A) \[48\pi {{R}^{2}}T\]
B) \[12\pi {{r}^{2}}T\]
C) \[96\pi {{R}^{2}}T\]
D) \[192\pi {{r}^{2}}T\]
Correct Answer: D
Solution :
Volume of big drop \[=64\times volume\text{ }of\text{ }tiny\text{ }drops\] Or \[\frac{4}{3}\pi {{R}^{3}}=64\times \frac{4}{3}\pi {{r}^{3}}\] Or \[R=4r\] So, the gain in surface energy = work done in splitting a liquid drop of radius R into n identical drops \[=4\pi T{{R}^{2}}({{n}^{1/3}}-1)\] \[=4\pi T{{(4r)}^{2}}({{64}^{1/3}}-1)=192\,\pi {{r}^{2}}T\]
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