In millikan oil drop experiment a charged drop falls with a terminal velocity V. If an electric field E is applied vertically upwards it moves with terminal velocity 2V in upward direction. If electric field reduces to E/2 then its terminal velocity will be -

Options

A. v/2
B. v
C. 3V/2
D. 2v

Answer

C

Explanation

When `E = 0 implies mg = 6pi eta rv" "...(i)`
When `E ne 0 implies mg + qE = 6pi eta r (2v) " "....(ii)`
When `E = E//2 implies mg +q(E/2) = 6pi eta r(v.) " "...(iii)`
By (i), (ii) and (iii) , `v.=3/2 v`