Let's break down the problem: **Given Data:** - Length of rod, \( L = 2 \) m \( = 2000 \) mm - Gap at end \( = 0.25 \) mm - Temperature rise, \( \Delta T = 100^\circ \)C - Coefficient of thermal expansion, \( \alpha = 18 \times 10^{-6} \) per \( ^\circ \)C - Modulus of elasticity, \( E = 1 \times 10^5 \) N/mm² **Step 1: Calculate thermal expansion if there were no restraint.** \[ \Delta L_{thermal} = \alpha \cdot L \cdot \Delta T \] \[ \Delta L_{thermal} = 18 \times 10^{-6} \times 2000 \times 100 \] \[ \Delta L_{thermal} = 18 \times 10^{-6} \times 200,000 \] \[ \Delta L_{thermal} = 3.6 \text{ mm} \] **Step 2: The rod can expand only by the gap (0.25 mm), the rest is restrained and creates compressive stress.** \[ \text{Extra expansion resisted by rod} = \Delta L_{thermal} - \text{gap} \] \[ = 3.6 \text{ mm} - 0.25 \text{ mm} = 3.35 \text{ mm} \] **Step 3: Calculate the compressive strain and stress.** \[ \text{Strain} (\epsilon) = \frac{\text{Extra expansion}}{L} = \frac{3.35}{2000} \] \[ \epsilon = 0.001675 \] \[ \text{Stress} (\sigma) = E \cdot \epsilon = 1 \times 10^5 \times 0.001675 \] \[ \sigma = 167.5 \text{ N/mm}^2 \] **Final Answer:** \[ \boxed{167.5~\text{N/mm}^2} \] So, the correct answer is **167.5 N/mm²** (Option 4 in your list).