Two small drops of mercury, each of radius $R$, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

Radius of one drop of mercury is $R$. $\therefore \quad$ The volume of one drop $=\frac{4}{3} \pi R^3$ $\therefore$ Total volume of the two drops, $$ V=2 \times \frac{4}{3} \pi R^3=\frac{8}{3} \pi R^3 $$ Let the radius of the large drop formed be $R^{\prime}$. The volume of the large drop is also $V$. $$ \therefore \quad \frac{4}{3} \pi R^{\prime 3}=\frac{8}{3} \pi R^3 \Rightarrow R^{\prime 3}=2 R^3 \Rightarrow R^{\prime}=2^{1 / 3} R \text {. } $$ Now the surface area of the two drops is $$ S_1=2 \times 4 \pi R^2=8 \pi R^2 $$ and the surface area of the resultant drop is $$ S_2=4 \pi R^{\prime 2}=4 \pi 2^{2 / 3} R^2 $$ Let $T$ be the surface tension of mercury. Therefore the surface energy of the two drops before coalescing is $$ U_1=S_1 T=8 \pi R^2 T $$ and the surface energy after coalescing, $$ \begin{gathered} U_2=S_2 T=2^{2 / 3} \times 4 \pi R^2 T \\ \therefore \quad \frac{U_1}{U_2}=\frac{8 \pi R^2 T}{2^{2 / 3} \times 4 \pi R^2 T}=\frac{2}{2^{2 / 3}}=2^{1 / 3} . \end{gathered} $$