JEE Main 2014 · 6 Apr · Shift 1, Physics question 7

The particles sit at the corners of a square of side $\sqrt2R$. On one particle:

• each of the two neighbours (distance $\sqrt2R$) pulls with $\dfrac{GM^2}{2R^2}$ at $45^\circ$ to the radius;
• the opposite particle (distance $2R$) pulls with $\dfrac{GM^2}{4R^2}$.

Net force towards the centre:
$$\frac{GM^2}{R^2}\left(\frac{2\cos45^\circ}2+\frac14\right)=\frac{GM^2}{4R^2}\left(1+2\sqrt2\right)=\frac{Mv^2}R,$$
so $v=\dfrac12\sqrt{\dfrac{GM}{R}\left(1+2\sqrt2\right)}$.