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

Two iron solid discs of negligible thickness have radii

R_{1}

and

R_{2}

and moment of intertia

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 =

____

Answer: 16

Solution

For a uniform solid disc of negligible thickness, mass

M = σ π R^{2}

, where

σ

is the same areal density for both discs since they are identical iron sheets. The moment of inertia about the central axis perpendicular to the disc is

I = \frac{1}{2} M R^{2} = \frac{1}{2} σ π R^{4}

I \propto R^{4}

. With

R_{2} = 2 R_{1}

\frac{I _{1}}{I _{2}} = (\frac{R _{1}}{R _{2}})^{4} = (\frac{1}{2})^{4} = \frac{1}{16}

Hence

x = 16

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

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