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Question

Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is I. If the same rod is bent into a ring and its moment of inertia about its diameter is I, then the ratio II is

A
3π2/2
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B
8π2/3
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C
2π2/3
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D
5π2/3
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Solution

The correct option is C 2π2/3
Let the length of the rod be L and mass be M.
Then, moment of inertia of the rod about a perpendicular axis through the center is given by:
I=ML212......(i)

Now, the rod is bent in the shape of a ring. Hence, the circumference of the ring equals the length of the rod.
L=2πR...............(ii) where R is the radius of the ring

Moment of inertia of ring about its diameter is given by:
I=MR22.....................(iii)

From (i), (ii) and (iii),
II=ML212MR22
=L26R2=(2πR)26R2=2π23

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