A large number of water drops, each of radius r, combine to have a drop of radius $R$ . If the surface tension is $T$  and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be:

$\text{R}$  is the radius of bigger drop is the radius of n water drops. Water drops are 
combined to make bigger drop. So, Volume of n drops = volume of bigger drop
$\Rightarrow n\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3\Rightarrow R=r{n}^{1/3}\Rightarrow n={\left(\frac{R}{r}\right)}^3$
$\text{ΔU}=T (Change in surface area)\Rightarrow \text{ΔU}=\text{T}(\text{n}4\pi {\text{r}}^2-4\pi {\text{R}}^2)$
$\Rightarrow \text{ΔU}=4\pi \text{T}[{\left(\frac{\text{R}}{\text{r}}\right)}^3{\text{r}}^2-{\text{R}}^2]\Rightarrow \frac{4\pi \text{T}(\frac{{\text{R}}^3}{\text{r}}-{\text{R}}^2)}{\text{J}}$
$\Rightarrow \frac{\text{ΔU}}{\text{V}}=\frac{4\pi \text{T}(\frac{{\text{R}}^3}{\text{r}}-{\text{R}}^2)}{\text{J}\times \frac{4}{3}\pi {\text{R}}^3}=\frac{3\text{ T}}{\text{ J}}\frac{1}{\text{r}}-\frac{1}{\text{R}}]$