Question:
A piece of metal having a square cross section of area 400 mm\(^2\) is pulled with 40 kN force, producing only elastic deformation. If the Young's modulus of the metal is \(40 \times 10^9~\text{N/m}^2\), then the strain is:
A piece of metal having a square cross section of area 400 mm\(^2\) is pulled with 40 kN force, producing only elastic deformation. If the Young's modulus of the metal is \(40 \times 10^9~\text{N/m}^2\), then the strain is:
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Strain in elastic materials can be calculated using \(\epsilon = \frac{\sigma}{Y}\), where \(\sigma = \frac{F}{A}\) and Y is Young's modulus.
Updated On: Jun 19, 2026
- \(1 \times 10^{-3}\)
- \(1.5 \times 10^{-3}\)
- \(2.5 \times 10^{-3}\)
- \(4.0 \times 10^{-3}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the problem.
We are given a metal bar with cross-sectional area \(A = 400~\text{mm}^2 = 400 \times 10^{-6}~\text{m}^2\), applied force \(F = 40~\text{kN} = 4 \times 10^4~\text{N}\), and Young's modulus \(Y = 40 \times 10^9~\text{N/m}^2\). We are asked to calculate the strain \(\epsilon\).
Step 2: Recall the relation between stress, strain, and Young's modulus.
\[ Y = \frac{\text{Stress}}{\text{Strain}} \Rightarrow \text{Strain } \epsilon = \frac{\text{Stress}}{Y} \] Stress \(\sigma = \frac{F}{A}\).
Step 3: Calculate stress.
\[ \sigma = \frac{F}{A} = \frac{4 \times 10^4}{400 \times 10^{-6}} = \frac{4 \times 10^4}{4 \times 10^{-4}} = 1 \times 10^8~\text{N/m}^2 \]
Step 4: Calculate strain.
\[ \epsilon = \frac{\sigma}{Y} = \frac{1 \times 10^8}{40 \times 10^9} = 0.0025 = 2.5 \times 10^{-3} \]
Step 5: Conclusion.
The strain in the metal bar is \(2.5 \times 10^{-3}\).
We are given a metal bar with cross-sectional area \(A = 400~\text{mm}^2 = 400 \times 10^{-6}~\text{m}^2\), applied force \(F = 40~\text{kN} = 4 \times 10^4~\text{N}\), and Young's modulus \(Y = 40 \times 10^9~\text{N/m}^2\). We are asked to calculate the strain \(\epsilon\).
Step 2: Recall the relation between stress, strain, and Young's modulus.
\[ Y = \frac{\text{Stress}}{\text{Strain}} \Rightarrow \text{Strain } \epsilon = \frac{\text{Stress}}{Y} \] Stress \(\sigma = \frac{F}{A}\).
Step 3: Calculate stress.
\[ \sigma = \frac{F}{A} = \frac{4 \times 10^4}{400 \times 10^{-6}} = \frac{4 \times 10^4}{4 \times 10^{-4}} = 1 \times 10^8~\text{N/m}^2 \]
Step 4: Calculate strain.
\[ \epsilon = \frac{\sigma}{Y} = \frac{1 \times 10^8}{40 \times 10^9} = 0.0025 = 2.5 \times 10^{-3} \]
Step 5: Conclusion.
The strain in the metal bar is \(2.5 \times 10^{-3}\).
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