If the ratio of lengths, radii and Young's modulus of steel and brass wires shown in the figure are a, b and c respectively, the ratio between the increase in lengths of brass and steel wires would be Brass and steel wire would be

From the figure,

$\frac{l_1}{l_2}=\mathrm{a}, \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\mathrm{b},\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\mathrm{c}$

${\mathrm{E}}_1=\frac{{\mathrm{F}}_1}{{\mathrm{A}}_1}.\frac{l_1}{∆l_1}$

${\mathrm{E}}_2=\frac{{\mathrm{F}}_2}{{\mathrm{A}}_2}.\frac{l_2}{∆l_2}$

thus,

$\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{F}}_1{\mathrm{A}}_2{l}_{\mathit{1}}∆l_{\mathit{2}}}{{\mathrm{F}}_2{\mathrm{A}}_1{l}_{\mathit{2}}∆l_{\mathit{1}}}$

Now the force acting on the steel wire is,

$T=F_1=2kg$

the force acting on the brass wire is,

$F_2=T_2=T+2kg=4kg$

$\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{2}{4}\frac{{{\mathrm{\pi }}_2}^2}{{{\mathrm{\pi }}_1}^2}\frac{l_1}{l_2}\frac{∆l_{\mathit{2}}}{∆l_1}$

$\mathrm{C}=\frac{1}{2}\frac{1}{{\mathrm{b}}_2}\mathrm{a}\frac{∆l_2}{∆l_1}$

$\frac{∆l_1}{∆l_2}=\frac{\mathrm{a}}{2{\mathrm{b}}^2\mathrm{c}}$

Hence the correct answer is $\frac{\mathrm{a}}{2{\mathrm{b}}^2\mathrm{c}}.$