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If I1 is the moment of inertia of a thin rod about an  axis perpendicular to its length and passing through its centre of mass and I2 is  the moment of inertia of ring about an axis perpendicular to the plane of ring and passing through its centre formed by bending the rod, then I1I2=

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a

3π2

b

2π2

c

π23

d

π22

answer is C.

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Detailed Solution

length of the rod  l=2πr   r=l2πI1=ml212  and I2=mr2=ml2π2=ml24π2I1I2=4π212=π23
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