Three solid spheres each of mass m and radius R are placed at three corners of an equilateral triangle of side ‘d’ released. The speed of any one sphere at the time of collision would be d>2R [Assume there is no external gravitational force acting on the system of three spheres].

From conservation of mechanical energy

Energy of system at the instance of release = Energy of system at the instance of Collision

$∴ 3 [- \frac{G m^{2}}{d}] = 3 [- \frac{G m^{2}}{2 R}] + 3 [\frac{1}{2} m v^{2}]$

and $v = \sqrt{G m (\frac{1}{R} - \frac{2}{d})}$

Therefore, the correct answer is (D).

Three solid spheres each of mass m and radius R are placed at three corners of an equilateral triangle of side ‘d’ released. The speed of any one sphere at the time of collision would be $(d > 2 R)$ [Assume there is no external gravitational force acting on the system of three spheres].