The modulus of elasticity is 44.4GPa
Explanation:
Given
original length L1=10cm
New length L2=10.036 cm
Force F=16000N
[tex]dimensions\ of\ the\ bar\ is\ 1 cm\times1cm\\Area=1cm\times1cm=1cm^2=0.0001m^2[/tex]
Stress of the bar σ=Force/Area
[tex]=16000/0.0001=16\times 10^7N/m^2[/tex]
Change in length ΔL=L2-L1=10.036-10=0.036 cm
Strain is obtained by dividing the change in length by original length
Strain ε=ΔL/L1
=0.036/10=0.0036
Modulus of elasticity=stress/strain
=σ/ε
[tex]=16\times10^7/0.0036\\=4.44\times10^\ 10[/tex]Pa
[tex]=44.4GPa[/tex]
