A steel rod bas a radius of 20 mm and a length of 2.0 m. A force of 62.8 kN stretches it along its length. Young's modulus of steel is 2.0 × 10^{11} N/m^2 . The longitudinal strain produced in the wire is _______ × 10^{-5}
Correct Answer: 25
Approach Solution - 1
Approach Solution -2
Young's Modulus and Strain Problem
Step 1: Young's Modulus and Strain
Young's modulus (\( Y \)) is defined as the ratio of stress to strain:
\( Y = \frac{\text{stress}}{\text{strain}} \)
Stress is defined as force (\( F \)) per unit area (\( A \)), and strain is the change in length (\( \Delta L \)) divided by the original length (\( L \)).
Step 2: Calculate Strain
We are given \( F = 62.8 \text{ kN} = 62.8 \times 10^3 \text{ N} \), \( r = 20 \text{ mm} = 20 \times 10^{-3} \text{ m} \), \( L = 2.0 \text{ m} \), and \( Y = 2.0 \times 10^{11} \text{ N/m}^2 \). The cross-sectional area of the rod is
\( A = \pi r^2 = \pi (20 \times 10^{-3} \text{ m})^2 = 400\pi \times 10^{-6} \text{ m}^2 \)
Strain is given by:
\( \text{strain} = \frac{\text{stress}}{Y} = \frac{F/A}{Y} = \frac{F}{AY} \)
\( \text{strain} = \frac{62.8 \times 10^3 \text{ N}}{(400\pi \times 10^{-6} \text{ m}^2)(2.0 \times 10^{11} \text{ N/m}^2)} = \frac{62.8 \times 10^3}{800\pi \times 10^5} = \frac{62.8}{800 \times 3.14} \times 10^{-2} \)
\( \text{strain} \approx \frac{62.8}{2512} \times 10^{-2} \approx 0.025 \times 10^{-2} = 25 \times 10^{-5} \)
Conclusion:
The longitudinal strain produced in the wire is \( \mathbf{25 \times 10^{-5}} \).
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