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. $\frac{3v_0}{2}$
• D. $2v_0$

Answer 1

Answer: (A)

$F_v=mg=6\pi nr{v}_0=mg\dots \text{ (1) }$
when field is applied vertically upward
$qE=mg+6\pi nr\left(2v_0\right)\dots (2)$
From (1) and (2)
$qE=18\pi \eta r{v}_0\dots (3)$
If field is reduced of half $mg+6\pi \eta r{v}^{\prime }=\frac{qE}{2}$
from (3) putting value of $qE$
$6\pi \eta r{v}_0+6\pi \eta r{v}^{\prime }=\frac{18\pi \eta r{v}_0}{2}$
$6\pi \eta r{v}^{\prime }=3\pi \eta r{v}_0$
$v^{\prime }=\frac{v_0}{2}$