A brass wire 1.8 m long at `27^(@)C` is held taut with a little tension b/w 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 ? `a=2.0xx10^(-5)""^(@)C^(-1)andY=0.91xx10^(11)Pa`?

Explanation

Given, `l_(2)=1.8m,theta_(1)=27^(@)C=-39^(@)C`,
`alpha=2.0xx10^(-5)""^(@)C^(-1),d=2xx10^(-3)m`
`Y=0.91xx10^(11)` Pa.
We know that `l_(2)=(alpha_(1)l_(1)t_(2)+l_(1))/(1+alphat_(1))" using "alpha=(l_(2)-l_(1))/(l_(1)t_(1)-l_(2)t_(2))`
i.e. `l_(2)=((-2xx10^(-5)xx1.8xx39+1.8))/(1+2xx10^(-5)xx27)=(1.7986)/(1.00054)=1.7976m`
`:.D_(1)=l_(1)-l_(2)=1.8-1.7976=0.00237m`
We know that `Y=(Fl)/(ADeltal)` where F=2T
`:.Y=(2Tlxx4)/(piD^(2)xxpil)=(8Tl)/(piD_(2)Deltal)`
or `T=(piD^(2)YDeltal)/(8l)`
`=(3.142xx(2xx10^(-3))^(2)xx0.91xx10^(11)xx0.00237)/(8xx1.8)=1.8829xx10^(-6+11)xx10^(-3)`
`T=1.883xx10^(2)N`.