A circular disc has radius R_1 and thickness | vedclass

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(C) The moment of inertia of a circular 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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$16$

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(C) The moment of inertia of a circular disc about its central axis is given by $I = \frac{1}{2} M R^2$.Since the discs are made of the same material, their density $ho$ is the same.The mass of a disc is $M = 	ext{Volume} 	imes 	ext{density} = (\pi R^2 T) ho$.Therefore, the moment of inertia is $I = \frac{1}{2} (\pi R^2 T ho) R^2 = \frac{1}{2} \pi ho R^4 T$.Given that $I_1 = I_2$, we have:$\frac{1}{2} \pi ho R_1^4 T_1 = \frac{1}{2} \pi ho R_2^4 T_2$$R_1^4 T_1 = R_2^4 T_2$$\frac{T_1}{T_2} = \frac{R_2^4}{R_1^4} = \left( \frac{R_2}{R_1} ight)^4$.Given $\frac{R_1}{R_2} = 2$, so $\frac{R_2}{R_1} = \frac{1}{2}$.$\frac{T_1}{T_2} = \left( \frac{1}{2} ight)^4 = \frac{1}{16}$.Comparing this with $\frac{T_1}{T_2} = \frac{1}{\alpha}$, we get $\alpha = 16$.

$A$ 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 . . . . . . .

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