$\begin{array}{1 1}(A)\;\large\frac{10\pi}{3}\normalsize cm&(B)\;\large\frac{20}{33}\normalsize cm\\(C)\;\large\frac{20\pi}{3}\normalsize cm&(D)\;\large\frac{12\pi}{5}\normalsize cm\end{array} $

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- $\theta=\large\frac{l}{r}$
- $\theta$=angle subtended by arc
- $l$=length of the arc
- $r$=radius of the circle
- $1^{\large\circ}=(\large\frac{\pi}{180})$ radian=0.01746(approx)
- Diameter=$2\times$ radius

Given :

Diameter=40cm

Length of the chord=20cm

Radius=20cm

Since radius=length of chord=20cm

Hence the formed $\Delta$le in the circle is equilateral $\Delta$le

$\Rightarrow \theta=60^{\large\circ}$

We know that $l=r\theta$

$l=20\times 60^{\large\circ}\times \large\frac{\pi}{180^{\large\circ}}$

$l=\large\frac{20\pi}{3}$

Thus length of the minor arc of the chord is $\large\frac{20\pi}{3}$

Hence (C) is the correct answer.

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