Question

In Millikon oil drop experiment a charged drop falls with a terminal velocity $v_{0}.$ If an electric field $E$ is applied vertically upwards it moves with terminal velocity $2v_{0}$ in upward direction. If electric field is reduced to $\frac{E}{2}$ then its terminal velocity will be

• A. $\frac{v_{0}}{2}$
• B. $v_{0}$
• C. $\frac{3 v_{0}}{2}$
• D. $2v_{0}$

Answer 1

Answer: (A)

$ F_{ v }=m g=6 \pi n r v_0=m g \ldots \text { (1) } $
when field is applied vertically upward
$ q E=m g+6 \pi n r\left(2 v_0\right) \ldots(2) $
From (1) and (2)
$ q E=18 \pi \eta r v_0 \ldots(3) $
If field is reduced of half $m g+6 \pi \eta r v^{\prime}=\frac{q E}{2}$
from (3) putting value of $q E$
$6 \pi \eta r v_0+6 \pi \eta r v^{\prime}=\frac{18 \pi \eta r v_0}{2}$
$6 \pi \eta r v^{\prime}=3 \pi \eta r v_0$
$ v^{\prime}=\frac{v_0}{2} $