As shown in the figure, forces of \(10^5\,N\) each are applied in opposite directions, on the upper and lower faces of a cube of side \(10\,cm\), shifting the upper face parallel to itself by \(0.5\,cm\) . If the side of another cube of the same material is, \(20\,cm\) then under similar conditions as above, the displacement will be......... \(cm\)

For first cube 

\(Stres{s_1} = \frac{{force}}{{force}} = \frac{{{{10}^5}}}{{\left( {{{0.1}^2}} \right)}}\)

\(Strai{n_1} = \frac{{change\,in\,lengt{h_1}}}{{original\,lengt{h_1}}} = \frac{{0.5 \times {{10}^{ - 2}}}}{{0.1}}\)

For second block,

\(Stres{s_2} = \frac{{forc{e_2}}}{{are{a_2}}} = \frac{{{{10}^5}}}{{\left( {{{0.2}^2}} \right)}}\)

\(Strai{n_2} = \frac{{change\,in\,lengt{h_2}}}{{original\,lengt{h_2}}} = \frac{x}{{0.2}}\)

\(x\) is the displacement for second block.

\(For\,same\,material,\frac{{Stress_1}}{{Strai{n_1}}} = \frac{{Stress_2}}{{Strai{n_2}}}\)

\(or,\frac{{\frac{{10.5}}{{{{\left( {0.1} \right)}^2}}}}}{{\frac{{0.5 \times {{10}^{ - 2}}}}{{0.1}}}} = \frac{{\frac{{{{10}^5}}}{{{{\left( {0.2} \right)}^2}}}}}{{\frac{x}{{0.2}}}}\)

\(Solving\,we\,get,\,x = 0.25\,cm\)