Question

A brass wire $1.8\, m$ 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 $2.0\, mm$ ? 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 $Y=\frac{F L}{A \Delta L}=\frac{F L}{\left(\pi D^{2} / 4\right) \Delta L}=\frac{4 F L}{\pi D^{2} \Delta L}$
F(tension in the wire) $=\frac{Y \pi D^{2} \Delta L}{4 L}$
$=\frac{\left(0.91 \times 10^{11}\right) \times 3.142 \times\left(2 \times 10^{-3}\right)^{2} \times\left(2.376 \times 10^{-3}\right)}{4 \times 1.8}$
$=3.8 \times 10^{2} N$