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Step-by-Step Solution
Step 1: Identify the known parameters
• Length of steel rail, $L = 5\,\text{m}$ (though not directly needed in the final formula for force).
• Cross-sectional area, $A = 40\,\text{cm}^2 = 40 \times 10^{-4}\,\text{m}^2$.
• Temperature rise, $\Delta \theta = 10\,^\circ \text{C}$.
• Coefficient of linear expansion of steel, $\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}$.
• Young’s modulus of steel, $Y = 2 \times 10^{11}\,\text{N}\,\text{m}^{-2}$.
Step 2: Write the formula for thermal stress-induced force
When expansion is completely prevented, the thermal stress developed is related to the strain, which in turn comes from the thermal expansion. The change in length $\Delta L$ would have been:
$\displaystyle \Delta L = \alpha L \,\Delta \theta$
But since it is not allowed to expand, this change in length is forcibly restricted, creating stress in the rail. Using the relation for Young’s modulus,
$\displaystyle Y = \frac{F/A}{\Delta L/L} \quad \Rightarrow \quad F = Y\,A \,\frac{\Delta L}{L}\,. $
Substituting $\Delta L / L = \alpha \,\Delta \theta,$ we get
$\displaystyle F = Y\,A\,\alpha\,\Delta \theta\,.$
Step 3: Substitute the values
$\displaystyle F \;=\; \bigl(2 \times 10^{11}\bigr)\,\bigl(40 \times 10^{-4}\bigr)\,\bigl(1.2\times 10^{-5}\bigr)\,\bigl(10\bigr)\,. $
Step 4: Simplify the expression
First compute the product in steps:
$40 \times 10^{-4} = 4.0 \times 10^{-3}$
$1.2 \times 10^{-5} \times 10 = 1.2 \times 10^{-4}$
Now multiply everything:
$\displaystyle F = \bigl(2 \times 10^{11}\bigr)\,\bigl(4.0 \times 10^{-3}\bigr)\,\bigl(1.2 \times 10^{-4}\bigr)\,. $
Combine terms carefully:
$2 \times 4.0 \times 1.2 = 9.6,$ and powers of 10 combine to $10^{11} \times 10^{-3} \times 10^{-4} = 10^{11-3-4} = 10^4.$
So,
$\displaystyle F = 9.6 \times 10^4 \,\text{N} \;\approx\; 1 \times 10^5\,\text{N}.$
Step 5: State the final answer
The force developed in the rail is approximately
$\displaystyle 1 \times 10^5\,\text{N}.$
