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

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

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

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

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

answer is D.

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Detailed Solution

When drops combine to form a single drop of radius R.
$\begin{array}{l} Δ E = T [4 π R^{2} - n \times 4 π r^{2}] \\ = 4 π T [R^{2} - \frac{R^{3}}{r^{3}} r^{2}] [∵ R^{3} = n r^{3}] \\ = 4 π T R^{3} [\frac{1}{R} - \frac{1}{r}] \end{array}$
So energy released, $| E | = 4 π T R^{3} [\frac{1}{r} - \frac{1}{R}]$
If this energy is converted into kinetic energy, then
$\begin{array}{l} \frac{1}{2} m V^{2} = 4 π R^{3} T [\frac{1}{r} - \frac{1}{R}] \\ \frac{1}{2} \times [\frac{4}{3} π R^{3} ρ] V^{2} = 4 π R^{3} T [\frac{1}{r} - \frac{1}{R}] \\ \Rightarrow V^{2} = \frac{6 T}{ρ} [\frac{1}{r} - \frac{1}{R}] \\ \Rightarrow V = \sqrt{\frac{6 T}{ρ} [\frac{1}{r} - \frac{1}{R}]} \end{array}$