Q 11-04-175JEE MainJEE Main 2021 (26 Aug, Shift 2)Easy
A particle of mass $m$ is suspended from a ceiling through a string of length $L$. The particle moves in a horizontal circle of radius $r$ such that $r = \frac{L}{\sqrt{2}}$. The speed of particle will be:
Answer: (A) $\sqrt{rg}$
For a conical pendulum, $\tan\theta = \dfrac{v^2}{rg}$, where $\theta$ is the angle of the string with the vertical.
$\sin\theta = \dfrac{r}{L} = \dfrac{1}{\sqrt{2}} \Rightarrow \theta = 45^\circ$, so $v^2 = rg\tan45^\circ = rg$ and $v = \sqrt{rg}$.
Solution by Sreeraj P, M.Sc Physics