Q 11-09-031JEE MainJEE Main 2026 (5 Apr, Shift 2)Easy
Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is ______ ($S$ is the surface tension of mercury).
Answer: (B) $16\pi r^2S$
Volume is conserved: $R^3 = 8r^3$, so $R = 2r$.
Area before: $8 \times 4\pi r^2 = 32\pi r^2$. After: $4\pi(2r)^2 = 16\pi r^2$.
Energy released $= S \times 16\pi r^2$.
Solution by Sreeraj P, M.Sc Physics