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MHT CET 2026 (13 May, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2026 (13 May, Shift 2)

Q25MHT CET 2026 · 13 MAY · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

A thin wire of length LL and uniform linear mass density ρ\rho is bent into a circular loop with centre at 'OO' as shown in figure. The moment of inertia of the loop about the axis XX′XX' is

Figure

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Answer: (d)

  1. Mass M=ρLM = \rho L, radius R=L2πR = \frac{L}{2\pi}.
  2. XX′XX' is a tangent in the plane of the loop: I=MR22+MR2=32MR2I = \frac{MR^2}{2} + MR^2 = \frac{3}{2}MR^2.
  3. I=32ρL⋅L24π2=3ρL38π2I = \frac{3}{2}\rho L \cdot \frac{L^2}{4\pi^2} = \frac{3\rho L^3}{8\pi^2}
  4. So the answer is (d) 3ρL38π2\frac{3\rho L^3}{8\pi^2}.
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