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JEE Main 2025 (8 Apr, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2025 (8 Apr, Shift 2)

Q55JEE Main 2025 · 8 APR · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

A rod of linear mass density 'λ' and length 'L' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is.

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Answer: (a)

  1. Mass M=λLM = \lambda L, radius R=L2πR = \frac{L}{2\pi}.
  2. About a diameter: I=MR22=λL2⋅L24π2=λL38π2I = \frac{MR^2}{2} = \frac{\lambda L}{2}\cdot\frac{L^2}{4\pi^2} = \frac{\lambda L^3}{8\pi^2}
  3. So the answer is (a) λL38π2\frac{\lambda L^3}{8\pi^2}.
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