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Concepts Of Physics MCQ Edition [Volume 2]PhysicsHeat Transfer

A composite slab is formed by pasting together two plates of thicknesses L₁ and L₂ and thermal conductivities K₁ and K₂ . The slabs possess equal cross-sectional area. Determine the equivalent conductivity of the composite slab.

Options

  1. AK₁ L₂ + K₂ L₁ L₁ + L₂
  2. BK₁ K₂ (L₁ + L₂) L₁ K₁ + L₂ K₂
  3. CK₁ L₁ + K₂ L₂ L₁ + L₂
  4. DK₁ K₂ (L₁ + L₂) L₁ K₂ + L₂ K₁

Correct answer

D. K₁ K₂ (L₁ + L₂) L₁ K₂ + L₂ K₁

Step-by-step solution

For a composite slab connected in series, the equivalent thermal resistance is the sum of the individual thermal resistances. R_ eq = R₁ + R₂ Using the formula for thermal resistance R = L KA , where A is the cross-sectional area: L₁ + L₂ K_ eq A = L₁ K₁ A + L₂ K₂ A L₁ + L₂ K_ eq = L₁ K₂ + L₂ K₁ K₁ K₂ K_ eq = K₁ K₂ (L₁ + L₂) L₁ K₂ + L₂ K₁

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