Question

A spherical liquid drop of radius $R=\frac{1}{\sqrt{\pi }}\text{m}$ is divided into ${}^{\prime }{n}^{\prime }$ equal droplets. If the surface tension is $2\text{N/m}$, and work done in the process is $16\text{J}$, then find the value of $n$?

• A. $3$
• B. $8$
• C. $9$
• D. $27$

Answer 1

The correct option is D $27$

We know,
Work done in splitting of bigger drop
$dW=4\pi R^{2}T(n^{1/3}-1)$
$\Rightarrow 16=4\times \pi \times {\left(\frac{1}{\sqrt{\pi }}\right)}^2\times 2(n^{1/3}-1)$
$\Rightarrow 3=n^{1/3}$
$\Rightarrow n=27$