Splitting & Merging of Drops question from MHT CET 2023 (10 May, Shift 2)

Answer: (d)

1. Energy released $= T\,\Delta A = 3VT\left(\frac{1}{r} - \frac{1}{R}\right)$, where $V = \frac{4}{3}\pi R^3$.
2. KE of big drop $= \frac{1}{2}(\rho V) v^2$
3. $\frac{1}{2}\rho V v^2 = 3VT\left(\frac{1}{r} - \frac{1}{R}\right) \Rightarrow v = \sqrt{\frac{6T}{\rho}\left[\frac{1}{r} - \frac{1}{R}\right]}$
4. So the answer is (d).