1. Given

A disc with central hole has:

• inner radius $a$
• outer radius $b$
• surface mass density varying as $\sigma(r)=\frac{\sigma_0}{r}$

We need the radius of gyration $k$ about the axis through the centre and perpendicular to the disc.

By definition, $k=\sqrt{\frac{I}{M}}$ where $M$ is total mass and $I$ is moment of inertia about the given axis.

2. Take a thin circular ring element

At radius $r$, thickness $dr$:

• area of ring $dA=2\pi r\,dr$
• mass of ring $dm=\sigma(r)\,dA=\frac{\sigma_0}{r}(2\pi r\,dr)=2\pi\sigma_0\,dr$

So the mass element is independent of $r$.

3. Find total mass

Integrate from $r=a$ to $r=b$: $M=\int_a^b dm=\int_a^b 2\pi\sigma_0\,dr=2\pi\sigma_0(b-a)$

4. Find moment of inertia

For a ring element, $dI=r^2\,dm$ Thus, $dI=r^2(2\pi\sigma_0\,dr)$

Integrating, $I=2\pi\sigma_0\int_a^b r^2\,dr$ $I=2\pi\sigma_0\left[\frac{r^3}{3}\right]_a^b$ $I=\frac{2\pi\sigma_0}{3}(b^3-a^3)$

Now use $b^3-a^3=(b-a)(b^2+ab+a^2)$ So, $I=\frac{2\pi\sigma_0}{3}(b-a)(a^2+ab+b^2)$

5. Compute radius of gyration

$k^2=\frac{I}{M}$

Substitute $I$ and $M$: $k^2=\frac{\frac{2\pi\sigma_0}{3}(b-a)(a^2+ab+b^2)}{2\pi\sigma_0(b-a)}$

$k^2=\frac{a^2+ab+b^2}{3}$

Therefore, $k=\sqrt{\frac{a^2+ab+b^2}{3}}$

6. Match with options

This matches: $\boxed{\sqrt{\frac{a^2+b^2+ab}{3}}}$ which is Option C.

7. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.