Question

The work done to get $n$ smaller equal size spherical drops from a bigger size spherical drop of water is proportional to:

• A. $\left(\frac{1}{n^{2/3}}\right)-1$
• B. $\left(\frac{1}{n^{1/3}}\right)-1$
• C. $n^{1/3}-1$
• D. $n^{4/3}-1$

Answer 1

The correct option is C $n^{1/3}-1$

Let us say $r$ is the radius of ${}^{\prime }{n}^{\prime }$ smaller droplets. Since volume of water remains the same, we have
$n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$\Rightarrow r=\frac{R}{n^{1/3}}$
We know that work done in splitting one bigger drop into ${}^{\prime }{n}^{\prime }$ smaller droplet = change in the surface energy

$\therefore dW=(n\times 4\pi r^2\times T)-(4\pi R^2\times T)$
where $R$ is the radius of bigger block.
$\Rightarrow dW=n\times 4\pi {\left(\frac{R}{n^{1/3}}\right)}^2\times T-4\pi R^2\times T$
$\Rightarrow dW=4\pi R^{2}T(n^{1/3}-1)$
So, work done is proportional to $(n^{1/3}-1)$