Question

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

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

Let us say $r$ is the radius of each eight of the bubble. Since volume of water remains the same, we have
$\frac{4}{3}\pi R^3=8\times \frac{4}{3}\pi r^3\Rightarrow r=\frac{R}{2}$
So the work done in this process will be = change in the surface energy i.e.
$(8\times 4\pi r^2\times T)-(4\pi R^2\times T)=(8\times 4\pi {\left(\frac{R}{2}\right)}^2\times T)-(4\pi R^2\times T)\because r=\frac{R}{2}=8\pi R^{2}T-4\pi R^{2}T=4\pi R^{2}T$