A steel rod of cross sectional area 16cm²  and two brass rods each of cross sectional area 10cm² together support a load of 500 kg as shown in the figure. (given ${\mathrm{Y}}_{\mathrm{steel}}=2\times {10}^{11}\frac{\mathrm{N}}{{\mathrm{m}}^2}$ and ${\mathrm{Y}}_{\mathrm{brass}}={10}^{11}\frac{\mathrm{N}}{{\mathrm{m}}^2}$). The stress in steel rod is given by $\frac{5000}{\mathrm{k}} {\mathrm{Nm}}^{-2}$. Then the value of $\mathrm{k}$ is 

Area of steel rod  ${\mathrm{A}}_{\mathrm{s}}=16{\mathrm{cm}}^2$
Area of two brass rods${\mathrm{A}}_{\mathrm{b}}=2\times 10=20{\mathrm{cm}}^2$
$\mathrm{F}=5000\mathrm{kg}{\mathrm{\sigma }}_{\mathrm{s}}=\mathrm{stress}\mathrm{in}\mathrm{steel}$
${\mathrm{\sigma }}_{\mathrm{B}}=\mathrm{stress}\mathrm{in}\mathrm{brass}$
$\frac{{\mathrm{\sigma }}_{\mathrm{s}}}{{\mathrm{y}}_{\mathrm{s}}}{\mathrm{L}}_{\mathrm{s}}=\frac{{\mathrm{\sigma }}_{\mathrm{B}}}{{\mathrm{y}}_{\mathrm{B}}}{\mathrm{L}}_{\mathrm{B}}\Rightarrow {\mathrm{\sigma }}_{\mathrm{s}}=\frac{{\mathrm{y}}_{\mathrm{s}}}{{\mathrm{y}}_{\mathrm{B}}}.\frac{{\mathrm{L}}_{\mathrm{B}}}{{\mathrm{L}}_{\mathrm{s}}}.{\mathrm{\sigma }}_{\mathrm{B}}$$\Rightarrow \left(\frac{2\times {10}^6}{{10}^6}\right)\left(\frac{20}{30}\right){\mathrm{\sigma }}_{\mathrm{B}}\Rightarrow {\mathrm{\sigma }}_{\mathrm{s}}=\frac{4}{3}{\mathrm{\sigma }}_{\mathrm{B}}\to \left(1\right)$
Now, $\mathrm{F}={\mathrm{\sigma }}_{\mathrm{s}}{\mathrm{A}}_{\mathrm{s}}+{\mathrm{\sigma }}_{\mathrm{B}}{\mathrm{A}}_{\mathrm{B}}$
$\begin{array}{l}5000={\mathrm{\sigma }}_{\mathrm{s}}\left(16\right)+{\mathrm{\sigma }}_{\mathrm{B}}\left(20\right)\to \left(2\right)\\ \mathrm{from}\left(\mathrm{i}\right)\&\left(\mathrm{ii}\right){\mathrm{\sigma }}_{\mathrm{B}}=120.9\mathrm{kg}/{\mathrm{cm}}^2\\ {\mathrm{\sigma }}_{\mathrm{s}}=\frac{5000}{31} \mathrm{kg}/{\mathrm{cm}}^2\end{array}$