Question

A brass wire $1.8$ m long at ${27}^{0}C$ is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of $-{39}^{o}C$, what is the tension developed in the wire, if its diameter is $2.0mm$? Co-efficient of linear expansion of brass $=2.0\times {10}^{-5}{K}^{-1}$; Young's modulus of brass $=0.91\times {10}^{11}$ Pa

Answer 1

Initial temperature, $T_1={27}^{o}C$
Length of the brass wire at $T_1,l=1.8m$
Final temperature, $T_2=39C$
Diameter of the wire, $d=2.0mm=2\times$${10}^{-3}m$
Tension developed in the wire $=F$
Coefficient of linear expansion of brass, $=2.0\times {10}^{-5}$$K^{-1}$
Youngs modulus of brass, $Y=0.91\times {10}^{11}$ Pa
Youngs modulus is given by the relation:
Y= Stress / Strain

$Y=\frac{F/A}{\Delta L/L}$
$\Delta L=F\times L/(A\times Y)$ ......(i)
Where,
$F=$ Tension developed in the wire
$A=$ Area of cross-section of the wire.
$\Delta L=$ Change in the length, given by the relation:
$\Delta L=$$\alpha L(T_2$ -$T_1$) .....(ii)
Equating equations (i) and (ii), we get:

$\alpha L(T_2-T_1)=\frac{FL}{\pi (d/2{)}^{2}Y}$
$F=\alpha (T_2-T_1)Y\pi (d/2{)}^2$

$F=2\times {10}^{-5}\times (-39-27)\times 3.14\times 0.91\times {10}^{11}\times (2\times {10}^{-3}/2{)}^2$

$F=-3.8\times {10}^{2}N$

(The negative sign indicates that the tension is directed inward.)
Hence, the tension developed in the wire is 3.8 $\times {10}^2$ N.