Gravitational Potential & Energy question from JEE Main 2022 (27 Jun, Shift 2)

Answer: (a)

1. Total PE = sum of $-\frac{Gm_im_j}{r_{ij}}$ over every pair (10 pairs).
2. 4 sides ($m$–$m$, distance $d$): $-\frac{4Gm^2}{d}$
3. 2 diagonals ($m$–$m$, distance $\sqrt{2}d$): $-\frac{2Gm^2}{\sqrt{2}d} = -\frac{\sqrt{2}Gm^2}{d}$
4. 4 pairs $M$–$m$ (distance $\frac{d}{\sqrt{2}}$): $-\frac{4\sqrt{2}GMm}{d}$
5. So the answer is (a) $U = -\frac{Gm}{d}\left[(4+\sqrt{2})m + 4\sqrt{2}M\right]$.