A thin ring has mass 0.25kg and radius 0.5m.its moment of inertia about an axis passing through its centre and perpendicular to its plane is
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I=MR²
Substitute the values and u will get the answer
The answer is 62.5×10^-3kgm²
Substitute the values and u will get the answer
The answer is 62.5×10^-3kgm²
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