Q 11-03-097JEE MainJEE Main 2023 (6 Apr, Shift 1)Medium
A particle is moving with constant speed in a circular path. When the particle turns by an angle $90^\circ$, the ratio of instantaneous velocity to its average velocity is $\pi:x\sqrt2$. The value of $x$ will be
Answer: (A) 2
For a $90^\circ$ turn the displacement is $\sqrt2R$ and the time is $\dfrac{\pi R}{2v}$, so the average velocity is $\dfrac{2\sqrt2v}{\pi}$.
Ratio $v:\dfrac{2\sqrt2v}{\pi}=\pi:2\sqrt2$, so $x=2$.
Solution by Sreeraj P, M.Sc Physics