A) V
B) 2 V
C) 4 V
D) 8 V
Correct Answer: C
Solution :
In Millikans oil drop experiment, in order that the drop stays stationary force due to electric field should equal to gravitational force. \[\therefore \] \[F=qE=mg\] ...(i) Also, \[mass=volume\times density\] \[m=\frac{4}{3}\pi {{r}^{3}}\rho \] and \[E=\frac{V}{d}\] Hence, Eq. (i), becomes \[\frac{4}{3}\pi {{r}^{3}}\rho g=q.\frac{V}{d}\] \[\therefore \] \[\frac{{{V}_{1}}}{{{V}_{2}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}\left( \frac{{{q}_{2}}}{{{q}_{1}}} \right)\] Given, \[{{r}_{2}}=2r,\,{{r}_{1}}=r,\,{{q}_{2}}=2q,\,{{q}_{1}}=q\] \[\therefore \] \[\frac{{{V}_{1}}}{{{V}_{2}}}={{\left( \frac{r}{2r} \right)}^{3}}\left( \frac{{{q}_{2}}}{{{q}_{1}}} \right)\] \[\therefore \] \[{{V}_{2}}=4{{V}_{1}}=4V\] (\[\therefore \]\[{{V}_{1}}=V\](given))You need to login to perform this action.
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