b
\(As\,Y = \frac{{FL}}{{A\Delta L}} = \frac{{4FL}}{{\pi {D^2}\Delta L}}\)

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

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

Where subscripts \('S'\) and \('C'\) refer to copper and steel respectively.

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

\({F_C} = 5mg\)

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

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