Dear Amrit pal
consider this hollow cone ,now cut a ring of thickness dx at a distance x from the origin ,
so radius of the ring will be r'=xtanΘ where Θ is the half angel of cone
so moment of inertia of this ring is dI =dm r'2
dm =[ M/πrL ]2πr'dL where L is slant hight
so dI =[2M/rL]r'3dx/cosΘ
I =o∫h tan3Θ [2M/rh]x3dx
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed solution very quickly.
We are all IITians and here to help you in your IIT JEE preparation.
All the best.