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MHT CET 2026 (16 Apr, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2026 (16 Apr, Shift 1)

Q7MHT CET 2026 · 16 APR · SHIFT 1Easy

Moment of Inertia & Radius of Gyration

A Solid sphere has mass M and radius R. Its moment of inertia about a parallel axis passing through a point at a distance R/3 from its centre is

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Answer: (c)

  1. Icm=25MR2I_{cm} = \frac{2}{5}MR^2 about a diameter.
  2. I=25MR2+M(R3)2=18+545MR2I = \frac{2}{5}MR^2 + M\left(\frac{R}{3}\right)^2 = \frac{18 + 5}{45}MR^2
  3. I=23MR245I = \frac{23MR^2}{45}
  4. So the answer is (c) 23MR2/4523MR^2/45.
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