Question

# A brass wire 1.8 m long at $$27^o C$$ is held taut with negligible tension between two rigid supports Diameter of the wire is 2mm, its coefficient of linear expansion,$$\alpha_{Brass}=2\times 10^{-5}\,^o\,C^{-1}$$ and its Young's modulus, $$Y_{Brass}=9\times 10^{10}N\,\, m^{-2}$$,If the wire is cooled to temperature $$-39^o C$$, tension developed in the wire is :

A
2.7×102N
B
3.7×102N
C
4.7×102N
D
5.7×102N

Solution

## The correct option is B $$3.7 \times 10^{2}N$$Let $$F$$ be tension developed in the wire.$$\therefore Y=\dfrac{F/A}{\triangle L/L}$$As $$\triangle L=L\alpha \triangle T$$$$\therefore Y=\dfrac{F}{A\alpha \triangle T}$$ or $$F=YA\alpha \triangle T$$Here , $$\alpha=2.0\times 10^{-5} \,^o C^{-1},Y=9\times 10^{10}N m^{-2}$$$$A=\pi(1\times 10^{-3})^2\,\,M^2$$  $$\triangle T=66^oC$$$$\therefore F=9 \times 10^{10}\times \pi(1\times 10^{-3})^2\times 2.0\times 10^{-5}\times 66$$$$\Rightarrow3.7 \times 10^2 N$$Physics

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