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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. `2 : 1`B. `1 : 2`C. `1 : 4`D. `4 : 1` |
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Answer» Correct Answer - B Using `v = sqrt((2QV)/(m))` `rArr v prop sqrt(Q) rArr (v_(A))/(v_(B)) = sqrt((Q_(A))/(Q_(B))) = sqrt((q)/(4q)) = (1)/(2)` |
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