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

(1.79:1) Given:
Length of the steel wire,$L_{1} = 4.7\; m$
Area of cross-section of the steel wire,$A_{1} = 3.0 \times 10^{-5}\; m^{2}$
Length of the copper wire,$L_{2} = 3.5\; m$
Area of cross-section of the copper wire,$A_{2} = 4.0 \times 10^{-5}\; m^{2}$
Change in length for both wires is the same,$\Delta L_{1} = \Delta L_{2} = \Delta L$
Force applied in both cases is the same,$F_{1} = F_{2} = F$
Young's modulus $Y$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta L}$.
For steel wire:
$Y_{1} = \frac{F \cdot L_{1}}{A_{1} \cdot \Delta L} = \frac{F \cdot 4.7}{3.0 \times 10^{-5} \cdot \Delta L} \quad \dots(i)$
For copper wire:
$Y_{2} = \frac{F \cdot L_{2}}{A_{2} \cdot \Delta L} = \frac{F \cdot 3.5}{4.0 \times 10^{-5} \cdot \Delta L} \quad \dots(ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{Y_{1}}{Y_{2}} = \frac{F \cdot 4.7}{3.0 \times 10^{-5} \cdot \Delta L} \times \frac{4.0 \times 10^{-5} \cdot \Delta L}{F \cdot 3.5}$
$\frac{Y_{1}}{Y_{2}} = \frac{4.7 \times 4.0}{3.0 \times 3.5} = \frac{18.8}{10.5} \approx 1.79$
The ratio of the Young's modulus of steel to that of copper is $1.79:1$.