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Question: Answered & Verified by Expert
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 )
PhysicsMechanical Properties of SolidsJEE Main
Solution:
2150 Upvotes Verified Answer
The correct answer is: 25

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

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