
Using parallel axis theorem, we can write total moment of inertia as, I=(52mR2+md2)×2
⇒I=2(52×2×(21)2+2×(43)2)=2053kgm2
Therefore, x=53.
JEE Main 2024 — Physics Mechanics
Two identical spheres each of mass 2kg and radius 50cm are fixed at the ends of a light rod so that the separation between the centers is 150cm. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is 20xkgm2, where the value of x is
Held on 31 Jan 2024 · Verified 6 Jul 2026.
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