To solve this problem, we need to find the compressive force developed in the metallic bar when it is heated from $0^\circ$C to $100^\circ$C$$ without allowing it to expand.$\\$Given:$\\$• Young's modulus $(Y) = 0.5 \times 10^{11}$ N/m$^2 \\$• Coefficient of linear thermal expansion $(\alpha) = 10^{-5}$ °C$^{-1} \\$• Initial length $(L) = 1$ m$\\$• Area of cross-section $(A) = 10^{-3}$ m$^2 \\$• Temperature change $(\Delta T) = 100^\circ$C$\\$When the bar is heated, it tends to expand by a length $\Delta L$ given by:$\\ \Delta L = \alpha L \Delta T \\$Substitute the known values:$\\ \Delta L = 10^{-5} \times 1 \times 100 = 10^{-3}$ m$\\$Since the bar is not allowed to expand, a compressive force $(F)$ is developed to prevent this expansion.$\\$The strain $(\epsilon)$ in the bar due to this force is given by:$\\ \epsilon = \frac{\Delta L}{L} = \frac{10^{-3}}{1} = 10^{-3} \\$Using Hooke's Law, the stress $(\sigma)$ developed in the bar is:$\\ \sigma = Y \epsilon \\$Substitute the known values:$\\ \sigma = 0.5 \times 10^{11} \times 10^{-3} = 0.5 \times 10^8$ N/m$^2 \\$The compressive force $(F)$ is given by:$\\ F = \sigma \times A \\$Substitute the known values:$\\ F = 0.5 \times 10^8 \times 10^{-3} = 0.5 \times 10^5$ N$\\$Therefore, the compressive force developed in the bar is $50 \times 10^3$ N.$\\$This corresponds to Option 4.