Question

A spherical liquid drop of radius $R$ is divided into eight equal droplets. If surface tension is $T$, then work done in the process will be

• A. $2\pi R^2T$
• B. $3\pi R^2T$
• C. $4\pi R^2T$
• D. $2\pi RT^2$

Answer 1

Answer: (C)

Let r be radius of each small droplets.
Since the volume remains the same
$\therefore \frac{4}{3}\pi\,R^{3}=8 \times \left(\frac{4}{3}\pi r^{3}\right)$
$r=\frac{1}{2}R$
Surface area of big droplet $=4\pi R^{2}$
Surface area of 8 small droplets $=8 \times 4 \pi r^{2}$
Increase in surface area $=8\times 4\pi r^{2}-4 \pi R^{2}$
$=4\pi (8r^{2}-R^{2})$
$=4\pi\left[8\left(\frac{1}{2}R\right)^{2}-R^{2}\right]=4\pi\left[R^{2}\right]$
$=4\pi R^{2}$
Work done = Surface tension x increase in surface area
$=T \times (4\pi R^{2})$
$=4 \pi R^{2}T$