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

Answer 1

The correct option is D $\sqrt{\frac{6T}{\rho }(\frac{1}{r}-\frac{1}{R})}$

$n\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3$

$\Rightarrow n=(\frac{R}{r}{)}^3$

Change in energy $=n4\pi r^{2}T-4\pi R^{2}T$

$=4\pi T(\frac{R^3}{r}-R^2)$

Now

$\frac{1}{2}m{v}^2=4\pi T(\frac{R^3}{r}-R^2)$

$\frac{1}{2}\times \frac{4}{3}\pi R^3\rho \times v^2=4\pi T{R}^3(\frac{1}{r}-\frac{1}{R})$

$\Rightarrow v=\sqrt{\frac{12s}{\rho }(\frac{1}{r}-\frac{1}{R})}$

$=\sqrt{\frac{6T}{\rho }(\frac{1}{r}-\frac{1}{R})}$