JEE Advanced 2025 · Paper 2 · Physics Q3

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PhysicsSingle correct+3 / -1Question 19 of 48

As shown in the figures, a uniform rod $OO'$ of length $l$ is hinged at the point $O$ and held in place vertically between two walls using two massless springs of same spring constant. The springs are connected at the midpoint and at the top-end ( $O'$ ) of the rod, as shown in Fig. 1 and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $f_1$ . On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2 and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $f_2$ . Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\dfrac{f_1}{f_2}$ is:

Fig. 1: a vertical rod OO′ of length l hinged at O on the floor between two walls; one horizontal spring joins the left wall to the rod's midpoint (height l/2) and another joins the rod's top end O′ (height l) to the right wall. Fig. 2: the same rod with both springs attached at the midpoint, one to each wall.

2

\sqrt2

\sqrt{\dfrac52}

\sqrt{\dfrac25}

Answer and worked solution

Official answer: C

For a small tilt $\theta$ , a spring attached at height $h$ stretches by $h\theta$ and gives restoring torque $kh^2\theta$ . The moment of inertia $I$ about O is the same in both cases.

Fig. 1: springs at $\frac l2$ and $l$ : $\tau=k\left(\frac{l^2}4+l^2\right)\theta=\frac54kl^2\theta$ .
Fig. 2: both springs at $\frac l2$ : $\tau=2k\frac{l^2}4\theta=\frac12kl^2\theta$ .

$\frac{f_1}{f_2}=\sqrt{\frac{5/4}{1/2}}=\sqrt{\frac52}.$

Topics:Simple and compound pendulum Torque and rotational dynamics

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JEE Advanced 2025 · Paper 2, Physics question 3