A) 10 V
B) 7.5 V
C) 5 V
D) 2.5 V
Correct Answer: C
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) Potential of one small drop \[(V')=\frac{q}{4\pi {{\varepsilon }_{0}}r}\] Similarly, potential of big drop \[(V)=\frac{Q}{4\pi {{\varepsilon }_{0}}r}\] Now, \[\frac{V'}{V}=\frac{q}{Q}\times \frac{R}{r}\] \[\Rightarrow \] \[\frac{V'}{20}=\frac{q}{8q}\times \frac{2\,r}{r}\] [from Eqs,(i) and (ii)] \[\therefore \] V'= 5 volt
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