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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 gram cm"^(2)`

B

`"100 gram cm"^(2)`

C

`"6 gram cm"^(2)`

D

`"1 gram cm"^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`I=mr^(2)+mr^(2)=2mr^(2)="6 gram - cm"^(2)`
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