A uniform ring of mass m is lying at a distance $1.73 a$ from the centre of a sphere of mass $M$ just over the sphere where $a$ is the small radius of the ring as well as that of the sphere. Then, the gravitational force exerted by one on the other is

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

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\frac{G M m}{(1.73 a)^{2}}

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

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1.73 \frac{G M m}{8 a^{2}}

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Solution

The correct option is A $1.73 \frac{G M m}{8 a^{2}}$
In this case the expression of gravitational force is $F = G M m \frac{r}{(r^{2} + R^{2})^{\frac{3}{2}}}$
here, $r = 1.73 a, R = a$
$∴ F = G M m \frac{1.73 a}{(2.99 a^{2} + a^{2})^{\frac{3}{2}}} = G M m \frac{1.73 a}{a^{3} (4)^{\frac{3}{2}}} = 1.73 \frac{G M m}{8 a^{2}} ($ take $2.99 \sim 3)$

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