NEET (UG) 2024 — Question 45

When the bar is heated but not allowed to expand, the thermal strain is constrained, and compressive stress develops. Thermal strain = coefficient of linear expansion × temperature change: $\alpha \Delta T = (10^{-5} \, \text{°C}^{-1})(100 \, \text{°C}) = 10^{-3}$. Since expansion is prevented, this is the longitudinal strain in the bar. Stress developed, $\sigma = Y \times \text{strain} = (0.5 \times 10^{11} \, \text{N m}^{-2}) \times 10^{-3} = 0.5 \times 10^{8} \, \text{N m}^{-2}$. Compressive force, $F = \sigma \times A = (0.5 \times 10^{8} \, \text{N m}^{-2}) \times (10^{-3} \, \text{m}^{2}) = 0.5 \times 10^{5} \, \text{N} = 5 \times 10^{4} \, \text{N}$. Express in terms of $\times 10^{3}$: $5 \times 10^{4} \, \text{N} = 50 \times 10^{3} \, \text{N}$. Thus, the compressive force is $50 \times 10^{3} \, \text{N}$, which corresponds to option B.