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

The Correct Option is A

Given that the strain produced is 0.2% (or 0.002) and the Young's modulus is \( 7 \times 10^9 \, \text{N/m}^2 \), calculate the area of the cross-section that can support a load of \( 104 \, \text{N} \).

Step 1: Young's Modulus and Stress Relationship

Young's modulus (\( E \)) is defined as the ratio of stress to strain:

\( E = \frac{\text{stress}}{\text{strain}} \)

Rearranging this equation to find stress:

\(\text{stress} = E \times \text{strain} \)

Step 2: Substituting the Given Values

The given strain is \( 0.002 \) (which is 0.2%), and the Young's modulus is \( E = 7 \times 10^9 \, \text{N/m}^2 \).

Substitute these values into the stress equation:

\(\text{stress} = (7 \times 10^9 \, \text{N/m}^2) \times 0.002 = 1.4 \times 10^7 \, \text{N/m}^2 \)

Step 3: Solving for the Area

The stress is the force per unit area. We can rearrange the equation \( \text{stress} = \frac{\text{force}}{\text{area}} \) to solve for the area:

\(\text{area} = \frac{\text{force}}{\text{stress}} \)

Substitute the given force of \( 104 \, \text{N} \) and the calculated stress \( 1.4 \times 10^7 \, \text{N/m}^2 \):

\(\text{area} = \frac{104 \, \text{N}}{1.4 \times 10^7 \, \text{N/m}^2} = 0.714 \times 10^{-3} \, \text{m}^2 \)

So, the area is:

\(\text{area} = 7.14 \times 10^{-4} \, \text{m}^2 \)

Therefore, the area of the cross-section that can support a load of 104 N is approximately: \( 7.14 \times 10^{-4} \, \text{m}^2 \), which corresponds to option (A) 7.1 x 10^-4 m².