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) \[\pi \sigma {{D}^{2}}\left( {{n}^{4/3}}-1 \right)\]
Correct Answer: A
Solution :
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 \[=n\times 4\pi {{r}^{2}}\sigma -4\pi {{R}^{2}}\sigma \] \[=n\times 4\pi \left( \frac{{{R}^{2}}}{{{n}^{2/3}}} \right)\sigma -4\pi {{R}^{2}}\sigma \] \[=4\pi {{R}^{2}}\sigma [{{n}^{1/3}}-1]\] \[=\pi {{D}^{2}}\sigma [{{n}^{1/3}}-1]\] \[\left[ \because \,R=\frac{D}{2} \right]\]
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