Physics · Transverse waves

JEE Main 2026 — 21 January, Morning Shift — Question 40

Two strings (A, B) having linear densities $\mu_{\mathrm{A}}=2 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ and $\mu_{\mathrm{B}}=4 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ and lengths $\mathrm{L}_{\mathrm{A}}=2.5 \mathrm{~m}$ and $\mathrm{L}_{\mathrm{B}}=1.5 \mathrm{~m}$ respectively are joined. Free ends of A and B are tied to two rigid supports C and D , respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time $\mathrm{t}_{1}$ and $\mathrm{t}_{2}$ , respectively, to reach the joint. The ratio $\mathrm{t}_{1} / \mathrm{t}_{2}$ is :

Option A:
1.08
Option B:
1.9
Option C:
1.67
Option D:
1.18
Correct

Answer: D

Step-by-step solution

Given $\mathrm{L}_{\mathrm{A}}=2.5 \mathrm{~m}$ ,

\begin{aligned} & L_{B}=1.5 \mathrm{~m} & T=500 \mathrm{~N} & v_{A}=\sqrt{\frac{T}{\mu_{A}}}=\sqrt{\frac{500}{2 \times 10^{-4}}}=5 \sqrt{10} \times 10^{2} \mathrm{~m} / \mathrm{s} & v_{B}=\sqrt{\frac{T}{\mu_{B}}}=\sqrt{\frac{500}{4 \times 10^{-4}}}=5 \sqrt{5} \times 10^{2} \mathrm{~m} / \mathrm{s} & t_{1}=\frac{L_{A}}{v_{A}}=\frac{2.5}{5 \sqrt{10}} \times 10^{-2} \mathrm{~s} & t_{2}=\frac{L_{B}}{v_{B}}=\frac{1.5}{5 \sqrt{5}} \times 10^{-2} \mathrm{~s} & \therefore \frac{t_{1}}{t_{2}}=\frac{2.5}{5 \sqrt{10}} \times \frac{5 \sqrt{5}}{1.5}=\frac{5}{3} \times \frac{1}{\sqrt{2}}=\frac{1.66}{1.41}=1.18 \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Morning Shift
Subject: Physics
Chapter: Transverse waves
Topic: super position, Reflection and Transmission of waves