Two wires of diameter $0.25 \;cm ,$ one made of steel and the other made of brass are loaded as shown in Figure. The unloaded length of steel wire is $1.5 \;m$ and that of brass wire is $1.0 \;m .$ Compute the elongations of the steel and the brass wires.

Elongation of the brass wire $=1.3 \times 10^{-4} m$ Diameter of the wires, $d=0.25 m$ Hence, the radius of the wires, $r=d / 2=0.125 cm$ Length of the steel wire, $L_{1}=1.5 m$ Length of the brass wire, $L_{2}=1.0 m$

Total force exerted on the steel wire:

$F_{1}=(4+6) g=10 \times 9.8=98 N$

Young's modulus for steel:

$Y_{1}=\frac{\left(\frac{F_{1}}{A_{1}}\right)}{\left(\frac{\Delta L_{1}}{L_{1}}\right)}$

$\therefore \Delta L_{1} =\frac{F_{1} \times L_{1}}{A_{1} \times Y_{1}}=\frac{F_{1} \times L_{1}}{\pi r_{1}^{2} \times Y_{1}}$

$=\frac{98 \times 1.5}{\pi\left(0.125 \times 10^{-2}\right)^{2} \times 2 \times 10^{11}}=1.49 \times 10^{-4} m$

Total force on the brass wire:

$F_{2}=6 \times 9.8=58.8 N$

Young's modulus for brass

$Y_{2}=\frac{\left(\frac{F_{2}}{A_{2}}\right)}{\left(\frac{\Delta L_{2}}{L_{2}}\right)}$

$\therefore \Delta L_{2}=\frac{F_{2} \times L_{2}}{A_{2} \times Y_{2}}=\frac{F_{2} \times L_{2}}{\pi r_{2}^{2} \times V_{2}}$

$=\frac{58.8 \times 1.0}{\pi \times\left(0.125 \times 10^{-2}\right)^{2} \times\left(0.91 \times 10^{11}\right)}$$=1.3 \times 10^{-4} m$

Elongation of the steel wire $=1.49 \times 10^{-4} m$

Elongation of the brass wire $=1.3 \times 10^{-4} m$