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

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

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

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

answer is A.

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Detailed Solution

$C o n s e r v i n g v o l u m e, N (\frac{4}{3} π r^{3}) = (\frac{4}{3} π R^{3}) \Rightarrow R^{3} = N r^{3}$

Loss of surface energy = $U_{i} - U_{f} = T A_{f} - T A_{i} = N (4 π r^{2} T) - 4 π R^{2} T$

$\Rightarrow$ heat produced $= \frac{4 π T}{J} (N r^{2} - R^{2})$
$\Rightarrow$ heat produced per unit volume $= \frac{4 π T (N r^{2} - R^{2})}{\frac{4}{3} π R^{3} J}$ $= \frac{3 T}{J} (\frac{N r^{2}}{R^{3}} - \frac{1}{R}) = \frac{3 T}{J} (\frac{1}{r} - \frac{1}{R})$