1.

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.

Answer»

SOLUTION :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)`


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