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`.

Explanation

Here, `L = 1.8m, T_(1)=27^(@)C, T_(2) = -39^(@)C, F=?, r=1 mm = 10^(-3)m`,
`alpha = 2xx10^(-05)^@C^(-1), Y = 0.91 xx 10^(11) N//m^2`
From, `Y = (F.L)/(A Delta L), Delta L= (FL)/(AY)`
Also, `Delta L = alpha L Delta T :. (FL)/(AY) = alpha L Delta T` or `F = alpha Delta T AY = alpha (T_2 - T_1) pi r^2 Y`
`=2xx10^(-5)xx (-39 - 27)xx(22)/(7) (10^(-3)xx 0.91xx10^(11) = -3.77xx10^2 N`
Negative sign indicates that the force is inwards due to contraction of the wire.