A) \[{{n}^{1/3}}C\]
B) \[{{n}^{2/3}}C\]
C) \[{{n}^{1/4}}C\]
D) \[nC\]
Correct Answer: A
Solution :
Let radius of small drop = r \[\Rightarrow \] Volume of n drops \[=n\frac{4}{3}\pi {{r}^{3}}\] Again, let R be the radius of bigger drop. \[\Rightarrow \] \[\frac{4}{3}\pi {{R}^{3}}=\frac{4}{3}\pi n{{r}^{3}}\] or \[R={{n}^{1/3}}r\] \[\therefore \] \[C=4\pi {{\varepsilon }_{0}}r\] and \[C'=4{{\varepsilon }_{0}}R\] Hence, \[C'=C{{n}^{1/3}}={{n}^{1/3}}C\]
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