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] $