A brass wire 1.8 m long at 27^(@)C is held taut with little tension between two rigid supports. If the wire 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 xx 10^(-5)//^(@)C, Young's modulus of brass = 0.91 xx 10^(11) Pa.

A brass wire 1.8 m long at 27^(@)C is held taut with little tension be

1.	A brass wire 1.8 m long at 27^(@)C is held taut with little tension between two rigid supports. If the wire 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 xx 10^(-5)//^(@)C, Young's modulus of brass = 0.91 xx 10^(11) Pa.
Answer» Solution :Here`I=1.8m,t_(1)=27^(@)C,t_(2)=-39^(@)C` `r=(2.0)/(2)=1.0 mm =1.0xx10^(-3)m` `r=2.0xx10^(-5)""^(@)C^(-1),Y=0.91xx10^(11)Pa` As `"" DeltaI=Ialpha(t_(2)-t_(1))` `:.` Strain `(DeltaI)/(I)=alpha(t_(2)-t_(1))` `"Stress" = "Strain" xx "Young’s modulus" =alpha(t_(2)-t_(1))xx`Y `=2.0xx10^(-5)xx(-39-27)xx0.91xx10^(11)=1.2xx10^(8)Nm^(-2)` [Numerically] Tension developed in the wire = Stress xx AREA of cross-section `= Strees xxPr^(2)=1.2xx10^(8)xx3.14xx(1.0xx10^(-3))^(2)=3.77xx10^(2)N`.