JEE Main 2018PhysicsWaves and SoundMediumMCQ

JEE Main 2018 — Waves and Sound Question with Solution

JEE Main 2018 (08 Apr)

Question

A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7×103 kg m-3 and its Young's modulus is 9.27×1010 Pa. What will be the fundamental frequency of the longitudinal vibrations?

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Show full solutionCorrect option: B
Correct answer
B5 kHz

Step-by-step explanation



L=⋋2⇒⋋=2L=1.2 m

v=yρ=5.86×103 m s-1

∴ Fundamental frequency=v⋋=4.88 kHz

≈5 kHz

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About this question

This is a previous-year question from JEE Main 2018, 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.