$\Delta L = \frac{{4FL}}{{\pi {D^2}Y}}$

$\therefore \frac{{\Delta {L_s}}}{{\Delta {L_c}}} = \frac{{{F_S}}}{{{F_C}}}\frac{{{L_S}}}{{{L_C}}}\frac{{D_C^2}}{{D_S^2}}\frac{{{Y_C}}}{{{Y_S}}}$

Where subscripts $'S'$ and $'C'$ refer to copper and steel respectively.

$Here,\,{F_S} = \left( {5m + 2m} \right)g = 7mg$

${F_C} = 5mg$

$\frac{{{L_S}}}{{{L_C}}} = q,\frac{{{D_S}}}{{{D_C}}} = p,\frac{{{Y_S}}}{{{Y_C}}} = s$

$\therefore \frac{{\Delta {L_S}}}{{\Delta {L_C}}} = \left( {\frac{{7mg}}{{5mg}}} \right)\left( q \right){\left( {\frac{1}{p}} \right)^2}\left( {\frac{1}{s}} \right) = \frac{{7q}}{{5{p^2}s}}$