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, T 1 = 27 ° C
Length of the brass wire at T 1 , l = 1.8 m
Final temperature, T 2 = –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 × 10 11 Pa
Young’s modulus is given by the relation:
F = Tension developed in the wire
A = Area of cross-section of the wire.
Δ L = Change in the length, given by the relation:
Δ L = α L ( T 2 – T 1 ) … ( 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 ×10 2 N.
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