A large number of liquid drops each of radius r coalesce to form a single drop of radius R. The energy released in the process is converted into the kinetic energy of the big drop so formed. The speed of the big drop is: (given surface tension of liquid is T,density of liquid is $ρ$ )

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

$\sqrt{\frac{4 T}{ρ} (\frac{1}{r} - \frac{1}{R})}$

$\sqrt{\frac{T}{ρ} (\frac{1}{r} - \frac{1}{R})}$

$\sqrt{\frac{6 T}{ρ} (\frac{1}{r} - \frac{1}{R})}$

answer is D.

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Detailed Solution

Energy released $= \frac{1}{2} m v^{2}$

$T 4 π R^{3} [\frac{1}{r} - \frac{1}{R}] = \frac{1}{2} \frac{4}{3} π R^{3} ρ v^{2}$

$v = \sqrt{\frac{6 T}{ρ} [\frac{1}{r} - \frac{1}{R}]}$

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