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Physics Question on Surface tension

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\rho)

A

Tρ(1r−1R)\sqrt{\frac{T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}

B

2Tρ(1r−1R)\sqrt{\frac{2T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}

C

T4ρ(1r−1R)\sqrt{\frac{T}{4\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}

D

6Tρ(1r−1R)\sqrt{\frac{6T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}

Answer

6Tρ(1r−1R)\sqrt{\frac{6T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}

Explanation

Solution

When drops combine to form a single drop of radius R.
Then energy released,
E=4πTR3[1r−1R]E = 4\pi TR^{3}\left[\frac{1}{r}-\frac{1}{R}\right]
If this energy is converted into kinetic energy then
12mv2=4πR3T[1r−1R]\frac{1}{2}mv^{2} = 4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]
12×[43πR3ρ]v2=4πR3T[1r−1R]\frac{1}{2}\times\left[\frac{4}{3}\pi R^{3}\rho\right]v^{2} = 4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]
v=6Tρ[1r−1R]v = \sqrt{\frac{6T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right]}