A uniform ring of mass $m$ and radius $r$ is placed directly above a uniform sphere of mass $M$ and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance $r \sqrt{3}$ as shown in the figure.
The gravitational force exerted by the sphere on the ring will be

\frac{G M m}{8 r^{2}}

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\frac{G M m}{4 r^{2}}

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

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

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Solution

The correct option is C $\sqrt{3} \frac{G M m}{8 r^{2}}$
The gravitational field due to the ring at a distance $\sqrt{3} r$ is given by
$E = \frac{G M (\sqrt{3} r)}{{[r^{2} + (\sqrt{3} r)^{2}]}^{3 / 2}} \Rightarrow E = \frac{\sqrt{3} G m}{8 r^{2}}$
Attractive force $= \frac{\sqrt{3} G m M}{8 r^{2}}$

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