Answer:
The stress is calculated as [tex]1.003\times 10^{9}\ Pa[/tex]
Solution:
As per the question:
Length of the wire, l = 75.2 cm = 0.752 m
Diameter of the circular cross-section, d = 0.560 mm = [tex]0.560\times 10^{- 3}\ m[/tex]
Mass of the weight attached, m = 25.2 kg
Elongation in the wire, [tex]\Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m[/tex]
Now,
The stress in the wire is given by:
[tex]Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}[/tex] (1)
Now,
Force is due to the weight of the attached weight:
F = mg = [tex]25.2\times 9.8 = 246.96\ N[/tex]
Cross sectional Area, A = [tex]\pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}[/tex]
Using these values in eqn (1):
[tex]\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa[/tex]
