Question

The work done in breaking a spherical drop of a liquid (surface tension T) of radius R into 8 equal drops is

• A. $\pi$$R^{2}T$
• B. $2\pi$$R^{2}T$
• C. $16\pi$$R^{2}T$
• D. $4\pi$$R^{2}T$

Answer 1

The correct option is D $4\pi$$R^{2}T$

$\begin{array}{c}Volumeofbiggerdrop=Totalvolumeof8dropeltes\\ \frac{4}{3}\times \pi \times R^3=8\times \frac{4}{3}\times \pi \times r^3\\ \Rightarrow r=\frac{R}{2}\\ Changeinsurfaceareainthisprocess=ds=4\times \pi \times \left[8\times r^2-R^2\right]=4\times \pi \times R^2\\ So,workdone=T.ds=T\times 4\pi \times R^2=4\pi R^{2}T\end{array}$

Hence,

option $\left(D\right)$ is correct answer.