Q.

A particle A has charge +q and particle I has charge + 4 q with each of them having the same mass m. When allowed to faII from rest through the same electric potential difference, the ratio of their speeds VA/VB will become

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a

2 : 1

b

1 : 2

c

1 : 4

d

4 : 1

answer is B.

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Detailed Solution

Acceleration = (Force/mass)∴a=Fm=qEmSo, aA=4aB Now, v2=u2+2ax, where u=0and x is same for both particle∴ VA2=2aAx and  VB2=2aBxor   VA2VB2=aAaB=14or   VAVB=12 or VA:VB=1:2
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