Answer:
Weight required = 194.51 N
Explanation:
The elongation is given by
[tex]\Delta L=\frac{PL}{AE}[/tex]
Length , L= 1.6 m
Diameter, d = 1.1 mm
Area
[tex]A=\frac{\pi d^2}{4}=\frac{\pi \times (1.1\times 10^{-3})^2}{4}=9.50\times 10^{-7}m^2[/tex]
Change in length, ΔL = 2.8 mm = 0.0028 m
Young's modulus of copper, E = 117 GPa = 117 x 10⁹ Pa
Substituting,
[tex]\Delta L=\frac{PL}{AE}\\\\0.0028=\frac{P\times 1.6}{9.50\times 10^{-7}\times 117\times 10^9}\\\\P=194.51N[/tex]
Weight required = 194.51 N
