A certain number of spherical liquid drops of radius \(r\) coalesce to form a single drop of radius \(R\) and volume \(V\). If \(T\) is the surface tension of the liquid, which one of the following statements is true for the energy \((E)\) in the process?

Change in surface energy is given by
\(E=T(\Delta A) \quad---(1)\)
The initial area is given by,
\(\mathrm{A}=\left(4 \pi r^2\right) n\)
The final area is given by,
\(\mathrm{a}=4 \pi R^2\)
Therefore, the change in area is given by,
\(\begin{aligned}
& \Delta A=a-A \\
& \Rightarrow \Delta A=4 \pi\left(n r^2-R^2\right)---(2)
\end{aligned}\)
Now, using volume conservation: \(\left(\frac{4}{3} \pi r^2\right) n=\frac{4}{3} R^3\)
\(\therefore n=\frac{R^3}{r^3} \quad---(3)\)
\(\Delta A=4 \pi\left[\frac{R^3}{r^3} \cdot r^2-R^2\right]=4 \pi\left[\frac{R^3}{r}-\frac{R^3}{R}\right]=\left(\frac{4}{3} \pi R^3\right) 3\) \(\left[\frac{1}{r}-\frac{1}{R}\right]=3 V\left[\frac{1}{r}-\frac{1}{R}\right]\)
Introducing above value in equation (1)
\(E=3 V T\left[\frac{1}{r}-\frac{1}{R}\right]\)