A steel rod of length $1 \mathrm{~m}$ and cross sectional area $10^{-4} \mathrm{~m}^{2}$ is heated from $0^{\circ} \mathrm{C}$ to $200^{\circ} \mathrm{C}$ without being allowed to extend or bend. The compressive tension produced in the rod is ___________ $\times 10^{4} \mathrm{~N}$. (Given Young's modulus of steel $=2 \times 10^{11} \mathrm{Nm}^{-2}$, coefficient of linear expansion $=10^{-5} \mathrm{~K}^{-1}$ )

<p>The change in length of the rod when it is heated is given by the equation:</p> <p>$\Delta L = L_0 \cdot \alpha \cdot \Delta T$</p> <p>where</p> <ul> <li>$\Delta L$ is the change in length,</li> <li>$L_0$ is the original length,</li> <li>$\alpha$ is the coefficient of linear expansion, and</li> <li>$\Delta T$ is the change in temperature.</li> </ul> <p>Substituting the given values:</p> <p>$\Delta L = 1 \, \text{m} \cdot 10^{-5} \, \text{K}^{-1} \cdot 200 \, \text{K} = 0.002 \, \text{m}$</p> <p>The rod is not allowed to extend or bend, so a stress is created in the rod. This stress can be calculated using Young&#39;s modulus (Y), which is the ratio of the stress (force per unit area, F/A) to the strain (change in length per unit length, $\Delta L / L_0$):</p> <p>$Y = \frac{F/A}{\Delta L / L_0}$</p> <p>Rearranging for F gives:</p> <p>$F = Y \cdot A \cdot \frac{\Delta L}{L_0}$</p> <p>Substituting the given values:</p> <p>$F = 2 \times 10^{11} \, \text{N/m}^2 \cdot 10^{-4} \, \text{m}^2 \cdot \frac{0.002 \, \text{m}}{1 \, \text{m}} = 4 \times 10^{4} \, \text{N}$</p> <p>So the compressive tension produced in the rod is $4 \times 10^{4} \, \text{N}$.</p>