Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: the radius of gyration of a disc of mass 50 gram and radius 2.5 cm,about an axis passing through its
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BALGANESH (1654)

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1/2mr^2 = mk^2 K = r/root2 = 2.5 / root 2 = 1.76 so the answer is c
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