Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' I₁ '. The same rod is bent into a ring and its moment of inertia about the diameter is ' I₂ '. Then I₁/I₂ is (a) (3π²)/(2) (b) (2π²)/(3) (c) (π²)/(3) (d) (π²)/(9)

Answer: (b). Rod: I₁ = (ML²)/(12). Ring of radius R = L/2π about a diameter: I₂ = MR²/2 = (ML²)/(8π²). (I₁)/(I₂) = (8π²)/(12) = (2π²)/(3) So the answer is (b) (2π²)/(3).

Moment of inertia of the rod about an axis passing through the centre… | MHT CET 2025 (26 Apr, Shift 1)

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Moment of Inertia & Radius of Gyration

Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' $I_1$ '. The same rod is bent into a ring and its moment of inertia about the diameter is ' $I_2$ '. Then $I_1/I_2$ is

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Answer: (b)

Rod: $I_1 = \frac{ML^2}{12}$ .
Ring of radius $R = \frac{L}{2\pi}$ about a diameter: $I_2 = \frac{MR^2}{2} = \frac{ML^2}{8\pi^2}$ .
$\frac{I_1}{I_2} = \frac{8\pi^2}{12} = \frac{2\pi^2}{3}$
So the answer is (b) $\frac{2\pi^2}{3}$ .

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Moment of Inertia & Radius of Gyration question from MHT CET 2025 (26 Apr, Shift 1)

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