Q 12-04-221JEE MainJEE Main 2019 (12 Jan, Shift 1)Easy
A proton and an $\alpha$-particle (with their masses in the ratio of $1 : 4$ and charges in the ratio of $1 : 2$) are accelerated from rest through a potential difference $V$. If a uniform magnetic field $(B)$ is set up perpendicular to their velocities, the ratio of the radii $r_p : r_\alpha$ of the circular paths described by them will be:
Answer: (B) $1 : \sqrt2$
$r = \dfrac{\sqrt{2mqV}}{qB} = \dfrac1B\sqrt{\dfrac{2mV}{q}} \propto \sqrt{\dfrac mq}$
$$\frac{r_p}{r_\alpha} = \sqrt{\frac{1}{1}\cdot\frac{2}{4}} = \frac{1}{\sqrt2}$$
Solution by Sreeraj P, M.Sc Physics