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 x volume of tiny 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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