A steel rod of cross-sectional area $16{ \mathrm{cm}}^2$ and two brass rods each of cross-sectional area $10{ \mathrm{cm}}^2$ together support a load of $5000 \mathrm{kg}$ as shown in the figure. (Given, $Y_{\text{steel }}=2\times {10}^6 \mathrm{kg}{ \mathrm{cm}}^{-2}and{Y}_{\text{brass }}={10}^6 \mathrm{kg}{ \mathrm{cm}}^{-2}$ ). Choose the correct option(s).

Area of steel rod, $A_s=16{ \mathrm{cm}}^2$

$$\begin{gathered} \text{ Area of two brass rods, } \\ {A}_B=2\times 10=20{ \mathrm{cm}}^2 \\ F=5000 \mathrm{kg} \\ {\sigma }_S=\text{ Stress in steel } \\ \text{ and } {\sigma }_B=\text{ Stress in brass } \end{gathered}$$

Decrease in length of steel rod = Decrease in length of brass rod

$\frac{{\sigma }_S}{Y_S}·L_S=\frac{{\sigma }_B}{Y_B}·L_B\Rightarrow {\sigma }_S=\frac{Y_S}{Y_B}·\frac{L_B}{L_S}·{\sigma }_B$

or ${\sigma }_S=\left(\frac{2\times {10}^6}{{10}^6}\right)\left(\frac{20}{30}\right){\sigma }_B$

$\Rightarrow {\sigma }_S=\frac{4}{3}{\sigma }_B......\left(i\right)$

Now,$F={\sigma }_S{A}_S+{\sigma }_B{A}_B$

or $5000={\sigma }_S\times 16+{\sigma }_B\times 20......\left(ii\right)$

From Eqs. (i) and (ii), we get 

${\sigma }_B=120.9 \mathrm{kg}{ \mathrm{cm}}^{-2}and{\sigma }_S=161.2 \mathrm{kg}{ \mathrm{cm}}^{-2}$