Q. 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 $E/2,$ then terminal velocity will becomes
A) $V/2$
B) $V$
C) $3V/2$
D) $2V$
Explanation
(c) In the absence of electric field ($i.e.$ $E = 0$)
$mg = 6\pi\eta rv$
…(i)
In the presence of Electric field
$mg + QE = 6\pi\eta r(2v)$
…(ii)
When electric field to reduced to $E/2$
$mg + Q(E/2) = 6\pi\eta r\left( v^{'} \right)$
…(iii)
After solving (i), (ii) and (iii)
We get $v^{'} = \frac{3}{2}v$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.