In a circle of diameter 40 cm, the length of a chord is 20 cm.

Diameter of the circle = 40 cm

$\therefore$ Radius $(r)$ of the circle $=\frac{40}{2} \mathrm{~cm}=20 \mathrm{~cm}$

Let $A B$ be a chord (length $=20 \mathrm{~cm}$ ) of the circle.

In $\triangle \mathrm{OAB}, \mathrm{OA}=\mathrm{OB}=$ Radius of circle $=20 \mathrm{~cm}$

Also, $A B=20 \mathrm{~cm}$

Thus, $\triangle \mathrm{OAB}$ is an equilateral triangle.

$\therefore \theta=60^{\circ}=\frac{\pi}{3}$ radian

We know that in a circle of radius $r$ unit, if an arc of length $/$ unit subtends an angle $\theta$ radian at the centre, then $\theta=\frac{l}{r}$.

$\frac{\pi}{3}=\frac{\overparen{\mathrm{AB}}}{20} \Rightarrow \overparen{\mathrm{AB}}=\frac{20 \pi}{3} \mathrm{~cm}$

Thus, the length of the minor arc of the chord is $\frac{20 \pi}{3} \mathrm{~cm}$.