1. Understand the geometry

A disc of radius $R=20\text{ cm}$ has a circular hole of radius $r=5\text{ cm}$ removed.

Since the hole touches the outer edge of the disc internally, the distance between the centers is $d=R-r=20-5=15\text{ cm}.$

So the center of the removed hole is at a distance $15\text{ cm}$ from the origin.

2. Use center of mass of a system with removed mass

Treat the removed circular hole as a negative mass.

Let surface mass density be uniform and equal to $\sigma$.

• Mass of full disc: $M=\sigma \pi R^2=\sigma\pi(20)^2=400\sigma\pi$

• Mass of removed hole: $m=\sigma \pi r^2=\sigma\pi(5)^2=25\sigma\pi$

Residual mass: $M_{\text{rem}}=M-m=400\sigma\pi-25\sigma\pi=375\sigma\pi$

3. Find the shift of center of mass

Take the origin at the center of the original disc. The original full disc has COM at the origin.

If the hole center is at $x=15\text{ cm}$, then for the remaining disc: $x_{\text{COM}}=\frac{M\cdot 0 - m\cdot 15}{M-m}$

Magnitude of shift is $|x_{\text{COM}}|=\frac{m\cdot 15}{M-m}$

Substitute values: $|x_{\text{COM}}|=\frac{25\sigma\pi\cdot 15}{375\sigma\pi}$

$|x_{\text{COM}}|=\frac{25\cdot 15}{375}=\frac{15}{15}=1\text{ cm}$

4. Match with options

The center of mass of the remaining disc is at a distance $\boxed{1.0\text{ cm}}$ from the origin.

So the correct option is D.

5. Compare with stored answer

Stored correct answer: D

Our derived answer: D

They agree.