Two wires of diameter 0.25 cm, 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 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Solution

Given,
Diameter of the wires = 0.25 cm
That is, d_s= d_B= 0.25 cm

Therefore,
Radius of the wires will be given by,

$r_{s} = r_{B} = 0.125 cm = 1.25 \times 10^{- 4} m$

Unloaded length of the steel wire, L_s=1.5 m
Unloaded length of the brass wire, L_B= 1 m

Young's modulus of steel, Y_s = 2 $\times 10^{11}$ Pa
Young's modulus of brass, Y_B = $0.91 \times 10^{11} P a$
m₁ = 4 kg and m₂ = 6 kg

Brass wire is under a tension of load 6 kg.

Therefore, increase in length is given by

$∆ L_{s} = \frac{F}{A_{s}} \frac{L_{B}}{Y_{B}} = \frac{6 \times 9.8 \times 10}{π (1.25 \times 10^{- 4})^{2} \times 0.91 \times 10^{11}} m$
Since, brass wire is under a tension of load 6 kg.

"
Steel wire is under a tension of 10 kg load.

Therefore, elongation of length of the wire,

$∴ ∆ L_{S} = \frac{F}{A_{s}} \frac{L_{s}}{Y_{s}} = \frac{10 \times 9.8 \times 1.5}{π (1.25 \times 10^{- 4})^{2} \times 10^{11}} m = 1.5 \times 10^{- 4} m$

Some More Questions From Mechanical Properties of Fluids Chapter

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Mechanical Properties Of Fluids

Two wires of diameter 0.25 cm, 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 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Some More Questions From Mechanical Properties of Fluids Chapter

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