A circular disc of radius R is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through an angle of 60° and released. Its angular velocity when it reaches the equilibrium position will be

PE at θ = 60° is MgR (1 - cos θ) where R is the radius of the disc. Gain in KE when the disc reaches the equilibrium position =12 Iω2 where I=ICM+Mx2=12MR2+MR2=32MR2. Here x is the distance between the centre of mass and the axis of rotation, i.e. x = R.Now PE = KE givesMgR1−cos⁡60∘=12Iω2=34MR2ω2  (∵h=R)which gives ω=2g3R, which is choice (b).