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. `1/2mr^(2)`
B. `mr^(2)`
C. `3/2mr^(2)`
D. `2mr^(2)`

Answer

c

Explanation

Because the plane of two ringns are mutually perpendicular and centres are coincident, hence `n` axis, which is passing through the centre of one of the rings and to its plane, will be along the diameter of other ring. Hence, moment of inertia of the system
`=I_(CM)+I_("diameter")=mr^(2)+(mr^(2))/2`
`=3/2mr^(2)`