Moment of inertia of a disc of mass m and radius R about the centre of mass and perpendicular to plane is I₁ and moment of inertia of disc at a distance 2R/3 from centre and perpendicular to plane is I₂ then I₁/I₂ is (a) 6/17 (b) 9/17 (c) 8/9 (d) 5/7

Answer: (b). I₁ = mR²/2 (centre, perpendicular to plane). Parallel axis at 2R/3: I₂ = mR²/2 + m4R²/9 = 17/18mR² (I₁)/(I₂) = (1/2)/(17/18) = 9/17 So the answer is (b) 9/17.

Moment of inertia of a disc of mass m and radius R about the centre… | MHT CET 2026 (15 May, Shift 2)

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Moment of Inertia & Radius of Gyration

Moment of inertia of a disc of mass $m$ and radius $R$ about the centre of mass and perpendicular to plane is $I_1$ and moment of inertia of disc at a distance $\frac{2R}{3}$ from centre and perpendicular to plane is $I_2$ then $I_1/I_2$ is

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Answer: (b)

$I_1 = \frac{mR^2}{2}$ (centre, perpendicular to plane).
Parallel axis at $\frac{2R}{3}$ : $I_2 = \frac{mR^2}{2} + m\frac{4R^2}{9} = \frac{17}{18}mR^2$
$\frac{I_1}{I_2} = \frac{1/2}{17/18} = \frac{9}{17}$
So the answer is (b) $\frac{9}{17}$ .

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Moment of Inertia & Radius of Gyration question from MHT CET 2026 (15 May, Shift 2)

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