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? Coefficient of linear expansion of brass = 2.0 x ${10}^{-5}{/}^0\mathrm{C}$, Young's modulus of brass = 0.91 x ${10}^{11}$ Pa.

As the wire is not free to contract, the thermal stress is developed at the wire.
The change in temperatwe,
$∆\mathrm{T} = {27}^0\mathrm{C}-(-{39}^0\mathrm{C}) = {66}^0\mathrm{C}$

Let $∆$L be the change in length of the wire.
$∆\mathrm{L} = \mathrm{\alpha L}∆\mathrm{T} = (2\times {10}^{-5})\times 1.8\times 66 = 2.376\times {10}^{-3} \mathrm{m}$

As  $\mathrm{Y} = \frac{\mathrm{FL}}{\mathrm{A}∆\mathrm{L}} = \frac{\mathrm{FL}}{{\displaystyle \frac{\left({\mathrm{\pi D}}^2\right)}{4}}∆\mathrm{L}} = \frac{4\mathrm{FL}}{{\mathrm{\pi D}}^2∆\mathrm{L}}$

F(tension in the wire) = $\frac{{\mathrm{Y\pi D}}^2∆\mathrm{L}}{4\mathrm{L}}$

$=\frac{(0.91\times {10}^{11})\times 3.142\times {(2\times {10}^{-3})}^2\times (2.376\times {10}^{-3})}{4\times 1.8}$

= $3.8 \times {10}^2\mathrm{N}$