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Physics MCQs for NEET — Practice Questions with Answers

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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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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]

$\frac{R}{n}$

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

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

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

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Explanation

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

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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

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Explanation

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

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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)

$\frac{1}{4} R$

$\frac{3}{8} R$

$\frac{R}{2}$

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Explanation

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

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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)

$- 2 m g R_{e}$

$2 m g R_{e}$

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

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

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Explanation

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

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

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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

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Explanation

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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

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Explanation

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

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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

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Explanation

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

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