If I_(1) is the moment of inertia of a thin' rod about an axis perpendicular to its length and passing through the centre of mass and I_(2) is the moment of inertia of the ring formed by bending this rod in the form of a ring, then (I_1)/(I_2) is

If I_(1) is the moment of inertia of a thin' rod about an axis perpend

1.	If I_(1) is the moment of inertia of a thin' rod about an axis perpendicular to its length and passing through the centre of mass and I_(2) is the moment of inertia of the ring formed by bending this rod in the form of a ring, then (I_1)/(I_2) is
Answer» `1 : 1` `PI^2 : 3` `pi : 4` `3 : 5` Solution :Let M and L be the mass and the length of the rod respectively . `therefore I_(1) = (1)/(12) ML^(2) ""because L = 2pi r or r = (L)/(2pi)` `therefore I_(2) = MR^(2) = M ((L)/(2pi))^(2)` THUS , `(I_(1))/(I_(2)) = (pi^2)/(3)`