JEE Main 2025 Jan 28 Shift 1, Physics Q22: Moment of Inertia

JEE Main2025Jan 28, Shift 1Physics

The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is $n$ times higher than the moment of inertia of the given ring. Here, $n =$ ______.

Consider all the bodies have equal masses.

Your answer

Solution

Use the standard moments of inertia, taking all axes through the centre and lying in the plane of the body for the disc and ring (diameter axis):

Disc (about diameter): $I_{1} = \frac{M R _{1}^{2}}{4}$ .
Ring (about diameter): $I_{2} = \frac{M R _{2}^{2}}{2}$ .
Solid sphere (about any diameter): $I_{3} = \frac{2 M R _{1}^{2}}{5}$ (same radius as the disc).

Step 1: use $I_{1} = 2.5 I_{2}$ to relate $R_{1}$ and $R_{2}$ .

$\frac{M R _{1}^{2} /4}{M R _{2}^{2} /2} = 2.5 \Rightarrow \frac{R _{1}^{2}}{2 R _{2}^{2}} = \frac{5}{2} \Rightarrow \frac{R _{1}^{2}}{R _{2}^{2}} = 5$ .

Step 2: compute $n = I_{3} / I_{2}$ .

$n = \frac{2 M R _{1}^{2} /5}{M R _{2}^{2} /2} = \frac{4 R _{1}^{2}}{5 R _{2}^{2}} = \frac{4 \cdot 5}{5} = 4$ .

$∴ n = 4$ .

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JEE Main 2025 Jan 28 Shift 1, Physics Q22: Moment of Inertia

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