Question

Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 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. Given youngs modulus of steel = $2\times {10}^{-11}Pa$ and that of brass is $0.9\times {10}^{-11}Pa$

Answer 1

Diameter of the wires, d $=0.25m$
Hence, the radius of the wires, $r=\frac{d}{2}=0.125cm$
Length of the steel wire, $L_1=1.5m$
Length of the brass wire, $L_2=1.0m$

Total force exerted on the steel wire:

$F_1=(4+6)g=10\times 9.8=98N$
Young’s modulus for steel:

$Y_1=\frac{\left(\frac{F_1}{A_1}\right)}{\left(\frac{△L_1}{L_1}\right)}$
Where,
$△L_1=$ Change in the length of the steel wire
$A_1=$ Area of cross-section of the steel wire $=\pi r_1^2$

Young’s modulus of steel, $Y_1=2.0\times {10}^{11}Pa$

$\therefore △L_1=\frac{F_1\times L_1}{(A_1\times Y_1)}$

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

Total force on the brass wire:

$F_2=6\times 9.8=58.8N$
Young’s modulus for brass:

$Y_2=\frac{\left(\frac{F_2}{A_2}\right)}{\left(\frac{△L_2}{L_2}\right)}$
Where,

$△L_2=$ Change in the length of the brass wire
$A_2=$ Area of cross-section of the brass wire $=\pi r_1^2$

$\therefore △L_2=\frac{F_2\times L_2}{(A_2\times Y_2)}$
$=(58.8\times 1)/\left[\right(\pi \times (0.125\times {10}^{-2}{)}^2\times (0.91\times {10}^{11}\left)\right]=1.3\times {10}^{-4}m$
Elongation of the steel wire $=1.5\times {10}^{–4}m$
Elongation of the brass wire $=1.3\times {10}^{-4}m$.