Given $Y_{\text {steel }}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ and
$Y_{\text {Aluminium }}=7 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$.
For steel wire $A, A_1=1 \mathrm{~mm}^2=1 \times 10^{-6} \mathrm{~m}^2$, $Y_1=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
For aluminium wire $B, A_2=2 \mathrm{~mm}^2=2 \times 10^{-6} \mathrm{~m}^2$, $Y_2=7 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$
$l_1=l_2=l$ (for both $A$ and $B$ )
(a) Let the mass $\mathrm{m}$ be suspended at a distance $x$ from $A$. Let $F_1$ and $F_2$ are tensions in $A$ and $B$ that produce equal stress in both.
$$
\begin{aligned}
&\therefore \quad \frac{F_1}{A_1}=\frac{F_2}{A_2} \\
&\Rightarrow \quad \frac{F_1}{F_2}=\frac{A_1}{A_2}=\frac{1 \times 10^{-6}}{2 \times 10^{-6}}=\frac{1}{2}
\end{aligned}
$$
For equilibrium of the rod, the moments of forces about the point of suspension of the mass should be equal.
$$
\begin{aligned}
&\therefore F_1 x=F_2(1.05-x) \\
&\Rightarrow \frac{1.05-x}{x}=\frac{F_1}{F_2}=\frac{1}{2} \\
&\Rightarrow 2.1-2 x=x \Rightarrow x=0.7 \mathrm{~m}
\end{aligned}
$$
(b) For equal strain in both the wires,
$$
\begin{aligned}
&\frac{F_1}{A_1 Y_1}=\frac{F_2}{A_2 Y_2} \Rightarrow \frac{F_1}{F_2}=\frac{A_1 Y_1}{A_2 Y_2} \\
&=\frac{1}{2} \times \frac{2 \times 10^{11}}{7 \times 10^{10}}=\frac{10}{7}
\end{aligned}
$$
Equating the moments again,
$$
\begin{aligned}
&F_1 x=F_2(1.05-x) \Rightarrow \frac{1.05-x}{x} \\
&=\frac{F_1}{F_2}=\frac{10}{7} \Rightarrow 10 x=7.35-7 x \\
&\Rightarrow x=0.4324 \mathrm{~m}
\end{aligned}
$$