Question

A cylindrical, 0.500-m rod has a diameter of 0.02 m. The rod is stretched to a length of 0.501 m by a force of 3000 N. What is the Young's modulus of the material?

Answer 1

Y = 4.775 x 10⁹ Pa = 4.775 GPa

Step-by-step explanation:

First, we calculate the stress on the rod:

`stress = (Force)/(Area) = (3000\ N)/(\pi r^2) \\\\stress = (3000\ N)/(\pi (0.01\ m)^2)\\\\stress = 9.55\ x\ 10^6\ Pa = 9.55 MPa\\`

Now, we calculate the strain:

`strain = (Change\ in Length)/(Original\ Length)\\\\strain = (0.501\ m - 0.5\ m)/(0.5\ m)\\\\strain = 0.002\\`

Now, we will calculate the Young's Modulus (Y):

`Y = (stress)/(strain)\\\\Y = (9.55\ x\ 10^6\ Pa)/(0.002) \\`

Y = 4.775 x 10⁹ Pa = 4.775 GPa