Question

A steel wire of length 4.7 m and cross-sectional area 3.0 × 10^-5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10^–5 m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Answer 1

Given

• Steel wire: length \( L_s = 4.7 \,\text{m} \), area \( A_s = 3.0 \times 10^{-5} \,\text{m}^2 \)
• Copper wire: length \( L_c = 3.5 \,\text{m} \), area \( A_c = 4.0 \times 10^{-5} \,\text{m}^2 \)
• Same load \( F \) is applied on both.
• Extension is the same: \( \Delta L_s = \Delta L_c \).

Formula

Young’s modulus: \[ Y = \frac{FL}{A \,\Delta L} \]

Use Equal Extensions

For steel: \[ Y_s = \frac{F L_s}{A_s \,\Delta L} \] For copper: \[ Y_c = \frac{F L_c}{A_c \,\Delta L} \] Ratio: \[ \frac{Y_s}{Y_c} = \frac{\dfrac{F L_s}{A_s \,\Delta L}}{\dfrac{F L_c}{A_c \,\Delta L}} = \frac{L_s}{A_s} \cdot \frac{A_c}{L_c} = \frac{L_s A_c}{L_c A_s} \]

Substitute Values

\[ \frac{Y_s}{Y_c} = \frac{4.7 \times 4.0 \times 10^{-5}}{3.5 \times 3.0 \times 10^{-5}} = \frac{4.7 \times 4.0}{3.5 \times 3.0} = \frac{18.8}{10.5} \approx 1.79 \]

Final Answer

\[ \boxed{\dfrac{Y_{\text{steel}}}{Y_{\text{copper}}} \approx 1.8} \]