$A$ rod of length $1.05 \; m$ having negligible mass is supported at its ends by two wires of steel (wire $A$) and aluminium (wire $B$) of equal lengths as shown in Figure. The cross-sectional areas of wires $A$ and $B$ are $1.0 \; mm^{2}$ and $2.0 \; mm^{2}$ respectively. At what point along the rod should a mass $m$ be suspended in order to produce $(a)$ equal stresses and $(b)$ equal strains in both steel and aluminium wires?

(A) Given:
Length of rod $L = 1.05 \; m$
Area of steel wire $A$ $(a_1)$ = $1.0 \; mm^2 = 1.0 \times 10^{-6} \; m^2$
Area of aluminium wire $B$ $(a_2)$ = $2.0 \; mm^2 = 2.0 \times 10^{-6} \; m^2$
Young's modulus for steel $(Y_1)$ = $2 \times 10^{11} \; N/m^2$
Young's modulus for aluminium $(Y_2)$ = $7.0 \times 10^{10} \; N/m^2$
$(a)$ For equal stresses:
Stress = $\frac{F}{a}$. If stresses are equal,$\frac{F_1}{a_1} = \frac{F_2}{a_2} \implies \frac{F_1}{F_2} = \frac{a_1}{a_2} = \frac{1.0}{2.0} = 0.5$.
Taking torque about the point of suspension at distance $y$ from wire $A$:
$F_1 y = F_2 (1.05 - y) \implies \frac{F_1}{F_2} = \frac{1.05 - y}{y} = 0.5$.
$1.05 - y = 0.5y \implies 1.5y = 1.05 \implies y = 0.7 \; m$.
$(b)$ For equal strains:
Strain = $\frac{\text{Stress}}{Y} = \frac{F}{aY}$. If strains are equal,$\frac{F_1}{a_1 Y_1} = \frac{F_2}{a_2 Y_2}$.
$\frac{F_1}{F_2} = \frac{a_1 Y_1}{a_2 Y_2} = \left(\frac{1.0}{2.0}\right) \times \left(\frac{2 \times 10^{11}}{7 \times 10^{10}}\right) = 0.5 \times \frac{20}{7} = \frac{10}{7}$.
Taking torque about the point of suspension at distance $y_1$ from wire $A$:
$\frac{F_1}{F_2} = \frac{1.05 - y_1}{y_1} = \frac{10}{7}$.
$7(1.05 - y_1) = 10y_1 \implies 7.35 - 7y_1 = 10y_1 \implies 17y_1 = 7.35 \implies y_1 \approx 0.432 \; m$.