Question

Young’s modulus of steel is $ 2 \times 10^{11} \, \text{N/m}^2 $ and strain at elastic limit is 0.15. The value of limiting stress will be

• A. \( 1.33 \times 10^{12} \, \text{N/m}^2 \)
• B. \( 1.33 \times 10^{11} \, \text{N/m}^2 \)
• C. \( 3 \times 10^{10} \, \text{N/m}^2 \)
• D. \( 3 \times 10^{11} \, \text{N/m}^2 \)

Hint:

Stress and strain are related through Young's modulus. Limiting stress can be found by multiplying the Young’s modulus by the strain at the elastic limit.

Answer 1

The Correct Option is C

The relationship between Young's modulus ($Y$), stress ($\sigma$), and strain ($\epsilon$) is given by:
\[ Y = \frac{\sigma}{\epsilon} \]  
We need to find the limiting stress ($\sigma$). Rearranging the formula, we get: \[ \sigma = Y \times \epsilon \] Substituting the given values:
\[ \sigma = (2 \times 10^{11} \, \text{N/m}^2) \times (0.15) \] \[ \sigma = 0.30 \times 10^{11} \, \text{N/m}^2 \] \[ \sigma = 3.0 \times 10^{-1} \times 10^{11} \, \text{N/m}^2 \] \[ \sigma = 3.0 \times 10^{(-1 + 11)} \, \text{N/m}^2 \] \[ \sigma = 3.0 \times 10^{10} \, \text{N/m}^2 \] The value of limiting stress is \(3 \times 10^{10} \, \text{N/m}^2\).