Q 11-09-107JEE MainJEE Main 2022 (26 Jul, Shift 1)Medium
A water drop of radius $1\ \text{cm}$ is broken into $729$ equal droplets. If the surface tension of water is $75\ \text{dyne cm}^{-1}$, then the gain in surface energy up to first decimal place will be [Given $\pi = 3.14$]
Answer: (C) $7.5\times10^{-4}\ \text{J}$
Volume conservation: $r = \dfrac{R}{729^{1/3}} = \dfrac R9$.
$$\Delta A = 4\pi(729r^2 - R^2) = 4\pi R^2(9 - 1) = 32\pi\ \text{cm}^2 = 100.48\ \text{cm}^2$$
$$\Delta E = 75\times100.48 = 7536\ \text{erg}\approx7.5\times10^{-4}\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics