A) 10 V
B) 7.5V
C) 5V
D) 2.5V
Correct Answer: A
Solution :
Volume of 8 drops = Volume of big drop \[\therefore \] \[\left( \frac{4}{3}\pi {{r}^{3}} \right)\times 8=\frac{4}{3}\pi {{R}^{3}}\] \[\Rightarrow \] \[2r=R\] ...(i) According to charge conservation \[8q=Q\] ...(ii) We have \[V\propto \frac{1}{r}\] Or \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{r}_{2}}}{n}\] \[\Rightarrow \] \[\frac{20}{{{V}_{2}}}=\frac{2r}{r}\] \[\Rightarrow \] \[{{V}_{2}}=10\,V\]
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