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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Thermal Properties of Matter

NTA Abhyas 2020 · Physics · Thermal Properties of Matter

A partition wall has two layers of different materials A and B in...

A partition wall has two layers of different materials AA and BB in contact with each other. They have the same thickness but the thermal conductivity of layer AA is twice that of BB. At steady state, if the temperature difference across the layer BB is 50K50K, then the corresponding temperature difference across the layer AA isA partition wall has two layers of different materials A and B in contact with each other. They have the same thickness

Options

  1. A50 K\text{50 K}
  2. B12.5 K\text{12}\text{.5 K}
  3. C25 K\text{25 K}
  4. D60 K\text{60 K}

Correct answer

C. 25 K\text{25 K}

Step-by-step solution

A partition wall has two layers of different materials A and B in contact with each other. They have the same thickness

Let TT be the junction temperature

Here, KA=2KB,T−TB=50K{K}_{A} = 2{K}_{B}, T - {T}_{B} = 50K

At the steady state HA=HB{H}_{A} = {H}_{B}

⇒KAA(TA−T)L2=KBA(T−TB)L2⇒ \dfrac{{K}_{A} A\left({T}_{A} - T\right)}{\dfrac{L}{2}} = \dfrac{{K}_{B} A\left(T - {T}_{B}\right)}{\dfrac{L}{2}}

⇒2KB(TA−T)=KB(T−TB)⇒ 2{K}_{B} \left({T}_{A} -T\right) = {K}_{B} \left(T - {T}_{B}\right)

⇒TA−T=T−TB2=502=25K⇒ {T}_{A} - T = \dfrac{T - {T}_{B}}{2} = \dfrac{50}{2} =25K
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