Dual Nature Of Light Question 71
Question: 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 $ \frac{E}{2} $ , then terminal velocity will become [CBSE PMT 1999]
Options:
A) $ \frac{V}{2} $
B) V
C) $ \frac{3V}{2} $
D) 2V
Show Answer
Answer:
Correct Answer: C
Solution:
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(v’) $ ?(iii) After solving (i), (ii) and (iii) We get $ v’=\frac{3}{2}v $
BETA
