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

Answer 1

Answer: (D)

$n\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$n=\frac{R^3}{r^3}$
$\Delta U=n4\pi r^{2}T-4\pi R^{2}T$
$=4\pi T\left(n{r}^2-R^2\right)$
$=4\pi T\left(\frac{R^3{r}^2}{r^3}-R^2\right)$
$\therefore$ Energy released,
$U=4\pi T{R}^3[\frac{1}{r}-\frac{1}{R}]$
If this energy is converted into kinetic energy then
$\frac{1}{2}m{v}^2=4\pi R^{3}T[\frac{1}{r}-\frac{1}{R}]$
$\frac{1}{2}\times [\frac{4}{3}\pi R^3\rho ]v^2=4\pi R^{3}T[\frac{1}{r}-\frac{1}{R}]$
$v=\sqrt{\frac{6T}{\rho }[\frac{1}{r}-\frac{1}{R}]}$