Answer:
The rope is stretched by 0.3373 cm
Explanation:
As we know
Youngs Modulus ([tex]5.0 * 10^9[/tex] N/m2) is Stress divided by strain
Stress = force/Area = (65*9.8)/(pi * (7.3*10^-3)^2)
Starin = change in length/original length of the rope = l/12 meters
Substituting the given values we get
[tex]5.0 * 10^9[/tex] N/m2 = {([tex]\frac{\frac{24*9.8}{\pi *(7.3*10^{-3})^2} }{\frac{\delta L}{L} }[/tex])
[tex]5.0 * 10^9[/tex] N/m2 = {([tex]\frac{\frac{24*9.8}{\pi *(7.3*10^{-3})^2} }{\frac{\delta L}{12} }[/tex])
[tex]\frac{\delta L}{12} = \frac{\frac{65*9.8}{pi * (7.3*10^-3)^2} }{5.0 * 10^9}[/tex]
The rope is stretched by 0.3373 cm
