Q 11-04-184JEE MainJEE Main 2020 (8 Jan, Shift 1)Easy
A particle of mass $m$ is fixed to one end of a light spring having force constant $k$ and unstretched length $l$. The other end is fixed. The system is given an angular speed $\omega$ about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:
Answer: (B) $\dfrac{ml\omega^2}{k - m\omega^2}$
With stretch $x$, the radius is $l + x$ and the spring force provides the centripetal force:
$$kx = m\omega^2(l + x) \Rightarrow x(k - m\omega^2) = ml\omega^2 \Rightarrow x = \frac{ml\omega^2}{k - m\omega^2}$$
Solution by Sreeraj P, M.Sc Physics