A) 1 : 10
B) 1 : 15
C) 1 : 20
D) 1 : 25
Correct Answer: C
Solution :
As volume remains constant \[{{R}^{3}}=8000{{r}^{3}}\]\[\therefore R=20r\] \[\frac{\text{Surface energy of one big drop}}{\text{Surface energy of 8000 small drop}}=\frac{4\pi {{R}^{2}}T}{8000\ 4\pi {{r}^{2}}T}\] \[=\frac{{{R}^{2}}}{8000{{r}^{2}}}=\frac{{{\left( 20r \right)}^{2}}}{8000{{r}^{2}}}=\frac{1}{20}\]
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