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ExamsJEE AdvancedPhysics

A composite slab has four layers of equal thickness made of two materials whose conductivities alternate as k, 2k, k, 2k. In steady state the face and interface temperatures from one side to the other are 20 degC, 10 degC, theta, -5 degC, -10 degC. Find theta. (Heat current is the same through every layer.)

  1. 5 degC
  2. 0 degC
  3. -2.5 degC
  4. -10 degC

Correct answer: 0 degC

Solution

Matching k*deltaT across layers, the drop across the third layer (conductivity k) must equal the first layer's drop pattern, giving theta = 0 degC.

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