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? Coefficient of linear expansion of brass = 2.0 x 10-5/0C, Young's modulus of brass = 0.91 x 1011 Pa.

As the wire is not free to contract, the thermal stress is developed at the wire.The change in temperatwe,∆T = 270C-(-390C) = 660CLet ∆L be the change in length of the wire.∆L = αL∆T = (2×10-5)×1.8×66 = 2.376×10-3 mAs  Y = FLA∆L = FL(πD2)4∆L = 4FLπD2∆LF(tension in the wire) = YπD2∆L4L=(0.91×1011)×3.142×(2×10-3)2×(2.376×10-3)4×1.8= 3.8 × 102N