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Thermal Properties of Matter

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Question : 9 of 22
Marks: +1, -0
A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39°C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass =2.0×10−5K−1= 2.0 \times 10^{-5} \mathrm{K}^{-1}; Young’s modulus of brass =0.91×1011Pa.= 0.91 \times 10^{11} \mathrm{Pa}.
Solution:  
Here l1=1.8 m, t1=27∘C,l_{1}=1.8 \ \text{m},\ t_{1}=27^{\circ}\mathrm{C},
t2=−39∘Ct_{2}=-39^{\circ}\mathrm{C}
∴t=t2−t1\therefore t=t_{2}-t_{1}
=39−27=−66∘C=39-27=-66^{\circ}\mathrm{C}
l2=l_{2}= length at t2∘Ct_{2}^{\circ}\mathrm{C}
For brass, α=2×10−5K−1\alpha=2 \times 10^{-5} \mathrm{K}^{-1}
∴Y=0.91×1011Pa\therefore Y=0.91\times 10^{11} \mathrm{Pa}
diameter of wire, d=2.0 mm=2.0×10−3 md=2.0 \ \text{mm}=2.0\times 10^{-3}\ \text{m}
If AA be the area of cross-section of the wire, then A=πd24=π(10−3)2A= \frac{\pi d^{2}}{4}=\pi (10^{-3})^{2}
If F be the tension developed in the wire, then using the relation
Y=FAΔll1,Y= \frac{\frac{F}{A}}{\frac{\Delta l}{l_{1}}}, we get Δl=FlAY\Delta l = \frac{F l}{A Y}
AlsoΔl=lαΔt ∴FlAY=αlΔT\Delta l = l \alpha \Delta t \ \therefore \frac{F l}{A Y} = \alpha l \Delta T
or F=αΔTAYF = \alpha \Delta T A Y
=α(T2−T1)πr2Y= \alpha (T_{2} - T_{1}) \pi r^{2} Y
=2×10−5(−66)×227(10−3)2×0.91×1011= 2 \times 10^{-5} (-66) \times \frac{22}{7} (10^{-3})^{2} \times 0.91 \times 10^{11}
=−3.8×102 N= -3.8 \times 10^{2} \ \mathrm{N}
Negative sign indicates that the force is inwards due to the contraction of the wire.
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