(a)

Given that:

Cross-sectional area of wire A, ${\mathrm{a}}_1$ = 1.0 mm${}^2$ = 1.0 × ${10}^{-6}$ m${}^2$

Cross-sectional area of wire B, ${\mathrm{a}}_2$ = 2.0 mm${}^2$ = 2.0 × 10${}^{-6}$ m${}^2$

Young’s modulus for steel, Y${}_1$ = 2.0 × 10${}^{11}$ Nm${}^{-2}$

Young’s modulus for aluminium, ${\mathrm{Y}}_2$ = 7.0 × ${10}^{10}$ Nm${}^{-2}$

Let a small mass m is suspended to the rod at a distance x from the end where wire A is attached.

$\mathrm{Stress}=\frac{\mathrm{Force}}{\mathrm{Area}}=\frac{\mathrm{F}}{\mathrm{a}}$

When the two wires have equal stresses,

$\frac{{\mathrm{F}}_1}{{\mathrm{a}}_1}=\frac{{\mathrm{F}}_2}{{\mathrm{a}}_2}$

$\frac{{\mathrm{F}}_1}{{\mathrm{F}}_2}=\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{1}{2}$...............(i)

Taking torque about the point of suspension,

${\mathrm{F}}_1\mathrm{y}={\mathrm{F}}_2(1.05-\mathrm{y})$
$\frac{{\mathrm{F}}_1}{{\mathrm{F}}_2}=\frac{(1.05-\mathrm{y})}{\mathrm{y}}..............\left(\mathrm{ii}\right)$

Using equations (i) and (ii),

$\frac{\left(1.05-\mathrm{y}\right)}{\mathrm{y}}=\frac{1}{2}$
$2(1.05-\mathrm{y})=\mathrm{y}$
$2.1-2\mathrm{y}$
$3\mathrm{y}=2.1$
$\Rightarrow \mathrm{y}=0.7\mathrm{m}$

In order to produce equal stress in both of the wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b)

Young's modulus = $\frac{\mathrm{Stress}}{\mathrm{Stain}}$

If the strain in the two wires is equal,

$\frac{{\mathrm{F}}_1}{{\displaystyle {\mathrm{a}}_1{\mathrm{Y}}_1}}=\frac{{\mathrm{F}}_2}{{\displaystyle {\mathrm{a}}_2{\mathrm{Y}}_2}}$
$\frac{{\mathrm{F}}_1}{{\mathrm{F}}_2}=\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\frac{{\mathrm{Y}}_1}{{\mathrm{Y}}_2}=\frac{1}{2}\times \frac{2\times {10}^{11}}{7\times {10}^{10}}=\frac{10}{7}...........\left(\mathrm{iii}\right)$

Let a mass m is suspended at a distance ${\mathrm{y}}_1$ from the side where wire A attached.

Taking torque about the point where mass m is suspended,

${\mathrm{F}}_1{\mathrm{y}}_1={\mathrm{F}}_2\left(1.05-{\mathrm{y}}_1\right)$
$\frac{{\mathrm{F}}_1}{{\mathrm{F}}_2}=\frac{\left(1.05-{\mathrm{y}}_1\right)}{{\mathrm{y}}_1}...............\left(\mathrm{iv}\right)$

Using equations (iii) and (iv),

$\frac{\left(1.05-{\mathrm{y}}_1\right)}{{\mathrm{y}}_1}=\frac{10}{7}$
$7(1.05-{\mathrm{y}}_1)=10{\mathrm{y}}_1$
$17{\mathrm{y}}_1=7.35$
$\Rightarrow {\mathrm{y}}_1=0.432 \mathrm{m}$

In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached.