Question

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 $\rho$)

• A. $\sqrt{\frac{T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}$
• B. $\sqrt{\frac{2T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}$
• C. $\sqrt{\frac{T}{4\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}$
• D. $\sqrt{\frac{6T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)}$

Answer 1

Answer: (D)

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} $
$= 4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]$
$v = \sqrt{\frac{6T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right]}$