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Two mercury drops (each of radius r ) merge to form a bigger drop. The surface energy of the bigger drop, if T is the surface tension, is

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$4 {πr}^{2} T$

$2^{\frac{5}{3}} {πr}^{2} T$

$2^{\frac{8}{3}} {πr}^{2} T$

$2 {πr}^{2} T$

answer is C.

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Detailed Solution

Volume of big drop = Total volume of small drops

$\Rightarrow \frac{4}{3} {πR}^{3} = 2 (\frac{4}{3} {πr}^{3}) \Rightarrow R = 2^{1 / 3} r$

Surface energy of big drop

$\begin{array}{r} U_{big} & = 4 {πR}^{2} T \\ = 4 π (2^{2 / 3} r^{2}) T \\ = 2^{8 / 3} {πr}^{2} T \end{array}$

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