Answer:
Total elongation will be 0.012 m
Explanation:
We have given diameter of the cylinder = 2.1 mm
Length of wire [tex]L=3.2\times 10^4mm[/tex]
So radius [tex]r=\frac{d}{2}=\frac{2.1}{2}=1.05mm=1.05\times 10^{-3}m[/tex]
Load F = 280 N
Elastic modulus = 207 Gpa
Area of cross section [tex]A=\pi r^2=3.14\times (1.05\times 10^{-3})^2=3.461\times 10^{-6}m^2[/tex]
We know that elongation in wire is given by [tex]\delta =\frac{FL}{AE}[/tex], here F is load, L is length, A is area and E is elastic modulus
So [tex]\delta =\frac{FL}{AE}=\frac{280\times 32}{3.461\times 10^{-6}\times 207\times 10^9}=0.012m[/tex]
