A liquid drop having surface energy ' \(E\) ' is spread into 512 droplets of same size. The final surface energy of the droplets is

Surface area of drop, \(\mathrm{A}_1=4 \pi \mathrm{R}^2\)
Surface area of 512 droplets, \(\mathrm{A}_2=512\left(4 \pi \mathrm{r}^2\right)\) volume of drop \(=\mathrm{n} \times\) (volume of droplet)
\(\therefore \quad \frac{4}{3} \pi \mathrm{R}^3=512 \times \frac{4}{3} \pi \mathrm{r}^3\)
\(\therefore \quad \mathrm{R}=8 \mathrm{r}\)
\(\therefore \quad \mathrm{A}_2=\frac{512\left(4 \pi \mathrm{R}^2\right)}{64}\)
\(\therefore \quad \mathrm{A}_2=8\left(4 \pi \mathrm{R}^2\right)\)
Surface energy \(\propto\) Area
\(\therefore \quad \frac{E_2}{E_1}=\frac{A_2}{A_1}=\frac{8\left(4 \pi R^2\right)}{4 \pi R^2}\)
\(\therefore \quad \mathrm{E}_2=8 \mathrm{E}_1=8 \mathrm{E}\)
\(\ldots .\left(\because E_1=E\right)\)