A) \[V{{N}^{1/3}}\]
B) \[V{{N}^{2/3}}\]
C) \[V\]
D) \[VN\]
Correct Answer: B
Solution :
Key Idea: Volume of N small drops is equal to volume of single big drop. Volume of N small drops = volume of one big drop i.e., \[\left( \frac{4}{3}\pi {{r}^{3}} \right)N=\frac{4}{3}\pi {{R}^{3}}\] \[\Rightarrow \] \[R={{N}^{1/3}}r\] Also electric potential on big drop is \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Nq}{R}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Nq}{{{N}^{1/3}}r}\] \[\Rightarrow \] \[V={{N}^{2/3}}V\] \[\left( as\,V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r} \right)\]
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