Q. 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
A) 2:1
B) 1:2
C) 1:4
D) 4:1
Explanation
(b) Using $v = \sqrt{\frac{2QV}{m}} \Rightarrow v \propto \sqrt{Q} \Rightarrow \frac{v_{A}}{v_{B}} = \sqrt{\frac{Q_{A}}{Q_{B}}} = \sqrt{\frac{q}{4q}} = \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.