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 small drop. i.e. \[\frac{4}{3}\pi {{r}^{3}}=n\frac{4}{3}\pi {{r}^{3}}\] \[\Rightarrow \]\[R={{n}^{1/3}}r\] ??(i) Now, potential of small drop \[V'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] or \[q=v\times 4\pi {{\varepsilon }_{0}}f\] ?..(ii) So, potential of big drop is given by \[V'=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{nq}{R}\] ?..(iii) Using equations (i), (ii) then equation (Hi) becomes \[V'=\frac{n4\pi {{\varepsilon }_{0}}\times V\times r}{4\pi {{\varepsilon }_{0}}\times {{n}^{1/3}}r}={{n}^{2/3}}V\]
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