A bar magnet of length `5.0cm` and breadth `1.2` cm is rotated about an axis passing through its center and perpendicular to its plane. Find its moment of inertia if the mass of the magnet is `200g`.

Explanation

Moment of inertia `I=(M(l^(2)+b^(2)))/(12)`
`=200xx10^(-3)(((5xx10^(-2))^(2)+(1.2xx10^(-2))^(2))/(12))`
`=4.407xx10^(-5)kgm^(2)`