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......... $cm$
JEE MAIN 2018, Diffcult
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For same material teh ratio of stress to strain is same
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.
$For\,same\,material,\frac{{Stress_1}}{{Strai{n_1}}} = \frac{{Stress_2}}{{Strai{n_2}}}$
$or,\frac{{\frac{{10.5}}{{{{\left( {0.1} \right)}^2}}}}}{{\frac{{0.5 \times {{10}^{ - 2}}}}{{0.1}}}} = \frac{{\frac{{{{10}^5}}}{{{{\left( {0.2} \right)}^2}}}}}{{\frac{x}{{0.2}}}}$
$Solving\,we\,get,\,x = 0.25\,cm$
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