Q 12-12-076JEE MainJEE Main 2023 (6 Apr, Shift 2)Medium
A small particle of mass $m$ moves in such a way that its potential energy $U=\dfrac12m\omega^2r^2$ where $\omega$ is constant and $r$ is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of $n^\text{th}$ orbit will be proportional to
Answer: (A) $\sqrt n$
The force $m\omega^2r$ supplies the centripetal force: $\dfrac{mv^2}{r}=m\omega^2r\Rightarrow v=\omega r$.
Bohr: $mvr=\dfrac{nh}{2\pi}\Rightarrow m\omega r^2=\dfrac{nh}{2\pi}\Rightarrow r\propto\sqrt n$.
Solution by Sreeraj P, M.Sc Physics