Question

A brass wire $1.8m$ long at $27^{\circ }C$ is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of $-39^{\circ }C$, what is the tension developed in the wire, if its diameter is $2.0mm$ ? Coefficient of linear expansion of brass $=2.0\times 10^{-5}/{}^{\circ }C$, Young's modulus of brass $=0.91\times 10^{11}Pa$

• A. $2.4\times 10^{2}N$
• B. $3.8\times 10^{2}N$
• C. $1.8\times 10^{2}N$
• D. $4.8\times 10^{2}N$

Answer 1

Answer: (B)

As the wire is not free to contract, the thermal stress is developed at the wire.
The change in temperature,
$\Delta T=27^{\circ }C-\left(-39^{\circ }C\right)=66^{\circ }C$
Let $\Delta L$ be the change in length of the wire.
$\Delta L=\alpha L\Delta T=\left(2\times 10^{-5}\right)\times 1.8\times 66$
$=2.376\times 10^{-3}m$
As $Y=\frac{FL}{A\Delta L}=\frac{FL}{\left(\pi D^2/4\right)\Delta L}=\frac{4FL}{\pi D^2\Delta L}$
F(tension in the wire) $=\frac{Y\pi D^2\Delta L}{4L}$
$=\frac{\left(0.91\times 10^{11}\right)\times 3.142\times {\left(2\times 10^{-3}\right)}^2\times \left(2.376\times 10^{-3}\right)}{4\times 1.8}$
$=3.8\times 10^{2}N$