Q 11-09-118JEE MainJEE Main 2021 (26 Feb, Shift 1)Medium
A large number of water drops, each of radius $r$, combine to have a drop of radius $R$. If the surface tension is $T$ and mechanical equivalent of heat is $J$, the rise in heat energy per unit volume will be:
Answer: (D) $\frac{3T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$
Volume is conserved: $n\cdot\frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 = V$.
Energy released $= T\,\Delta A = T\left(n\,4\pi r^2 - 4\pi R^2\right) = 3TV\left(\frac{1}{r} - \frac{1}{R}\right)$, since $n\,4\pi r^2 = \dfrac{3V}{r}$ and $4\pi R^2 = \dfrac{3V}{R}$.
Heat per unit volume:
$$\frac{Q}{V} = \frac{3T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$$
Solution by Sreeraj P, M.Sc Physics