Physics - Elasticity Question with Solution | TestHub

PhysicsElasticityMiscellaneousHard2 minPYQ_2022
PhysicsHardnumerical

A uniform heavy rod of mass20 kg. Cross sectional area0.4 m2and length20 mis hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight isx×10-9 m. The value ofxis _____ .
(Given. Young's modulusY=2×1011Nm-2andg=10 m s-2)

Answer:
25.00
Solution:

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

Stream:JEESubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous
⏱ 2mℹ️ Source: PYQ_2022

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