Blog Log in Get Started Free

Gravitation MCQs for NEET — Physics Questions with Answers

Practice free Gravitation (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

The depth d at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface, is [R = radius of the earth]

$\frac{R}{n}$

$R [\frac{n - 1}{n}]$

$\frac{R}{n^{2}}$

$R [\frac{n}{n + 1}]$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) $g^{1} = g [1 - \frac{d}{R}] \Rightarrow \frac{g}{n} = g [1 - \frac{d}{R}] \Rightarrow d = [\frac{n - 1}{n}] R$

● ● ● ● ●

The diameters of two planets are in the ratio 4 : 1 and their mean densities in the ratio 1 : 2. The acceleration due to gravity on the planets will be in ratio

1 : 2

2 : 3

2 : 1

4 : 1

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c) $g = \frac{4}{3} G πRp \Rightarrow \frac{g_{1}}{g_{2}} = \frac{p_{1} R_{1}}{p_{2} R_{2}} = \frac{1}{2} \times \frac{4}{1} = \frac{2}{1}$

● ● ● ● ●

At what altitude in metre will the acceleration due to gravity be 25% of that at the earth's surface (Radius of earth = R metre)

$\frac{1}{4} R$

$\frac{3}{8} R$

$\frac{R}{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

b) $g' = g {[\frac{R}{R + h}]}^{2} \Rightarrow \frac{g}{4} = g {[\frac{R}{R + h}]}^{2} \Rightarrow \frac{1}{2} = \frac{R}{R + h} \Rightarrow R + h = 2 R S o, h = R$

● ● ● ● ●

The gravitational potential energy of a body of mass ‘m’ at the earth’s surface $- m g R_{e}$ . Its gravitational potential energy at a height $R_{e}$ from the earth’s surface will be (Here $R_{e}$ is the radius of the earth)

$- 2 m g R_{e}$

$2 m g R_{e}$

$\frac{1}{2} m g R_{e}$

$- \frac{1}{2} m g R_{e}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) $∆ U = U_{2} - U_{1} = \frac{m g h}{1 + \frac{h}{R_{e}}} = \frac{m g R_{e}}{1 + \frac{R_{e}}{R_{e}}} = \frac{m g R_{e}}{2}$

$\Rightarrow U_{2} + (m g R_{e}) = \frac{m g R_{e}}{2} \Rightarrow U_{2} = - \frac{1}{2} m g R_{e}$

● ● ● ● ●

If the angular speed of the earth is doubled, the value of acceleration due to gravity (g) at the north pole

Doubles

Becomes half

Remains same

Becomes zero

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

● ● ● ● ●

Escape velocity of a body of 1 kg mass on a planet is 100 m/sec. Gravitational Potential energy of the body at the planet is -

– 5000 J

– 1000 J

– 2400 J

5000 J

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) $V_{e} = \sqrt{\frac{2 G M}{R}} = 100 \Rightarrow \frac{G M}{R} = 5000 P o t e n t i a l e n e r g y U = - \frac{G M m}{r} = - 5000 J$

● ● ● ● ●

At the surface of a certain planet, acceleration due to gravity is one-quarter of that on earth. If a brass ball is transported to this planet, then which one of the following statements is not correct ?

The mass of the brass ball on this planet is a quarter of its mass as measured on earth

The weight of the brass ball on this planet is a quarter of the weight as measured on earth

The brass ball has the same mass on the other planet as on earth

The brass ball has the same volume on the other planet as on earth

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) Mass of the ball always remain constant. It does not depend upon the acceleration due to gravity

● ● ● ● ●

A body of mass m is placed on the earth’s surface. It is taken from the earth’s surface to a height 3R . The change in gravitational potential energy of the body is -

$\frac{2}{3} m g R$

$\frac{3}{4} m g R$

$\frac{m g R}{2}$

$\frac{m g R}{4}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) $∆ U = \frac{m g h}{1 + \frac{h}{R}} = \frac{m g \times 3 R}{1 + \frac{3 R}{R}} = \frac{3}{4} m g R$

● ● ● ● ●

Weight of 1 kg becomes 1/6 kg on moon. If radius of moon is $1.768 \times 10^{6} m$ , then the mass of moon will be -

$1.99 \times 10^{30} k g$

$7.56 \times 10^{22} k g$

$5.98 \times 10^{24} k g$

$7.65 \times 10^{22} k g$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) $g_{m} = \frac{G M_{m}}{{R_{m}}^{2}}$ and $g_{m} = \frac{g_{e}}{6} = \frac{9.8}{6} m / s^{2} = 1.63 m / s^{2}$

● ● ● ● ●

A geo-stationary satellite is orbiting the earth at a height of 6 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 2.5 R from the surface of earth is

10 hr

$(6 / \sqrt{2}) h r$

6 hr

$6 \sqrt{2} h r$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) Distances of the satellite from the centre are 7R and 3.5R respectively.

$\frac{T_{2}}{T_{1}} = {(\frac{R_{2}}{R_{1}})}^{\frac{3}{2}} \Rightarrow T_{2} = 24 {(\frac{3.5 R}{7 R})}^{\frac{3}{2}} = 6 \sqrt{2} h r$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Gravitation question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.