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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)

Question diagram: A uniform heavy rod of mass 20 kg . Cross sectional area 0 .
Answer:
25.00
Solution:

Tension at a distance x from lower end =mglx

If the elongation in the element is taken as dl then, using Hooke's law,

Y=mgxdxAldldl=maxdxAlY0ldl=0lmgxdxAlYl=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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