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A long wire PQR is made by joining two wires $PQ$ and $QR$ of equal radii. $PQ$ has length 4.8 m and mass $0.06 kg .$ $QR$ has length $2.56 m$ and mass $0.2 kg .$ The wire PQR is under a tension of $80 N .$ A sinusoidal wave-pulse of amplitude 3.5 cm is sent along the wire $PQ$ from the end P. Calculate the time taken by the wave-pulse to reach the other end R of the wire.

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0.36 s

$0.14 s$

2.54 s

1.14 s

answer is D.

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Detailed Solution

$v = \sqrt{\frac{T}{m}}$

Mass per unit length of wire $PQ$ is $m_{1} = \frac{0.06 kg}{4.8 m}$ $= 1.25 \times 10^{- 2} {kgm}^{- 1}$

$∴$ Speed of wave-pulse in wire $PQ$ is

$v_{1} = {(\frac{80}{1.25 \times 10^{- 2}})}^{1 / 2} = 80 {ms}^{- 1}$

Mass per unit length of wire $QR$ is $m_{2} = \frac{0.2 kg}{2.56 m} = \frac{0.2}{2.56} {kgmm}^{- 1}$

$∴$ speed of wave-pulse in wire $QR$ is

$v_{2} = {(\frac{80}{0.2 / 2.56})}^{1 / 2} = 32 {ms}^{- 1}$

(a)Time taken by the wave-pulse to travel from $P t o R$ is

$\begin{aligned} t = t_{1} + t_{2} = \frac{l_{1}}{v_{1}} + \frac{l_{2}}{v_{2}} = \frac{4.8}{80} + \frac{2.56}{32} \\ = 0.06 + 0.08 = 0.14 s \end{aligned}$

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