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PhysicsHeat & ThermoConductionHard2 minPYQ_2016
PhysicsHardsingle choice

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 at10oC.Now the end P is maintained at10oC,while the end S is heated and maintained at400oC.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 is1.2×10-5 K-1,the change in length of the wire PQ is

Question diagram: The ends Q and R of two thin wires, PQ and RS, are soldered

Options:

Answer:
A
Solution:


Heat flow from P to Q
dQdt=2KAT-101
Heat flow from Q to S
dQdt=KA (400-T)1
At steady state heat flow is same in whole combination
2KAT-101=KA400-T
2T-20=400-T
3T=420
T=140o

Temp of junction is140oC
Temp at a distance x from end P
IsTx=130x+10o
Change in length dx is dy
dy=αdxTx-10
0ydy=01αdx130x+10-10
y=αx22×13001
y=1.2×10-5×65
y=78.0×10-5m=0.78 mm

Stream:JEE_ADVSubject:PhysicsTopic:Heat & ThermoSubtopic:Conduction
2mℹ️ Source: PYQ_2016

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