A) \[{{n}^{2/3}}V\]
B) \[{{n}^{1/3}}V\]
C) \[nV\]
D) \[{{V}^{n/3}}\]
Correct Answer: A
Solution :
Volume of big drop\[=n\] \[\times \] volume of a small drop \[\Rightarrow \] \[\frac{4}{3}\pi {{R}^{3}}=n\frac{4}{3}\pi {{r}^{3}}\] \[\Rightarrow \] \[R={{n}^{1/3}}r\] ... (i) Potential of small drop \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] \[\Rightarrow \] \[q=V4\pi {{\varepsilon }_{0}}r\] ... (ii) So, potential of big drop\[=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{nq}{R}\] \[\Rightarrow \] \[V=\frac{4\pi {{\varepsilon }_{0}}\times V\times r}{4\pi {{\varepsilon }_{0}}\times {{n}^{1/3}}r}={{n}^{2/3}}V\]
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