Physics - Fluid Question with Solution | TestHub

Let $\mathrm{R}$ be the radius of the bigger drop, then Volume of bigger drop $=2\times$ volume of small drop $$\frac{4}{3}\pi R^3=2\times \frac{4}{3}\pi r^3\Rightarrow \mathrm{R}=2^{1/3}r$$ Surface energy of bigger drop, $$\mathrm{E}=4\pi R^{2}T=4\times 2^{2/3}\pi r^{2}T=2^{8/3}\pi r^{2}T$$