Question

A spherical liquid drop of diameter D breaks up to n identical spherical drops. If the surface tension of [lie liquid is $\sigma$ , the change in energy in this process is

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

Answer 1

Answer: (A)

Suppose, the radius of big and small drops are R and r respectively. Volume of big drop = Volume of n small drops $\frac{4}{3}\pi R^3=n\frac{4}{3}\pi r^3$ $R=n^{1/3}r$ $r=\frac{R}{n^{1/3}}$ Change energy in this process E = Surface energy of n small drops - Surface energy of big drop $=n\times 4\pi r^2\sigma -4\pi R^2\sigma$ $=n\times 4\pi \left(\frac{R^2}{n^{2/3}}\right)\sigma -4\pi R^2\sigma$ $=4\pi R^2\sigma [n^{1/3}-1]$ $=\pi D^2\sigma [n^{1/3}-1]$ $\left[\because R=\frac{D}{2}\right]$