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`.

Options

A. `2.4 xx 10^(2) N`
B. `3.8 xx 10^(2) N`
C. `1.8 xx 10^(2) N`
D. `4.8 xx 10^(2) N`

Answer

B

Explanation

As the wire is not free to contract, the thermal stress is developed at the wire.
The change in temeperature,
`Delta T = 27^(@)C -(-39^(@)C) = 66^(@)C`
Let `Delta L` be the change in length of the wire.
`Delta L = alphaL Delta T = (2xx10^(-5))xx1.8xx66 = 2.376xx10^(-3)m`
As, `Y = (FL)/(A Delta L) = (FL)/((piD^(2)//4)Delta L) = (4FL)/(pi D^(2)Delta L)`
`F` (tension in the wire) = `(Y pi D^(2)Delta L)/(4L)`
`=((0.91xx10^(11))xx3.142xx(2xx10^(-3))^(2)xx(2.376xx10^(-3)))/(4xx1.8)`
`=3.8 xx 10^(2)N`.