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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

`10gm cm^(2)`

B

`100 gm cm^(2)`

C

`6gm cm^(2)`

D

`1gm cm^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`I = 2 MR^(2)=2xx3xx(1)^(2)=6 "gm cm"^(2)`.
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