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JEE Advanced 2016 · Paper 2, Physics question 4

The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1 m at 10 °C. Now the end P is maintained at 10 °C, while the end S is heated and maintained at 400 °C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2×10−5 K−11.2\times10^{-5}\ \mathrm K^{-1}, the change in length of the wire PQ is

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Answer and worked solution

Official answer: A

Junction temperature. The wires are in series with equal lengths and the same heat current:
2K(T−10)=K(400−T)  ⟹  T=140 ∘C.2K(T-10)=K(400-T)\implies T=140\ ^\circ\mathrm C.

Expansion of PQ. The temperature along PQ rises linearly from 10 °C to 140 °C, so its average rise is 65 °C:
ΔL=αL ΔTavg=1.2×10−5×1×65=0.78 mm.\Delta L=\alpha L\,\Delta T_{avg}=1.2\times10^{-5}\times1\times65=0.78\text{ mm}.

(A)

Topics:ConductionTemperature and thermal expansion
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