Splitting & Merging of Drops question from MHT CET 2026 (13 Apr, Shift 1)

Answer: (b)

1. Volume conserved: $64\, r^3 = R^3 \Rightarrow r = R/4$.
2. Final area $= 64 \cdot 4\pi \frac{R^2}{16} = 16\pi R^2$; initial area $= 4\pi R^2$.
3. $W = T\,\Delta A = T(16\pi R^2 - 4\pi R^2) = 12\pi R^2 T$
4. So the answer is (b) $12\pi R^2 T$.