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.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

0.36 s

$0.14 s$

2.54 s

1.14 s

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

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

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET