# 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.

2
by surajiop

Log in to add a comment

by surajiop

Log in to add a comment

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

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,

⇒

⇒

⇒

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,

⇒

⇒

⇒