A bar magnet of length 5.0 cm 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 200 g.

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)`