Q 11-09-100JEE MainJEE Main 2023 (1 Feb, Shift 1)Medium
A mercury drop of radius $10^{-3}$ m is broken into $125$ equal size droplets. Surface tension of mercury is $0.45\ \text{N m}^{-1}$. The gain in surface energy is
Answer: (A) $2.26\times10^{-5}\ \text{J}$
Each droplet has radius $r=\dfrac R5=2\times10^{-4}$ m.
$$\Delta A=4\pi(125r^2-R^2)=4\pi(5\times10^{-6}-10^{-6})=16\pi\times10^{-6}\ \text{m}^2$$
$\Delta E=T\Delta A=0.45\times16\pi\times10^{-6}\approx2.26\times10^{-5}$ J.
Solution by Sreeraj P, M.Sc Physics