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Question No. 2
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−1, the change in length of the wire PQ is
(A) 0.78 mm
(B) 0.90 mm
(C) 1.56 mm
(D) 2.34 mm

Solution:

Let’s denote:

  • Wire PQ: length L1=1 m at 10°C, thermal conductivity k1=2k (since it is twice that of RS), coefficient of linear expansion α1=1.2×10−5 K−1.

  • Wire RS: length L2=1 m at 10°C, thermal conductivity k2=k.

The system is soldered at Q and R, and thermally insulated. End P is at 10°C, end S is at 400°C.

Step 1: Find the temperature at the junction (Q/R)
Since the wires are in series, the heat current through both is the same.
Let Tj be the junction temperature.

Heat current through PQ:

H1=k1A(Tj−10)L1

Heat current through RS:

H2=k2A(400−Tj)L2

Set H1=H2:

k1(Tj−10)L1=k2(400−Tj)L2

Given k1=2k, k2=k, L1=L2=1 m:

2k(Tj−10)=k(400−Tj)

Divide both sides by k:

2(Tj−10)=400−Tj2Tj−20=400−Tj2Tj+Tj=400+203Tj=420Tj=140∘C

Step 2: Determine the average temperature of wire PQ
The temperature varies linearly along PQ from 10°C at P to 140°C at Q.
So, the average temperature of PQ is:

Tavg, PQ=10+1402=75∘C

Step 3: Calculate the change in length of PQ due to thermal expansion
The initial temperature is 10°C, and the average temperature is 75°C.
So, the change in temperature is:

ΔT=75−10=65∘C

The change in length is given by:

ΔL=α1L1ΔT=(1.2×10−5)×1×65ΔL=1.2×10−5×65=78×10−5 m=0.78×10−3 m=0.78 mm

Final Answer:

0.78 mm

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