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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 $2.0 \;mm$ ? Co-efficient of Itnear expansion of brass $=2.0 \times 10^{-5}\; K ^{-1} ;$ Young's modulus of brass $=0.91 \times 10^{11} \;Pa$
$3.8 \times 10^{2}\; N$
$9.4 \times 10^{4}\; N$
$2.7 \times 10^{1}\; N$
$8.4 \times 10^{2}\; N$
Solution
Initial temperature, $T_{1}=27^{\circ} C$
Length of the brass wire at $T_{1}, l=1.8\; m$
Final temperature, $T_{2}=-39^{\circ} C$
Diameter of the wire, $d=2.0 mm =2 \times 10^{-3} m$
Tension developed in the wire $=F$
Coefficient of linear expansion of brass, $\alpha=2.0 \times 10^{-5} K ^{-1}$
Young's modulus of brass, $Y=0.91 \times 10^{11} Pa$ p://wuw tiwariacademy.com/
Young's modulus is given by the relation:
$Y=\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$
$\Delta L=\frac{F \times L}{A \times Y}$
$F=$ Tension developed in the wire
$A=$ Area of cross-section of the wire.
$\Delta L=$ Change in the length, given by the relation:
$\Delta L=\alpha L\left(T_{2}-T_{1}\right)$
$\alpha L\left(T_{2}-T_{1}\right)=\frac{F L}{\pi\left(\frac{d}{2}\right)^{2} \times Y}$
$F=\alpha\left(T_{2}-T_{1}\right) \pi Y\left(\frac{d}{2}\right)^{2}$
$F=2 \times 10^{-5} \times(-39-27) \times 3.14 \times 0.91 \times 10^{11} \times\left(\frac{2 \times 10^{-3}}{2}\right)^{2}$
$=-3.8 \times 10^{2} N$
(The negative sign indicates that the tension is directed inward.)
Hence, the tension developed in the wire is $3.8 \times 10^{2}\; N$
