JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

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 _____.

Your answer

Solution

For a uniform disc about its central symmetry axis, $I = \frac{1}{2} m R^{2}$ . With $m = π R^{2} T ρ$ (mass = volume $\times$ density):

$I = \frac{1}{2} (π R^{2} T ρ) R^{2} = \frac{π ρ}{2} R^{4} T$ .

Same density (same material), so $I_{1} = I_{2}$ gives $R_{1}^{4} T_{1} = R_{2}^{4} T_{2}$ .

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

So $α = 16$ .

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