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Thermal Properties of Matter

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Question : 10 of 22
Marks: +1, -0
A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod arefree to expand (Coefficient of linear expansion of brass =2.0×10−5 K−1,= 2.0 \times 10^{-5} \,\mathrm{K}^{-1},steel =1.2×10−5 K−1).= 1.2 \times 10^{-5} \,\mathrm{K}^{-1}).
Solution:  
For brass rod : α=2.0×10−5 K−1\alpha=2.0 \times 10^{-5} \,\mathrm{K}^{-1}
l1=50 cm;l_{1}=50 \,\mathrm{cm} ;
ΔT=250−40=210∘ C\Delta T=250-40=210^{\circ} \,\mathrm{C}
The length of the brass rod at 250°C is given by l2=l1(1+αΔT)l_2 = l_1 (1+ \alpha \Delta T)
=50(1+2.0×10−5×210)= 50(1 + 2.0 \times 10^{-5} \times 210)
=50.21 cm= 50.21 \,\mathrm{cm}
For steel rod: α′=1.2×10−5 K−1\alpha'=1.2 \times 10^{-5} \,\mathrm{K}^{-1}
l1′=50 cm;l_{1}'=50 \,\mathrm{cm} ;
ΔT′=250−40=210∘ C\Delta T'=250-40=210^{\circ} \,\mathrm{C}
The length of the steel rod at 250°C is given by
l2′=l1′(1+α′ΔT′)l_{2}'=l_{1}'(1+\alpha' \Delta T')
=50(1+1.2×10−5×210)=50 (1+1.2 \times 10^{-5} \times 210)
=50.126 cm=50.126 \,\mathrm{cm}
Therefore, the length of the combined rod at 250°C
=l2+l2′=l_{2}+l_{2}'
=50.21+50.126=50.21+50.126
=100.336 cm=100.336 \,\mathrm{cm}
As the length of the combined rod at 40∘ C=50+50=10040^{\circ} \,\mathrm{C} = 50 + 50 = 100 cm
The change in length of the combined rod at 250°C
=100.336−100.0=0.336 cm= 100.336 - 100.0 = 0.336 \,\mathrm{cm}
No thermal stress is developed at the junction since the rod freely expand.
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