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ExamsJEE AdvancedPhysics

A composite block is assembled from slabs A, B, C, D and E having different thermal conductivities (expressed as multiples of a constant K) and different sizes (expressed in terms of a length), with all slabs of the same width, as shown in the figure. Heat Q flows only from the left face to the right face through the arrangement (A first, then B, C, D in parallel, then E). In the steady state, which statement is correct?

  1. Heat flow through A and E are equal, and heat flow through C equals the sum of heat flows through B and D.
  2. Heat flow through slab E is the maximum of all slabs.
  3. The temperature difference across slab E is the smallest.
  4. Heat flow through slab A is smaller than that through slab E.

Correct answer: Heat flow through A and E are equal, and heat flow through C equals the sum of heat flows through B and D.

Solution

This is the JEE Advanced 2011 composite-slab problem. A is the single entry slab and E the single exit slab, so the full heat current Q passes through both: heat flow through A = heat flow through E. The intermediate slabs B, C, D are in parallel and split that same current. With the given geometry C carries half the total while B and D together carry the other half, so heat through C = heat through B + heat through D.

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