Eight small drops of mercury each of radius ' \(r\) ', coalesce to form a large single drop. The ratio of total surface energy before and after the change is

Let \(\mathrm{R}\) be the radius of the coalesced drop.
\(
\begin{aligned}
\therefore & \frac{4}{3} \pi R^3=8 \times \frac{4}{3} \pi r^3 \\
& R^3=8 \pi r^3 \\
& R=2 r
\end{aligned}
\)
Surface Energy \(\mathrm{E}=\mathrm{T}\).dA
\(
\begin{aligned}
& E_1=8 \times T \times d A=8 \times T \times 4 \pi r^2 \\
& E_2=T \times d A=T \times 4 \pi R^2
\end{aligned}
\)
Where \(E_1\) and \(E_2\) are the surface energies before and after coalescing.
Dividing equation (i) by (ii),
\(
\begin{aligned}
& \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{8 \mathrm{r}^2}{\mathrm{R}^2} \\
& \text { but } \mathrm{R}=2 \mathrm{r} \\
& \therefore \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{8 \mathrm{r}^2}{4 \mathrm{r}^2}=\frac{2}{1}\end{aligned}
\)