Question

A water drop of radius $r$ is divided into eight equal droplets. What is the work done in the process? (Given surface tension $=\sigma)$

• A. $2 \pi pr ^{2} \sigma$
• B. $3 \pi r^{2} \sigma$
• C. $4 \pi r^{2} \sigma$
• D. $24 \pi r^{2} \sigma$

Answer 1

Answer: (C)

$\frac{4}{3} \pi r^{3}=8\left(\frac{4}{3} \pi r^{'3}\right)$
$ \Rightarrow r'=\frac{r}{2}$
$\Delta A=8\left(4 \pi r^{2}\right)-4 \pi r^{2}=4 \pi\left[8\left(\frac{r}{2}\right)^{2}-r^{2}\right]$
$=4 \pi\left(2 r^{2}-r^{2}\right)=4 \pi r^{2}$
Workdone $=\sigma \cdot \Delta A=\sigma\left(4 \pi r^{2}\right)$