Electrostatics Question 359

Question: A particle $ A $ has charge $ +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 $ \frac{v _{A}}{v _{B}} $ will become [BHU 1995; MNR 1991; UPSEAT 2000; Pb PET 2004]

Options:

A) $ 2:1 $

B) $ 1:2 $

C) $ 1:4 $

D) $ 4:1 $

Show Answer

Answer:

Correct Answer: B

Solution:

Using $ v=\sqrt{\frac{2QV}{m}} $

therefore $ v\propto \sqrt{Q} $

therefore $ \frac{v _{A}}{v _{B}}=\sqrt{\frac{Q _{A}}{Q _{B}}}=\sqrt{\frac{q}{4q}}=\frac{1}{2} $



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