Correct Answer: 30 cm

Solution :
Given: O is the centre of a circle.
AB = 16 cm is a chord of the given circle and AB is at a distance of 15 cm from O. Another chord CD is at a distance of 8 cm from O.
We join OA and draw a perpendicular ON to AB.
So, $\angle$ONA = 90°
So, OA is the radius of the circle.
Now AB = 2AN, since the perpendicular, drawn from the centre of a circle to the chord, bisects the chord.
So, AN = $\frac{16}{2}=8$ cm
Given, ON = 15 cm
By using Pythagoras's theorem, we get,
OA = $\sqrt{AN^2+ON^2}=\sqrt{8^2+15^2}=17$ cm
Now, in ΔOMC, OM = 8 cm and OC = OA = 17 cm (radii of the same circle).
Also, $\angle$OMC= 90°, since OM is the perpendicular distance of CD from O.
Using Pythagoras theorem, we get,
CM = $\sqrt{OC^2-OM^2}=\sqrt{17^2-8^2}=15$ cm.
M bisects CD in half.
So, CD = 2CM = 2 × 15 = 30 cm
So, the length of the chord is 30 cm.
Hence, the correct answer is 30 cm.