Option B: $10\sqrt{2}$ cm
The chord subtends 90° at the centre. Using the right triangle formed by two radii (each 10 cm) and the chord: chord $= \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}$ cm.
Since the angle at the centre is 90°, the two radii form a right-angled isosceles triangle with the chord as hypotenuse. Apply Pythagoras theorem: $\text{chord} = \sqrt{r^2 + r^2} = r\sqrt{2} = 10\sqrt{2}$. This is a standard geometry result often tested alongside sector/segment area problems (see Q.4, Exercise 11.1).