JEE Main & Advanced
Physics
Photo Electric Effect, X- Rays & Matter Waves
Question Bank
Cathode Rays and Positive Rays
question_answer
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]
A) \[\frac{V}{2}\]
B) V
C) \[\frac{3V}{2}\]
D) 2V
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 \[mg+Q\,\left( E/2 \right)=6\pi \eta r(v')\] ?(iii) After solving (i), (ii) and (iii) We get \[v'=\frac{3}{2}v\]