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(B) The moment of inertia of a uniform disc about its central axis is given by $I = \frac{1}{2} M R^{2}$.Since the discs are made of the same material

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$ 2 $

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(B) The moment of inertia of a uniform disc about its central axis is given by $I = \frac{1}{2} M R^{2}$.Since the discs are made of the same material and have the same thickness $t$,the mass $M$ can be expressed as $M = \rho V = \rho (\pi R^{2} t)$,where $\rho$ is the density.Substituting this into the formula for $I$,we get $I = \frac{1}{2} (\rho \pi R^{2} t) R^{2} = \frac{1}{2} \rho \pi t R^{4}$.Since $\rho$,$\pi$,and $t$ are constant for both discs,we have $I \propto R^{4}$.Given the ratio $\frac{I_{1}}{I_{2}} = \frac{1}{16}$,we can write $\frac{R_{1}^{4}}{R_{2}^{4}} = \frac{1}{16}$.Taking the fourth root of both sides,we get $\frac{R_{1}}{R_{2}} = \frac{1}{2}$.Given $R_{1} = R$ and $R_{2} = \alpha R$,we have $\frac{R}{\alpha R} = \frac{1}{2}$,which implies $\frac{1}{\alpha} = \frac{1}{2}$.Therefore,$\alpha = 2$.

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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