Q 11-09-048JEE MainJEE Main 2025 (24 Jan, Shift 1)Medium
The amount of work done to break a big water drop of radius $R$ into 27 small drops of equal radius is $10\ \text{J}$. The work done required to break the same big drop into 64 small drops of equal radius will be
Answer: (A) $15\ \text{J}$
Volume is conserved: $n\cdot\tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi R^3 \Rightarrow r = R/n^{1/3}$.
Work $= T\,\Delta A = T\cdot4\pi(nr^2 - R^2) = 4\pi TR^2(n^{1/3} - 1)$.
- $n = 27$: $n^{1/3} - 1 = 2$
- $n = 64$: $n^{1/3} - 1 = 3$
$$W_{64} = 10\times\frac{3}{2} = 15\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics