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A particle A has charge +q and particle B has charge +4q with each of them having the same mass m. When allowed to fall from rest through same electrical potential difference, the ratio of their speed VA : VB will become A. 2:1 B. 1:2 C. 1:4 D. 4:1

A particle A has charge +q and particle B has charge +4q with each of them having the same mass m. When allowed to fall from rest through same electrical potential difference, the ratio of their speed VA : VB will become
A. 2:1 B. 1:2 C. 1:4 D. 4:1

Grade:Upto college level

2 Answers

Archana
11 Points
3 years ago
Ka/Kb=q×v/4q×v =1/2mVa^2/1/2mVb^2 Va^2/Vb^2=1/4 Va/VB=1/2 or Va:Vb=1:2 It is the ratio between the kinectic energies between the two particles.... Hope this is useful for u guys and this is the correct answer for this question....
Rishi Sharma
askIITians Faculty 646 Points
10 months ago
Dear Student,
Please find below in the attached image the solution to your problem.

Thanks and Regards
645-734_4.PNG

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