Two wires of diameter $0.25 \; cm,$ one made of steel and the other made of brass,are loaded as shown in the figure. The unloaded length of the steel wire is $1.5 \; m$ and that of the brass wire is $1.0 \; m.$ Compute the elongations of the steel and the brass wires. (Given: Young's modulus of steel $Y_s = 2.0 \times 10^{11} \; Pa,$ Young's modulus of brass $Y_b = 0.91 \times 10^{11} \; Pa$)

(N/A) Given:
Diameter of wires,$d = 0.25 \; cm = 0.25 \times 10^{-2} \; m$
Radius of wires,$r = d/2 = 0.125 \times 10^{-2} \; m$
Length of steel wire,$L_s = 1.5 \; m$
Length of brass wire,$L_b = 1.0 \; m$
Young's modulus of steel,$Y_s = 2.0 \times 10^{11} \; Pa$
Young's modulus of brass,$Y_b = 0.91 \times 10^{11} \; Pa$
$1$. For the steel wire:
The steel wire supports both the $4.0 \; kg$ and $6.0 \; kg$ masses.
Total force,$F_s = (4.0 + 6.0) \times 9.8 = 98 \; N$
Elongation,$\Delta L_s = \frac{F_s L_s}{A Y_s} = \frac{F_s L_s}{\pi r^2 Y_s}$
$\Delta L_s = \frac{98 \times 1.5}{\pi \times (0.125 \times 10^{-2})^2 \times 2.0 \times 10^{11}} \approx 1.49 \times 10^{-4} \; m$
$2$. For the brass wire:
The brass wire supports only the $6.0 \; kg$ mass.
Total force,$F_b = 6.0 \times 9.8 = 58.8 \; N$
Elongation,$\Delta L_b = \frac{F_b L_b}{A Y_b} = \frac{F_b L_b}{\pi r^2 Y_b}$
$\Delta L_b = \frac{58.8 \times 1.0}{\pi \times (0.125 \times 10^{-2})^2 \times 0.91 \times 10^{11}} \approx 1.3 \times 10^{-4} \; m$