JEE Main 2025 Jan 28 Shift 1, Physics Q24: Moment of Inertia

JEE Main2025Jan 28, Shift 1Physics

Two iron solid discs of negligible thickness have radii $R_{1}$ and $R_{2}$ and moment of inertia $I_{1}$ and $I_{2}$ , respectively. For $R_{2} = 2 R_{1}$ , the ratio of $I_{1}$ and $I_{2}$ would be $1/ x$ , where $x =$ ______

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Solution

Both discs are cut from the same iron sheet (same surface mass density $σ$ and negligible thickness). The mass of a thin disc is $M = σ π R^{2}$ , so for the same material:

$M_{1} = σ π R_{1}^{2}$ , $M_{2} = σ π R_{2}^{2} = σ π (2 R_{1})^{2} = 4 M_{1}$ .

Moment of inertia (about the central axis): $I = \frac{1}{2} M R^{2}$ .

$\frac{I _{1}}{I _{2}} = \frac{M _{1} R _{1}^{2}}{M _{2} R _{2}^{2}} = \frac{M _{1}}{4 M _{1}} \cdot \frac{R _{1}^{2}}{( 2 R _{1} ) ^{2}} = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$ .

$∴ x = 16$ .

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JEE Main 2025 Jan 28 Shift 1, Physics Q24: Moment of Inertia

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