Physics Thermometry, Thermal Expansion and Calorimetry JEE (Advanced) / IIT-JEE Problems (Previous Years) MCQ (Single Correct)

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 1m 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

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Text Solution

Verified by Experts
The correct answer is:
A

d λ = dx α 1 ( θ – 10)

Δ λ =

⇒ θ = 10 + 130x

Δ λ =

Δ λ = 130 α 1

Δ λ = 130 × 1.2 × 10 –5 × = 78 × 10 –5 = 0.78mm

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