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

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

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

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

answer is A.

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Detailed Solution

As the volume remains the same, we have
$\frac{4}{3} {πR}^{3} = n (\frac{4}{3} {πr}^{3}) or R^{3} = {nr}^{3} Work done W = T \times ΔA = T [4 {πr}^{2} n - 4 {πR}^{2}] or W = T 4 π [{nr}^{2} - R^{2}] Further, W = \frac{1}{2} {mv}^{2} = \frac{1}{2} (\frac{4}{3} {πR}^{3} ρ) v^{2} From eqs . (2) and (3), we have \frac{1}{2} (\frac{4}{3} {πR}^{3} ρ) v^{2} = T 4 π [{nr}^{2} - R^{2}] v^{2} = \frac{6 T}{ρ} [\frac{{nr}^{2}}{R^{3}} - \frac{R^{2}}{R^{3}}] V = \sqrt{\frac{6 T}{ρ} [\frac{{nr}^{2}}{{nr}^{3}} - \frac{1}{R}]} V = \sqrt{\frac{6 T}{ρ} [\frac{1}{r} - \frac{1}{R}]}$