A wire PQ has length $4.8 \mathrm{~m}$ and mass $0.06 \mathrm{~kg}$. Another wire QR has length $2.56…
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)
0.1 s
0.12 s
0.14 s
0.16 s
Solution
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:
$\mathrm{t}=0.06+0.08=0.14 \mathrm{~s}$