Question

A metallic rod breaks when strain produced is 0.2%. The Young's Modulus of the material is 7 x 109 N/m2 . The area of cross section to support a load of 104 N is

• A. 7.1 x 10-4 m2
• B. 7.1 x 10-8 m2
• C. 7.1 x 10-2 m2
• D. 7.1 x 10-6 m2

Answer 1

Answer: (A)

7.1 x 10-4 m2

Answer 2

Young's modulus (E) is defined as the ratio of stress to strain:
E = stress / strain
Rearranging the equation, we get:
stress = E * strain
Given that the strain produced is 0.2% (or 0.002) and the Young's modulus is 7 x 109 N/m2, we can calculate the stress:
stress = (7 x 109 N/m2) * 0.002
= 1.4 x 107 N/m2
The stress is the force per unit area, so we can rearrange the equation to solve for the area:
stress = force / area
area = force / stress
Substituting the given force of 104 N and the calculated stress, we can find the area:
area = (104 N) / (1.4 x 107 N/m2)
= 0.714 x 10-3 m2
= 7.14 x 10-4 m2
Therefore, the area of the cross section that can support a load of 104 N is approximately 7.14 x 10-4 m2. The closest option is (A) 7.1 x 10-4 m2.