A) 100 : 1
B) 1000 : 1
C) 10 : 1
D) 1 : 100
Correct Answer: D
Solution :
Volume of liquid remain same ie, volume of 1000 small drops will be equal to volume of one big drop\[n\frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}\] \[1000{{r}^{3}}={{R}^{3}}\] \[\Rightarrow \]\[R=10r\] \[\therefore \]\[v=\sqrt{\mu rg}\] \[\frac{\text{Surface energy of one small drop}}{\text{Surface energy of one big drop}}=\frac{4\pi {{r}^{2}}T}{4\pi {{R}^{2}}T}\] \[=\frac{1}{100}\]
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