JEE Advanced 2025 Paper 2 · Q03 · Torque

Question

A uniform rod $OO'$ of length $l$ is hinged at the point $O$ and held vertically between two walls using two massless springs of same spring constant $k$. In Fig. 1 one spring is attached at the midpoint and the other at the top end $O'$ of the rod; the rod is made to oscillate by a small angular displacement, giving frequency $f_1$. In Fig. 2 both springs are attached at the midpoint of the rod, giving frequency $f_2$. Ignoring gravity and assuming motion only in the plane of the diagram, the value of $f_1/f_2$ is:

SolveKar AI · Accepted Answer

Moment of inertia about the hinge $O$: $I = (1/3) M l^2$. Fig. 1: restoring torques from the two springs (one at $l/2$, one at $l$) about $O$ are $-k(l/2)^2	heta$ and $-k l^2 	heta$. Total $-k l^2 	heta(1/4 + 1) = -(5/4) k l^2 	heta$. $\omega_1^2 = (5 k l^2/4)/(M l^2/3) = 15 k/(4 M)$. Fig. 2: both springs at midpoint, restoring torque $= -2k(l/2)^2	heta = -(k l^2/2)	heta$. $\omega_2^2 = (k l^2/2)/(M l^2/3) = 3 k/(2 M)$. $f_1^2/f_2^2 = \omega_1^2/\omega_2^2 = (15/4)(2/3) = 5/2$. So $f_1/f_2 = \sqrt{5/2}$. Answer (C).

JEE Advanced 2025 Paper 2 · Q03 · Torque

Solution