1. Identify the axes

A thin circular disk lies in the $xy$-plane.

• The axis $z$ is the axis through the center of the disk and perpendicular to its plane.
• The axis $z'$ is another axis perpendicular to the plane, passing through a point on the rim of the disk.

So we need the ratio: $I_z : I_{z'}$

2. Moment of inertia of a disk about its central perpendicular axis

For a thin circular disk of mass $M$ and radius $R$, $I_z = \frac{1}{2}MR^2$

3. Use the parallel axis theorem for the axis through the rim

The distance between the central axis $z$ and the rim axis $z'$ is $R$.

By parallel axis theorem, $I_{z'} = I_z + MR^2$

Substitute $I_z = \frac{1}{2}MR^2$: $I_{z'} = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$

4. Find the ratio

$I_z : I_{z'} = \frac{1}{2}MR^2 : \frac{3}{2}MR^2 = 1:3$

5. Check options

• A: $1:3$ ✅
• B: $1:4$ ❌
• C: $1:5$ ❌
• D: $1:2$ ❌

Therefore, the correct option is A.