question_answer 35)

                  The Young's modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?                

Answer:

                  \[{{\text{Y}}_{\text{e}}}\text{=}\,\frac{\text{(strees)}}{\text{longitudinal}\,\,\text{strain}}\]                 \[\therefore \] \[\,\frac{{{Y}_{s}}}{{{Y}_{r}}}=\,\frac{{{(strees)}_{s}}}{{{(strees)}_{r}}}.\] Since, \[{{Y}_{s}}>\,{{Y}_{r}}\]                 \[\therefore \] \[{{(\text{strees})}_{s}}\,>\,{{(\text{strees})}_{r}}\]