then they will reach the earth together. It shows that objects travel the same distance in same interval of time.

He also concluded that the velocity of a freely falling object does not remain constant instead it increases at a constant rate. In other words, the motion of a freely falling object is a uniformly accelerated motion. This acceleration is same for all objects.

The acceleration of a freely falling object does not depend on the mass, size and shape of the object. This acceleration is known as acceleration due to gravity.

The rate at which the velocity ofthe freely falling objectincreases is called acceleration due to gravity. It is a vector quantity.

It is represented by g.

Unit

The S.I. unit of acceleration due to gravity is m/s².

The value of g changes from place to place. On equator, it is slightly less as compared to that at poles. At altitudes above the earth’s surface or at depth below the earth surface value of g decreases. The value of g is zero at the centre of the earth. The mean value of g on the earth surface is taken to be 9.8 m/s². The value of acceleration due to gravity of the earth is 9.8 m/s²

Relationship between g and G

Let g = acceleration due to gravity at a planet

M = mass

R = radius

m = mass of object

By Newton’s law of motion, force on a body due to gravity on its surface

F = mass × acceleration due to gravity

= m g

By Newton’s gravitational law, force is

F = \(\frac{G M m}{R^2}\)

Therefore,

\(\frac{G M m}{R^2}=m g\)

∴ Acceleration due to gravity

g = \(\frac{G M}{R^2}\)

i.e. acceleration due to gravity on the surface of a planet depends on its mass (M) and its radius (R).

Variation in the Value of g with Depth below the Surface of the Earth

Let us consider a body of mass m at a depth h below the surface of earth. Than radius of the inner solid sphere of the earth = R – h

Volume of the inner solid sphere of the earth = \(\frac{4}{3} \pi(R-h)^3\)

If d is the average density of the earth, then

Mass of inner solid sphere of earth = \(\frac{4}{3} \pi(R-h)^3 d\)

According to the law of gravitation,

\(m g_d=G \times \frac{4}{3} \pi(R-h)^5 d \times \frac{m}{(R-h)^2}\)

This gives

\(g_{\mathrm{d}}=G \times \frac{4}{3} \pi(R-h) d\)

On the surface of earth,

g = \(\frac{G M}{R^2}\)

= \(G \times \frac{4}{3} \pi R^3 \frac{d}{R^2}\)

= \(G \times \frac{4}{3} \pi R d\)

From equation (1) and (2)

\(\frac{g_d}{g}=\frac{\left(G \times \frac{4}{3} \pi(R-h) d\right)}{\left(G \times \frac{4}{3} \pi R d\right)}\)

= \(\frac{(R-h)}{R}\)

\(g_{\mathrm{d}}=g\left(\frac{1-h}{R}\right) \quad \text { or } \quad \frac{(R-h)}{R}<1\)

So, gd < g

Thus, the value of g at a depth inside the earth is less than that on the surface of the earth.

The value \(\frac{(R-h)}{R}\) decreases with the value of h, i.e., depth below the surface of earth. So the value of g decreases as we go down below the surface of earth.

Chapter 3 Gravitation Mass

The mass of an object is the quantity of matter it contains. In other words, it is the measure of its inertia. It is a scalar quantity. The S.I. unit of mass is kilogram (kg). The mass of an object is constant, no matter if it is on earth or any other planet. If mass of a body is more, then its inertia will be more (Fig. 3.5).

Chapter 3 Gravitation Weight

The weight of an object is the force with which the earth attracts it towards its centre. Weight is a vector quantity. The S.I. unit of weight is Newton (N) and its C.G.S. unit is dyne 1 N = 105 dyne The weight of an object is also known as the force of gravity on that object.

Weight of an Object on the Moon

Weight of an object on the earth is the force with which the earth attracts the object. Similarly, the weight of an object on the moon is the force with which the moon attracts the object. Moon exerts less force of attraction than the earth because the mass of moon is lesser than the earth.

Mass of object = m

Weight of mass on moon = W_m

Mass of moon = M_m

Radius = R_m

As per universal law of gravitation, the weight of the object on the moon

Wm = \(G \frac{M_m m}{R_m{ }^2}\) (1)

Let the weight of same object on earth = W_e

Mass of earth = M

Radius of earth = R

Then We = \(G \frac{M m}{R^2}\) (2)

On substituting the known value in Eq (1) and (2), and comparing them, we find that

Weight of object on moon = 1/6 Weight of object on earth

Relationship between Weight and Mass

Weight = mass × Acceleration due to gravity

W = m × g

The gravitational unit of weight in M.K.S. system = Kilogram force (kgf)

C.G.S. unit is gramme force (gf)

1 kgf = 9.8 N

980 Dyne = 9.8 × 10^-3 N

Chapter 3 Gravitation Thrust and Pressure

Thrust

A force can be applied on a surface in any direction. If a force is applied on a surface in a direction normal or perpendicular to the surface, it is called thrust. The effect of thrust depends on the area of surface on which it acts. The effect of a thrust is less on a large area and vice versa.

Thrust exerted by an object on a surface is the force that object exerts on the surface which is equal to the weight of that object. Thrust is a vector quantity.

Unit

The S.I. unit of thrust = Newton (N)

C.G.S. unit of thrust = Dyne

Gravitational unit of thrust in M.K.S. system = kgf

Gravitational unit of thrust in C.G.S. system = gf

1 kgf = 9.8 N

1 gf = 980 dyne

Pressure

Pressure is the thrust per unit area of the surface. Pressure is a scalar quantity.

Pressure = \(\frac{\text { thrust }}{\text { Area }}\)

\(P=\frac{F}{A}\)

Factors Affecting the Pressure

The pressure exerted on a surface depends on two factors

1. Thrust
2. Area on which the thrust is applied

Unit

The S.I. unit of pressure = \(\frac{\text { Newton }}{\text { metre }^2}=\frac{\mathrm{N}}{\mathrm{m}^2}=\text { pascal }\)

Symbol of pascal = Pa

C.G.S. unit = dyne/cm²

1 dyne/cm² = 0.1 N/m²

1 N/m² = 10 dyne/cm²

1 bar = 105 N/m²

1 millibar = 10^-3 bar = 10² N/m²

Pressure in Fluids

Any substance that can flow is called a fluid. Therefore, all liquids and gases are fluids. As we know solid exerts pressure on a surface due to its weight. Similarly, a fluid exerts pressure due to its weight.

We have studied that a solid exerts pressure on the surface but liquid exerts pressure at all points and in all directions. It exerts pressure at the bottom and also on the sides of the container.

Factors Affecting the Pressure at a Point in a Liquid

The pressure at a point inside the liquid depends directly on the following factors:

1. Depth of the point below the free surface (h)
2. Density of liquid (d)
3. Acceleration due to gravity (g) at that place.

So, pressure at a point due to a fluid is given by

P = h × d × g

Laws of Liquid Pressure

1. Pressure at a point inside liquid increases with the depth from its free surface.
2. In a stationary liquid, pressure is same at all the points on a horizontal plane.
3. Pressure is same in all the directions about a point in liquid.
4. Pressure at same depth is different in different liquids. It increases with the increase in density of liquid.
5. A liquid seeks its own level.

Buoyancy

When an object is partially or completely immersed in water, an upward force acts on it. This upward force is known as buoyancy. Another name of this force is buoyant force or upthrust. It is denoted by the symbol FB.

The property of a fluid to exert an upward force on a body immersed in it is called buoyancy.

Why Objects Float or Sink when Placed on the Surface of Water?

To understand it let us take an example. Take an empty closed bottle and dip it in a vessel filled with water. You will notice that most of the part of the bottle is floating above the water level. Now push the bottle, an upwardforce will be felt. As we try to push the bottle inside water, we need to apply more and more force to dip it completely inside the water.

Even after it is completely dipped in water, force is required to let it remain there. As soon as we remove the force, the bottle will come up again and starts floating. This is due to upthrust of water.

Characteristic Properties of Upthrust

1. Larger the volume of an object submerged in fluid, greater is the upthrust.
2. For same volume inside the fluid, more the density of fluid, greater is the upthrust.
3. The upthrust acts on an object in upward direction at the centre of buoyancy.

Factors Affecting the Upthrust

The magnitude of upthrust on an object due to liquid depends on the following factors

1. Volume of object submerged in the liquid (V).
2. Density of the liquid in which the object is submerged (dl) and
3. Acceleration due to gravity (g) at that place.

Chapter 3 Gravitation Archimedes’ Principle

When an object is immersed in a liquid, it occupies the space that was earlier occupied by the liquid. Volume of liquid displaced by the object = Volume of submerged part of object

Therefore, the object experiences an upthrust which is equal to the weight of the liquid displaced.

Archimedes’s Principle States

When an object is immersed in a liquid partially or completely, it experiences an upthrust, which is equal to the weight of liquid displaced by it.

Illustration

In the a metallic piece is suspended with the help of a thread, from the hook of a spring balance. Note its weight. Now, take a vessel and fill it with water. Keep a measuring cylinder next to the vessel. Now dip the metallic piece inside the water.

Some water will get collected in the measuring cylinder. Now note the weight of the solid and also measure the volume of water in the measuring cylinder.

The density of water is 1 g/cm³, the volume of collected water in the measuring ­cylinder is equal to the mass in g or weight in gf of the water displaced by the solid when solid is completely dipped inside water.

It is found that the weight of water displaced by the solid is equal to the loss in weight of the solid in water. Loss in weight of the body is due to upthrust. So, upthrust is equal to weight of water displaced by the body. This verifies Archimedes’ Principle.

Chapter 3 Gravitation Relative Density

We have already studied that density of a substance is its mass per unit volume.

\(\text { Density }=\frac{\text { mass of the substance }}{\text { volume of the substance }}\)

Relative Density (R.D.) of a substance is the ratio of the density of that substance to the density of water at 4 °C.

\(\text { Relative Density }=\frac{\text { density of substance }}{\text { density of water at } 4^{\circ} \mathrm{C}}\)

= \(\frac{\text { mass of unit volume of substance }}{\text { mass of unit volume of water at } 4^{\circ} \mathrm{C}}\)

= \(\frac{\text { mass of substance }}{\text { mass of equal volume of water at } 4^{\circ} \mathrm{C}}\)

Relative density of a substance is defined as the ratio of mass of the substance to the mass of an equal volume of water at 4°C.

Unit

As it is a pure ratio it has no unit.

Chapter 3 Gravitation Fill In The Blanks

Question 1. The upward force acting onthe object immersed in liquid is known as ________.
Answer. Buoyancy

Question 2. ___________ is the force acting on a body normal to the surface.
Answer. Thrust

Question 3. A fluid contained in a vessel exerts a pressure at all points and in all directions. (True/False)
Answer. True

Question 4. The pressure experienced by a large surface area will be _________.
Answer. Less

Question 5. The Earth’s speed is ______ when it is closer to the Sun.
Answer. More

Question 6. The orbit of Earth around the Sun is ______.
Answer. Elliptical

Question 7. The acceleration due to gravity depends on mass and ______ of the planet.
Answer. Radius

Question 8. The weight of object in water is called its ______.
Answer. Apparent weight

Question 9. Weight of a body on Earth = ______ times the weight of a body on Moon.
Answer. Six

Question 10. To increase the gravitational force between two bodies, the distance between them must be ______.
Answer. Decreased

Question 11. A stone goes up in the air due to ______.
Answer. Kinetic energy

Question 12. ‘G’ is the ______ gravitational constant.
Answer. Universal

Question 13. Water exerts force on a submerged stone in the ______ direction.
Answer. Upward

Question 14. Weight of a body is ______ when measured underwater.
Answer. Less

Question 15. The value of acceleration due to gravity __________ if we go down below the surface of earth.
Answer. Decreases

Question 16. The acceleration due to gravity depends on the mass of the object. (True/False)
Answer. False

Question 17. _____________ is the point in body where its whole mass is assumed to be concentrated.
Answer. Centre of Mass

Question 18. The value of acceleration due to gravity of the earth is ____________.
Answer. 9.8 m/s²

Chapter 3 Gravitation Match the Columns

Question 1. Choose the correct unit.

Select the correct option:

1. A-1, B-2, C-3, D-4
2. A-4, B-3, C-2, D-1
3. A-2, B-1, C-3, D-4
4. A-2, B-3, C-1, D-4

Answer. 3. A-2, B-1, C-3, D-4

Question 2. Choose the correct unit.

Select the correct option:

1. A-1, B-2, C-3, D-4
2. A-4, B-3, C-2, D-1
3. A-2, B-1, C-3, D-4
4. A-2, B-3, C-1, D-4

Answer. 2. A-4, B-3, C-2, D-1

Question 3. Choose the correct nature of the physical quantities.

Select the correct option:

1. A-1, B-2, C-3, D-4
2. A-4, B-3, C-2, D-1
3. A-2, B-1, C-3, D-4
4. A-2, B-3, C-1, D-4

Answer. 1. A-1, B-2, C-3, D-4

Question 4. Choose the correct equation.

Select the correct option:

1. A-1, B-2, C-3, D-4
2. A-4, B-3, C-2, D-1
3. A-2, B-1, C-3, D-4
4. A-3, B-1, C-2, D-4

Answer. 4. A-3, B-1, C-2, D-4

Chapter 3 Gravitation Assertion Reasoning

Direction: For the following questions the options will remain the following:

• Both A and R are correct and R is correct explanation of A.
• Both A and R are correct but R is not a logical explanation of A.
• A is correct but R is incorrect.
• R is correct but A is incorrect.

1. Assertion: Relative Density has no unit.
   Reason: It is a unit.
2. Assertion: The liquid will flow out from the holes of the vessel.
   Reason: This is because liquid exerts pressure on the wall of the vessel.
3. Assertion: Weight is a scalar quantity.
   Reason: Force can be applied towards earth and in opposite direction.

Chapter 3 Gravitation Comprehension Passage

Sir Issac Newton was born in 1642 in Woolsthorpe near Grantham, England. He belonged to a poor family. His interest was not into farming, Instead, he went for studies to Cambridge University in the year 1661.

In the year 1665, the incident of falling apple happened. This incident made him think about the gravity of force which intact moon in its orbit. This gave birth to the invention of universal law of gravitation. He was the first person to discover about gravity.

Newton gave laws of motion. Along with it he worked on light and colour. To do observations on astrology, he designed an astronomical telescope. He was good in mathematics so and invented calculus.

Question 1. In which year Issac Newton was born?

1. 1662
2. 1642
3. 1622
4. 1602

Answer. 2. 1642

Question 2. In which year did apple fall on Newton’s head?

1. 1642
2. 1662
3. 1665
4. 1675

Answer. 3. 1665

Question 3. Who invented universal law of gravitation?

1. Issac
2. Newton
3. Issac Newton
4. Alberto

Answer. 3. Issac Newton

Question 4. In which country was Issac newton born?

1. England
2. Germany
3. London
4. India

Answer. 1. England

Question 5. Which branch of mathematics was invented by Issac Newton?

1. Integer
2. Algebra
3. Boolean
4. Calculus

Answer. 4. Calculus

Question 6. Centre of mass becomes the centre of gravity

1. At the core of the earth
2. On the surface of the earth
3. Force of gravity varies
4. Never happens

Answer. 1. At the core of the earth