1. Given surface density

   The circular plate is thin, so we treat it as a lamina with surface mass density $\sigma(r)=\rho_0 r$ where $r$ is the distance from the centre.

2. Find total mass $M$

   Consider a thin ring of radius $r$ and thickness $dr$.

   Its area is $dA=2\pi r\,dr$

   Hence its mass is $dm=\sigma(r)\,dA=\rho_0 r(2\pi r\,dr)=2\pi\rho_0 r^2\,dr$

   So total mass: $M=\int_0^R dm=2\pi\rho_0\int_0^R r^2\,dr$ $M=2\pi\rho_0\left[\frac{r^3}{3}\right]_0^R$ $M=\frac{2\pi\rho_0 R^3}{3}$

3. Moment of inertia about central perpendicular axis

   For the same ring element, $dI_O=r^2\,dm$

   Therefore, $I_O=\int_0^R r^2\,dm=\int_0^R r^2(2\pi\rho_0 r^2\,dr)$ $I_O=2\pi\rho_0\int_0^R r^4\,dr$ $I_O=2\pi\rho_0\left[\frac{r^5}{5}\right]_0^R$ $I_O=\frac{2\pi\rho_0 R^5}{5}$

   Now use $M=\frac{2\pi\rho_0 R^3}{3}$

   So, $I_O=\frac{2\pi\rho_0 R^5}{5}=\frac{3}{5}MR^2$

4. Moment of inertia about axis through edge

   The required axis is perpendicular to the plate and passes through the edge. By parallel axis theorem, $I_{\text{edge}}=I_O+MR^2$

   Hence, $I_{\text{edge}}=\frac{3}{5}MR^2+MR^2=\frac{8}{5}MR^2$

5. Compare with $I=aMR^2$

   Therefore, $a=\frac{8}{5}$

6. Option check

   • A: $\frac12$ ✗
   • B: $\frac32$ ✗
   • C: $\frac85$ ✓
   • D: $\frac35$ ✗

Therefore, the correct option is C.