Question

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 become
A. `1:2`
B. `2:1`
C. `1:4`
D. `4:1`

Answer 1

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)`