Question

A spherical water drop of radius $R$ is split up into $8$ equal droplets. If $T$ is the surface tension of water, then the work done in this process is-

• A. $4\pi R^{2}T$
• B. $8\pi R^{2}T$
• C. $48\pi R^{2}T$
• D. $2\pi R^{2}T$

Answer 1

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

Let smaller drop has radius $r$

Volume will remain same.

$\frac{4}{3}\pi R^3=8\times \frac{4}{3}\pi r^3⟹r=\frac{R}{2}$

Change in surface energy-

$A_{2}T-A_{1}T=8\times 4\pi r^{2}T-4\pi R^{2}T=4\pi [2R^2-R^2]T=4\pi R^{2}T$

Work done = change in surface energy = $4\pi R^{2}T$