**Solution:**

Let the two concentric circles with centre O.

AB be the chord of the larger circle which touches the smaller circle at point P.

∴ AB is tangent to the smaller circle to the point P.

⇒ OP ⊥ AB

By Pythagoras theorem in ΔOPA,

OA^{2} = AP^{2} + OP^{2}

⇒ 5^{2} = AP^{2} + 3^{2}

⇒ AP^{2} = 25 - 9

⇒ AP = 4

In ΔOPB,

Since OP ⊥ AB,

AP = PB (Perpendicular from the center of the circle bisects the chord)

AB = 2AP = 2 × 4 = 8 cm

**∴ The length of the chord of the larger circle is 8 cm.**