Question

A steel wire of length $4.7\text{m}$ and cross-sectional area $3.0\times {10}^{-5}{\text{m}}^2$ stretches by the same amount as a copper wire of length $3.5\text{m}$ and cross-sectional area $4.0\times {10}^{-5}{\text{m}}^2$ under a given load. What is the ratio of the Young's modulus of steel to that of copper?
[Answer upto two decimal places]

Answer 1

For steel wire : $L_S=4.7\text{m},A_S=3.0\times {10}^{-5}{\text{m}}^2$
For copper wire: $L_C=3.5\text{m},A_C=4.0\times {10}^{-5}{\text{m}}^2$
Applied force $F$ and extension $\left(\Delta L\right)$ are same for both the wires.
Young's modulus of steel can be written as,
$Y_S=\frac{F{L}_S}{A_S\Delta L}=\frac{F\times 4.7}{3.0\times {10}^{-5}\times \Delta L}...\left(i\right)$
Similarly for copper,
$Y_C=\frac{F{L}_C}{A_C\Delta L}=\frac{F\times 3.5}{4.0\times {10}^{-5}\times \Delta L}...\left(ii\right)$
From Eq.$\left(i\right)\&\left(ii\right)$,
$\frac{Y_S}{Y_C}=\frac{F\times 4.7}{3.0\times {10}^{-5}\times \Delta L}\times \frac{4\times {10}^{-5}\times \Delta L}{F\times 3.5}$
$\Rightarrow \frac{Y_S}{Y_C}=\frac{4.7\times 4}{3\times 3.5}=1.79$