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Asked in JEE Main 28th June 2nd Shift 2022 · Rotational energy and released bodies
For the disc, I=1/2 MR², so effective inertia I/(R²)=/M2=2 kg.
a=(mg)/(m+M/2)=(2(10))/(2+2)=5 m/s².
T=m(g-a)=2(10-5)=10 N.
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