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JEE Advanced 2016Paper 2
Exam
JEE Advanced
Year
2016
Paper
Paper 2
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Advanced 2016 · Paper 2 · Physics · Thermal Properties of Matter

The ends Q and R of two thin wires, PQ and RS, are soldered...

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 10oC.{10}^{o}C . Now the end P is maintained at 10oC,{10}^{o}C, while the end S is heated and maintained at 400oC.{400}^{o}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 1.2×10−5K−1,1.2 \times {10}^{- 5}{K}^{- 1}, the change in length of the wire PQ isThe ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a lengthThe ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length

Options

  1. A0.78 mm
  2. B0.90 mm
  3. C1.56 mm
  4. D2.34 mm

Correct answer

A. 0.78 mm

Step-by-step solution

The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length
Heat flow from P to Q
dQdt=2KA(T−10)1\dfrac{dQ}{dt} = \dfrac{2KA\left(T - 10\right)}{1}
Heat flow from Q to S
dQdt=KA(400−T)1\dfrac{dQ}{dt} = \dfrac{KA ( 400 - T )}{1}
At steady state heat flow is same in whole combination
2KA(T−10)1=KA(400−T)\dfrac{2KA\left(T - 10\right)}{1} = KA\left(400 - T\right)
2T−20=400−T2T - 20 = 400 - T
3T=4203T = 420
T=140oT = {140}^{o}
The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length
Temp of junction is 140oC{140}^{o}C
Temp at a distance x from end P
Is Tx=(130x+10o){T}_{x} = \left(130x + {10}^{o}\right)
Change in length dx is dy
dy=\alphadx(Tx−10)dy = \alphadx\left({T}_{x} - 10\right)
∫0∆ydy=∫01\alphadx(130x+10−10)\mathop{∫}_{0}^{∆ y}dy = \mathop{∫}_{0}^{1}\alphadx\left(130x + 10 - 10\right)
∆y=[αx22×130]01∆ y = {\left[\dfrac{\alpha{x}^{2}}{2} \times 130\right]}_{0}^{1}
∆y=1.2×10−5×65∆ y = 1.2 \times {10}^{- 5} \times 65
∆y=78.0×10−5m=0.78mm∆ y = 78.0 \times {10}^{- 5}m = 0.78mm
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