JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Fluid Mechanics

Volume is conserved when the drops coalesce, $8\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$, so the large drop has radius $R=2r$. Surface energy equals surface tension times surface area. The initial surface energy of the eight drops is $8(4\pi r^2)S=32\pi r^{2}S$, and the final surface energy of the single drop is $4\pi R^{2}S=4\pi (2r{)}^{2}S=16\pi r^{2}S$. The energy released is the decrease in surface energy, $32\pi r^{2}S-16\pi r^{2}S=16\pi r^{2}S$. The surface energy released in the process is $16\pi r^{2}S$.