Answer is (b) 5 cm
AM = BM = \(\frac { 1 }{ 2 }\) × AM = \(\frac { 1 }{ 2 }\) × 24 = 12 cm
In right ∆OAM
OA2 = AM2 + OM2
(13)2 = (12)2 + (OM)2
OM2 = (13)2 – (12)2 = 169 – 144
OM2 = 25
OM = \(\sqrt { 25 }\) = 5 cm
Thus, distance of chord from center of circle = 5 cm.