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MHT CET 2026 (13 May, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2026 (13 May, Shift 1)

Q26MHT CET 2026 · 13 MAY · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

Moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is I. If the same rod is bent in to a ring and its moment of inertia about its diameter is I1I_1, then the ratio I/I1I/I_1 is

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Answer: (a)

  1. Rod: I=ML212I = \frac{ML^2}{12}.
  2. Ring: R=L2πR = \frac{L}{2\pi}, about a diameter I1=MR22=ML28π2I_1 = \frac{MR^2}{2} = \frac{ML^2}{8\pi^2}.
  3. II1=8π212=2π23\frac{I}{I_1} = \frac{8\pi^2}{12} = \frac{2\pi^2}{3}
  4. So the answer is (a) 2π23\frac{2\pi^2}{3}.
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