Q 11-09-069JEE MainJEE Main 2024 (29 Jan, Shift 2)Medium
A small liquid drop of radius $R$ is divided into 27 identical liquid drops. If the surface tension is $T$, then the work done in the process will be:
Answer: (A) $8\pi R^2T$
Volume is conserved: $27\cdot\frac43\pi r^3 = \frac43\pi R^3 \Rightarrow r = \dfrac R3$.
Increase in surface area:
$$\Delta A = 27\times4\pi\frac{R^2}{9} - 4\pi R^2 = 12\pi R^2 - 4\pi R^2 = 8\pi R^2$$
$$W = T\Delta A = 8\pi R^2T$$
Solution by Sreeraj P, M.Sc Physics