Physics - Fluid Question with Solution | TestHub

As volume remains constant $R^3=8000r^3$ $\therefore R=20r$

$\frac{\text{Surface energy of one big drop}}{\text{Surface energy of 8000 small drop}}=\frac{4\pi R^{2}T}{80004\pi r^{2}T}$

$=\frac{R^2}{8000r^2}=\frac{{\left(20r\right)}^2}{8000r^2}=\frac{1}{20}$