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JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Rotational Mechanics

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.

Answer: 4

Solution

For rotation about a diameter, with all masses equal to M:

I_{d i sc} = \frac{M R _{d}^{2}}{4}, I_{r in g} = \frac{M R _{r}^{2}}{2}, I_{s p h er e} = \frac{2 M R _{d}^{2}}{5}

(the sphere shares the disc's radius

R_{d}

).

Given

I_{d i sc} = 2.5 I_{r in g}

:

\frac{M R _{d}^{2} /4}{M R _{r}^{2} /2} = 2.5 ⟹ \frac{R _{d}^{2}}{2 R _{r}^{2}} = 2.5 ⟹ R_{d}^{2} = 5 R_{r}^{2}

Then

n = \frac{I _{s p h er e}}{I _{r in g}} = \frac{2 M R _{d}^{2} /5}{M R _{r}^{2} /2} = \frac{4 R _{d}^{2}}{5 R _{r}^{2}} = \frac{4}{5} \times 5 = 4

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Integer answer

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