
Tension at a distance x from lower end =lmgx
If the elongation in the element is taken as d(Δl) then, using Hooke's law,
Y=Ald(Δl)mgxdx⇒d(Δl)=AlYmaxdx⇒∫0Δld(Δl)=∫0lAlYmgxdx⇒Δl=2AYmgl
⇒Δl=2×0.4×2×101120×10×20
⇒Δl=25×10−9
⇒x=25
A uniform heavy rod of mass 20kg. Cross sectional area 0.4m2 and length 20m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x×10−9m. The value of x is _____ .
(Given. Young's modulus Y=2×1011Nm−2 and g=10ms−2 )
Held on 26 Jul 2022 · Verified 6 Jul 2026.
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