(c) $p = \sqrt{2mE_{k}}$ or $p \propto \sqrt{m}\ \ \ $[$\because E_{k}$ is given to be constant] $\therefore\ \ \frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.