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) $16\pi r^2S$
Concept. Volume is conserved when drops merge, but total surface area falls. Surface energy released $=S\times$ decrease in area.
Step 1. New radius. $8\times\tfrac43\pi r^3=\tfrac43\pi R^3\Rightarrow R=2r$.
Step 2. Areas. Before: $8\times4\pi r^2=32\pi r^2$. After: $4\pi(2r)^2=16\pi r^2$.
Step 3. Energy released. $S(32-16)\pi r^2=16\pi r^2S$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
