Splitting & Merging of Drops question from JEE Main 2021 (20 Jul, Shift 2)

Answer: (a)

1. Volume is conserved: $2 \cdot \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R'^3 \Rightarrow R' = 2^{1/3}R$.
2. Surface energy $E = 4\pi r^2 T$, so $E_{before} = 2 \cdot 4\pi R^2 T$ and $E_{after} = 4\pi (2^{1/3}R)^2 T = 2^{2/3} \cdot 4\pi R^2 T$.
3. $\frac{E_{before}}{E_{after}} = \frac{2}{2^{2/3}} = 2^{1/3}$
4. So the answer is (a) $2^{1/3} : 1$.