Consider two uniform discs of the same thickness and different radii R1=R and R2=αR made of the same material. If the ratio of their moments of inertia I1 and I2, respectively, about their axes is I1:I2=1:16 then the value of α is:

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Consider two uniform discs of the same thickness and different radii $R_{1} = R$ and $R_{2} = α R$ 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 $α$ is:

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$2 \sqrt{2}$

$\sqrt{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} Moment of Inertia , I ∝ {mR}^{2} \\ \Rightarrow I \propto (σ π R^{2} t) R^{2} [Thickness and material is same] \\ \Rightarrow I \propto R^{4} \\ ∴ \frac{R_{2}}{R_{1}} = {(\frac{I_{2}}{I_{1}})}^{1 / 4} = (16)^{1 / 4} = 2 \\ \Rightarrow α = 2 \end{array}$