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 × 10 ¹¹ Pa.

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:

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:

(The negative sign indicates that the tension is directed inward.)

Hence, the tension developed in the wire is 3.8 ×10² N.