Practice portal › Rotational Motion › Moment of Inertia
Asked in JEE Main 24th Jan 1st Shift 2023 · Radius of gyration and numericals
Work in centimetres throughout, since the answer is asked for in cm.
Sphere about its own diameter: I_cm=2/5MR²=2/5M(25)=10M.
Parallel-axis theorem, with d=10 cm: I=10M+M(100)=110M.
Radius of gyration: k=√I/M=√110 cm.
x=110.
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