(b) \(5\sqrt3\) cm
Given, OP = 10 cm and OA = 5 cm Since, the tangent to a circle at a point is perpendicular to the radius through the point of contact
OA ⊥ PA ⇒ ∠OAP = 90°
∴ In ΔOAP, PA2 = OP2 – OA2
⇒ PA2 = 100 – 25 = 75
⇒ PA = \(\sqrt{75}\) = \(5\sqrt3\) cm.