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A solid cylinder ical steel coloumn is 4m long nd 9.0cm in diameter.wat will be its decrease its length wen carryin a lod of 80000 kg?Y=1.9*10^11pa.



We first find
  cross sectional area of the coloumn =πr² =(0.045m)²=6.36*10⁻³m²
then  from Y=(F/A)(ΔL/L)we have
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Length of cylinder (L) = 4m
diameter = 9 cm = 0.09 m
radius = 0.09/2 = 0.045 m
cross sectional area(A) = πr² = π(0.045)² = 0.0063585 m²
weight (F)= 80000 kgf = 80000×9.8 = 7.84×10⁵ N
Young's modulus(Y) = 1.9×10¹¹ Pa
Let change in length = ΔL

According to hooke's law,

Y= \frac{stress}{strain} = \frac{F/A}{ \Delta L/L } = \frac{F.L}{A. \Delta L }

 \Delta L= \frac{F.L}{A.Y}

\Delta L =  \frac{7.84*10^{5}*4}{0.0063585*1.9*10^{11}}

\Delta L=0.002596m=2.596mm

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