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Students can solve these MCQ Questions for Class 11 Physics with Answers and evaluate their preparation level. Class 11 Physics MCQ Questions of Gravitation with Answer are set up according to the Latest Exam Pattern and syllabus. These Class 11 MCQ Questions of Gravitation with answers clear for a speedy correction of the Chapter.

Gravitation is one of the key powers of attraction that exists between objects. Newton is the unit of gravitation force. To help you in an exam, here could be a list of MCQ questions for class 11 Gravitation with Answers.

Practice MCQ Questions for class 11 Physics Chapter-Wise

1. A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energy is

(a) Positive
(b) Negative
(c) Zero
(d) may be positive or negative

2. There is no atmosphere on the moon because

(a) it is closer to the earth
(b) it revolves round the earth
(c) it gets light from the sun
(d) the escape velocity of gas molecules is less than their root mean square velocity

3. The value of g at a particular point is 9.8 m/sec² suppose the earth suddenly shrink uniformly to half its present size without losing any mass. The value of g at the same point (assuming that the distance of the point from the centre of the earth does not shrink) will become

(a) 9.8 m/sec²
(b) 4.9 m/sec²
(c) 19.6 m/sec²
(d) 2.45 m/sec²

4. Who among the following first gave the experimental velocity of G?

(a) Cavendish
(b) Copernicus
(c) Brook Taylor
(d) none of these

5. A satellite S is move in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth .

(a) The acceleration of S is always directed towards the centre of the earth
(b) The angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant
(c) The total mechanical energy of S varies periodically with time
(d) The linear momentum of S remains constant in magnitude

6. In planetary motion

(a) the total angular momentum remains constant
(b) the linear speed remains constant
(c) neither the angular momentum nor angular speed remains constant
(d) the angular speed remains constant

7. If the earth is at one-fourth of its present distance from the sun, the duration of the year will be

(a) one-eighth the present year
(b) one-sixth the present year
(c) one-tenth the present year
(d) half the present year

8. Newton's universal law of gravitation applies to

(a) both small and big bodies
(b) only valid for solar system
(c) planets only
(d) small bodies only

9. Force of gravitational attraction is least

(a) at the equator
(b) at the poles
(c) at a point in between equator and any pole
(d) None of these

10. Assertion: If earth suddenly stops rotating about its axis, then the value of acceleration due to gravity will become same at all the places.

Reason: The value of acceleration due to gravity depends upon the rotation of the earth.

(a) Assertion is correct, reason is incorrect
(b) Assertion is incorrect, reason is correct
(c) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(d) Assertion is correct, reason is correct; reason is a correct explanation for assertion

11. Where will it be profitable to purchase one kilogram sugar?

(a) At equator
(b) At 40° latitude
(c) At poles
(d) At 45° latitude 

12. Earth is flattened at poles, bulged at the equator. This is due to

(a) the centrifugal force is more at the equator than at the poles
(b) the angular velocity of spinning about its axis is more at equator
(c) the angular velocity of spinning about its axis is less at equator
(d) earth revolves round the sun in an elliptical orbit

13. Average density of the earth

(a) is directly proportional to g
(b) is inversely proportional to g
(c) does not depend on g
(d) is a complex function of g

14. If the earth were to rotates faster than its present speed, the weight of an object will

(a) decrease at the equator but remain unchanged at the poles
(b) remain unchanged at the equator but decrease at the poles
(c) remain unchanged at the equator but increase at the poles
(d) increase at the equator but remain unchanged at the poles

15. At what height from the ground will the value of g be the same as that in 10 km deep mine below the surface of earth?

(a) 5 km
(b) 15 km
(c) 10 km
(d) 20 km

16. Consider Earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist B goes high up in a balloon. The gravitational field measured by

(a) each decreases at different rates
(b) each decreases at the same rate
(c) B goes on decreasing and that by A goes on increasing
(d) A goes on decreasing and that by B goes on increasing

17. In some region, the gravitational field is zero. The gravitational potential in this region.

(a) must be constant
(b) may be zero
(c) must be variable
(d) cannot be zero

18. In a gravitational field, at a point where the gravitational potential is zero

(a) the gravitational field is necessarily zero
(b) the gravitational field is not necesarily zero
(c) any value between one and infinite
(d) None of these

19. A planet is moving in an elliptical orbit. If T, V, E, and L stand, respectively, for its kinetic energy, gravitational potential energy, total energy and angular momentum about the center of orbit, then

(a) T is conserved
(b) V is always positive
(c) E is always negative
(d) None of these

20. If 'A' is areal velocity of a planet of mass M, its angular momentum is

(a) M/A
(b) 2MA
(c) A²M
(d) AM²

21. Kepler's second law regarding constancy of the areal velocity of a planet is a consequence of the law of conservation of:

(a) Energy
(b) Linear momentum
(c) Angular momentum
(d) Mass 

22. Weightlessness experienced while orbiting the earth in space-ship, is the result of

(a) Inertia 
(b) Acceleration
(c) Zero gravity
(d) Freefall towards the earth

23. The time period of a simple pendulum on a freely moving artificial satellite is

(a) Zero
(b) 2 sec
(c) 3 sec
(d)  Infinite

24. A body of mass m is taken to the bottom of a deep mine. Then

(a) Its mass increases
(b) Its mass decreases
(c) Its weight increases
(d) Its weight decreases

25. If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct ?

(a) g' on the Earth will not change.
(b) Raindrops will fall faster
(c) Time period of a simple pendulum on the Earth would decrease.
(d) Walking on the ground would become more difficult.

Answer:  

1. Answer: (b) Negative

Explanation: The sum of kinetic energy and potential energy is negative. This is because for the velocity less than escape velocity the missile is bounded due to gravitational field of the earth.  

2. Answer: (d) the escape velocity of gas molecules is less than their root mean square velocity

Explanation: There is no atmosphere at moon because escape velocity is less than the root mean square velocity of the molecules at moon. Hence all molecules escape.

3. Answer: (a) 9.8 m/sec² 

Explanation: Acceleration of gravitation, a = \(\frac{GM}{R^2}\)

Where, M is mass of earth and R is separation between both masses. If volume is reduce, keeping mass and separation between them same then.

4. Answer: (a) Cavendish

Explanation: .The value of G was not experimentally determined until nearly a century later (1798) by Lord Henry Cavendish using a torsion balance.

5. Answer: (a) The acceleration of S is always directed towards the centre of the earth

Explanation: Force on satellite is always towards earth, therefore, acceleration of satellite S is always directed towards centre of the earth. 

6. Answer: (a) the total angular momentum remains constant 

Explanation: Kepler's laws of planetary motion can be stated as follows: 

(i) All planets move about the Sun in elliptical orbits, having the Sun as one of the foci. 

(ii) A radius vector joining any planet to the Sun sweeps out equal areas in equal lengths of time.

7. Answer: (a) one-eighth the present year

Explanation: From Kepler's law,

T²∝ r³

\((\frac{T_1}{T_2})^2=\frac{(r_1}{r_2})^3\) \(=(\frac{r}{r/4})^3\)

\(\frac{T_1}{T_2}=(4)^{3/2}\)

\(T_2=\frac{T_1}{4^{3/2}}\)

T₂ = T₁/8

8. Answer: (a) both small and big bodies

Explanation: Newton's law of universal gravitation states that any two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

9. Answer: (a) at the equator

Explanation: Since the strength of gravity weakens as you get farther away from a gravitational body, the points on the equator are farther and have weaker gravity than the poles.

10. Answer: (b) Assertion is incorrect, reason is correct

Explanation: The value of acceleration due to gravity is maximum at poles and minimum at the equator because the centrifugal force for the earth's spinning is vanishes at poles and maximum at the equator. If the earth suddenly stops rotating about its axis, there is no effect of centrifugal force on the gravitational acceleration. So, the value of gravitational will be same at all places.

11. Answer: (a) At equator

Explanation: As the value of g is minimum at the equator for the same value of mass because of spherical shape of earth.  A weight of 1 kg sugar is measured less at the equator. 

12. Answer: (a) the centrifugal force is more at the equator than at the poles

Explanation: Earth is bulged out at equator and flattered at poles because of centripetal force. The earth spins at constant rate but rate of movement is different the equator is moving fastest and poles are not moving (ignoring the fact that earth is orbiting the sun).

13. Answer: (a) is directly proportional to g

Explanation: As g is directly proportional to average density hence inversely average density is proprotional to g.

14. Answer: (a) decrease at the equator but remain unchanged at the poles

Explanation: As the earth rotates faster, the centrifugal force on a body at the equator increases, thus decreasing the weight of the body. But since the poles lie on the axis of rotation, the centrifugal force on a body at the poles is 0. 

15. Answer: (d) 20 km 

Explanation: Same changes in value of g can be observe at a depth x and height 2x 

′d = x = 10km 

∴ the same charge can be observed at a height 2x = 20km

16. Answer: (a) each decreases at different rates

Explanation: The variation of 'g' with altitude is: \(g'=g(1-\frac{2h}{R})\)

The variation of 'g' with depth is: \(g'=g(1-\frac{d}{R})\)

So, from the above equations we can deduce that if a person goes deep into a mine or to a height above the Earth's surface, the gravitational field decreases at different rates.

17. Answer: (b) may be zero

Explanation: E⁰_g= −dv/dx 

If E_g = 0, then V= Constant and this constant may also be zero.

18. Answer: (b) 

Explanation: point where gravitational potential is zero, gravitational field has to be zero. as I = – (dv / dx) i.e. potential is zero then field intensity = 0

19. Answer: (c) E is always negative

Explanation: For the planet to move in any orbit, its total mechanical energy must be bound. Thus E must be negative always. T will not be conserved because v varies as r changes.  potential energy can never  be positive and L is always conserved with direction of L always in one direction.

20. Answer: (b) 

Explanation: dA/dt = L/2m

21. Answer: (c) Angular momentum

Explanation: Kepler’s second law, is a consequence of the fact that the force of gravity is a central force, which in turn implies that angular momentum is constant. 

22. Answer: (c) Zero gravity

Explanation: Weightlessness means that there is no reaction on a body from the floor. Since both the artificial satellite & the astronaut have same centripetal acceleration (as in a lift; which is falling freely, we does not feel any weight, because both lift & we fall with same acceleration).

23. Answer: (d) Infinite

Explanation: Time period of simple pendulum \(T = 2\pi\sqrt{\frac{1}{g'}}\)

In artificial satellite g =0

 ∴ T=infinite.

24. Answer: (d) Its weight decreases

Explanation: A body of mass mm is taken to the bottom of a deep mine. Then its weight decreases because acceleration due to gravity decreases. 

25. Answer: (a) g' on the Earth will not change.

Explanation: \(g= \frac{GM}{R^2}\) 

g is depends in 'G', hence 'g' will change.
Hence raindrops will fall faster and walking on ground will more difficult.

Time period \(\propto\frac{1}{\sqrt g}\), hence 'T' decreases.

Click here to practice MCQ Questions for Gravitation class 11

Correct option is A) 4 sec

B) spring balance

Weight of an object is the force with which it is attracted towards the earth.

The quantity of the matter present in an object is called the mass and it is related with the object’s inertia whereas weight is written as W = m × g, where g is acceleration due to gravity.

Hence, option D is correct.

Correct option is C) 5 m

D) radius of the earth

A) acceleration due to gravity

The equation is valid for spherical bodies.

The gravitational equations are valid for point masses. For spherical bodies the center of mass can be conveniently chosen at the center and mass is evenly distributed about the center.

Hence, option B is correct.

Correct option is B) 2G

Correct option is C) zero

Correct option is C) ma

Correct option is C) – g

If the distance between two objects is doubled the gravitational force becomes one fourth since,

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

Where R is the distance between the bodies.

So, if R’ = 2R, then

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

F' = G\(\frac{m_1\times m_2}{4\times R^2}\) = \(\frac{F}{4}\)

Hence, option D is correct.

Correct option is A) T² ∝ R³ 

Kepler’s law of periods states that: The cube of the mean distance of a planet from the sun is directly proportional to the square of time it takes to move around the sun.

School bags have wide straps so that their weight may spread over a large area of shoulder producing less pressure on the shoulder.

A piece of glass sinks in water but it floats in mercury because density of glass is more than that of water but less than that of mercury.

An iron nail sinks in water but it floats in mercury because density of iron is more than that of water but less than that of mercury.

A piece of steel sinks in water because steel is denser than water. However, a steel ship is a hollow object made of steel and contains a lot of air in it. Due to presence of a lot of air in it, the average density of the ship becomes less than the density of water. Hence a ship floats in water.

The sheet of tin sinks in water because the density of tin is higher than that of water. When the same sheet of tin is converted into a box or a boat, then due to the trapping of lot of ‘light’ air in the box or boat, the average density of the box or boat made of tin sheet becomes lower than that of water and hence it floats in water.

The formation of tides in the sea is because of the gravitational force

Due to gravitational effect of Moon there is high tides and low tides in the sea.

The value of g is equal to zero in a hollow sphere, hence, the weight (mg) will also be zero.

Geo-stationary Satellite

A satellite placed at a definite height directly above the Earth’s equator and revolves in the same direction as the Earth rotates; so that its orbital time period is same as the Earth’s rotation period (24 hours), is called a Geo-stationary satellite. The observer at the equator views the satellite as stationary, hence such types of satellites are also called geosynchronous satellites. These are used for communication, radio broadcasting, universe related studies & researches and gathering weather information. Communication satellite is also a Geo-stationary satellite.
We know the time period of a satellite,

T = \(\frac{2 \pi r^{3 / 2}}{\sqrt{G M}}\)
∴ r = \(\left[\frac{G M T^{2}}{4 \pi^{2}}\right]^{1 / 3}\) ……………. (1)
Putting G = 6.67 × 10^-11 Nm²/kg², mass of Earth M = 6 × 10²⁴ kg, time period T = 24 hours in above equation.

Orbital radius of the satellite from the equation (1) by putting all values;
r = 4.2 × 10⁴ = 42000 km
but r = R + h
Hence, height of the satellite from the surface of the Earth;
h = r – R
h = 42000 km – 6400 km
or h ≈ 36000 km
Hence, the height of a geostationary satellite from the Earth’s surface is approximately 36000 km.

Angular Velocity of Geo-stationary Satellite
Since, angular velocity ω = \(\frac{2 \pi}{T}=\frac{2 \times 3.14}{24 \times 60 \times 60}\)
ω = 7.3 × 10^-5 rad/s
Orbital velocity v = ωr
∴ v = 3.1 km/s.
For communication system, a satellite transmits signals to only \(\frac{1}{3}\) part of the Earth. Hence, to broadcast information throughout the Earth minimum three satellites are necessary.

Any communication satellite (geostationary) can not be placed over India’s capital New Delhi because it will not be in equatorial line.

(d) will depend on direction of projection.

EXPLANATION: 

Its minimum escape speed will be vₑ when projected vertically upward, but if projected otherwise the speed will be greater. So the escape speed will depend on the direction of projection. 

(d) will depend on direction of projection.

EXPLANATION: 

Its minimum escape speed will be vₑ when projected vertically upward, but if projected otherwise the speed will be greater. So the escape speed will depend on the direction of projection. 

The time period of the satellite,
T = \(2 \pi \sqrt{\frac{r^{3}}{G M}}\)
where, r = radius of the orbit,
M mass of Earth.
In above formula, the mass of the satellite (m) is not used. Therefore the time period T will not depend on m. Hence, on doubling the mass of the satellite, its time period will remain unchanged.

Time-period related with them T² ∝ a³.

Gravitational force provides the necessary centripetal force, which at right angles to motion of the Earth, so by the relation W = Ed cos θ
∴ W = Fd cos 90 = 0
hence no work is done by gravitational force.

For satellite,
Required centripetal force = Gravitational attraction between planet and satellite.
or F_c - F_g
This gravitational force will be the central force.

(c) Mass of the projectile