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Gravitation MCQs for NEET — Physics Questions with Answers

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A body of mass m rises to height h = R/5 from the earth's surface, where R is earth's radius. If g is acceleration due to gravity at earth's surface, the increase in potential energy is

mgh

$\frac{4}{5} m g h$

$\frac{5}{6} m g h$

$\frac{6}{7} m g h$

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Explanation

Substituting $R = 5 h w e g e t ∆ U = \frac{m g h}{1 + 1 / 5} = \frac{5}{6} m g h$

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Spot the wrong statement :

The acceleration due to gravity ‘g’ decreases if

We go down from the surface of the earth towards its centre

We go up from the surface of the earth

We go from the equator towards the poles on the surface of the earth

The rotational velocity of the earth is increased

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Explanation

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Which of the following statements is true?

g is less at the earth's surface than at a height above it or a depth below it

g is the same at all places on the surface of the earth

g has its maximum value at the equator

g is greater at the poles than at the equator

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Explanation

(d)

The value of 'g' that is gravity is greater at the poles because the gravitational pull is maximum at the poles and decreases as it comes down toward the equator.

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A spring balance is graduated on sea level. If a body is weighed with this balance at consecutively increasing heights from earth's surface, the weight indicated by the balance

Will go on increasing continuously

Will go on decreasing continuously

Will remain same

Will first increase and then decrease

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Explanation

(b) Because value of g decreases with increasing height

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The gravitational field due to a mass distribution is $E = K / x^{3}$ in the x-direction. (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is

K/x

K/2x

$K / x^{2}$

K/2 $x^{2}$

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Explanation

(d) Gravitational potential $= \int I d x = \int_{x}^{\infty} \frac{K}{x^{3}} d x = K {(\frac{x^{- 3 + 1}}{- 3 + 1})}_{x}^{\infty} = |\frac{- K}{2 x^{2}}| = \frac{K}{2 x^{2}}$

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The change in potential energy, when a body of mass m is raised to a height nR from the earth's surface is (R = Radius of earth)

$m g R \frac{n}{n - 1}$

nmgR

C.
mgR $\frac{n^{2}}{n^{2} + 1}$
D.

$m g R \frac{n}{n + 1}$

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Explanation

(d) $∆ U = \frac{m g h}{1 + \frac{h}{R}} = \frac{m g n R}{1 + \frac{n R}{R}} = \frac{n m g R}{n + 1}$

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If the earth suddenly shrinks (without changing mass) to half of its present radius, the acceleration due to gravity will be

g/2

g/4

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Explanation

(b) $g = \frac{G M}{R^{2}}$ . If radius shrinks to half of its present value then g will becomes four time

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The masses and radii of the earth and moon are $M_{1,} R_{1}$ and $M_{2}, R_{2}$ respectively. Their centres are distance d apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it escapes to infinity is

$2 \sqrt{\frac{G}{D} (M_{1} + M_{2})}$

$2 \sqrt{\frac{2 G}{D} (M_{1} + M_{2})}$

$2 \sqrt{\frac{G M}{D} (M_{1} + M_{2})}$

$2 \sqrt{\frac{G M (M_{1} + M_{2})}{D (R_{1} + R_{2})}}$

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Explanation

(a) Gravitational potential at mid point

$V = \frac{- G M_{1}}{d / 2} + \frac{- G M_{2}}{d / 2} N o w, P E = m \times V = \frac{- 2 G M}{d} (M_{1} + M_{2}) [m = m a s s o f p a r t i c l e] S o, f o r p r o j e c t i n g p a r t i c l e f r o m m i d p o i n t t o i n f i n i t y$

$K E = |P E| \Rightarrow \frac{1}{2} m v^{2} = \frac{2 G m}{d} (M_{1} + M_{2}) \Rightarrow v = 2 \sqrt{\frac{G (M_{1} + M_{2})}{d}}$

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If mass of earth is M, radius is R and gravitational constant is G, then work done to take 1 kg mass from earth surface to infinity will be

$\sqrt{\frac{G M}{2 R}}$

$\frac{G M}{R}$

$\sqrt{\frac{2 G M}{R}}$

$\frac{G M}{2 R}$

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Explanation

(b) Potential energy of the 1 kg mass which is placed at the earth surface= $- \frac{G M}{R}$

its potential energy at infinite = 0

Work done = change in potential energy = $\frac{G M}{R}$

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If $v_{e}$ and $v_{o}$ represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius R, then

$v_{e} = v_{o}$

$\sqrt{2} v_{o} = v_{e}$

$v_{e} = v_{o} / \sqrt{2}$

$v_{e}$ and $v_{o}$ are not related

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Explanation

(b) $v_{e} = \sqrt{2 g R} a n d v_{o} = \sqrt{g R} ∴ \sqrt{2} v_{o} = v_{e}$

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