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:

We are given two uniform circular discs having the same thickness and made of the same material. Their radii are $$R_1 = R$$ and $$R_2 = \alpha R$$, respectively. We denote their moments of inertia about the central axes (perpendicular to the plane of the discs) by $$I_1$$ and $$I_2$$.

First we recall the standard formula for the moment of inertia of a uniform disc about its central axis. The formula is

$$I = \dfrac{1}{2} M R^2,$$

where $$M$$ is the mass of the disc and $$R$$ is its radius.

Because the two discs have the same thickness and are made of the same material, their mass per unit volume (density) is the same. Let us call the mass density $$\rho$$ and the common thickness $$t$$. For any disc, the mass equals the product of density, volume, and area. The volume of a disc is the area of its face times the thickness, so the mass of a disc of radius $$R_i$$ is

$$M_i = \rho\,(\text{volume}) = \rho \, (\text{area}) \times (\text{thickness}) = \rho \,(\pi R_i^2)\,t.$$

Now we calculate each moment of inertia separately.

For the first disc, having radius $$R_1 = R$$, its mass is

$$M_1 = \rho\,\pi R^2\,t.$$

The corresponding moment of inertia is

$$I_1 = \dfrac{1}{2} M_1 R^2 = \dfrac{1}{2}\bigl(\rho \pi R^2 t\bigr) R^2 = \dfrac{1}{2}\,\rho\,\pi\,t\,R^4.$$

For the second disc, whose radius is $$R_2 = \alpha R$$, its mass becomes

$$M_2 = \rho\,\pi\,(\alpha R)^2\,t = \rho\,\pi\,\alpha^2 R^2\,t.$$

Using the same moment-of-inertia formula, we write

$$I_2 = \dfrac{1}{2} M_2 (\alpha R)^2 = \dfrac{1}{2} \bigl(\rho\,\pi\,\alpha^2 R^2\,t\bigr) (\alpha^2 R^2) = \dfrac{1}{2}\,\rho\,\pi\,t\,\alpha^4 R^4.$$

We are told that the ratio of the moments of inertia satisfies

$$I_1 : I_2 = 1 : 16.$$

Therefore we set up the equation

$$\dfrac{I_1}{I_2} = \dfrac{1}{16}.$$

Substituting the explicit expressions for $$I_1$$ and $$I_2$$, we have

$$\dfrac{\dfrac{1}{2}\,\rho\,\pi\,t\,R^4}{\dfrac{1}{2}\,\rho\,\pi\,t\,\alpha^4 R^4} = \dfrac{1}{16}.$$

Notice that the common factors $$\dfrac{1}{2}\,\rho\,\pi\,t\,R^4$$ cancel from numerator and denominator, leaving

$$\dfrac{1}{\alpha^4} = \dfrac{1}{16}.$$

Cross-multiplying, we obtain

$$\alpha^4 = 16.$$

Taking the positive fourth root (since radius is positive), we get

$$\alpha = 16^{1/4} = 2.$$

The numerical value of $$\alpha$$ is therefore $$2$$, which corresponds to Option C.

Hence, the correct answer is Option C.