two rings of same radius and same mass are placed such that their centers are at a common point and their planes are perpendicular to each other. THE MOMENT OFINERTIA ABOUT THE CENTER PASSING THROUGH CENTER AND PERPENDICULAR TO PLANE OF ONE OF THE RINGS ?

The moment of Inertia of such a system would be (3/2)MR² .

The moment of Inertia of the system = The moment of Inertia of Ring rotating about an axis perpendicular to its plane + The moment of Inertia of Ring rotating about an axis in its own plane.

The moment of Inertia of Ring rotating about an axis perpendicular to its plane= MR²

The moment of Inertia of Ring rotating about an axis in its own plane.= MR²/2

So the moment of inertia of the system is (3/2)MR²

The moment of Inertia of such a system would be (3/2)MR².
The moment of Inertia of the system = The moment of Inertia of Ring rotating about an axis perpendicular to its plane + The moment of Inertia of Ring rotating about an axis in its own plane.
The moment of Inertia of Ring rotating about an axis perpendicular to its plane= MR²
The moment of Inertia of Ring rotating about an axis in its own plane.= MR²/2
So the moment of inertia of the system is (3/2)MR²
5 years ago