JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Gravitation

Each mass sits at distance $R$ from the centre at the corners of a square, so adjacent masses are $\sqrt{2}R$ apart and opposite masses $2R$ apart. The two adjacent masses each pull with $\frac{G{M}^2}{2R^2}$, and at right angles their resultant is $\sqrt{2}\cdot \frac{G{M}^2}{2R^2}=\frac{G{M}^2}{\sqrt{2}{R}^2}$ toward the centre. The opposite mass adds $\frac{G{M}^2}{(2R{)}^2}=\frac{G{M}^2}{4R^2}$. Setting the net inward force equal to $\frac{M{V}^2}{R}$ gives $\frac{G{M}^2}{R^2}\left(\frac{1}{\sqrt{2}}+\frac{1}{4}\right)=\frac{M{V}^2}{R}$, so $V=\frac{1}{2}\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}$.