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JEE • PHYSICS • GRAVITATION

Gravitation practice · Easy

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 53 questions
1 Easy
1 of 53
Application +4 | -1
The ratio of the accelerations due to gravity on two planets P1&P2P_1 \& P_2 is K1K_1. The ratio of their respective radii is K2K_2. The ratio of their respective escape velocities is
View answer and explanation
Correct option: A
k1k2\sqrt{k_1 k_2}
Option A (k1k2\sqrt{{k_1 k_2}}) is correct because the escape velocity is proportional to the square root of the product of the acceleration due to gravity and the radius.
2 Easy
2 of 53
Analyzing +4 | -1
Statement-1: A particle is at a height R{R} from the surface of earth (Here R{R} is radius of earth). If mass of particle is m{m} then its gravitational potential energy is mgR{mgR}
Statement-2: If a particle is slowly lifted above the surface of earth then work is done by external agent. Work done by external agent is wasted in the form of heat energy.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is false
Option D is correct because both statements are false. The gravitational potential energy at a height R from the Earth's surface is -mgR/2, and the work done by an external agent is stored as potential energy, not wasted as heat.
3 Easy
3 of 53
Understanding +4 | -1
The gravitational field due to a uniform solid sphere of mass MM and radius aa at the centre of the sphere is
View answer and explanation
Correct option: D
zero
Option D (zero) is correct because the gravitational field at the center of a uniform solid sphere is zero due to symmetry.
4 Easy
4 of 53
Understanding +4 | -1
Statement-1: There is almost no effect of rotation of earth at poles.
Statement-2: Because rotation of earth is about polar axis.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true
Option C (Increases) is correct because a refrigerator transfers heat from the inside to the outside, so opening the door releases heat into the room.
5 Easy
5 of 53
Analyzing +4 | -1
A satellite is moving in a circular orbit round the earth with a diameter of orbit 2R2 R. At a certain point a rocket fixed to the satellite is fired such that it increases the velocity of the satellite tangentially. The resulting orbit of the satellite would be
View answer and explanation
Correct option: C
elliptical orbit with minimum distance from the earth equal to RR
Option C (elliptical orbit with minimum distance from the earth equal to RR) is correct because increasing the tangential velocity at a point in a circular orbit results in an elliptical orbit with that point as the perigee.
6 Easy
6 of 53
Application +4 | -1
A satellite of mass mm is circulating around the earth with constant angular velocity. If the radius is R0R_0 and mass of earth is MM, then the angular momentum about the centre of the earth is
View answer and explanation
Correct option: C
mGMR0m \sqrt{G M R_0}
Option C (mGMR0m \sqrt{G M R_0}) is correct because it is derived from the definition of angular momentum and the orbital velocity of a satellite.
7 Easy
7 of 53
Understanding +4 | -1
A planet is revolving round the sun in an elliptical orbit, If v\mathbf{v} is the velocity of the planet when its position vector from the sun is r\mathbf{r}, then areal velocity of the planet is
View answer and explanation
Correct option: C
∣12(r×v)∣\left|\frac{1}{2}(\mathbf{r} \times \mathbf{v})\right|
Option C (∣12(r×v)∣\left|\frac{{1}}{{2}}(\mathbf{{r}} \times \mathbf{{v}})\right|) is correct because the areal velocity is given by half the magnitude of the cross product of the position vector and the velocity vector.
8 Easy
8 of 53
Application +4 | -1
A body of mass mm is taken from earth surface to the height h\mathrm{h} equal to radius of earth, the increase in potential energy will be
View answer and explanation
Correct option: B
12mgR\frac{1}{2} \mathrm{mgR}
Option B (12mgR\frac{1}{2} \mathrm{mgR}) is correct because the change in potential energy is calculated using the formula ΔU=mgh1+h/R\Delta U = \frac{mgh}{1 + h/R} with h = R.
9 Easy
9 of 53
Application +4 | -1
The acceleration due to gravity on a planet is same as that on earth and its radius is 16 times that of earth. What will be the value of escape velocity on that planet if it is Ve{V_e} on earth.
View answer and explanation
Correct option: C
4Ve4{V_e}
Option C (4Ve4{V_e}) is correct because the escape velocity is proportional to the square root of the radius, and since the radius is 16 times that of Earth, the escape velocity is 4 times that of Earth.
10 Easy
10 of 53
Understanding +4 | -1
Which of the following statement is correct? (Take, earth as a sphere of uniform density)
View answer and explanation
Correct option: D
Acceleration due to gravity decreases with increasing depth.
Option D (Acceleration due to gravity decreases with increasing depth) is correct because the effective mass pulling the object towards the Earth's center decreases as depth increases.
11 Easy
11 of 53
Understanding +4 | -1
When you move from equator to pole, the value of acceleration due to gravity ( g\mathrm{g} )
View answer and explanation
Correct option: A
increases
Option A is correct because the acceleration due to gravity increases as you move from the equator to the poles due to the Earth's shape and rotation.
12 Easy
12 of 53
Understanding +4 | -1
A uniform spherical shell gradually shrinks maintaining its shape. The gravitational potential at the centre
View answer and explanation
Correct option: A
increases
Option A (increases) is correct because as the spherical shell shrinks, its radius decreases, and since the gravitational potential is inversely proportional to the radius, the potential increases (becomes less negative).
13 Easy
13 of 53
Understanding +4 | -1
Statement-1: The force of attraction due to a hollow spherical shell of uniform density, on a point mass situated inside it is zero.
Statement-2: The gravitation field due to the shell inside the shell will be zero.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true
Option A (Statement-1 is true, Statement-2 is true) is correct because the gravitational force and field inside a hollow spherical shell of uniform density are zero.
14 Easy
14 of 53
Understanding +4 | -1
If the speed of rotation of earth about its axis increases, then the weight of the body at the equator will
View answer and explanation
Correct option: B
decrease
Option B (decrease) is correct because the effective weight decreases with increasing rotational speed due to centrifugal force.
15 Easy
15 of 53
Understanding +4 | -1
The velocity with which a projectile must be fired so that it escapes Earth's gravitation does not depend on
View answer and explanation
Correct option: B
mass of the projectile
Option B (mass of the projectile) is correct because the escape velocity formula does not include the mass of the projectile.

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16 Easy
16 of 53
Understanding +4 | -1
Assertion: If a particle projected horizontally just above the surface of earth with a speed greater than escape speed, then it will escape from gravitational influence of earth.
Reason: Escape velocity is independent of its direction.
View answer and explanation
Correct option: A
Assertion is True, Reason is True; Reason is a correct explanation for Assertion
Option A is correct because both the assertion and reason are true, and the reason correctly explains why a particle with escape speed escapes Earth's gravity regardless of direction.
17 Easy
17 of 53
Application +4 | -1
The imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero is equal to
(Take, g=10 ms−2g=10 \mathrm{~ms}^{-2} for the acceleration due to gravity, if the earth were at rest, radius of earth equal to 6400 km6400 \mathrm{~km} and λ=60\lambda=60 )
View answer and explanation
Correct option: B
2.50×10−3 rad s−12.50 \times 10^{-3} \mathrm{~rad} \mathrm{~s}^{-1}
Option B (2.50×10−3 rad s−12.50 \times 10^{-3} \mathrm{~rad} \mathrm{~s}^{-1}) is correct because it is derived from setting the effective gravity at the equator to zero and solving for the angular velocity.
18 Easy
18 of 53
Understanding +4 | -1
Statement-1: A planet is moving around the sun in an elliptical orbit. Then from aphelion point to perihelion point the speed of the planet continuously increases.
Statement-2: According to Kepler's second law, the areal velocity of the line joining planet to the sun is constant.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true
Option A is correct because Kepler's second law states that areal velocity is constant, implying that the planet speeds up as it approaches the sun (perihelion) and slows down as it moves away (aphelion).
19 Easy
19 of 53
Analyzing +4 | -1
Statement-1: Orbiting satellite or body has K.E. of always less than that of Potential energy.
Statement-2: For any bound state, the magnitude of potential energy is always twice that of kinetic energy (K.E.)
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true
Option A (Statement-1 is true, Statement-2 is true) is correct because for an orbiting satellite, the kinetic energy is always less than the potential energy, and for a bound state, the magnitude of potential energy is twice the kinetic energy.
20 Easy
20 of 53
Analyzing +4 | -1
Earth orbiting satellite will escape, if
View answer and explanation
Correct option: C
Both (a) and (b) are correct
Option C (Both (a) and (b) are correct) is correct because increasing the speed by 41% or doubling the kinetic energy will allow the satellite to reach escape velocity.
21 Easy
21 of 53
Understanding +4 | -1
A body is projected from earth's surface to become its satellite, its time period of revolution will not depend upon
View answer and explanation
Correct option: D
radius of orbit
Option B is correct because the time period of a satellite's revolution does not depend on its own mass.
22 Easy
22 of 53
Understanding +4 | -1
The weight of the body at the centre of the earth is
View answer and explanation
Correct option: A
zero
Option A (zero) is correct because the acceleration due to gravity (g) at the center of the Earth is zero, thus the weight (mg) is also zero.
23 Easy
23 of 53
Understanding +4 | -1
According to Newton's law of gravitation, the magnitude of the force FF on point masses m1m_1 and m2m_2 (say) is given by ∣F∣=Gm1m2r2|F|=\frac{G m_1 m_2}{r^2}.
I. This equation is not directly applicable for the gravitational force between extended object (like the earth) and a point mass.
II. This equation can be applied to find the gravitational force between an extended object and a point mass provided rr is the distance of the point mass from the geometric centre of the extended objects.
Which of the following statement(s) about this equation is/are correct?
View answer and explanation
Correct option: A
Only I
Option A is correct because Newton's law of gravitation in its simple form applies directly only to point masses, not extended objects without further calculation.
24 Easy
24 of 53
Remembering +4 | -1
Period of revolution of a geostationary satellite around the earth is
View answer and explanation
Correct option: B
24 hours
Option B (24 hours) is correct because a geostationary satellite has the same period as the Earth's rotation, which is 24 hours.
25 Easy
25 of 53
Understanding +4 | -1
Assertion: Smaller the orbit of a planet around the sun, shorter is the time it takes to complete.
Reason: According to Kepler's third law of planetary motion, square of time period is proportional to cube of mean distance from the sun.
View answer and explanation
Correct option: A
Assertion is True, Reason is True; Reason is a correct explanation for Assertion
Option A is correct because both the assertion and reason are true, and the reason (Kepler's third law) correctly explains why planets closer to the sun have shorter orbital periods.
26 Easy
26 of 53
Understanding +4 | -1
Assertion: Earth does not retain hydrogen molecules and helium atoms in its atmosphere, but does retain much heavier molecules, such as oxygen and nitrogen.
Reason: Lighter molecules in the atmosphere have translational speed that is greater or closer to escape speed of earth.
View answer and explanation
Correct option: A
Assertion is True, Reason is True; Reason is a correct explanation for Assertion
Option A (Assertion is True, Reason is True; Reason is a correct explanation for Assertion) is correct because lighter molecules have higher average speeds, making it easier for them to reach escape velocity and leave the atmosphere.
27 Easy
27 of 53
Application +4 | -1
A ring of mass m1m_1 and radius RR is fixed in space at some location. An external agent brings a point mass m2m_2 from infinity to centre of the ring. Work done by the external agent will be
View answer and explanation
Correct option: A
−Gm1m2R-\frac{G m_1 m_2}{R}
Option A (−Gm1m2R-\frac{G m_1 m_2}{R}) is correct because the work done by the external agent is equal to the change in potential energy, which is negative in this case.
28 Easy
28 of 53
Understanding +4 | -1
I. The geo-stationary satellite always has circular orbit.
II. The time period of satellite depends on mass of the satellite, mass of the earth and height of satellite from the surface of earth.
Which of the following statement(s) is/are incorrect?
View answer and explanation
Correct option: A
Only I
Option A (Only I) is correct because geostationary satellites can have elliptical orbits, and the time period of a satellite depends on the mass of the Earth and the height of the satellite, not the satellite's mass.
29 Easy
29 of 53
Remembering +4 | -1
Gravitational potential energy of a body of mass mm at a height of hh above the surface of earth (M=(M= mass of earth R=R= radius of earth) is
View answer and explanation
Correct option: D
−GMm(R+h)-\frac{G M m}{(R+h)}
Option D is correct because the gravitational potential energy is negative and inversely proportional to the distance from the center of the Earth.
30 Easy
30 of 53
Understanding +4 | -1
Statement-1: It is possible to put an artificial satellite into orbit in such a way that it will always remain directly over New Delhi.
Statement-2: To put an artificial satellite into orbital such that it remains directly over head at a fixed location, its time period should be related to that of earth in the equatorial plane.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Option D (Statement-1 is false, Statement-2 is true) is correct. A geostationary satellite must be above the equator, so it cannot remain directly over New Delhi. However, the time period must match Earth's rotation.
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Key Takeaways

  • ✓ This set contains 53 single-correct-answer MCQs on GRAVITATION.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
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