Three uniform spheres each of mass m and diameter D are kept in such a way that each touches the other two, then magnitude of the gravitational force on any one sphere due to the other two is

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$\frac{\sqrt{3 {Gm}^{2}}}{4 D^{2}}$

$\frac{3 {Gm}^{2}}{D^{2}}$

$\frac{2 \sqrt{3} {Gm}^{2}}{D^{2}}$

$\frac{\sqrt{3} {Gm}^{2}}{D^{2}}$

answer is D.

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Detailed Solution

Gravitational force on one sphere due to the other two is $F = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} Cos θ} = \sqrt{3} F_{1}$
But $F_{1} = \frac{{Gm}^{2}}{D^{2}} \Rightarrow F = \frac{\sqrt{3} {Gm}^{2}}{D^{2}} (∵ F_{1} = F_{2})$

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