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Gravitation MCQs for NEET — Physics Questions with Answers

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Earth is revolving around the sun, if the distance of the Earth from the Sun is reduced to 1/4^th of the present distance then the present year length reduced to -

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{8}$

$\frac{1}{6}$

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Explanation

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Two satellite are revolving around the earth with velocities $v_{1}$ and $v_{2}$ and in radii $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively. Then -

$v_{1} = v_{2}$

$v_{1} > v_{2}$

$v_{1} < v_{2}$

$\frac{v_{1}}{r_{1}} = \frac{v_{2}}{r_{2}}$

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Explanation

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If orbital velocity of planet is given by $v = G^{a} M^{b} R^{c}$ , then -

$a = 1 / 3, b = 1 / 3, c = ‐ 1 / 3$

$a = 1 / 2, b = 1 / 2, c = ‐ 1 / 2$

$a = 1 / 2, b = ‐ 1 / 2, c = 1 / 2$

$a = 1 / 2, b = ‐ 1 / 2, c = ‐ 1 / 2$

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Explanation

(b) $v = \sqrt{\frac{G M}{R}} = G^{1 / 2} M^{1 / 2} R^{‐ 1 / 2}$

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The mass of a planet that has a moon whose time period and orbital radius are T and R respectively can be written as -

$4 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

$8 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

$12 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

$16 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

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Explanation

(a) $m ω^{2} R = \frac{G M m}{R^{2}} \Rightarrow {(\frac{2 π}{T})}^{2} R = \frac{G M}{R^{2}} \Rightarrow M = \frac{4 π^{2} R^{3}}{G T^{2}}$

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The eccentricity of earth's orbit is 0.0167. The ratio of its maximum speed in its orbit to its minimum speed is

2.507

1.033

8.324

1.000

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Explanation

(b) $\frac{v_{m a x}}{v_{m i n}} = \frac{1 + e}{1 - e} = \frac{1 + 0.0167}{1 - 0.0167} = 1.033$

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The radius of orbit of a planet is two times that of the earth. The time period of planet is -

4.2 years

2.8 years

5.6 years

8.4 years

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Explanation

(b) $T_{2} = T_{1} {[\frac{R_{2}}{R_{1}}]}^{3 / 2} = 1 \times {(2)}^{3 / 2} = 2.8 y e a r s$

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The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance is increased to 16r, then the new angular momentum will be -

16 L

64 L

$\frac{L}{4}$

4 L

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Explanation

(d) $L = m v r = m \sqrt{\frac{G M}{r} r} = m \sqrt{G M r} ∴ L \propto \sqrt{r}$

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The condition for a uniform spherical mass m of radius r to be a black hole is [G= gravitational constant and g= acceleration due to gravity] .

${(2 G m / r)}^{1 / 2} \leq c$

${(2 G m / r)}^{1 / 2} = c$

${(2 G m / r)}^{1 / 2} \geq c$

${(g m / r)}^{1 / 2} \geq c$

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Explanation

$v_{e}$ should be more than or equal to speed of light.

i.e. $\sqrt{\frac{2 G m}{r}} \geq c$

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There are two bodies of masses 100 kg and 10000 kg separated by a distance 1 m. At what distance from the smaller body, the intensity of gravitational field will be zero ?

$\frac{1}{9} m$

$\frac{1}{10} m$

$\frac{1}{11} m$

$\frac{10}{11} m$

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Explanation

(c) $\frac{G \times 100}{x^{2}} = \frac{G \times 10000}{(1 - x^{2})} \Rightarrow \frac{10}{x} = \frac{100}{1 - x} \Rightarrow x = \frac{1}{11} m$

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What is the intensity of gravitational field of the centre of a spherical shell ?

$G m / r^{2}$

Zero

None of these

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Explanation

so, work done = 0

Hence, intensity of gravitational field of the centre of spherical shell is zero.

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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.