A rod is 2.40 m long and has a diameter of 2.50 mm. A force of 2000 N is applied to the end to stretch the rod by 1.00 mm. What is the tensile modulus for this rod?
7.50 × 1011 N/m2
8.70 × 1011 N/m2
9.78 × 1011 N/m2
10.2 × 1011 N/m2
10.8 × 1011 N/m2
To find the tensile modulus, we can use Hooke's Law, which states that the strain (change in length divided by the original length) is proportional to the stress (force divided by the cross-sectional area).
First, let's calculate the cross-sectional area of the rod. The diameter is given as 2.50 mm, so we can find the radius by dividing it by 2: 2.50 mm / 2 = 1.25 mm = 0.00125 m.
Now, we calculate the area by using the formula for the area of a circle: A = π * r^2. Plugging in the value we found for the radius, we get: A = π * (0.00125 m)^2.
Next, let's calculate the strain. The change in length is given as 1.00 mm, and the original length of the rod is 2.40 m. So, the strain is: (1.00 mm / 2.40 m) = 0.00041667.
Using Hooke's Law, we can now find the tensile modulus. The tensile modulus (also called Young's modulus) is given by the formula: stress / strain.
The stress can be found by dividing the force applied (2000 N) by the cross-sectional area (π * (0.00125 m)^2).
Finally, we divide this stress by the strain to get the tensile modulus: tensile modulus = stress / strain.
Calculating this, we get:
tensile modulus = (2000 N) / (π * (0.00125 m)^2) / (0.00041667).
Simplifying this expression, we find that the tensile modulus is approximately 7.50 × 10^11 N/m^2.
Therefore, the correct option is:
7.50 × 10^11 N/m^2.