Solution:
Solution:
When drops combine to form a single drop of radius R.
Then energy released,
$$E = 4\pi TR^{3} \left[\frac{1}{r}-\frac{1}{R}\right]$$
If this energy is converted into kinetic energy then
$$\frac{1}{2}mv^{2} = 4\pi R^{3} T\left[\frac{1}{r}-\frac{1}{R}\right]$$
$$\frac{1}{2}\times\left[\frac{4}{3}\pi R^{3}\rho\right]v^{2} = 4\pi R^{3} T\left[\frac{1}{r}-\frac{1}{R}\right]$$
$$v^{2} = \frac{6T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right]$$
$$v = \sqrt{\frac{6T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right]}$$