A) \[V/n\]
B) \[Vn\]
C) \[V{{n}^{1/3}}\]
D) \[V{{n}^{2/3}}\]
Correct Answer: D
Solution :
Volume of big drop\[=n\times \]volume of small drops \[\frac{4}{3}\pi {{R}^{3}}=n\frac{4}{3}\pi {{r}^{3}}\] \[R={{n}^{1/3}}r\] Potential of each small drop \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] Potential of big drop \[V'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{nq}{R}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{nq}{{{n}^{1/3}}r}={{n}^{2/3}}V\]
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