\(4\pi {R^3}T\left[ {\frac{1}{r} - \frac{1}{R}} \right]\)

Where, \(R= radius\,of\,big\,drop\)

\(r=radius\,of\,small\,drop\)

\(T=surface\,tension\)

According to question

\(\frac{1}{2}m{v^2} = 4\pi {R^3}T\left[ {\frac{1}{r} - \frac{1}{R}} \right]\)

\( \Rightarrow \frac{1}{2}\left[ {\frac{4}{3}\pi {R^3}\rho } \right]{v^2} = 4\pi {R^3}T\left[ {\frac{1}{r} - \frac{1}{R}} \right]\)

\( \Rightarrow {V^2} = \frac{{6T}}{\rho }\left[ {\frac{1}{r} - \frac{1}{R}} \right] \Rightarrow V = {\left[ {\frac{{6T}}{\rho }\left( {\frac{1}{r} - \frac{1}{R}} \right)} \right]^{\frac{1}{2}}}\)