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Rotational Motion question

2025 · Shift 1 · Q33
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  5. /2025 · Shift 1 · Q33

Rotational Motion question

2025 · Shift 1 · Q33

JEE AdvancedPhysicsRotational MotionMCQ+3 / −1
The center of a disk of radius rrr and mass mmm is attached to a spring of spring constant kkk, inside a ring of radius R>rR>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=\frac{2 \pi}{\omega}T=ω2π​. The correct expression for ω\omegaω is (ggg is the acceleration due to gravity): JEE Advanced 2025 Paper 1 Online Physics - Rotational Motion Question 1 English
  1. A
    23(gR−r+km)\sqrt{\frac{2}{3} \left( \frac{g}{R - r} + \frac{k}{m} \right)}32​(R−rg​+mk​)​
  2. B
    2g3(R−r)+km\sqrt{\frac{2g}{3(R - r)} + \frac{k}{m}}3(R−r)2g​+mk​​
  3. C
    16(gR−r+km)\sqrt{\frac{1}{6} \left( \frac{g}{R - r} + \frac{k}{m} \right)}61​(R−rg​+mk​)​
  4. D
    14(gR−r+km)\sqrt{\frac{1}{4} \left( \frac{g}{R - r} + \frac{k}{m} \right)}41​(R−rg​+mk​)​
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