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

`10g-cm^(2)`

`100g-cm^(2)`

`6g-cm^(2)`

`1g-cm^(2)`

Text Solution

Verified by Experts

The correct Answer is:

MOI
`I=2MR^(2)=2xx3xx(1)^(2)=6"g-cm"^(2)`

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