Answer:
Let \[{{r}_{1}}\] and \[{{r}_{2}}\] be the radii of the two wires. (i) \[Stress=\frac{F}{A}=\frac{F}{\pi {{r}^{2}}}.\]For same load \[F,\frac{{{(stress)}_{1}}}{{{(stress)}_{2}}}=\frac{r_{2}^{2}}{r_{1}^{2}}\] (ii) Strain, \[\frac{\Delta l}{l}=\frac{F}{A\Upsilon }=\frac{F}{\pi {{r}^{2}}\Upsilon }\] For the two wires F and \[\Upsilon \] are same, so \[\frac{{{(strain)}_{1}}}{{{(strain)}_{2}}}=\frac{r_{2}^{2}}{r_{1}^{2}}\] (iii) Extension, \[\Delta l=\frac{F}{A}.\frac{l}{\Upsilon }=\frac{F}{\pi {{r}^{2}}}.\frac{l}{\Upsilon }\] For the two wires F, L and \[\Upsilon \] are same, so\[\frac{{{(\Delta l)}_{1}}}{{{(\Delta l)}_{2}}}=\frac{r_{2}^{2}}{r_{1}^{2}}\] Hence stress, strain and extension are all different for the two wires.
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