Question

If a number of little droplets of water of surface tension $\sigma$, all of the same radius $r$ combine to form a single drop of radius $R$ and the energy released is converted into kinetic energy, find the velocity acquired by the bigger drop.

Answer 1

No. of drops
Surface energy initially
$n\left(4\pi r^2\right)\sigma$
Surface energy find $n$
$4\pi R^2\sigma$
$n\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3$
$⟹R=n^{\frac{1}{2}}r$
$\Delta$ (Surface energy) $=n\left(4\pi r^2\right)\sigma -4\sigma R^2\sigma$
$=4\pi \sigma (n{r}^2-R^2)$
$=4\pi \sigma (n{r}^2-n^{\frac{2}{3}}{r}^2)$
$=4\pi \sigma r^2(n-n^{\frac{2}{3}})$
$=\Delta \left(ICE\right)$
$=\frac{1}{2}m{v}^2$
$⟹v=\sqrt{\frac{\delta \pi \sigma }{m}{r}^2(n-n^{\frac{2}{3}})}$