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The moment of inertia of a ring of mass ...

The moment of inertia of a ring of mass 5 gram and radius 1 cm about and axis passing through its edge and parallel to its natural axis is

A

`5 g - cm"^(2)`

B

`"2.5 g - cm"^(2)`

C

`"20 g - cm"^(2)`

D

`"10 g - cm"^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`I_(t)=I_(cm)+MR^(2)=MR^(2)+MR^(2)=2MR^(2)`
`=2xx5xx(1)^(2)="10 g-cm^(2)`
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