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 , the change in length of the wire PQ is
- (A)0.78 mm✓ Correct
- (B)0.90 mm
- (C)1.56 mm
- (D)2.34 mm
Answer: (A)
Solution
Since thermal conductivity of PQ is twice that of RS, the thermal resistance ratio is .
In steady state, heat flow is same through both wires, so:
This gives: , so C
Temperature varies linearly along PQ: where is in meters.
Expansion of element dx:
Total expansion: mm
In steady state, heat flow is same through both wires, so:
This gives: , so C
Temperature varies linearly along PQ: where is in meters.
Expansion of element dx:
Total expansion: mm
JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Single correct option