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Question

# A brass wire 1.8 m long at 270C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of −39oC, what is the tension developed in the wire, if its diameter is 2.0mm? Co-efficient of linear expansion of brass =2.0×10−5K−1; Young's modulus of brass =0.91×1011 Pa

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Solution

## Initial temperature, T1=27oCLength of the brass wire at T1,l=1.8mFinal temperature, T2=39C Diameter of the wire, d=2.0mm=2×10−3mTension developed in the wire =FCoefficient of linear expansion of brass, =2.0×10−5K−1Youngs modulus of brass, Y=0.91×1011 PaYoungs modulus is given by the relation:Y= Stress / Strain Y=F/AΔL/LΔL=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)=FLπ(d/2)2YF=α(T2−T1)Yπ(d/2)2F=2×10−5×(−39−27)×3.14×0.91×1011×(2×10−3/2)2F=−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.

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