A spherical liquid drop of radius \(\mathrm{R}\) is divided into 8 equal droplets. If surface tension is \(\mathrm{S}\), then the work done in this process will be

Work done, \(\mathrm{W}=\mathrm{S} \times \Delta \mathrm{A}\)
Where, \(\Delta \mathrm{A}\) is a change in surface area. Since, the radius of the big drop be \(\mathrm{R}\) \(\mathrm{A}_{\text {initial }}=4 \pi \mathrm{R}^2\)
Let the radius of the small drops be \(\mathrm{r}\) \(\mathrm{A}_{\text {final }}=8 \times 4 \pi \mathrm{r}^2\)
The volume is constant
\(
\begin{array}{ll}
\therefore & \frac{4}{3} \pi \mathrm{R}^3=8 \times \frac{4}{3} \pi \mathrm{r}^3 \\
\therefore & \mathrm{R}^3=8 \mathrm{r}^3 \\
\therefore & \mathrm{R}=2 \mathrm{r}
\end{array}
\)
Substituting the values in the formula.
\(
\begin{aligned}
& \mathrm{W}=\mathrm{S} \times \Delta \mathrm{A} \\
& \mathrm{W}=\mathrm{S}\left(\mathrm{A}_{\text {final }}-\mathrm{A}_{\text {initial }}\right)
\end{aligned}
\)
\(\begin{aligned} & \mathrm{W}=\mathrm{S}\left(8 \times 4 \pi \mathrm{r}^2-4 \pi \mathrm{R}^2\right) \\ & \mathrm{W}=\mathrm{S}\left(32 \pi\left(\frac{\mathrm{R}}{2}\right)^2-4 \pi \mathrm{R}^2\right) \\ & \mathrm{W}=\mathrm{S}\left(8 \pi \mathrm{R}^2-4 \pi \mathrm{R}^2\right) \\ & \mathrm{W}=4 \pi \mathrm{R}^2 \mathrm{~S}\end{aligned}\)