For steel wire $A,\ l_{1}=l\ ;\ A_{1}=1\ \mathrm{mm}^{2}\ ;$ $Y_{1}=2 \times 10^{11}\ \mathrm{N}\ \mathrm{m}^{-2}$ For aluminium wire $B,\ l_{2}=1\ ;\ A_{2}=2\ \mathrm{mm}^{2}\ ;$ $Y_{2}=7 \times 10^{10}\ \mathrm{N}\ \mathrm{m}^{-2}$ (a) Let mass m be suspended from the rod at distance x from theend where wire A is connected, Let $F_1 \text{ and } F_2$ be the tensions in two wires and there is equal stress in two wires, then$\frac{F_1}{A_1} = \frac{F_2}{A_2}$ or $\frac{F_1}{F_2} = \frac{A_1}{A_2} = \frac{1}{2} \ldots (i)$Taking moment of forces about thepoint of suspension of mass from the rod, we have$F_1 x = F_2 (1.05 - x)$or $\frac{1.05 - x}{x} = \frac{F_1}{F_2} = \frac{1}{2}$ or $2.10 - 2x = x$ or $x = 0.70\ \mathrm{m} = 70\ \mathrm{cm}$(b) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let and $F_1 F_2$ be the tension in the wires and there is equal strain in the two wires i.e. $\frac{F_1}{A_1 Y_1} = \frac{F_2}{A_2 Y_2}$or $\frac{F_1}{F_2} = \frac{A_1 Y_2}{A_2 Y_2}$$= \frac{1}{2} \times \frac{2 \times 10^{11}}{7 \times 10^{10}} = \frac{10}{7}$As the rod is stationary, so$F_1 x = F_2 (1.05 - x)$or $\frac{1.05 - x}{x} = \frac{F_1}{F_2} = \frac{10}{7}$ or $10x = 7.35 - 7x$or $x = 0.4324\ \mathrm{m} = 43.2\ \mathrm{cm}$