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A particle `A` has chrage `+q` and a particle `B` has charge `+4q` with each of them having the same mass `m`. When allowed to fall from rest through the same electric potential difference, the ratio of their speed `(v_(A))/(v_(B))` will becomeA. `1:2`B. `2:1`C. `1:4`D. `4:1` |
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Answer» Correct Answer - A `(1)/(2)mv_(A)^(2) = qV` `(1)/(2)mv_(B)^(2) = 4qV` Dividing, `(v_(A)^(2))/(v_(B)^(2)) = (qV)/(4qV)` or `(v_(A)^(2))/(v_(B)^(2)) = (1)/(4)` or `(v_(A))/(v_(B)) = (1)/(2)` |
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