Question

Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. 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.

Answer 1

Elongation of the steel wire = 1.49 × 10^–4 m
Elongation of the brass wire = 1.3 × 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, L₁ = 1.5 m
Length of the brass wire, L₂ = 1.0 m
Total force exerted on the steel wire: 

F₁ = (4 + 6) g = 10 × 9.8 = 98 N 

Young’s modulus for steel : 

\(Y1 =\frac{ (\frac{F_1 }{ A_1}) }{ (\frac{ΔL_1 }{ L_1})}\)

Where, ΔL₁ = Change in the length of the steel wire

A₁ = Area of cross-section of the steel wire = \(πr_1^2\)

Young’s modulus of steel, Y₁ = 2.0 × 10 ¹¹ Pa

∴ ΔL1 =\(\frac{ F_1 × L_1 }{A_1 × Y_1} = \frac{F_1 × L_1 }{ πr_1^2 × Y_1 }\)

= \(\frac{98 × 1.5 }{ π(0.125 × 10 ^{- 2})^2 × 2 × 10 ^{11}}\) = 1.49 × 10 - 4 m

Total force on the brass wire : F2 = 6 × 9.8 = 58.8 N

Young’s modulus for brass : 

\(Y2 =\frac{ (\frac{F2 }{ A2})}{(\frac{ΔL_2 }{ L_2})}\)

Where, 

ΔL2 = Change in length

A2 = Area of Cross - section of the brass wire

∴ ΔL2 = \(\frac{F_2 × L_2 }{ A_2 × Y_2 }=\frac{ F_2 × L_2 }{ πr_2^2 × Y_2 }\)

= \(\frac{58.8 × 1.0 }{ π × (0.125 × 10 ^{- 2})^2 × (0.91 × 10^{ 11}) }\)= 1.3 × 10 ^{- 4} m

Elongation of the steel wire = 1.49 × 10 ^{- 4} m

Elongation of the brass wire = 1.3 × 10 ^{- 4} m