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JEE Mains · Physics · STD 11- 8. mechanical properties of solids
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......... \(cm\)
- A \(1.00\)
- B \(0.25\)
- C \(0.37\)
- D \(0.75\)
Answer & Solution
Correct Answer
(B) \(0.25\)
Step-by-step Solution
Detailed explanation
For same material teh ratio of stress to strain is same For first cube \(Stres{s_1} = \frac{{force}}{{force}} = \frac{{{{10}^5}}}{{\left( {{{0.1}^2}} \right)}}\) \(Strai{n_1} = \frac{{change\,in\,lengt{h_1}}}{{original\,lengt{h_1}}} = \frac{{0.5 \times {{10}^{ - 2}}}}{{0.1}}\)…
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