JAMB 2001 · UME · Q25

The chord and the two radii to $S$ and $T$ are all $r$, so that triangle is equilateral and the arc makes $60^\circ$ at the centre.

Arc length is $\frac{\theta}{360^\circ} \times 2\pi r$, so the arc is $\frac{60}{360} \times 2\pi r$.

That simplifies to $\frac{\pi r}{3}$, option B.