A thin circular plate of mass $$M$$ and radius $$R$$ has its density varying as $$\rho(r) = \rho_0 r$$ with $$\rho_0$$ as constant and $$r$$ is the distance from its centre. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is $$I = aMR^{2}$$. The value of the coefficient $$a$$ is:

We have a thin circular plate of radius $$R$$ in which the surface-mass density depends on distance from the centre according to $$\rho(r)=\rho_0\,r$$, where $$\rho_0$$ is a constant and $$r$$ varies from $$0$$ to $$R$$. Because the plate is thin we treat it as a two-dimensional lamina, so an elemental ring of radius $$r$$ and infinitesimal width $$dr$$ has area $$dA = 2\pi r\,dr$$.

The mass of this elemental ring is therefore

$$dm = \rho(r)\,dA = \rho_0\,r \;(2\pi r\,dr) = 2\pi\rho_0\,r^{2}\,dr.$$

To find the total mass $$M$$ of the plate we integrate $$dm$$ from the centre $$r=0$$ out to the rim $$r=R$$:

$$\displaystyle M = \int_{0}^{R} 2\pi\rho_0\,r^{2}\,dr = 2\pi\rho_0 \int_{0}^{R} r^{2}\,dr = 2\pi\rho_0 \left[\frac{r^{3}}{3}\right]_{0}^{R} = \frac{2\pi\rho_0 R^{3}}{3}.$$

Now we calculate the moment of inertia about the axis perpendicular to the plate and passing through the centre. For an elemental ring the distance of every point from this axis is exactly $$r$$, so

$$dI_{\text{centre}} = r^{2}\,dm = r^{2}\;(2\pi\rho_0\,r^{2}\,dr) = 2\pi\rho_0\,r^{4}\,dr.$$

Integrating gives

$$\displaystyle I_{\text{centre}} = \int_{0}^{R} 2\pi\rho_0\,r^{4}\,dr = 2\pi\rho_0 \int_{0}^{R} r^{4}\,dr = 2\pi\rho_0 \left[\frac{r^{5}}{5}\right]_{0}^{R} = \frac{2\pi\rho_0 R^{5}}{5}.$$

To obtain the moment of inertia about an axis perpendicular to the plate but passing through the edge, we use the Parallel-Axis Theorem. The theorem states:

$$I_{\text{edge}} = I_{\text{centre}} + M\,d^{2},$$

where $$d$$ is the distance between the two parallel axes. Here the centre axis is at the centre and the new axis is at the rim, so $$d = R$$. Substituting we get

$$I_{\text{edge}} = \frac{2\pi\rho_0 R^{5}}{5} + M R^{2}.$$

We now express everything in terms of the total mass $$M$$ instead of $$\rho_0$$. From the earlier mass formula we have

$$2\pi\rho_0 R^{3} = 3M \; \Longrightarrow \; \frac{2\pi\rho_0 R^{5}}{5} = \frac{3M R^{2}}{5}.$$

Substituting this into the expression for $$I_{\text{edge}}$$ gives

$$I_{\text{edge}} = \frac{3M R^{2}}{5} + M R^{2} = M R^{2}\left(\frac{3}{5} + 1\right) = M R^{2}\left(\frac{3}{5} + \frac{5}{5}\right) = M R^{2}\left(\frac{8}{5}\right).$$

Thus the moment of inertia can be written in the required form $$I = a M R^{2}$$ with

$$a = \frac{8}{5}.$$

Hence, the correct answer is Option C.