A) 10 V
B) 100 V
C) 1000 V
D) \[\frac{V}{100}\]
Correct Answer: B
Solution :
Let R be radius of big drop and r of small drop, then volume of big drop = 1000 \[\times \] volume of small drop \[\frac{4}{3}\pi {{R}^{3}}=1000\times \frac{4}{3}\pi {{r}^{3}}\] \[\Rightarrow \] R = 10 r Potential of small drop \[V=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}.\frac{q}{r}\] ... (i) Potential of bigger drop \[V=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}\frac{1000\,q}{R}\] \[V=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}\times 1000\times \frac{q}{10\,r}\]...(ii) From Eqs. (i) and (ii), we get V = 100 V.
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