Chapter 7: Problem 845
A stress of \(3.18 \times 10^{8} \mathrm{Nm}^{2}\) is applied to steel rod of length \(1 \mathrm{~m}\) along its length. Its young's modulus is \(2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) Then what is the elongation produced in the rod in \(\mathrm{mm}\) ? (A) \(3.18\) (B) \(6.36\) (C) \(5.18\) (D) \(1.59\)
Short Answer
Step by step solution
Write down Hooke's Law formula
Calculate the strain
Solve for strain
Calculate elongation
Convert elongation to millimeters
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hooke's Law
- \(\sigma = E \epsilon\), where \( \sigma\) is stress, \( E\) is Young's modulus, and \( \epsilon \) is strain.
Strain
- \( \epsilon = \frac{\Delta L}{L} \), where \( \Delta L \) is the change in length and \( L \) is the original length.
Stress
- \( \sigma = \frac{F}{A} \), where \( F \) is the force applied and \( A \) is the cross-sectional area.
Elongation
- \( \Delta L = \epsilon \times L \), where \( \Delta L \) is the elongation, \( \epsilon \) is the strain, and \( L \) is the original length.