
Using parallel axis theorem,
IPQ=ICOM +Md2
=52MR2+Md2
=M[52×52+102]
=M[110]=Mk2
⇒k=110cm
Therefore, x=110.
Solid sphere A is rotating about an axis PQ. If the radius of the sphere is 5cm, then its radius of gyration about PQ will be xcm. The value of x is _____.

Held on 24 Jan 2023 · Verified 6 Jul 2026.
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Work through every JEE Main Mechanics PYQ, year by year.