Question

A large number of droplets, each of radius $a$, coalesce to form a bigger drop of radius $b$. Assume that the energy released in the process is converted into the kinetic energy of the drop. The velocity of the drop is ($\sigma =$surface tension, $\rho =$density)

• A. ${\left[\frac{\sigma }{\rho }(\frac{1}{a}-\frac{1}{b})\right]}^{1/2}$
• B. ${\left[\frac{2\sigma }{\rho }(\frac{1}{a}-\frac{1}{b})\right]}^{1/2}$
• C. ${\left[\frac{3\sigma }{\rho }(\frac{1}{a}-\frac{1}{b})\right]}^{1/2}$
• D. ${\left[\frac{6\sigma }{\rho }(\frac{1}{a}-\frac{1}{b})\right]}^{1/2}$

Answer 1

The correct option is D ${\left[\frac{6\sigma }{\rho }(\frac{1}{a}-\frac{1}{b})\right]}^{1/2}$

Energy released=decrease in surface energy of the droplets $=[n\times 4\pi a^2-4\pi b^2]\sigma$
Since the volume of the water in the droplets remains same, we have
$n\times \frac{4}{3}\pi a^3=\frac{4}{3}\pi b^3\Rightarrow n=\frac{b^3}{a^3}$
$\therefore$ Energy released $=[\frac{b^3}{a^3}\times 4\pi a^2-4\pi b^2]\sigma =4\pi b^2[\frac{b}{a}-1]\sigma$
The energy released in the process is converted into the kinetic energy of the drop, so we have
$\frac{1}{2}m{v}^2=4\pi b^2[\frac{b}{a}-1]\sigma$
$\Rightarrow \frac{1}{2}\left(\frac{4}{3}\pi b^3\right)\rho v^2=4\pi b^2[\frac{b}{a}-1]\sigma$
$\Rightarrow \frac{\rho }{6}{v}^2=[\frac{1}{a}-\frac{1}{b}]\sigma \Rightarrow v^2=\frac{6\sigma }{\rho }[\frac{1}{a}-\frac{1}{b}]\Rightarrow v={\left[\frac{6\sigma }{\rho }[\frac{1}{a}-\frac{1}{b}]\right]}^{1/2}$