Q 11-09-157JEE MainJEE Main 2025 (2 Apr, Shift 2)Medium
Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension, the surface energy released in this process is:
Answer: (A) $4\pi r^2T\left[2 - 2^{2/3}\right]$
Volume is conserved: $2\cdot\tfrac43\pi r^3 = \tfrac43\pi R^3 \Rightarrow R = 2^{1/3}r$.
Surface energy before: $2\times4\pi r^2T$. After: $4\pi R^2T = 4\pi r^2T\cdot2^{2/3}$.
$$\text{Energy released} = 4\pi r^2T\left[2 - 2^{2/3}\right]$$
Solution by Sreeraj P, M.Sc Physics