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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

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2 :1

$2^{1 / 3} : 1$

1: 2

$1 : 2^{1 / 3}$

answer is A.

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Detailed Solution

Let R' be radius of common drop. Using volume conservation,

$\frac{4}{3} {πR}^{3} \times 2 = \frac{4}{3} πR'^{3} \Rightarrow R' = 2^{1 / 3} R$

We know, $U = TA = T 4 {πR}^{2}$

$\begin{array}{l} \frac{U_{1}}{U_{2}} = \frac{A_{1}}{A_{2}} = \frac{2 (4 {πR}^{2})}{4 πR'^{2}} = \frac{2 R^{2}}{{(2^{1 / 3})}^{2} R^{2}} = \frac{2^{1 / 3}}{1} \end{array}$

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