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A large no. of water drops each of radius r combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be

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$\frac{2 T}{J} (\frac{1}{r} - \frac{1}{R})$

$\frac{2 T}{r J}$

$\frac{3 T}{r J}$

$\frac{3 T}{J} (\frac{1}{r} - \frac{1}{R})$

answer is D.

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Detailed Solution

Total engage in surface energy =T $[n \times 4 π r^{2} - 4 π R^{2}]$

But $n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$

$∴$ Change in surface energy = $4 π T [n r^{2} - R^{2}]$
$∴$ Heat energy per unit volume = $\frac{4 π T [n r^{2} - R^{2}]}{J \times \frac{4}{3} π R^{3}}$

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