A large number of water drops, each of radius $r$, combine to have a drop of radius $R$. If the surface tension is $\mathrm{T}$ and mechanical equivalent of heat is $J$, the rise in heat energy per unit volume will be:
Correct Option: , 4
$\mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \mathrm{R}^{3}$
$\therefore n^{1 / 3} r=R$
$\therefore$ Total change in surface energy
$=\left(\mathrm{n}\left(4 \pi \mathrm{r}^{2}\right)-4 \pi \mathrm{R}^{2}\right) \mathrm{T}$
$\Rightarrow 4 \pi \mathrm{T}\left(\mathrm{nr}^{2}-\mathrm{R}^{2}\right)$
$\therefore$ Heat energy
$=\frac{4 \pi \mathrm{T}\left(\mathrm{nr}^{2}-\mathrm{R}^{2}\right)}{\mathrm{J} \times \frac{4}{3} \pi \mathrm{R}^{3}}=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left(\frac{\mathrm{nr}^{2}}{\mathrm{R}^{3}}-\frac{1}{\mathrm{R}}\right)$
Put $n r^{3}=R^{3}$
$\therefore \frac{3 \mathrm{~T}}{\mathrm{~J}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$