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MHT CET 2025 (20 Apr, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2025 (20 Apr, Shift 1)

Q54MHT CET 2025 · 20 APR · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

Two discs A and B of same material and thickness have radii RR and 3R3R respectively. Their moments of inertia about their axis will be in the ratio

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Answer: (c)

  1. Same material and thickness: M=ρπR2t∝R2M = \rho \pi R^2 t \propto R^2, so I=MR22∝R4I = \frac{MR^2}{2} \propto R^4.
  2. IAIB=(R3R)4=181\frac{I_A}{I_B} = \left(\frac{R}{3R}\right)^4 = \frac{1}{81}
  3. So the answer is (c) 1:811 : 81.
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