# Consider two uniform discs of the same thickness and different radii R1=R and $R_{2}=\alpha R$ made of the same material. If the ratio of their moments of the interia I1 and I2, respectively, about their axes is $I_{1}:I_{2} = 1:16$ then the value of $\alpha$ is: Option: 1 Option: 2 Option: 3 2 Option: 4 4

$\text { Moment of inertia of disc is given by } 1=\frac{M R^{2}}{2}=\frac{\left[\rho\left(\pi R^{2}\right) t\right] R^{2}}{2}$

$\begin{array}{l} \mathrm{I} \propto \mathrm{R}^{4} \\ \frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{4} \\ \\ \frac{16}{1}=\alpha^{4} \\ \\ \alpha=2 \end{array}$

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