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

Options

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

Answer

B

Explanation

The geometrical axes of the rings are perpendicular to each other. For the first ring, the M.I. about an axis passing through its centre and perpendicular to its axis is `mr^(2)`. But the other ring is perpendicular to the first. Hence its axis becomes the diameter of the first ring and the M.I. of a ring about its diameter `=(1)/(2)mr^(2)`
`therefore" Total M.I. "=mr^(2)+(mr^(2))/(2)=(3)/(2)mr^(2)`