Answer
(a) $ \mathrm{Strain_C} = 2.06 \times 10^{-4} $
$ \mathrm{Strain_S} = 1.13 \times 10^{-4} $
(b) $ \Delta l_{\mathrm{C}} = 154 \, \mathrm{µm} $
$ \Delta l_{\mathrm{S}} = 84.9 \, \mathrm{µm} $
Work Step by Step
$\mathrm{C}$ for copper, $\mathrm{S}$ for steel
(a) $$ \mathrm{Strain_C} = \frac{\mathrm{Stress}}{Y_\mathrm{C}} = \frac{F_\perp}{AY_\mathrm{C}} = $$
$$ \frac{4000 \, \mathrm{N}}{\pi(0.0075 \, \mathrm{m})^2(11 \times 10^{10} \, \mathrm{Pa})} = 2.0577... \times 10^{-4} = 2.06 \times 10^{-4} $$
$$ \mathrm{Strain_S} = \frac{\mathrm{Stress}}{Y_\mathrm{S}} = \frac{F_\perp}{AY_\mathrm{S}} = $$
$$ \frac{4000 \, \mathrm{N}}{\pi(0.0075 \, \mathrm{m})^2(20 \times 10^{10} \, \mathrm{Pa})} = 1.1317... \times 10^{-4} = 1.13 \times 10^{-4} $$
(b) $$ \Delta l_{\mathrm{C}} = l_0 \times \mathrm{Strain_C} = (0.750 \, \mathrm{m})(2.058 \times 10^{-4}) = 1.54 \times 10^{-4} \, \mathrm{m} = 154 \, \mathrm{µm} $$
$$ \Delta l_{\mathrm{S}} = l_0 \times \mathrm{Strain_S} = (0.750 \, \mathrm{m})(1.132 \times 10^{-4}) = 8.49 \times 10^{-5} \, \mathrm{m} = 84.9 \, \mathrm{µm} $$
