A wire PQ has length 4.8 ~m and mass 0.06 ~kg . Another wire QR has length 2.56 ~m and mass 0.2 ~kg . Both wires have same radii and are joined as a single wire. This wire is under tension of 80 ~N . A wave pulse of amplitude 3.5 ~cm is sent along the wire PQ from end P . The time taken by the wave to reach the other end of single wire is (No power is dissipated during propagation)

Step-by-step solution

Mass per unit length for wire P Q , m _ PQ = M L = 0.06 4.8 = 1 80 ~kg / m Similarly for wire QR , m _ QR = 0.2 2.56 = 5 64 ~kg / m Velocity of the wave in PQ wire: y= T m_ PQ = 80 1 / 80 =80 ~m / s Time taken to travel PQ wire, t = d v = 4.8 80 =0.06 ~s Velocity of wave in QR wire, aligned v & = T _ net ^ m _ QR = (80+0.6) 64 5 & =32.12 ~m / s aligned Time taken to travel QR wire, t = 2.56 32.12 =0.08 ~s So, total time taken will be: t =0.06+0.08=0.14 ~s