Moment of inertia of a uniform disc about its own axis

For a uniform disc, $I = \frac{1}{2}MR^2$

Mass of each disc

Since both discs are made of the same material and have the same thickness, their masses are proportional to their volumes.

Volume of a disc: $V = \pi R^2 t$ where $t$ is the common thickness.

Hence, $M \propto R^2$

Dependence of moment of inertia on radius

Using $I = \frac{1}{2}MR^2$ and $M \propto R^2$, we get $I \propto R^2 \cdot R^2 = R^4$

Apply to the two discs

Given: $R_1 = R, \qquad R_2 = \alpha R$

Therefore, $I_1 : I_2 = R^4 : (\alpha R)^4 = 1 : \alpha^4$

But it is given that $I_1 : I_2 = 1 : 16$

So, $\alpha^4 = 16$

$\alpha = 2$

(Taking the positive value since radius is positive.)

Check options

• A: $\sqrt{2}$
• B: $2$
• C: $2\sqrt{2}$
• D: $4$

Hence the correct option is B.