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Physics MCQs for NEET — Practice Questions with Answers

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The escape velocity of a particle of mass m varies as

$m^{2}$

$m^{0}$

$m^{- 1}$

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Explanation

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Acceleration due to gravity is ‘g’ on the surface of the earth. The value of acceleration due to gravity at a height of 32 km above earth’s surface is (Radius of the earth = 6400 km)

0.9 g

0.99g

0.8 g

1.01 g

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Explanation

(b) h=32 km, R=6400k, so h $< < R$

$g^{1} = g (1 - \frac{2 h}{R}) = g (1 \frac{2 \times 32}{6400}) \Rightarrow g^{1} = \frac{99}{100} g = 0.99 g$

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

Zero

2 sec

3 sec

Infinite

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Explanation

d) Time period of simple pendulum T=2 $π$ $\sqrt{\frac{L}{g_{e f f}}}$

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For the moon to cease to remain the earth's satellite, its orbital velocity has to increase by a factor of -

$\sqrt{2}$

1/ $\sqrt{2}$

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Explanation

(b) $v_{e} = \sqrt{2} v_{0,}$ i.e. if the orbital velocity of moon is increased by factor of $\sqrt{2}$ then it will escape out from the gravitational field of earth

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The height of the point vertically above the earth’s surface, at which acceleration due to gravity becomes 1% of its value at the surface is (Radius of the earth = R)

10r

20r

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Explanation

(b)

$g = \frac{G M}{R^{2}} g' = \frac{G M}{(R + h)^{2}} & g' = 1 % o f g' = \frac{1}{100} \times g \frac{g'}{g} = {(\frac{R}{R + h})}^{2} \Rightarrow \frac{1}{100} = {(\frac{R}{R + h})}^{2} \Rightarrow h = 9 r$

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Escape velocity on a planet is $v_{e}$ . If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes

$4 v_{d}$

$2 v_{d}$

$v_{d}$

$\frac{1}{2} v_{d}$

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Explanation

(b) $v_{e} = \sqrt{\frac{2 G M}{R}} \Rightarrow v_{e} \propto \sqrt{M} i f R = c o n s \tan t$

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The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon. The ratio of the escape velocity on the surface of earth to that on the surface of moon will be

0.2

2.57

4.81

0.39

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Explanation

$v_{e} = \sqrt{\frac{2 G M}{R}} s o, \frac{v_{e}}{v_{m}} = \sqrt{\frac{M_{e} R_{m}}{M_{m} R_{e}}} = \sqrt{\frac{81}{3.5}} = 4.81$

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If radius of earth is R then the height h’ at which value of ‘g’ becomes one-fourth is

$\frac{R}{4}$

$\frac{3 r}{4}$

$\frac{R}{8}$

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Explanation

By solving h = R

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The escape velocity from the surface of earth is $V_{e}$ . The escape velocity from the surface of a planet whose mass and radius are 3 times those of the earth will be

$V_{d}$

$3 V_{d}$

$9 V_{d}$

$27 V_{d}$

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Explanation

(a) $v_{e} = \sqrt{\frac{2 G M}{R}} s o, v_{e} \propto \sqrt{\frac{M}{R}}$

If mass and radius of the planet are three times than that of earth then escape velocity will be same

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How much energy will be necessary for making a body of 500 kg escape from the earth ?

[g = 9.8 m / s^{2}, r a d i u s o f e a r t h = 6.4 \times 10^{6} m]

$A b o u t 9.8 \times 10^{6} J$

$A b o u t 6.4 \times 10^{8} J$

$A b o u t 3.1 \times 10^{10} J$

$A b o u t 27.4 \times 10^{12} J$

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Explanation

$P E = - \frac{G M m}{R} = \frac{g R^{2} m}{R} = - m g R = - 500 \times 9.8 \times 6.4 \times 10^{6} = - 3.1 \times 10^{10} J$

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Frequently Asked Questions

Yes. You can attempt every Physics 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.