Answers

  • Brainly User
2014-10-31T22:22:00+05:30
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
              ΔL=FL/AY=8.00*10⁴*9.81N*4.0m/6.36*13⁻³m²*1.9*10¹¹Pa=2.6*10⁻³=2.6mm
0
2014-11-01T12:13:13+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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

0