A copper wire of length 1.0 m and a steel wire of length 0.5 m having equal cross-sectional areas are joined end to end. The composite wire is stretched  by a certain load which stretches the copper wire by 1 mm. If the Young's moduli of copper and steel are 1.0 x 10¹¹ Nm^-2 and 2.0 x 10¹¹ Nm^-2  , the total extension (in mm) of the composite wire is

When a wire of length L, cross-sectional area A and Young's modulus Y is stretched with a force F, the extension I in the wire is given by

                $l=\frac{FL}{AY}$

Since F and A are the same for the two wires, we have

$For copper wire l_c=\frac{F L_c}{A{Y}_c}$

$$\begin{gathered} For steel wire, l_s=\frac{F L_s}{A{Y}_s} \\ l_s= l_c\frac{L_s}{L_c}.\frac{Y_c}{Y_s} \\ =1 mm\times \left(\frac{0.5m}{1.0m}\right)\times \left(\frac{1.0 \times {10}^{11}N{m}^{-2}}{2.0\times {10}^{11}N{m}^{-2}}\right) \\ =0.25 mm \end{gathered}$$

Total extension = 1 + 0.25 = 1.25 mm.