A uniform ring of mass $m$ and radius $a$ is placed directly above a uniform sphere of mass ( $M$ ) and of equal radius. The centre of the ring is at a distance $\sqrt{3}$ a from the centre of the sphere. The gravitational force (F) exerted by the sphere on the ring is

\frac{\sqrt{3} G M m}{8 a^{2}}

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\frac{2 G M m}{3 a^{2}}

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\frac{7 G M m}{\sqrt{2} a^{2}}

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\frac{3 G M m}{a^{2}}

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Solution

The correct option is D $\frac{\sqrt{3} G M m}{8 a^{2}}$
Given data :
Mass of the ring - $m$
Radius of the ring - $a$
Mass of the sphere - $M$
Radius of sphere - $a$
Distance between center of ring and center of sphere - $r = \sqrt{3} a$
The gravitational field, $F^{'}$ of the sphere is given by:
$F^{'} = \frac{G M r}{(r^{2} + a^{2})^{\frac{3}{2}}}$ ........(1)
where, $G$ is gravitational constant.
Using the value of $r$ in equation (1), we get:
$F^{'} = \frac{G M (\sqrt{3} a)}{((\sqrt{3} a)^{2} + a^{2})^{\frac{3}{2}}}$

$F^{'} = \sqrt{3} \frac{G M a}{(3 a^{2} + a^{2})^{\frac{3}{2}}}$

$F^{'} = \sqrt{3} \frac{G M a}{(4 a^{2})^{\frac{3}{2}}}$

$F^{'} = \sqrt{3} \frac{G M a}{(2^{3}) (a^{3})}$

$F^{'} = \frac{\sqrt{3} G M}{8 (a^{2})}$

Now, the force, $F$ exerted by the sphere on the ring is,
$F = F^{'} \times m$

$F = (\frac{\sqrt{3} G M}{8 (a^{2})}) m$

$F = \frac{\sqrt{3} G M m}{8 (a^{2})}$

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