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MHT CET 2025 (21 Apr, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2025 (21 Apr, Shift 2)

Q52MHT CET 2025 · 21 APR · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

Moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is ' II '. If the same rod is bent in the form of ring, its moment of inertia about the diameter is ' I1I_1 '. If I1=xII_1 = xI, then the value of ' x ' is

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Answer: (b)

  1. Rod: I=ML212I = \frac{ML^2}{12}; ring (R=L2πR = \frac{L}{2\pi}) about diameter: I1=ML28π2I_1 = \frac{ML^2}{8\pi^2}.
  2. x=I1I=128π2=32π2x = \frac{I_1}{I} = \frac{12}{8\pi^2} = \frac{3}{2\pi^2}
  3. So the answer is (b) 32π2\frac{3}{2\pi^2}.
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