Physics - Fluid Question with Solution | TestHub

Let R be the radius of the bigger drop, then Volume of bigger drop$=2\times$volume of small drop

$\frac{4}{3}\pi R^3=2\times \frac{4}{3}\pi r^3\Rightarrow R=2^{1/3}r$

Surface energy of bigger drop,

$E=4\pi R^{2}T=4\times 2^{2/3}\pi r^{2}T=2^{8/3}\pi r^{2}T$