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.

Answer 2

We are given the following information: 

• Maximum strain before breaking, \( \mathbf{\epsilon_{max} = 0.2\%} \)
• Young's Modulus of the material, \( \mathbf{Y = 7 \times 10^9 \text{ N/m}^2} \)
• Load (Force) to be supported, \( \mathbf{F = 10^4 \text{ N}} \)

We need to find the minimum area of cross-section \( A \) required to support this load without the rod breaking.

First, convert the strain percentage to a dimensionless value: \[ \epsilon_{max} = 0.2\% = \frac{0.2}{100} = \mathbf{0.002} \]

The rod breaks when the strain reaches \( \epsilon_{max} \). This corresponds to the ultimate tensile stress (\( \sigma_{max} \)) the material can withstand. Young's Modulus relates stress and strain: \[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\sigma}{\epsilon} \] So, the maximum stress the material can handle is: \[ \sigma_{max} = Y \times \epsilon_{max} \] \[ \sigma_{max} = (7 \times 10^9 \text{ N/m}^2) \times 0.002 \] \[ \sigma_{max} = (7 \times 10^9) \times (2 \times 10^{-3}) \text{ N/m}^2 \] \[ \sigma_{max} = 14 \times 10^{6} \text{ N/m}^2 \]

Stress is also defined as the force applied per unit area: \[ \sigma = \frac{F}{A} \] To support the load \( F \) without breaking, the stress produced in the rod must not exceed the maximum stress \( \sigma_{max} \). Therefore, the minimum required area \( A \) is given by: \[ \frac{F}{A} = \sigma_{max} \] \[ A = \frac{F}{\sigma_{max}} \]

Substitute the values for \( F \) and \( \sigma_{max} \): \[ A = \frac{10^4 \text{ N}}{14 \times 10^6 \text{ N/m}^2} \] \[ A = \frac{1}{14} \times 10^{4-6} \text{ m}^2 \] \[ A = \frac{1}{14} \times 10^{-2} \text{ m}^2 \] Calculating the value of \( \frac{1}{14} \): \[ \frac{1}{14} \approx 0.07143 \] So, \[ A \approx 0.07143 \times 10^{-2} \text{ m}^2 \] Converting to standard scientific notation: \[ \mathbf{A \approx 7.143 \times 10^{-4} \text{ m}^2} \]

Comparing this result with the given options:

• 7.1 x 10^-4 m²
• 7.1 x 10^-8 m²
• 7.1 x 10^-2 m²
• 7.1 x 10^-6 m²

The calculated area matches the first option.