Two rings of same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass the ring `= m`, radius `= r`)

Options

A. `1/2 mr^2`
B. `mr^2`
C. `3/2mr^2`
D. `2mr^2`

Answer

C

Explanation

Moment of inertia of a ring about its axis is, `I_2 = mr^2`
Moment of inertia of a ring about an axis passing through the center and lying in its plane is, `I_1 = (mr^2)/2`
We know that `I_1 = I_1 + I_2`
`implies I = (mr^2)/2 + mr^2`
`implies I =3/2 mr^2`