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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
$\frac{{5q}}{{7{p^2}s}}$
$\;\frac{{7q}}{{5{p^2}s}}$
$\;\frac{{2q}}{{5sp}}$
$\;\frac{{7q}}{{5sp}}$
Solution
$As\,Y = \frac{{FL}}{{A\Delta L}} = \frac{{4FL}}{{\pi {D^2}\Delta L}}$
$\Delta L = \frac{{4FL}}{{\pi {D^2}Y}}$
$\therefore \frac{{\Delta {L_s}}}{{\Delta {L_c}}} = \frac{{{F_S}}}{{{F_C}}}\frac{{{L_S}}}{{{L_C}}}\frac{{D_C^2}}{{D_S^2}}\frac{{{Y_C}}}{{{Y_S}}}$
Where subscripts $'S'$ and $'C'$ refer to copper and steel respectively.
$Here,\,{F_S} = \left( {5m + 2m} \right)g = 7mg$
${F_C} = 5mg$
$\frac{{{L_S}}}{{{L_C}}} = q,\frac{{{D_S}}}{{{D_C}}} = p,\frac{{{Y_S}}}{{{Y_C}}} = s$
$\therefore \frac{{\Delta {L_S}}}{{\Delta {L_C}}} = \left( {\frac{{7mg}}{{5mg}}} \right)\left( q \right){\left( {\frac{1}{p}} \right)^2}\left( {\frac{1}{s}} \right) = \frac{{7q}}{{5{p^2}s}}$
