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MHT CET 2024 (10 May, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2024 (10 May, Shift 1)

Q68MHT CET 2024 · 10 MAY · SHIFT 1Easy

Moment of Inertia & Radius of Gyration

The moment of inertia of uniform circular disc is maximum about an axis perpendicular to the disc and passing through point

Figure

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Answer: (b)

  1. Parallel axis theorem: I=MR22+Md2I = \frac{MR^2}{2} + Md^2, which grows with distance dd from the centre.
  2. BB is on the rim (d=Rd = R), the farthest of the marked points; AA is the centre, CC and DD lie inside.
  3. IB=32MR2I_B = \frac{3}{2}MR^2 is the largest.
  4. So the answer is (b) B.
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