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In the Millikan's experiment, the oil dr...

In the Millikan's experiment, the oil drop is subjected to a horizontal electric field of `2` N//S and the drop moves with a constant velocity making an angle of `45^(0)`with the horizontal . If the weight of the drop is `W`, then the electric charge, in coulomb, on the drop is

A

`W`

B

`W//2`

C

`W//4`

D

`w//8`

Text Solution

Verified by Experts

The correct Answer is:
B
`Tan45^(0)=(Eq)/w` `Eq=w`
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