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Gravitation MCQs for NEET — Physics Questions with Answers

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Four particles each of mass M, are located at the vertices of a square with side L. The gravitational potential due to this at the centre of the square is

$- \sqrt{32} \frac{G M}{L}$

$- \sqrt{64} \frac{G M}{L^{2}}$

zero

$\sqrt{32} \frac{G M}{L}$

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Explanation

$P o t e n t i a l a t t h e c e n t r e d u e t o \sin g l e m a s s = \frac{- G M}{L / \sqrt{2}} P o t e n t i a l a t t h e c e n t r e d u e t o a l l f o u r m a s s e s = - 4 \frac{G M}{L / \sqrt{2}} = - 4 \sqrt{2} \frac{G M}{L} = - \sqrt{32} \times \frac{G M}{L} .$

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The mass of the moon is $7.34 \times 10^{22} k g$ and the radius is $1.74 \times 10^{6} m$ . The value of gravitation field will be

$1.45 N / k g$

$1.55 N / k g$

$1.75 N / k g$

$1.62 N / k g$

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Explanation

(d) $g = \frac{G M}{R^{2}} = \frac{6.67 \times 10^{‐ 11} \times 7.34 \times 10^{22}}{{(1.74 \times 10^{6})}^{2}} = 1.62 N / k g$

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The centripetal force acting on a satellite orbiting round the earth and the gravitational force of earth acting on the satellite both equal F. The net force on the satellite is

Zero

$F \sqrt{2}$

2 F

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Explanation

(b) Actually gravitational force provides the centripetal force.

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Reason of weightlessness in a satellite is:

Zero gravity

Centre of mass

Zero reaction force by satellite surface

None

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Explanation

(c)

The reason of weightlessness in a satellite is a zero reaction force by satellite surface.

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Mass M is divided into two parts xM and $(1 - x) M$ . For a given separation, the value of x for which the gravitational attraction between the two pieces becomes maximum is

$\frac{1}{2}$

$\frac{3}{5}$

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Explanation

(a)

$F \propto xm \times (1 - x) m = {xm}^{2} (1 - x) = {xm}^{2} - x^{2} m^{2} For maximum force; \frac{dF}{dx} = 0 \frac{dF}{dx} = m^{2} - 2 {xm}^{2} = 0 \Rightarrow x = 1 / 2$

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If earth is supposed to be a sphere of radius R, if $g_{30 °}$ is value of acceleration due to gravity at latitude of 30^o and g at the pole, the value of $g - g_{30^{ο}}$ is

$\frac{1}{4} ω^{2} R$

$\frac{3}{4} ω^{2} R$

$ω^{2} R$

$\frac{1}{2} ω^{2} R$

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Explanation

(b) Acceleration due to gravity at latitude $λ$ is given by

$g' = g - R ω^{2} \cos^{2} λ a t 30^{ο}, g_{30^{ο}} = g - R ω^{2} \cos^{2} 30^{ο} = g - \frac{3}{4} R ω^{2} s o g - g_{30 °} = \frac{3}{4} ω^{2} R$

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The radii of two planets are respectively $R_{1}$ and $R_{2}$ and their densities are respectively $ρ_{1}$ and $ρ_{2}$ . The ratio of the accelerations due to gravity at their surfaces is

$g_{1} : g_{2} = \frac{ρ_{1}}{{R_{1}}^{2}} : \frac{ρ_{2}}{{R_{2}}^{2}}$

$g_{1} : g_{2} = R_{1} R_{2} : ρ_{1} ρ_{2}$

$g_{1} : g_{2} = R_{1} ρ_{2} : R_{2} ρ_{1}$

$g_{1} : g_{2} = R_{1} ρ_{1} : R_{2} ρ_{2}$

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Explanation

(d) $g = \frac{4}{3} π ρ G R S o \frac{g_{1}}{g_{2}} = \frac{R_{1} ρ_{1}}{R_{2} ρ_{2}}$

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If M the mass of the earth and R its radius, the ratio of the gravitational acceleration and the gravitational constant is

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

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

$M R^{2}$

$\frac{M}{R}$

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Explanation

(b) Acceleration due to gravity g= $\frac{G M}{R^{2}}$

$S o, \frac{g}{G} = \frac{M}{R^{2}}$

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Two sphere of mass m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be

$\frac{F}{3}$

$\frac{F}{9}$

3 F

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Explanation

(a) Gravitational force does not depend on the medium.

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Earth binds the atmosphere because of:

Gravity

Oxygen between earth and atmosphere

Both (a) and (b)

None of these

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Explanation

(a)

Various atom of gases present in the earth atmosphere has masses and a gravitational force acts between theses atoms and earth.

also this is the force that holds the various gases near the surface and the atmosphere is present.

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