Question 9: Properties of Matter

Question

A brass wire of length 2 m and radius 1 mm at $27^{\circ} \mathrm{C}$ is held taut between two rigid supports. Initially it was cooled to a temperature of $-43^{\circ} \mathrm{C}$ creating a tension $T$ in the wire. The temperature to which the wire has to be cooled in order to increase the tension in it to $1.4 T$, is $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$.

Accepted Answer

A metal rod of original length $L$ that is free to move. If we heat this rod by a temperature difference $\Delta T$, it will naturally expand.
The change in length ( $\Delta \mathrm{L}$ ) is given by the law of linear thermal expansion :
$$ \Delta \mathrm{L}=\mathrm{L} \cdot \alpha \cdot \Delta \mathrm{~T} $$
Where $\alpha=$ Coefficient of Linear Expansion
When a wire fixed at both ends is cooled, it contracts. Since the supports are rigid, they prevent this contraction, creating a tension in the wire.
$$ 	ext { Strain } \varepsilon=\frac{\Delta \mathrm{L}}{\mathrm{~L}}=\alpha \Delta \mathrm{T} $$
According to Hooke's Law, Stress is proportional to Strain within the elastic limit. The constant of proportionality is Young's Modulus (Y).
$$ Y=\frac{	ext { Stress }}{	ext { Strain }} $$
$$\Rightarrow $$ Stress $(\sigma)=\mathrm{Y} 	imes$ Strain
$$ \sigma=\mathrm{Y} \cdot(\alpha \Delta \mathrm{~T}) $$
Stress is defined as Force (F) per unit Area (A).
$$ 	ext { Stress }=\frac{\mathrm{F}}{\mathrm{~A}}=\mathrm{Y} \cdot(\alpha \Delta \mathrm{~T}) $$
$$\Rightarrow $$ $\mathrm{F}=(\mathrm{Y} \alpha \Delta \mathrm{T}) 	imes \mathrm{A}$
$$ \Rightarrow \mathrm{F} \propto \Delta \mathrm{~T} \Rightarrow \frac{\mathrm{~F}_1}{\mathrm{~F}_2}=\frac{\Delta \mathrm{T}_1}{\Delta \mathrm{~T}_2} $$
Initial Temperature $=\mathrm{t}_0=27^{\circ} \mathrm{C}$
Final Temperature $=\mathrm{t}_1=-43^{\circ} \mathrm{C}$
$$ \Delta \mathrm{T}_1=27-(-43)=70^{\circ} \mathrm{C} $$
Initial tension in the wire is, $\mathrm{F}_1=\mathrm{T}$
Final tension in the wire is $\mathrm{F}_2=1.4 \mathrm{~T}$
Let the new final temperature be $t_2$.
$$ \Delta \mathrm{T}_2=27-\mathrm{t}_2 $$
$$\Rightarrow $$ $\frac{\mathrm{F}_2}{\mathrm{~F}_1}=\frac{1.4 \mathrm{~T}}{\mathrm{~T}}=\frac{27-\mathrm{t}_2}{70}$
$$\Rightarrow $$ $1.4=\frac{27-t_2}{70}$
$$\Rightarrow $$ $27-t_2=1.4 	imes 70$
$$\Rightarrow $$ $27-t_2=98$
$$\Rightarrow $$$
t_2=27-98 $
$$\Rightarrow $$$ t_2=-71^{\circ} \mathrm{C}
$

Answer

-71

Answer

-80

Answer

-65

Answer

-86

Properties of Matter

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