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 \[v'=\frac{3}{2}v\]