As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is 120 cm² for each plate. The equivalent thermal conductivity of the compound plate is ( 1 + 5/α ) K, then the value of α will be ______________.

Solution

Physics sir · 4 steps

The plates are in series, so add their thermal resistances and match the total to one plate of 6.5 cm.

1. Series plates. Heat flows from 100°C through A and then through B, so the same heat current passes through both. The thermal resistances L/KA add, like resistors in series.

   (L₁ + L₂)/K_(eq)A = L₁/(K₁ A) + L₂/(K₂ A)

2. Put in numbers. The area is the same for both plates and cancels. With lengths in cm, 6.5/K_(eq) = 4/K + 2.5/2K = 5.25/K.
3. Solve for Keq. K_(eq) = 6.5/5.25K = 26/21K = (1 + 5/21)K.
4. So. Comparing with (1 + 5/α)K gives α = 21.

The common slip: 13. Averaging the conductivities weighted by thickness, (4K + 2.5 × 2K)/6.5, is the parallel rule. It gives (1 + 5/13)K and so 13, not 21.

The idea to keep. Slabs placed one after another in the heat path are in series. Add L/KA for each slab, and the equivalent conductivity is total length divided by Σ L/K.

Answer from the NTA final answer key. Solution by PrepWiser.