A) \[\frac{Q}{4}\]
B) \[\frac{Q}{2}\]
C) Q
D) \[\frac{3Q}{2}\]
Correct Answer: B
Solution :
Force on charge,\[F=QE\] or \[F=\frac{QV}{d}\] \[(\because \,\,V=Ed)\] For drop to be stationary, weight of drop = force due to charge \[i.e.,\] \[mg=\frac{QV}{d}\] For two drops, \[\frac{{{Q}_{1}}}{{{Q}_{2}}}\cdot \frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\] or \[\frac{{{Q}_{1}}}{{{Q}_{2}}}\cdot \frac{{{V}_{1}}}{{{V}_{2}}}=\frac{\frac{4}{3}\pi r_{1}^{3}\rho }{\frac{4}{3}\pi r_{2}^{3}\rho }\] or \[\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{r_{2}^{3}}{r_{1}^{3}}\times \frac{{{V}_{1}}}{{{V}_{2}}}\] \[\therefore \] \[\frac{{{Q}_{2}}}{Q}=\frac{{{(r/2)}^{3}}}{{{r}^{3}}}\times \frac{2400}{600}\] or \[{{Q}_{2}}=\frac{Q}{2}\]
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