A rod of length $l$ and negligible mass is suspended at its two ends by two wires of steel (wire $A$) and aluminium (wire $B$) of equal lengths (figure). The cross-sectional areas of wires $A$ and $B$ are 1.0 mm² and 2.0 mm², respectively

$(Y_{Al}=70\times {10}^9 {\mathrm{Nm}}^{-2}$ and $Y_{steel}=200\times {10}^9 {\mathrm{Nm}}^{-2})$

Let the mass is placed at $x$ from the end $B$

Let $T_{A }$and $T_B$ be the tensions in wire $A$ and wire $B$, respectively.

For the rotational equilibrium of the system,

                                 $\sum _{ }^{ }\tau =0$           (Total torque = 0 )

$\Rightarrow$            $T_{B}x-T_A(l-x)=0$

$\Rightarrow$                               $\frac{T_B}{T_A}=\frac{l-x}{x}$                          …(i)

             Stress in wire $A$$=S_A=\frac{T_A}{a_A}$

             Stress in wire $B=S_B=\frac{T_B}{a_B}$

where, $a_{A }$and $a_B$ are cross-sectional areas of wire $A$ and $B$,respectively 

By question, $a_B=2a_A$

Now, for equal stress              $S_A=S_B$

$\Rightarrow$                   $\frac{T_A}{a_A}=\frac{T_B}{a_B} \Rightarrow \frac{T_B}{T_A}=\frac{a_B}{a_A}=2$

$\Rightarrow$                      $\frac{l-x}{x}=2 \Rightarrow \frac{l}{x}-1=2$

$\Rightarrow$                             $x=\frac{l}{3}$

$\Rightarrow$                        $l-x=l-l/3=\frac{2l}{3}$

Hence mass $m$ should be placed closer to $B$.

For equal strain,     (strain)_A = (strain)_B

$\Rightarrow$  $\frac{\left(Y_A\right)}{S_A}=\frac{Y_B}{S_B}$     (where, $Y_A$ and $Y_B$ are Young modulii)

$\Rightarrow$            $\frac{Y_{steel}}{Y_{Al}}=\frac{T_A}{T_B}\times \frac{a_B}{a_A}=\left(\frac{x}{l-x}\right)\left(\frac{2a_A}{a_A}\right)$

$\Rightarrow$      $\frac{200\times {10}^9}{70\times {10}^9}=\frac{2x}{l-x}$

$\Rightarrow$                    $\frac{20}{7}=\frac{2x}{l-x}$

$\Rightarrow$                    $\frac{10}{7}=\frac{x}{l-x}$

$\Rightarrow$           $10l-10x=7x$

$\Rightarrow$                     $17x=10l$

$\Rightarrow$                          $x=\frac{10l}{17}$

                           $l-x=l-\frac{10l}{17}=\frac{7l}{17}$

Hence, mass $m$ should be placed closer to wire $A$