A hanging object cannot be considered as a point mass and it oscillates about a  fixed axis which does not pass through its centre of mass uniform  rod of mass M  and length L. The rod is pivoted at one end and hangs vertically in equilibrium with  its centre of mass vertically below the point of suspension.  The rod is slightly at  the lower end and  released. It then oscillates in a vertical plane in a simple  harmonic manner, at any instant angular displacement of the rod from its vertical  position is $\theta$ assuming small $\theta$, time period of oscillation of the rod is (Here I is the  moment of inertia of the rod about the axis about which the rod oscillates )

detailed solution

Correct option is C

For an object executing angular simple harmonic motion. τ=cθ  Where, τ=Iα , Therefore, Iα=−cθ Where d2θdt2=α and Restoring torque acting on a rod after a small  displacement θ , about an axis passing through point of contact O with the  curved path, τ=−Mgrsinθ , for small angles sinθ≈θ  Therefore,  τ=−cθ=−Mgl2θ………………….(1)Therefore Id2θdt2=−Mgl2θ  is the equation of angular simple harmonic motion.Here,  α=d2θdt2=−ω2θSubstituting this we get  −Iω2θ=−Mgl2θi.e. −Iω2=−Mgl2 Moment of inertia of a rod oscillating about one of its end is given by I=Ml23  Substituting this we get  Ml23ω2=Mgl2i.e.  l3ω2=g12simplifying this we get  ω2=3g2l⇒ω=3g2lWe know that  ω=2πTUsing this we get  2πT=3g2lHence,  T=2π2l3g