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 $=2.0 \times 10^{-5} /{ }^{\circ} C$ and Young's modulus of brass $=0.91 \times 10^{11} Pa$ )

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

Answer 1

Answer: (C)

Thermal stress that is produced in an elastic wire is $Y \alpha \theta$ per unit area, where $Y$ is Young's modulus, $\alpha$ coefficient of linear expansion and $\theta$ change of temperature. Thus, tension developed in the wire is
$T =$ Thermal stress $ \times $ area of cross section
$=Y \alpha \times(\Delta T)\left[\pi r^{2}\right] N $
$=0.91 \times 10^{11} \times 2 \times 10^{-5} \times[27-(-33)]\left[\pi \times 1 \times 10^{-6}\right] N $
$=0.91 \times 2 \times 6 \times 3.14 \times 10 \,N$
$=34.4 \times 10 N =0.34 \,kN$