Answer:
[tex]3978873.58\ \text{Pa}[/tex]
[tex]0.00002122\ \text{m}[/tex]
Explanation:
F = Force = 5000 N
r = Radius of circular cross section = 2 cm
l = Length of disk = 0.8 cm
A = Area = [tex]\pi r^2[/tex]
Y = Young's modulus = [tex]1.5\times 10^9\ \text{N/m}^2[/tex]
Pressure is given by
[tex]P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{5000}{\pi (2\times 10^{-2})^2}\\\Rightarrow P=3978873.58\ \text{Pa}[/tex]
The pressure on the cross section is [tex]3978873.58\ \text{Pa}[/tex]
The change in length of the cross section is given by
[tex]\Delta L=\dfrac{PL}{Y}\\\Rightarrow \Delta L=\dfrac{3978873.58\times 0.8\times 10^{-2}}{1.5\times 10^9}\\\Rightarrow \Delta L=0.00002122\ \text{m}[/tex]
The deformation produced is [tex]0.00002122\ \text{m}[/tex]
