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Moment of inertia of a ring of mass m = ...

Moment of inertia of a ring of mass m = 3 gm and radius r = 1 cm about an axis passing through its edge and parallel to its natural axis is

A

`10 g- cm^(2)`

B

`100 g- cm^(2)`

C

`6 g- cm^(2)`

D

`1 g- cm^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
MOI
`I= 2MR^(2)=2 xx 3 xx (1)^(2)= 6 g-cm^(2)`
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