Q 11-09-026NEETJEE MainHard
A liquid drop of radius $R$ and surface tension $T$ is split into $1000$ identical small drops. The work done is
Answer: (D) $36\pi R^2T$
Volume is conserved: $1000r^3 = R^3$, so $r = \dfrac{R}{10}$.
New area $= 1000 \times 4\pi\dfrac{R^2}{100} = 40\pi R^2$. The increase is $40\pi R^2 - 4\pi R^2 = 36\pi R^2$.
$W = T \times \Delta A = 36\pi R^2T$.
Solution by Sreeraj P, M.Sc Physics