An aluminium rod with Young's modulus Y=7.0×1010 N/m2 undergoes elastic strain of 0.04%. The energy per unit volume stored in the rod in SI unit is:
- 2800
- 5600
- 8400
- 11200
The Correct Option is B
Solution and Explanation
Step 1: Recall the Energy Density Formula
The energy per unit volume \( u \) is given by: \[ u = \frac{1}{2} \cdot \text{stress} \cdot \text{strain} \] We also know: \[ \text{stress} = Y \cdot \text{strain} \]
Step 2: Substitute the Values
Given: \[ Y = 7.0 \times 10^{10} \, \text{N/m}^2, \quad \text{strain} = \frac{0.04}{100} = 0.0004 \] Substitute these values into the formula: \[ u = \frac{1}{2} \cdot Y \cdot (\text{strain})^2 \] \[ u = \frac{1}{2} \cdot 7.0 \times 10^{10} \cdot (0.0004)^2 \] \[ u = 5600 \, \text{J/m}^3 \]
Final Answer
The energy per unit volume is: \[ u = 5600 \, \text{J/m}^3 \]
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