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.
Initial temperature, T1=27oC Length of the brass wire at T1,l=1.8m Final temperature, T2=39C Diameter of the wire, d=2.0mm=2×10−3m Tension developed in the wire =F Coefficient of linear expansion of brass, =2.0×10−5K−1 Youngs modulus of brass, Y=0.91×1011 Pa Youngs modulus is given by the relation: Y= Stress / Strain Y=ΔL/LF/A ΔL=F×L/(A×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)=π(d/2)2YFL F=α(T2−T1)Yπ(d/2)2 F=2×10−5×(−39−27)×3.14×0.91×1011×(2×10−3/2)2 F=−3.8×102N (The negative sign indicates that the tension is directed inward.) Hence, the tension developed in the wire is 3.8 ×102 N.
