A copper wire and an aluminium wire have lengths in the ratio 3 : 2, diameters in the ratio 2 : 3 and forces applied in the ratio 4 : 5. Find the ratio of the increase in length of the two wires. $({\mathrm{Y}}_{\mathrm{Cu}}=1.1 \times {10}^{11}{\mathrm{Nm}}^2, {\mathrm{Y}}_{\mathrm{AI}}=0.70 \times {10}^{11}{\mathrm{Nm}}^{-2})$

Given,

${\mathrm{Y}}_{\mathrm{Cu}}=1.1 \times {10}^{11}{\mathrm{Nm}}^2 \& {\mathrm{Y}}_{\mathrm{AI}}=0.70 \times {10}^{11}{\mathrm{Nm}}^{-2}$

${\mathrm{L}}_{\mathrm{cu}}:{\mathrm{L}}_{\mathrm{m}}=3:2 \& {\mathrm{d}}_{\mathrm{cu}}:{\mathrm{d}}_{\mathrm{m}}=2:3$

$$\begin{gathered} {\mathrm{A}}_{\mathrm{cu}}:{\mathrm{A}}_{\mathrm{Al}}=\frac{{\displaystyle \frac{\mathrm{\pi }}{4}{\left({\mathrm{d}}_{\mathrm{cu}}\right)}^2}}{{\displaystyle \frac{\mathrm{\pi }}{4}{\left({\mathrm{d}}_{\mathrm{Al}}\right)}^2}} \\ \Rightarrow {\mathrm{A}}_{\mathrm{cu}}:{\mathrm{A}}_{\mathrm{Al}}={\left(\frac{2}{3}\right)}^2 \end{gathered}$$

${\mathrm{F}}_{\mathrm{cu}}:{\mathrm{F}}_{\mathrm{Al}}=4:5$

The ratio of wire length increases

$\mathrm{Y}=\frac{{\displaystyle \frac{\mathrm{F}}{{\mathrm{\pi r}}^2}}}{{\displaystyle \frac{∆\mathrm{l}}{\mathrm{l}}}}\Rightarrow \mathrm{Y}=\frac{\mathrm{Fl}}{{\mathrm{\pi r}}^2∆\mathrm{l}}$

$\Rightarrow ∆\mathrm{l}=\frac{\mathrm{Fl}}{{\mathrm{Y\pi r}}^2}$

$\frac{∆{\mathrm{l}}_{\mathrm{Cu}}}{∆{\mathrm{l}}_{\mathrm{Al}}}=\frac{\frac{{\mathrm{F}}_{\mathrm{Cu}}{\mathrm{l}}_{\mathrm{Cu}}}{{\mathrm{Y}}_{\mathrm{Cu}}{{\mathrm{\pi r}}_{\mathrm{Cu}}}^2}}{\frac{{\mathrm{F}}_{\mathrm{Al}}{\mathrm{l}}_{\mathrm{Al}}}{{\mathrm{Y}}_{\mathrm{Al}}{{\mathrm{\pi r}}_{\mathrm{Al}}}^2}}\Rightarrow \frac{∆{\mathrm{l}}_{\mathrm{Cu}}}{∆{\mathrm{l}}_{\mathrm{Al}}}=\frac{{\mathrm{F}}_{\mathrm{Cu}}{\mathrm{l}}_{\mathrm{Cu}}{\mathrm{Y}}_{\mathrm{Al}}{{\mathrm{\pi r}}_{\mathrm{Al}}}^2}{{\mathrm{F}}_{\mathrm{Al}}{\mathrm{l}}_{\mathrm{Al}}{\mathrm{Y}}_{\mathrm{Cu}}{{\mathrm{\pi r}}_{\mathrm{Cu}}}^2}$

$\Rightarrow \frac{∆{\mathrm{l}}_{\mathrm{Cu}}}{∆{\mathrm{l}}_{\mathrm{Al}}}=\frac{{\mathrm{F}}_{\mathrm{Cu}}}{{\mathrm{F}}_{\mathrm{Al}}}\times \frac{{\mathrm{l}}_{\mathrm{Cu}}}{{\mathrm{l}}_{\mathrm{Al}}}\times \frac{{\mathrm{Y}}_{\mathrm{Al}}}{{\mathrm{Y}}_{\mathrm{Cu}}}\times {\left(\frac{{\mathrm{\pi r}}_{\mathrm{Al}}}{{\mathrm{\pi r}}_{\mathrm{Cu}}}\right)}^2$

$\Rightarrow \frac{∆{\mathrm{l}}_{\mathrm{Cu}}}{∆{\mathrm{l}}_{\mathrm{Al}}}=\frac{4}{5}\times \frac{3}{2}\times \frac{0.70\times {10}^{11}}{1.1\times {10}^{11}}\times {\left(\frac{3}{2}\right)}^2$

$\Rightarrow \frac{∆{\mathrm{l}}_{\mathrm{Cu}}}{∆{\mathrm{l}}_{\mathrm{Al}}}=\frac{4}{5}\times \frac{3}{2}\times \frac{7}{11}\times \frac{9}{4}=\frac{189}{110}$

Hence the correct answer is 189:110.