JEE Main 2024PhysicsWaves and SoundMediumNumerical

JEE Main 2024Waves and Sound Question with Solution

JEE Main 2024 (30 Jan Shift 1)

Question

In a closed organ pipe, the frequency of fundamental note is 30 Hz. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz. If the organ pipe has a cross-sectional area of 2 cm2, the amount of water poured in the organ tube is ________g. (Take speed of sound in air is 330 m s-1 )

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Show full solutionCorrect answer: 400
Correct answer
400

Step-by-step explanation

Fundamental frequency for closed organ pipe in the first part:

V4l1=30l1=114 m

Similarly, for second part:

V4l2=110l2=34 m

Therefore, change in length Δl=2 m

Change in volume =l=400 cm3

M=400 g;ρ=1 g cm-3

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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.