JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Rotational Mechanics

A circular disc has radius

R_{1}

and thickness

T_{1}

. Another circular disc made of the same material has radius

R_{2}

and thickness

T_{2}

. If the moment of inertia of both discs are same and

\frac{R _{1}}{R _{2}} = 2

then

\frac{T _{1}}{T _{2}} = \frac{1}{α}

. The value of

α

is ____ .

Answer: 16

Solution

For a disc

I = \frac{1}{2} m R^{2}

with

m = ρ π R^{2} T

, so

I = \frac{1}{2} ρ π R^{4} T

. Equal moments of inertia for the same material require

R_{1}^{4} T_{1} = R_{2}^{4} T_{2}

, hence

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

. Comparing with

\frac{1}{α}

, the value of

α

16

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Integer answer

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