A thin wire of length L and uniform linear mass density p is bent into a circular loop with centre O as shown in fig. The moment of inertia of the loop about the axis XX' is

The moment of inertia of the loop about the diameter = M R2/ 2. So, the moment of inertia of the loop about XX' is given byIXX=MR22+MR2=32MR2where M = mass of the loop and .R = radius of the loop NowM=Lρ  and  R=L/2π∴ IXX′=32(Lρ)(L/2π)2=3L3ρ/8π2