In Millikan's oil drop experiment, a charged drop falls with terminal velocity $V$. If an electric field $E$ is applied in vertically upward direction then it starts moving in upward direction with terminal velocity $2V$.If magnitude of electric field is decreased to $\frac{E}{2}$, then terminal velocity will become
In Millikan's oil drop experiment, a charged drop falls with terminal velocity $V$. If an electric field $E$ is applied in vertically upward direction then it starts moving in upward direction with terminal velocity $2V$.If magnitude of electric field is decreased to $\frac{E}{2}$, then terminal velocity will become
- A
$\frac{V}{2}$
- B
$V$
- C
$\frac{{3V}}{2}$
- D
$2V$
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