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Step-by-Step Solution
Step 1: Identify the Relevant Physical Quantities
• Area of cross-section of the track, A = 0.01 \text{ m}^2
• Temperature change, \Delta T = 10^\circ \text{C}
• Coefficient of linear expansion, \alpha = 10^{-5} \text{ per }^\circ\text{C}
• Young's modulus of the material, Y = 10^{11} \text{ N/m}^2
Step 2: Write the Formula for Thermal Stress
When the track is not allowed to expand freely, it develops stress (denoted by \sigma ). The relation between thermal stress and expansion is
\sigma = Y \, \alpha \, \Delta T.
Step 3: Express the Strain Energy Density
Strain energy per unit volume, u , stored in the material due to stress is given by
u = \frac{1}{2} \cdot \frac{\sigma^2}{Y}.
Substitute \sigma = Y \,\alpha \,\Delta T into this expression:
u = \frac{1}{2}\,\frac{\left(Y \,\alpha \,\Delta T\right)^2}{Y}
= \frac{1}{2} \,Y \,\alpha^2 \,\Delta T^2.
Step 4: Find the Energy Stored per Unit Length
Energy stored per unit length of the track is the strain energy density multiplied by the cross-sectional area A :
\text{Energy per unit length}
= A \times u
= A \times \left(\frac{1}{2} Y\,\alpha^2\,\Delta T^2\right).
Step 5: Substitute the Numerical Values
Plug in the given values:
A = 0.01 \text{ m}^2 = 10^{-2}\,\text{m}^2, \quad
Y = 10^{11}\,\text{N/m}^2, \quad
\alpha^2 = (10^{-5})^2 = 10^{-10},\quad
\Delta T^2 = (10)^2 = 100.
Thus,
\text{Energy per unit length}
= \frac{1}{2} \times 10^{-2} \times 10^{11} \times 10^{-10} \times 100.
Step 6: Simplify the Expression
Combine the powers of 10 carefully:
10^{-2} \times 10^{11} \times 10^{-10} = 10^{-2} \times 10^{1} = 10^{-1} = 0.1,
so the product before multiplying by \frac{1}{2} and 100 is
0.1 \times 100 = 10.
Finally, multiply by \frac{1}{2} :
\text{Energy per unit length} = \frac{1}{2} \times 10 = 5 \text{ J/m}.
Final Answer
The energy stored per meter in the track is 5 J/m.
