Let ' \(n\) ' is the number of liquid drops, each with surface energy ' \(E\) '. These drops join to form single drop. In this process

Let \(r=\) radius of each small drop and \(\mathrm{R}=\) radius of a big single drop.
Then, \(n \times \frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi \mathrm{R}^3\)
\(\therefore \quad \mathrm{R}=\mathrm{n}^{1 / 3} \mathrm{r}\)...(i)
Initial surface energy,
\(\mathrm{E}_1=\mathrm{n} \times 4 \pi \mathrm{r}^2 \times \mathrm{T}=\mathrm{nE}\)
Final surface energy,
\(\begin{aligned}
\mathrm{E}_2 & =4 \pi \mathrm{R}^2 \times \mathrm{T}=4 \pi \mathrm{r}^2 \mathrm{n}^{2 / 3} \times \mathrm{T} \quad \ldots[\operatorname{From}(\mathrm{i})] \\
& =\mathrm{n}^{2 / 3} \mathrm{E}
\end{aligned}\)
Energy released \(=E_1-E_2=\left[E\left(n-n^{2 / 3}\right)\right]\)