If the ratio of lengths, radii and Young's modulus of steel and...

Free body diagram of the two blocks are Given, $\quad \frac{l_1}{l_2}=a, \frac{r_1}{r_2}=b, \frac{Y_1}{Y_2}=c$ Let Young's modulus of steel is $Y_1$ and of brass is $Y_2$ $\therefore \quad Y_1=\frac{F_1 \cdot l_1}{A_1 \cdot \Delta l_1}$...(i)$and \quad Y_2=\frac{F_2 \cdot l_2}{A_2 \Delta l_2}$...(ii) Dividing Eq. (i) by Eq. (ii), we get $\frac{Y_1}{Y_2}=\frac{\frac{F_1 \cdot l_1}{A_1 \cdot \Delta l_1}}{\frac{F_2 \cdot l_2}{A_2 \cdot \Delta l_2}}$ or $\quad \frac{Y_1}{Y_2}=\frac{F_1 \cdot A_2 \cdot l_1 \cdot \Delta l_2}{F_2 \cdot A_1 \cdot l_2 \cdot \Delta l_1}$...(iii) Force on steel wire from free body diagram $T=F_1=2 g \mathrm{~N}$ Force on brass wire from free body diagram $F_2=T^{\prime}=T+2 g=4 \mathrm{~g} \mathrm{~N}$ Now, putting the values of $F_1, F_2$ in Eq. (iii), we get $\frac{Y_1}{Y_2}=\left(\frac{2 g}{4 g}\right) \cdot\left(\frac{\pi r_2^2}{\pi r_1^2}\right) \cdot\left[\frac{l_1}{l_2}\right] \cdot\left(\frac{\Delta l_2}{\Delta l_1}\right)$ or $\quad c=\frac{1}{2}\left(\frac{1}{b^2}\right) \cdot a\left(\frac{\Delta l_2}{\Delta l_1}\right)$ or \quad \frac{\Delta l_1}{\Delta l_2}=\left(\frac{a}{2 b^2 c}\right)