A) 9V
B) 3V
C) 27V
D) \[\frac{1}{3}V\]
Correct Answer: A
Solution :
Suppose the radius of small drop is r and charge is q and the radius of big drop is R and charge is Q. Then \[\frac{4}{3}\pi {{R}^{3}}=27.\frac{4}{3}\pi {{r}^{3}}\] \[\Rightarrow \] \[R=3r\] Potential at small drop \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] \[\therefore \]Potential at big drop\[V'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{R}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{27q}{3r}\] \[=9V\]
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