A circular disc has radius R₁ and thickness T₁. Another circular disc made of the same material has radius R₂ and thickness T₂. If the moment of inertia of both discs are same and (R₁)/(R₂) = 2 then (T₁)/(T₂) = (1)/(α). The value of α is ______. Numerical answer type. Work it out, then open the answer.

Answer: 16. I = 1/2(ρπ R² T)R² ∝ TR⁴. T₁R₁⁴ = T₂R₂⁴ ⇒ (T₁)/(T₂) = ((R₂)/(R₁))⁴ = 1/16 So α = 16.

A circular disc has radius R₁ and thickness T₁. Another circular disc… | JEE Main 2026 (22 Jan, Shift 1)

Q30JEE Main 2026 · 22 JAN · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

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 $\frac{T_1}{T_2} = \frac{1}{\alpha}$ . The value of $\alpha$ is ______.

Numerical answer type. Work it out, then open the answer.

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Answer: 16

$I = \frac{1}{2}(\rho\pi R^2 T)R^2 \propto TR^4$ .
$T_1R_1^4 = T_2R_2^4 \Rightarrow \frac{T_1}{T_2} = \left(\frac{R_2}{R_1}\right)^4 = \frac{1}{16}$
So $\alpha = 16$ .

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