Question

Two wires of diameter $0.25cm$. one made of steel and the other made of brass are loaded as shown in Fig. $9.13$. The unloaded length of steel wire is $1.5m$ and that of load wire is $1.0m$. Compute the elongation of the steel and the brass wires.

Answer 1

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

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 = $\frac{d}{2}$ = 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:

$Y_1$ = $\frac{F_1/A_1}{\Delta L_1/L_1}$

Where, $\Delta L_1$ = Change in the length of the steel wire

$A_1$= Area of ​​cross-section of steel wire = $\pi r_1^2$

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

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

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

= $1.49\times {10}^{-4}$ m