As shown in the figure, forces of 10^5 ~N each are applied in...

For same material the ratio of stress to strain is same For first cube \text{ Stress }_1=$\frac{\text{ force }_1}{\text{ area }_1}=\frac{10^5}{\left(0.1^2\right)}$$For second block,$$\begin{aligned}&\text{ stress }_2=\frac{\text{ force }_2}{\text{ area }_2}=\frac{10^5}{\left(0.2^2\right)}\\ &\text{ strain }_2=\frac{\text{ change in length }{ }_2}{\text{ original length }}=\frac{x}{0.2} \end{aligned}$$x is the displacement for second block. For same material,$\frac{\text{ stress }_1}{\text{ strain }_1}=\frac{\text{ stress }_2}{\text{ strain }_2}\text{ or, } \frac{\frac{10.5}{(0.1)^2}}{\frac{0.5 \times 10^{-2}}{0.1}}=\frac{\frac{10^5}{(0.2)^2}}{\frac{x}{0.2}} \text{ Solving we get, }$x=0.25$\mathrm{~cm}$