Physics · Sound Waves

JEE Main 2024 — 27 January, Shift 2 — Question 52

A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of length 350 cm , both vibrating in fundamental mode. The velocity of sound is ___________\_\_\_\_\_\_\_\_\_\_\_ m/s\mathrm{m} / \mathrm{s}.

Answer: 294

Numerical answer — enter this value.

Step-by-step solution

fc=v4ℓ1fo=v2ℓ2\mathrm{f}_{\mathrm{c}}=\frac{\mathrm{v}}{4 \ell_{1}} \quad \mathrm{f}_{\mathrm{o}}=\frac{\mathrm{v}}{2 \ell_{2}}

∣fc−f0∣=7\left|\mathrm{f}_{\mathrm{c}}-\mathrm{f}_{0}\right|=7

v4×150−v2×350=7\frac{v}{4 \times 150}-\frac{v}{2 \times 350}=7

v600 cm−v700 cm=7\frac{\mathrm{v}}{600 \mathrm{~cm}}-\frac{\mathrm{v}}{700 \mathrm{~cm}}=7

v6m−v7m=7\frac{v}{6 m}-\frac{v}{7 m}=7

v(142)=7\mathrm{v}\left(\frac{1}{42}\right)=7

v=42×7\mathrm{v}=42 \times 7

=294 m/s=294 \mathrm{~m} / \mathrm{s}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Physics
Chapter
Sound Waves
Topic
Vibrations in rod and Air Columns - Organ pipes
A closed organ pipe 150 cm long gives 7 beats per second with an open… | JEE Main 2024 PYQ with Solution · DhiX AI