Q 11-04-150JEE MainJEE Main 2022 (27 Jun, Shift 2)Medium
One end of a massless spring of spring constant $k$ and natural length $l_0$ is fixed while the other end is connected to a small object of mass $m$ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $\omega$ about an axis passing through fixed end, then the elongation of the spring will be
Answer: (C) $\dfrac{m\omega^2 l_0}{k - m\omega^2}$
With elongation $x$ the radius is $l_0 + x$, and the spring force provides the centripetal force:
$$kx = m\omega^2(l_0 + x)$$
$$x(k - m\omega^2) = m\omega^2 l_0 \Rightarrow x = \frac{m\omega^2 l_0}{k - m\omega^2}$$
Solution by Sreeraj P, M.Sc Physics