Two rings of the 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 of the ring = m, radius = r)

Options

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

Answer

C

Explanation

Because the planes of two rings are mutually perpendicular and centres are coincident , hence an axis , which is passing through centre of one of the rings and perpendicular 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)/(2) mr^(2)`