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.


Given,

Initial temperature, T1 = 27°C = 27 + 273.15 K = 300.15 K


Length of the brass wire at T1, L = 1.8 m


Final temperature, T2 = –39°C = -39 + 273.15 K = 234.15 K


Diameter of the wire, d = 2.0 mm = 2 × 10-3 m


Co-efficient of linear expansion of brass, α = 2.0 × 10–5 K–1


Young’s modulus of brass, Y = 0.91 × 1011 Pa


Let the tension developed in the wire be T.


Young’s modulus is given by


Y = Stress/Strain


Y = (F/A)/(∆L/L)


∆L = FL/AY


So, for tension T,


ΔL = TL/AY


Where, T = Tension developed in the wire


A = Area of cross-section of the wire


ΔL = Change in the length


But, ΔL = αL(T2 – T1)


αL(T2 – T1) = TL/[π(d/2)2 × Y]


T = απ(T2 – T1)Y(d/2)2


T = (2.0 × 10–5 K–1)×3.14×(234.15 K - 300.15 K)×( 0.91 × 1011 Pa)×( 2 × 10-3 m/2)2


T = -377.37 N


The negative sign indicates that the direction of tension is inwards.


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