

MCQOPTIONS
Saved Bookmarks
1. |
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 |
Answer» D. 2V | |