Splitting & Merging of Drops question from JEE Main 2014 (19 Apr, Shift 1)

Answer: (d)

1. Volume is conserved: $n \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow n = \frac{R^3}{r^3}$.
2. Energy released $= T\,\Delta A = T \cdot 4\pi(nr^2 - R^2) = 4\pi R^3 T\left(\frac{1}{r} - \frac{1}{R}\right)$.
3. This becomes KE: $\frac{1}{2}\left(\frac{4}{3}\pi R^3 \rho\right)v^2 = 4\pi R^3 T\left(\frac{1}{r} - \frac{1}{R}\right)$.
4. $v^2 = \frac{6T}{\rho}\left(\frac{1}{r} - \frac{1}{R}\right)$
5. So the answer is (d) $\sqrt{\frac{6T}{\rho}\left(\frac{1}{r} - \frac{1}{R}\right)}$.