11. To place three squares with side lengths of $1 \mathrm{~cm}$ inside a circular dish. The squares must not have any part outside the dish and must not overlap: what is the minimum radius of the circular dish?

To place three squares with side lengths of $1 \mathrm{~cm}$ inside a circular dish. The squares must not have any part outside the dish and must not overlap: what is the minimum radius of the circular dish?
11. $\frac{5 \sqrt{17}}{16}$.

The placement of three squares can be considered in the following scenarios:
First scenario: Placed side by side to form a $3 \mathrm{~cm} \times 1 \mathrm{~cm}$ rectangle, as shown in Figure 5. The diameter of the circular disc must be at least the diagonal of the rectangle, which is $\sqrt{10} \mathrm{~cm}$. The radius must be at least $\frac{\sqrt{10}}{2} \mathrm{~cm}$.

Second scenario: As shown in Figure 6. In this case, the radius of the circular disc must be at least the diagonal of the square, which is $\sqrt{2} \mathrm{~cm}$.

Third scenario: As shown in Figure 7. In this case, the diameter of the circular disc must be at least $2 \sqrt{2} \mathrm{~cm}$, and the radius must be at least $\sqrt{2} \mathrm{~cm}$.

In the third scenario, the upper square gradually moves to the right until it reaches the middle position (as shown in Figure 8).

Fourth scenario: As shown in Figure 8. In this case, the radius $r=O B=O .4<\sqrt{2} \mathrm{~cm}$. It is easy to see that the distance from any point in the figure to $O$ is no greater than $O A$. Let $O C$ $=x$, then $O D=2-x, O B^{2}=r^{2}=\frac{1}{4}+x^{2}, O A^{2}=r^{2}=1$ $+(2-x)^{2}$, which can be calculated to give $x=\frac{19}{16}$. Thus, $r=\frac{5 \sqrt{17}}{16}$.