Question

A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of the strains developed. Y of steel = 2 × 10¹¹N m⁻². Y of copper = 1.3 × 10 11 N m⁻². 

Answer 1

Given:
Young's modulus of steel = 2 × 10¹¹ N m⁻²
Young's modulus of copper = 1.3 × 10 11 N m⁻²
Both wires are of equal length and equal cross-sectional area. Also, equal tension is applied on them.
 As per the question :

\[L_{\text{ steel}} = L_{\text{Cu}} \]
\[ A_{\text{ steel}} = A_{\text{Cu}} \] 
\[ F_{\text{Cu}} = F_{\text{Steel }} \]

Here: L_steel and L_Cu denote the lengths of steel and copper wires, respectively.
           A_steel and A_Cu denote the cross-sectional areas of steel and copper wires, respectively.
           F_steel and F_Cu  denote the tension of steel and cooper wires, respectively.

\[\frac{\text{Strain of Cu}}{\text{Strain of steel}} = \frac{\frac{∆ L_{\text{Steel}}}{L_{\text{Steel}}}}{\frac{∆ L_{\text{cu}}}{L_{\text{cu}}}} = \frac{F_{\text{Steel}} L_{\text{Steel}} A_{\text{cu}} Y_{\text{cu}}}{A_{\text{Steel}} Y_{\text{Steel}} F_{\text{cu}} L_{\text{cu}}}\]

 

                                 \[ \left( \text{Using } \frac{∆ L}{L} = \frac{F}{AY} \right)\]

\[ \Rightarrow \frac{\text{ Strain of Cu }}{\text{ Strain of steel }} = \frac{Y_{\text{ cu }}}{Y_{\text{ Steel } }} = \frac{1 . 3 \times {10}^{11}}{2 \times {10}^{11}}\]

\[ \Rightarrow \frac{\text{ Strain of Cu} }{ \text{ Strain of steel }} = \frac{13}{20}\]

\[ \Rightarrow \frac{\text{ Strain of steel }}{\text{Strain of Cu }} = \frac{20}{13}\]

Hence, the required ratio is 20 : 13.