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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
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B
3.7×102N
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C
4.7×102N
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D
5.7×102N
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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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