The masses m each are placed at the corners of an equilateral triangle of side a and a mass $\frac{m}{3}$ is placed at the centre of the centre of triangle. The 3 masses are in uniform circular motion with velocity v on the circumcircle of the triangle, the centripetal force being provided by their mutual attraction. Then

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$v = \sqrt{\frac{GM}{a} \frac{(\sqrt{3} + 1)}{\sqrt{3}}}$

$v = \sqrt{\frac{GM}{a} \frac{(\sqrt{3} - 1)}{\sqrt{3}}}$

The mass $\frac{m}{3}$ experiences no force

$If \frac{m}{3} is absent, v = \sqrt{\frac{GM}{a}}$

answer is A, C, D.

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Detailed Solution

The result of all forces on each m acts in the direction towards centre.
$\begin{array}{l} F_{r} = F^{'} + 2 Fcos 30^{\circ} = F^{'} + \sqrt{3} F \\ = \frac{Gm \frac{M}{3}}{a^{2} / 3} + \frac{\sqrt{3} Gmm}{a^{2}} = \frac{{Gm}^{2}}{a^{2}} [1 + \sqrt{3}] \end{array}$
Since this provides the centripetal force:
$\begin{array}{l} \frac{{Gm}^{2}}{a^{2}} [1 + \sqrt{3}] = \frac{{MV}^{2}}{[\frac{a}{\sqrt{3}}]} \\ ∴ v = \sqrt{\frac{Gm}{a} [\frac{1 + \sqrt{3}}{\sqrt{3}}]} \Rightarrow If \frac{m}{3} \end{array}$
is absent, then
$F_{r} = \sqrt{3} F \Rightarrow∴ \sqrt{3} \frac{Gmm}{a^{2}} = \frac{{mv}^{2}}{\frac{a}{\sqrt{3}}} \Rightarrow v = \sqrt{\frac{Gm}{a}}$