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. $2\pi R^{2}T$
• B. $3\pi R^{2}T$
• C. $4\pi R^{2}T$
• D. $2\pi R{T}^2$

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
$=T\times (4\pi R^2)$
$=4\pi R^{2}T$