A) 1
B) 2
C) 4
D) 6
Correct Answer: C
Solution :
As volume remains constant therefore \[R={{n}^{1/3}}r\] \[\frac{\text{Energy of big drop}}{\text{Energy of small drop}}=\frac{4\pi {{R}^{2}}T}{4\pi {{r}^{2}}T}=\frac{{{R}^{2}}}{{{r}^{2}}}\]\[={{(8)}^{2/3}}=4\]
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