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A particle A has chrage +q and a particl...

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

`2:1`

B

`1:2`

C

`1:4`

D

`4:1`

Text Solution

Verified by Experts

The correct Answer is:
B
`K=qV implies1/2mv^2=qV implies v=sqrt((2q)/m)`
v = Velocity
V = Potential
`v_A/v_B=sqrt(q_1/q_2)=sqrt(q/(4q))=1/2`
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