A copper wire and an aluminium wire have lengths in the ratio 5 : 2, diameters in the ratio 4 : 3 and forces applied in the ratio 4 : 5. Find the ratio of increase in length of the copper wire to that of aluminium wire.

$(\mathrm{Let} {\mathrm{Y}}_{\mathrm{Cu}}=1.1\times {10}^{11} {\mathrm{Nm}}^{-2}, {\mathrm{Y}}_{\mathrm{Al}}=0.7\times {10}^{11} {\mathrm{Nm}}^{-2} )$

Given the ratio of copper to aluminum wire length is $l_1:l_2=5:2$

and the wire's diameter ratio is ${\mathrm{d}}_1:{\mathrm{d}}_2=4:3$

The increase in length of a wire under tension can be calculated using Hooke's Law:

copper wire,

$∆l_1=\frac{{\mathrm{F}}_1{l}_1}{{\mathrm{A}}_1{\mathrm{Y}}_1}\mathrm{\_\_\_\_\_\_\_\_}\left(1\right)$

aluminum wire,

$∆l_2=\frac{{\mathrm{F}}_2{l}_2}{{\mathrm{A}}_2{\mathrm{Y}}_2}\mathrm{\_\_\_\_\_\_\_\_}\left(2\right)$

from equations 1 and 2,

$\frac{∆l_1}{∆l_2}=\frac{l_1}{l_2}.\frac{{\mathrm{A}}_2}{{\mathrm{A}}_1}.\frac{{\mathrm{Y}}_2}{{\mathrm{Y}}_1}.\frac{{\mathrm{F}}_1}{{\mathrm{F}}_2}\left[\frac{{\mathrm{Y}}_2}{{\mathrm{Y}}_1}=\frac{{\mathrm{Y}}_{\mathrm{Al}}}{{\mathrm{Y}}_{\mathrm{Cu}}}=\frac{0.7}{1.1}\right]$

$\frac{∆l_1}{∆l_2}=\frac{5}{2}.\frac{{{\mathrm{\pi r}}_2}^2}{{{\mathrm{\pi r}}_1}^2}.\frac{0.7\times {10}^{11}}{1.1\times {10}^{11}}.\frac{4}{5}$

$\frac{∆l_1}{∆l_2}=\frac{5}{2}\times \frac{{\left({\displaystyle \frac{3}{2}}\right)}^2}{{\left(2\right)}^2}\times \frac{7}{11}\times \frac{4}{5}$

$\frac{∆l_1}{∆l_2}=\frac{63}{88}$

Hence the correct answer is $\frac{63}{88}.$