' \(n\) ' 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 [T = surface tension of liquid, \(\rho=\) density of liquid.]

\(\begin{aligned}
& \frac{4}{3} \pi R^3=n \times \frac{4}{3} \pi r^3... (given) \\
\therefore \quad & R^3=n r^3
\end{aligned}\)
Energy released,
\(\begin{aligned}
\Delta \mathrm{U} & =\mathrm{T} \times 4 \pi \mathrm{r}^2 \times \mathrm{n}-\mathrm{T} \times 4 \pi \mathrm{R}^2 \\
& =\mathrm{T} \times 4 \pi \frac{\mathrm{R}^3}{\mathrm{r}}-\mathrm{T} \times 4 \pi \mathrm{R}^2
\end{aligned}\)
This energy is converted into K.E
\(\begin{array}{ll}
\therefore \quad & \frac{1}{2} m v^2=T \times 4 \pi R^3\left[\frac{1}{r}-\frac{1}{R}\right] \\
\Rightarrow & \frac{1}{2} \rho \times \frac{4}{3} \pi R^3 \times v^2=T \times 4 \pi R^3\left[\frac{1}{r}-\frac{1}{R}\right] \\
& v^2=\frac{6 T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right] \\
\therefore \quad v & =\sqrt{\frac{6 T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right]}
\end{array}\)