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 $$ \frac{R_{1}}{R_{2}}=2 \text { then }\frac{T_{1}}{T_{2}}=\frac{1}{\alpha} $$. The value of $$\alpha$$ is__________.

Two discs of the same material have the same moment of inertia. Given $$R_1/R_2 = 2$$, find $$\alpha$$ where $$T_1/T_2 = 1/\alpha$$.

$$I = \frac{1}{2}MR^2$$

The mass of a disc is $$M = \rho \cdot \pi R^2 \cdot T$$ (density times volume), where $$T$$ is thickness. Substituting:

$$I = \frac{1}{2}(\rho \pi R^2 T) R^2 = \frac{1}{2}\rho \pi R^4 T$$

$$\frac{1}{2}\rho\pi R_1^4 T_1 = \frac{1}{2}\rho\pi R_2^4 T_2$$

$$R_1^4 T_1 = R_2^4 T_2$$

$$\frac{T_1}{T_2} = \frac{R_2^4}{R_1^4} = \left(\frac{R_2}{R_1}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$$

$$\frac{T_1}{T_2} = \frac{1}{16} = \frac{1}{\alpha}$$, so $$\alpha = 16$$.

The correct answer is 16.