Question

A spherical liquid drop of radius $R$ is divided into 8 equal droplets. If the surface tension is $T$, then the 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

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

Let us say $r$ is the radius of eight smaller droplets. Since volume of water remains the same, we have
$8\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$\Rightarrow r=\frac{R}{2}$
We know that work done in splitting one bigger drop into ${}^{\prime }{n}^{\prime }$ smaller droplet = change in the surface energy

$\therefore dW=(8\times 4\pi r^2\times T)-(4\pi R^2\times T)$
where $R$ is the radius of bigger block.
$\Rightarrow dW=8\times 4\pi {\left(\frac{R}{2}\right)}^2\times T-4\pi R^2\times T$
$\Rightarrow dW=4\pi R^{2}T$

Alternative:
$dW=4\pi R^{2}T(n^{1/3}-1)$
$dW=4\pi R^{2}T(8^{1/3}-1)=4\pi R^{2}T(2-1)$
$dW==4\pi R^{2}T$