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If l(1) is the moment of inertia of a th...

If `l_(1)` is the moment of inertia of a thin rod about an axis perpendicular to its length and passing thorugh its centre of mass and `l_(2)` the moment of inertia of the ring formed by the same rod about an axis passing through the centre of mass of the ring and perpendicular to the plane of the ring. then find the ratio `(l_(1))/(l_(2))`.

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
`I_(1) = (Ml^(2))/(12)`
If `r` is radius of ring, then `l = 2 pi r, r = l//2 pi`
`l_(2) = M r^(2) = M(l//2pi)^(2) = (Ml^(2))/(4pi^(2))`
`(I_(1))/(I_(2)) = (Ml^(2))/(12).(4pi^(2))/(Ml^(2)) = (pi^(2))/(3)`
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