Consider a small element `dx` of the rod which is at a distance `x` from the end `P` if `theta` is the inclination of rod w.r.t the axis of rotation, the radius of the circle in which the element rotates is given by `sin theta=r/xrArrr=x sin theta`
`M.I.` of the element about the axis of rotation is `dI=dm.r^(2)`
where `dm `is the mass of element `dm =m/L dx`
`dI=m/Ldx(x sin theta)^(2).` Total `M.I.` of the rod is given by `I=int dI=int_(0)^(L)m/Lsin^(2) thetax^(2)dx, I=(mL^(2))/3 sin^(2) theta`
