Moment of inertia of uniform rod about an axis perpendicular to it and passing from COM is `I_1`. The rod is bent in the ring then the moment of inertia of ring about axis is `I_2` then `I_1/I_2 = `(`n pi^2`)/3 then value of n is

The moment of inertia of a uniform rod about a perpendicular axis passing through one end is I_(1) . The same rod is bent into a ring and its moment of inertia about a diameter is I_(2) . Then I_(1)//I_(2) is

The M.I. of a uniform rod about a perpendicular axis passing through one of its ends is I_(1) . The same rod is bent into a ring and its moment of inertia about a diameter is I_(2) . Then (I_(1))/(I_(2)) is

If I_(1) is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mas and I_(2) is the moment of inertia of the ring about an axis perpendicular to plane of ring and passing through its centre formed by bending the rod, then

The moment of inertia of a uniform ring about an axis passing through its centre and perpendicular to its plane is 100kgm^(2) . What is the moment of inertia of the ring about its diameter ?

A wire of mass m and length l is bent in the form of circular ring. The moment of inertia of the ring about its axis is

If l_(1) is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and l_(2) is the moment of inertia (about central axis) of the ring formed by bending the rod, then the ratio of I_(1) to I_(2) is