Question

A particle A has charge +q and a particle B has charge +4q. each having the same m . allowed to fall from rest through the same electric potential difference. the ratio of the speed of A to that of B will e

• A. 2 :1
• B. 1 :2
• C. 4 :1
• D. 1 :4

Answer 1

The correct option is B 1 :2

Particle A	Particle B
Charge =q	Charge =4q
mass=m	mass=m
potential=$V_A$	potential=$V_B$

Now, kinetic energy, $K=$charge $\times$potential

$=qv$

and kinetic energy $=\frac{1}{2}m{v}^2$

$\begin{array}{c}\therefore \frac{K_A}{K_B}=\frac{qv}{4qv}=\frac{\frac{1}{2}m{v}_A^2}{\frac{1}{2}m{v}_B^2}\\ or,\frac{V_A}{V_B}=\frac{1}{2}\end{array}$

Hence Option $B$ is correct .