A) 1000 : 1
B) 1 : 1000
C) 10 : 1
D) 1 : 10
Correct Answer: D
Solution :
As volume remain constant therefore \[R={{n}^{1/3}}r\] \[\frac{\text{surface energy of one big drop}}{\text{surface energy of }n\text{ drop}}=\frac{4\pi {{R}^{2}}T}{n\times 4\pi {{r}^{2}}T}\] \[\frac{{{R}^{2}}}{n{{r}^{2}}}=\frac{{{n}^{2/3}}{{r}^{2}}}{n{{r}^{2}}}\]= \[\frac{1}{{{n}^{1/3}}}=\frac{1}{{{(1000)}^{1/3}}}=\frac{1}{10}\]
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