Question:
Consider two uniform discs of the same thickness and different radii \(R_1 = R\) and \(R_2 = \alpha R\) made of the same material. If the ratio of their moments of inertia \(I_1\) and \(I_2\), respectively, about their axes is \(I_1 : I_2 = 1 : 16\) then the value of \(\alpha\) is :
Consider two uniform discs of the same thickness and different radii \(R_1 = R\) and \(R_2 = \alpha R\) made of the same material. If the ratio of their moments of inertia \(I_1\) and \(I_2\), respectively, about their axes is \(I_1 : I_2 = 1 : 16\) then the value of \(\alpha\) is :
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Mass is proportional to \(R^2\) (same thickness and density), and \(I = \frac{1}{2}MR^2\), so \(I \propto R^4\).
Updated On: Nov 5, 2026
- \(\sqrt{2}\)
- \(2\)
- \(2\sqrt{2}\)
- \(4\)
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The Correct Option is B
Solution and Explanation
Mass of a disc \(M = \rho \pi R^2 t\), so \(M \propto R^2\).
\(I = \frac{1}{2}MR^2 = \frac{1}{2}\rho\pi t R^4\), so \(I \propto R^4\).
\(\frac{I_2}{I_1} = \left(\frac{R_2}{R_1}\right)^4 = \alpha^4 = 16 \Rightarrow \alpha = 2\).
Answer: (2) 2
\(I = \frac{1}{2}MR^2 = \frac{1}{2}\rho\pi t R^4\), so \(I \propto R^4\).
\(\frac{I_2}{I_1} = \left(\frac{R_2}{R_1}\right)^4 = \alpha^4 = 16 \Rightarrow \alpha = 2\).
Answer: (2) 2
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