JEE Challenger
More from Thermal Properties of Matter
PhyMedAdv 2016 • Paper 2

Thermal Expansion and Junction Temperature of Composite Wires

The ends QQ and RR of two slender wires, PQPQ and RSRS, are soldered together. At 10 ∘C10 \text{ }^\circ\text{C}, each wire has an initial length of 1 m1 \text{ m}. The free end PP is maintained at 10 ∘C10 \text{ }^\circ\text{C} while end SS is kept at 400 ∘C400 \text{ }^\circ\text{C}. The lateral surface of the system is thermally insulated from the surroundings. If the thermal conductivity of wire PQPQ is twice that of wire RSRS, and the linear thermal expansion coefficient of PQPQ is 1.2×10−5 K−11.2 \times 10^{-5} \text{ K}^{-1}, determine the expansion in length of wire PQPQ.

Options

A

0.78 mm0.78 \text{ mm}

Correct
B

0.90 mm0.90 \text{ mm}

C

1.56 mm1.56 \text{ mm}

D

2.34 mm2.34 \text{ mm}

Topics & Concepts

Thermal Properties of MatterThermal ExpansionHeat Transfer

Step-by-Step Solution

To determine the expansion in length of the wire PQPQ, we proceed through two main steps: finding the junction temperature under steady-state heat conduction, and then calculating the expansion due to the resulting linear temperature gradient.

Step 1: Find the Junction Temperature (TJT_J)

Under steady-state thermal conditions, the rate of heat flow (HH) through wire PQPQ must equal the rate of heat flow through wire RSRS because the lateral surfaces of the wires are thermally insulated.

Let TJT_J be the temperature at the junction Q−RQ-R. The ends are maintained at TP=10 ∘CT_P = 10\text{ }^\circ\text{C} and TS=400 ∘CT_S = 400\text{ }^\circ\text{C}.

The heat flow equation for each wire is: H=KPQA(TJ−TP)L=KRSA(TS−TJ)LH = \frac{K_{PQ} A (T_J - T_P)}{L} = \frac{K_{RS} A (T_S - T_J)}{L}

Given that KPQ=2KRSK_{PQ} = 2 K_{RS} and both wires have the same cross-sectional area AA and initial length L=1 mL = 1\text{ m}: 2KRS(TJ−10)=KRS(400−TJ)2 K_{RS} (T_J - 10) = K_{RS} (400 - T_J)

Canceling KRSK_{RS} from both sides gives: 2(TJ−10)=400−TJ2(T_J - 10) = 400 - T_J 2TJ−20=400−TJ2T_J - 20 = 400 - T_J 3TJ=420  ⟹  TJ=140 ∘C3T_J = 420 \implies T_J = 140\text{ }^\circ\text{C}

Step 2: Calculate the Expansion of Wire PQPQ

Since the heat flow is steady and the lateral surface is insulated, the temperature varies linearly along the length of wire PQPQ from TP=10 ∘CT_P = 10\text{ }^\circ\text{C} to TJ=140 ∘CT_J = 140\text{ }^\circ\text{C}.

The average temperature of wire PQPQ is: Tavg=TP+TJ2=10 ∘C+140 ∘C2=75 ∘CT_{\text{avg}} = \frac{T_P + T_J}{2} = \frac{10\text{ }^\circ\text{C} + 140\text{ }^\circ\text{C}}{2} = 75\text{ }^\circ\text{C}

The average temperature increase relative to the initial temperature T0=10 ∘CT_0 = 10\text{ }^\circ\text{C} is: ΔTavg=Tavg−T0=75 ∘C−10 ∘C=65 K\Delta T_{\text{avg}} = T_{\text{avg}} - T_0 = 75\text{ }^\circ\text{C} - 10\text{ }^\circ\text{C} = 65\text{ K}

Using the formula for linear thermal expansion: ΔL=L0⋅α⋅ΔTavg\Delta L = L_0 \cdot \alpha \cdot \Delta T_{\text{avg}}

Substituting the given values (L0=1 mL_0 = 1\text{ m} and α=1.2×10−5 K−1\alpha = 1.2 \times 10^{-5}\text{ K}^{-1}): ΔL=1 m×(1.2×10−5 K−1)×65 K\Delta L = 1\text{ m} \times (1.2 \times 10^{-5}\text{ K}^{-1}) \times 65\text{ K} ΔL=78×10−5 m=0.78×10−3 m=0.78 mm\Delta L = 78 \times 10^{-5}\text{ m} = 0.78 \times 10^{-3}\text{ m} = 0.78\text{ mm}

Thus, the expansion in length of wire PQPQ is 0.78 mm0.78\text{ mm}, which corresponds to Option A.

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