Q 12-04-096JEE MainJEE Main 2024 (9 Apr, Shift 2)Easy
A proton and a deutron ($q = +e$, $m = 2.0\ \text{u}$) having same kinetic energies enter a region of uniform magnetic field $\vec B$, moving perpendicular to $\vec B$. The ratio of the radius $r_d$ of deutron path to the radius $r_p$ of the proton path is:
Answer: (A) $\sqrt2:1$
$r = \dfrac{\sqrt{2mK}}{qB}$. Same $q$ and $K$, so $r \propto \sqrt m$:
$$\frac{r_d}{r_p} = \sqrt{\frac{2}{1}} = \sqrt2 : 1$$
Solution by Sreeraj P, M.Sc Physics