JEE Main 2020PhysicsMechanical Properties of SolidsMediumNumerical

JEE Main 2020 — Mechanical Properties of Solids Question with Solution

JEE Main 2020 (02 Sep Shift 2)

Question

A wire of density 9×10–3 kg cm–3 is stretched between two clamps 1 m apart. The resulting strain in the wire is  4.9 × 10–4. The lowest frequency of the transverse vibrations in the wire (Young's modulus of wire  Y = 9×1010 Nm–2 ), (to the nearest integer),_______

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Show full solutionCorrect answer: 35
Correct answer
35

Step-by-step explanation

Fundamental frequency in the string

f=12ℓTμ=12ℓTρA=12ℓYΔℓρℓ
f=12ℓYΔℓρℓ
=129×1010×4.9×10-49×10-3×10-6=35

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About this question

This is a previous-year question from JEE Main 2020, covering the Mechanical Properties of Solids 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.