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The M.I. of a uniform rod about a perpen...

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

A

`(pi^(2))/(3)`

B

`(2pi^(2))/(3)`

C

`(4pi^(2))/(3)`

D

`(8pi^(2))/(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
`I_(1)= (ML^(2))/(3)` ....(i)
`2pi R = L rArr R= (L)/(2pi)`
`I_(2)= (MR^(2))/(2)= (M)/(2) (L^(2))/(4pi^(2))= (ML^(2))/(8pi^(2))` ...(ii)
From (i) & (ii)
`(I_(1))/(I_(2))= (1)/(3) xx 8pi^(2)= (8pi^(2))/(3)`
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