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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
6 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
3 years 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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