A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa.
Answers
Initial temperature, T1 = 27°C
Length of the brass wire at T1, l = 1.8 m
Final temperature, T2 = –39°C
Diameter of the wire, d = 2.0 mm = 2 × 10–3 m
Tension developed in the wire = F
Coefficient of linear expansion of brass, α = 2.0 × 10–5 K–1
Young’s modulus of brass, Y = 0.91 × 1011 Pa
Young’s modulus is given by the relation:
γ = Stress / Strain = (F/A) / (∆L/L)
∆L = F X L / (A X Y) ......(i)
Where,
F = Tension developed in the wire
A = Area of cross-section of the wire.
ΔL = Change in the length, given by the relation:
ΔL = αL(T2 – T1) … (ii)
Equating equations (i) and (ii), we get:
αL(T2 - T1) = FL / [ π(d/2)2 X Y ]
F = α(T2 - T1)πY(d/2)2
F = 2 × 10-5 × (-39-27) × 3.14 × 0.91 × 1011 × (2 × 10-3 / 2 )2
= -3.8 × 102 N
(The negative sign indicates that the tension is directed inward.)
Hence, the tension developed in the wire is 3.8 ×102 N.
☑️ Initial temperature, T1 = 27°C
☑️ Length of the brass wire at T1, l = 1.8 m
☑️ Final temperature, T2 = –39°C
☑️ Diameter of the wire, d = 2.0 mm = 2 × 10–3 m
☑️ Tension developed in the wire = F
☑️ Coefficient of linear expansion of brass, α = 2.0 × 10–5 K–1
Young’s modulus of brass, Y = 0.91 × 1011 Pa
Young’s modulus is given by the relation:
γ = Stress / Strain = (F/A) / (∆L/L)
∆L = F X L / (A X Y) ……(i)
Where,
F = Tension developed in the wire
A = Area of cross-section of the wire.
ΔL = Change in the length, given by the relation:
ΔL = αL(T2 – T1) … (ii)
Equating equations (i) and (ii), we get:
αL(T2 – T1) = FL / [ π(d/2)2X Y ]
F = α(T2 – T1)πY(d/2)2
F = 2 × 10-5 × (-39-27) × 3.14 × 0.91 × 1011 × (2 × 10-3 / 2 )2
= -3.8 × 102 N
(The negative sign indicates that the tension is directed inward.)
Hence, the tension developed in the wire is 3.8 ×102 N.