A uniform ring of mass M and radius r is placed directly above a uniform sphere of mass 8 M and of same radius R. The centre of the ring is at a distance of d = $\sqrt{3}$ R from the centre of the sphere. The gravitational attraction between the sphere and the ring is

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

$\frac{3 {GM}^{2}}{2 R^{2}}$

$\frac{2 {GM}^{2}}{\sqrt{2} R^{2}}$

$\frac{\sqrt{3} {GM}^{2}}{R^{2}}$

answer is D.

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Detailed Solution

From the figure, the gravitational intensity due to the ring at a distance $d = \sqrt{3} R$ on its axis is
$I = \frac{GM}{{(d^{2} + R^{2})}^{3 / 2}} = \frac{GM \times \sqrt{3} R}{{(3 R^{2} + R^{2})}^{3 / 2}} = \frac{\sqrt{3} GM}{8 R^{2}}$
Force on sphere
$= (8 M) I = (8 M) \times \frac{\sqrt{3} M}{8 R^{2}} = \frac{\sqrt{3} {GM}^{2}}{R^{2}}$