Q 11-09-117JEE MainJEE Main 2021 (20 Jul, Shift 2)Easy
Two small drops of mercury each of radius $R$ coalesce to form a single large drop. The ratio of total surface energy before and after the change is
Answer: (A) $2^{\frac13} : 1$
Volume conservation: $R'^3 = 2R^3 \Rightarrow R' = 2^{1/3}R$. Surface energy $\propto$ area:
$$\frac{E_1}{E_2} = \frac{2\times4\pi R^2}{4\pi(2^{1/3}R)^2} = \frac{2}{2^{2/3}} = 2^{1/3}$$
Solution by Sreeraj P, M.Sc Physics