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)

$4\pi R^2T$

Answer 2

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