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

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