JEE Main 2024PhysicsWaves and SoundHardNumerical

JEE Main 2024 — Waves and Sound Question with Solution

JEE Main 2024 (27 Jan Shift 2)

Question

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-1.

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Show full solutionCorrect answer: 294
Correct answer
294

Step-by-step explanation

The formula to calculate the fundamental frequency in a closed pipe is given by

fc=v4l1   ...1

And, that in an open pipe is given by

fo=v2l2   ...2

Given that

fc-fo=7   ...3

From equations (1), (2) and (3), it follows that

v4×150-v2×350=7⇒v600 cm-v700 cm=7⇒v6m-v7m=7⇒v142=7⇒v=42×7=294 m s-1

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About this question

This is a previous-year question from JEE Main 2024, covering the Waves and Sound chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.