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

Initial temperature,` T_(1) = 27^(@)C`
Final temperature, `T_(2) = –39^(@)C`
Diameter of the wire, d = 2.0 mm `= 2 × 10^(–3) m`
Tension developed in the wire = F
Coefficient of linear expansion of brass, `α= 2.0 × 10^(–5) K^(–1)`
Young’s modulus of brass, `Y = 0.91 × 10^(11)` Pa
Young’s modulus is given by the relation:
`Y=("Stress")/("Strain")=((F)/(A))/((DeltaL)/(L))`
`DeltaL=(FxxL)/(AxxY)`...(i)
Where,
F = Tension developed in the wire
A = Area of cross-section of the wire.
ΔL = Change in the length, given by the relation:
`DeltaL=aL(T_(2)-T_(1))`...(ii)
Equating equations (i) and (ii), we get:
`aL(T_(2)-T_(1))=(FL)/(pi((d)/(2))^(2)xxY)`
`F=a(T_(2)-T_(1))piY((d)/(2))^(2)`
`F=2xx10^(-5)xx(-39-27)xx3.14xx0.91xx10^(11)xx((2xx10^(-3))/(2))^(2)`
`= -3.8xx10^(2)N`
(The negative sign indicates that the tension is directed inward.) Hence, the tension developed in the wire is `3.8 ×10^(2) `N.