Splitting & Merging of Drops question from MHT CET 2023 (13 May, Shift 1)

Answer: (c)

1. Energy released $= 3VT\left(\frac{1}{r} - \frac{1}{R}\right)$ (area change $\times T$, with $V = \frac{4}{3}\pi R^3$).
2. Heat $= \frac{3VT}{J}\left(\frac{1}{r} - \frac{1}{R}\right) = m s \Delta\theta = V \rho s \Delta\theta$
3. For water in CGS units $\rho = 1$, $s = 1$: $\Delta\theta = \frac{3T}{J}\left(\frac{1}{r} - \frac{1}{R}\right)$
4. So the answer is (c).