In Milikan'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 become

Options

A. `v_0/2`
B. `v_0`
C. `(3v_0)/2`
D. `2v_0`

Answer

A

Explanation

`F_(v) = mg = 6 pi nrv_0 = mg …(1)`
when field is applied vertically upward
`qE = mg + 6pinr(2v_0) ... (2)`
From (1) & (2)
`qE = 18 pi eta r v_0 ... (3)`
If field is reduced of half
` mg+ 6 pi eta rv. = (qE)/2`
from (3) putting value of qE
`6pietarv_0=6pietarv.=(18pietarv_0)/2`
`6pietarv.=3pietarv_0`
`v.=v_0/2`