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← JEE Advanced 2025 · Paper 1
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JEE Advanced 2025 · Paper 1, Physics question 1

The center of a disk of radius rr and mass mm is attached to a spring of spring constant kk, inside a ring of radius R>rR > r as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke's law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as T=2πωT = \dfrac{2\pi}{\omega}. The correct expression for ω\omega is (gg is the acceleration due to gravity):

A large ring of radius R (radius drawn as a dashed arrow from its centre) in a vertical plane. A small disk of radius r rests at the bottom of the ring, touching its inner surface; a short double arrow by the disk shows it can move along the ring. A coiled spring of constant k runs along the inside of the ring from the disk's centre up the lower-left arc to a fixed mount on the ring.
Choose one option
Answer and worked solution

Official answer: A

Let ss be the arc displacement of the disk's centre, which moves on a circle of radius R−rR-r. The spring stretches by ss.

Kinetic energy. Rolling without slipping, with I=12mr2I=\frac12mr^2:
KE=12mv2+12⋅12mr2(vr)2=34ms˙2.KE=\tfrac12mv^2+\tfrac12\cdot\tfrac12mr^2\left(\tfrac vr\right)^2=\tfrac34m\dot s^2.

Potential energy. For small ss the height rise is s22(R−r)\frac{s^2}{2(R-r)}, so
PE=mg2(R−r)s2+12ks2.PE=\frac{mg}{2(R-r)}s^2+\frac12ks^2.

Frequency.
ω2=mgR−r+k32m=23(gR−r+km).\omega^2=\frac{\frac{mg}{R-r}+k}{\frac32m}=\frac23\left(\frac g{R-r}+\frac km\right).