(c) As , $q\ V = \frac{1}{2}\ m\ v^{2}$ Or $v = \ \sqrt{2q\ V/m}$; So, $v\ \propto \ \sqrt{q\ /m}$ $\therefore\frac{v_{2}}{v_{1}}\ \propto \ \ \sqrt{\frac{2q/4m}{q/m}}$ = $\frac{1}{\sqrt{2}}$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.