JEE Main 2022PhysicsMechanical Properties of SolidsHardNumerical

JEE Main 2022 — Mechanical Properties of Solids Question with Solution

JEE Main 2022 (26 Jul Shift 2)

Question

A uniform heavy rod of mass 20 kg. Cross sectional area 0.4 m2 and length 20 m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x×10-9 m. The value of x is _____ .
(Given. Young's modulus Y=2×1011Nm-2 and g=10 m s-2 )

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Show full solutionCorrect answer: 25
Correct answer
25

Step-by-step explanation

Tension at a distance x from lower end =mglx

If the elongation in the element is taken as d∆l then, using Hooke's law,

Y=mgxdxAld∆l⇒d∆l=maxdxAlY⇒∫0∆ld∆l=∫0lmgxdxAlY⇒∆l=mgl2AY

⇒Δl=20×10×202×0.4×2×1011

⇒Δl=25×10-9

⇒x=25

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About this question

This is a previous-year question from JEE Main 2022, 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.