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Question

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 (about central axis) of the ring formed by bending the rod, then

A
I1:I2=1:1
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B
I1:I2=π2:3
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C
I1:I2=π:4
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D
I1:I2=3:5
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Solution

The correct option is B I1:I2=π2:3
I1=Ml212
I2=MR2
l=2πR
R=l2π
I=Ml24π2
I1I2=Ml212×4π2Ml2=π23
I1:I2=π2:3.
Hence, the answer is I1:I2=π2:3.

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