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Physics MCQs for NEET — Practice Questions with Answers

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The velocity with which a projectile must be fired so that it escapes earth’s gravitation does not depend on:

Mass of the earth

Mass of the projectile

Radius of the projectile’s orbit

Gravitational constant

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Explanation

(b) We know that, $V_{e} = \sqrt{\frac{2 G M}{r}}$

From above expression we can see, mass of projectile does not matter.

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The gravitational force between two stones of mass 1 kg each separated by a distance of 1 metre in vacuum is

Zero

6.675 $\times 10^{‐ 5} n e w t o n$

$6.675 \times 10^{‐ 11} n e w t o n$

$6.675 \times 10^{‐ 8} n e w t o n$

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Explanation

(c) $F = G \frac{m_{1} m_{2}}{r^{2}} = 6.675 \times \frac{1 \times 1}{1^{2}} \times 10^{‐ 11} = 6.675 \times 10^{‐ 11} N$

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The radius of a planet is $\frac{1}{4}$ of earth’s radius and its acceleration due to gravity is double that of earth’s acceleration due to gravity. How many times will the escape velocity at the planet’s surface be as compared to its value on earth’s surface ?

$\frac{1}{\sqrt{2}}$

$\sqrt{2}$

$2 \sqrt{2}$

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Explanation

$(a) v = \sqrt{2 g R} \Rightarrow \frac{v_{p}}{v_{e}} = \sqrt{\frac{g_{p}}{g_{e}} \times \frac{R_{e}}{R_{p}}} = \sqrt{2 \times \frac{1}{4}} = \frac{1}{\sqrt{2}} s o, v_{p} = \frac{v_{e}}{\sqrt{2}}$

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The escape velocity for the earth is $V_{d}$ . The escape velocity for a planet whose radius is four times and density is nine times that of the earth, is

$36 v_{d}$

$12 v_{d}$

$6 v_{d}$

$20 v_{d}$

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Explanation

(b) $v \propto R \sqrt{ρ} so, \frac{v_{p}}{v_{e}} = \frac{R_{p}}{R_{e}} \times \sqrt{\frac{ρ_{p}}{ρ_{e}}} = 4 \sqrt{9} = 12 \Rightarrow v_{p} = 12 v_{e}$

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Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

$ν = \frac{1}{2 R} \sqrt{\frac{1}{G m}}$

$ν = \sqrt{\frac{G m}{2 R}}$

$ν = \frac{1}{2} \sqrt{\frac{G m}{R}}$

$ν = \sqrt{\frac{4 G m}{R}}$

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Explanation

i.e. $\frac{m v^{2}}{R} = \frac{G m \times m}{{(2 R)}^{2}}$

$\Rightarrow ν = \frac{1}{2} \sqrt{\frac{G m}{R}}$

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The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is (g = acceleration due to gravity at the surface of the earth)

$\frac{g}{9}$

$\frac{g}{3}$

$\frac{g}{4}$

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Explanation

(a) $g' = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{R + 2 R})}^{2} = \frac{g}{9}$

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If V, R, and g denote respectively the escape velocity from the surface of the earth, the radius of the earth, and acceleration due to gravity, then the correct equation is:

$V = \sqrt{g R}$

V= $\sqrt{\frac{4}{3} g R^{3}}$

V = R $\sqrt{g}$

V= $\sqrt{2 g R}$

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Explanation

$v = \sqrt{\frac{2 G M}{R}} \sin c e, g = \frac{G M}{R^{2}} \Rightarrow v = \sqrt{2 g R}$

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The depth at which the effective value of acceleration due to gravity is $\frac{g}{4}$ is

$\frac{3 R}{4}$

$\frac{R}{2}$

$\frac{R}{4}$

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Explanation

b) $g' = g (1 - \frac{d}{R}) \Rightarrow \frac{g}{4} = g (1 - \frac{d}{R}) \Rightarrow d = \frac{3 R}{4}$

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The value of ‘g’ at a particular point is $9.8 m / s^{2}$ . Suppose the earth suddenly shrinks 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 earth does not shrink) will now be

$4.9 m / s e c^{2}$

$3.1 m / s e c^{2}$

$9.8 m / s e c^{2}$

$19.6 m / s e c^{2}$

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Explanation

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The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is $V_{d}$ on earth -

$V_{d}$

2 $V_{d}$

$4 V_{d}$

$\frac{V_{d}}{2}$

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Explanation

$(b) v = \sqrt{2 g R} \Rightarrow \frac{v_{p}}{v_{e}} = \sqrt{\frac{g_{p}}{g_{e}} \times \frac{R_{e}}{R_{p}}} = \sqrt{1 \times 4} = 2 s o, v_{p} = 2 v_{e}$

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