A spherical liquid drop of radius $\mathrm{R}$ is divided into 8 equal droplets. If surface tension is…
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
$2 \pi R^2 S$
$3 \pi R^2 S$
$4 \pi R^2 S$
$2 \pi \mathrm{RS}^2$
Solution
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.
$\begin{aligned}
& \mathrm{W}=\mathrm{S} \times \Delta \mathrm{A} \\
& \mathrm{W}=\mathrm{S}\left(\mathrm{A}_{\text {final }}-\mathrm{A}_{\text {initial }}\right)
\end{aligned}$
$\begin{aligned} & \mathrm{W}=\mathrm{S}\left(8 \times 4 \pi \mathrm{r}^2-4 \pi \mathrm{R}^2\right) \\ & \mathrm{W}=\mathrm{S}\left(32 \pi\left(\frac{\mathrm{R}}{2}\right)^2-4 \pi \mathrm{R}^2\right) \\ & \mathrm{W}=\mathrm{S}\left(8 \pi \mathrm{R}^2-4 \pi \mathrm{R}^2\right) \\ & \mathrm{W}=4 \pi \mathrm{R}^2 \mathrm{~S}\end{aligned}$
.