Q 11-04-113JEE MainJEE Main 2023 (24 Jan, Shift 2)Medium
A body of mass $200\ \text{g}$ is tied to a spring of spring constant $12.5\ \text{N m}^{-1}$, while the other end of spring is fixed at point $O$. If the body moves about $O$ in a circular path on a smooth horizontal surface with constant angular speed $5\ \text{rad s}^{-1}$, then the ratio of extension in the spring to its natural length will be
Answer: (C) $2:3$
The spring force provides the centripetal force: $kx=m\omega^2(l_0+x)$.
$$12.5x=0.2\times25(l_0+x)=5l_0+5x\ \Rightarrow\ 7.5x=5l_0\ \Rightarrow\ \frac{x}{l_0}=\frac23$$
Solution by Sreeraj P, M.Sc Physics