Question

A brass wire $2 \, m$ long at $27 \,{}^\circ C$ is held taut with negligible tension between two rigid supports. If the wire is cooled to a temperature of $ \, - \, 33 \,{}^\circ C$ , then the tension developed in the wire, its diameter being $2 \, mm$ , will be (coefficient of linear expansion of brass = $2.0 \, \times \, 10-5 \,{}^\circ C^{- 1} \, $ and Young's modulus of brass = $0.91 \, \times \, 10^{11} \, Pa$ )

• A. $3400N$
• B. $34kN$
• C. $0.34kN$
• D. $6800N$

Answer 1

Answer: (C)

Thermal stress that is produced in an elastic wire is Y $ \, α \, \theta $ per unit area, wher Y is Young's modulus, $\alpha $ coefficient of linear expansion and $\theta $ change of temperature.
Thus, tension developed i the wire is
T = Thermal stress x area of cross section
$= \text{Y} \alpha \times \left(\Delta \text{T}\right) \left[\pi \left(\text{r}\right)^{2}\right] \text{N}$
$= \text{0.91} \times 1 0^{1 1} \times 2 \times 1 0^{- 5} \times \left[2 7 - \left(- 3 3\right)\right] \left[\pi \times 1 \times 1 0^{- 6}\right] \text{N}$
$= \text{0.91} \times 2 \times 6 \times \text{3.14} \times 1 0 N$
$= \text{34.4} \times 1 0 N = \text{0.34} kN$