A steel wire of `2.0 mm^2` cross- section is held straight(but under no tension) by attaching it firmly to two points a distance `1.50 m` apart at `30^@ C`. If the temperature now decreases to `-10^@ C` and if the two points remain fixed, what will be the tension in the wire ? For steel, `Y = 20.0000 MPa`.

Explanation

KEY IDEA
If free to do so, the wire would contract but since we have tied its ends, it will not contract and maintain the original length .
Till now we have seen that when the length of a wire is changed it produces strain and hence stress . This situation is different as strain will be produced because of wire maintaining its length . At a lower temperature, the wire would have a unstrained length smaller than the original length . However, since its ends are tied, it will maintain its length but develop strain . Or in other words it has longer length than it would have at this temperature of not tied at its ends.
Calculations : If free to do so , the wire would contract a distance `DeltaL` as it cooled , where
`DeltaL=alphaL_0DeltaT`
`=(1.2xx10^(-5)//""^@C)(1.5m)(40^@C)`
`= 7.2 xx10^(-4)m`
But the ends are fixed , As a result , forces at the ends must, in effect , stretch the wire this same length , `DeltaL` . Therefore from `Y=(F//A)//(DeltaL//L)` , we have tension
`F=(YA DeltaL)/(L)`
`=((2xx10^(11)N//m^(2))(2xx10^(-6)m^2)(7.2xx10^(-4)m))/(1.5m)`
Note : Strictly , we should have substituted `(1.5-7.2xx10^(-4))m` for L in the expression for tension . However , the error incurred in not doing so, is negligible.