In a circle of diameter $40$ cm, the length of a chord is $20$ cm. Find the length of minor arc of the chord.

Diameter of the circle $=40$ cm
$\therefore$ Radius (r) of the circle $=\frac{40}{2}=20$ cm
Let AB be a chord (length = 20 cm) of the circle.
In $△OAB$, OA = OB = Radius of circle = 20 cm
Also, AB = 20 cm
Thus, $△OAB$ 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 I unit subtends an angle $\theta$ radian at the centre, then $\theta =\frac{l}{r}$
$\therefore \frac{\pi }{3}=\frac{arcAB}{20}⟹arcAB=\frac{20}{3}\pi$ cm