Q79 / 90 · JEE Main 2021 (20 Jul Shift 2) · Physics

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

Solution

$\frac{4}{3} π R^{3} + \frac{4}{3} π R^{3} = \frac{4}{3} π R^{' 3}$

$R^{'} = 2^{\frac{1}{3}} R \dots (i)$

$A_{i} = 2 [4 π R^{2}]$

$A_{f} = 4 π R^{' 2}$

$\frac{U_{i}}{U_{f}} = \frac{A_{i}}{A_{f}} = \frac{2 R^{2}}{2^{\frac{2}{3}} R^{2}} = 2^{\frac{1}{3}}$

Hence, the correct answer is option: 'A'.