The center of a disk of radius and mass is attached to a spring of spring constant , inside a ring of radius 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 . The correct expression for is ( is the acceleration due to gravity):
← JEE Advanced 2025 · Paper 1
PhysicsSingle correct+3 / -1Question 17 of 48
JEE Advanced 2025 · Paper 1, Physics question 1

Answer and worked solution
Official answer: A
Let be the arc displacement of the disk's centre, which moves on a circle of radius . The spring stretches by .
Kinetic energy. Rolling without slipping, with :
Potential energy. For small the height rise is , so
Frequency.