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Q.

If n identical small droplets each of radius r merge to form a big drop of radius R and the energy released in this process is absorbed by the big drop so as to increase its kinetic energy then the speed of the big drop (T be the surface tension and d be the density of the liquid droplets )

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a

$\sqrt{\frac{6 T}{d} [\frac{1}{R} - \frac{1}{r}]}$

b

$\sqrt{\frac{6 T}{d} [\frac{1}{r} - \frac{1}{R}]}$

c

$\sqrt{\frac{6 d}{T} [\frac{1}{R} - \frac{1}{r}]}$

d

$\sqrt{\frac{6 d}{T} [\frac{1}{r} - \frac{1}{R}]}$

answer is B.

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Detailed Solution

$E n e r g y r e l e a s e d = 4 π n r^{2} T - 4 π R^{2} T n = \frac{R^{3}}{r^{3}} \frac{1}{2} {mv}^{2} = E = 4 {πR}^{3} T (\frac{1}{r} - \frac{1}{R}) \frac{1}{2} \frac{4}{3} {πR}^{3} {dv}^{2} = 4 {πR}^{3} T (\frac{1}{r} - \frac{1}{R}) v^{2} = \frac{6 T}{d} (\frac{1}{r} - \frac{1}{R})$

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