Physics - Fluid Question with Solution | TestHub

As volume of 125 droplets = Volume of the drop $\therefore 125\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$ or $125r^3=R^3$ or $r=\frac{R}{5}$ The surface area of the drop $=4\pi R^2$ and the surface area of 125 droplets $=125\left(4\pi r^2\right)=125\left(4\pi {\left(\frac{R}{5}\right)}^2\right)=20\pi R^2$ $\begin{array}{rl}\therefore \text{ The increase in surface area }& =20\pi R^2-4\pi R^2\\ & =16\pi R^2\end{array}$ The change in surface energy $=$ Surface tension $\times$ increase in surface area $=T\times 16\pi R^2=16\pi R^{2}T$