Question

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.​

Answer 1

Explanation:

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.

Answer 2

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