Physics - Fluid Question with Solution | TestHub

When a drop of radius $R$ split into $n$ smaller drops (each of radius $r$ ), then surface area of liquid increases. Hence, the work is to be done against surface tension. Since, the volume of liquid remains constant $$\begin{array}{l}\therefore \frac{4}{3}\pi R^3=n\frac{4}{3}\pi r^3\therefore R^3=n{r}^3\\ \text{ Work done }=T\times \Delta A\\ =T\left[n4\pi r^2-4\pi R^2\right]\\ =4\pi R^{2}T\left(n^{1/3}-1\right)\end{array}$$