If the ratio of diameters, lengths and Young's modulus of steel and copper wires shown in the figure are p, q and s respectively, then the corresponding ratio of increase in their lengths would be

$$\begin{gathered} \text{ As }Y=\frac{FL}{A\Delta L}=\frac{4FL}{\pi D^2\Delta L} \\ \Delta L=\frac{4FL}{\pi D^{2}Y} \\ \therefore \frac{\Delta L_S}{\Delta L_C}=\frac{F_S}{F_C}\frac{L_S}{L_C}\frac{D_C^2}{D_S^2}\frac{Y_C}{Y_S} \end{gathered}$$

$$\begin{gathered} \text{ where subscripts }S\text{ and }C\text{ refer to copper and steel respectively.} \\ Here, \\ {F}_S=(5m+2m)g=7mg \\ {F}_C=5mg \\ \frac{L_S}{L_C}=q,\frac{D_S}{D_C}=p,\frac{Y_S}{Y_C}=s \\ \therefore \frac{\Delta L_S}{\Delta L_C}=\left(\frac{7mg}{5mg}\right)\left(q\right){\left(\frac{1}{p}\right)}^2\left(\frac{1}{s}\right)=\frac{7q}{5p^{2}s} \end{gathered}$$