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COMEDK2023PhysicsThermal Properties of Matter

Two slabs are of the thicknesses d₁ and d₂ . Their thermal conductivities are K₁ and K₂ , respectively. They are in series. The free ends of the combination of these two slabs are kept at temperatures ₁ and ₂ . Assume ₁> ₂ . The temperature of their common junction is

Options

  1. AK₁ ₁+K₂ ₂ ₁+ ₂
  2. BK₁ ₁ d₁+K₂ ₂ d₂ K₁ d₂+K₂ d₁
  3. CK₁ ₁ d₂+K₂ ₂ d₁ K₁ d₂+K₂ d₁
  4. DK₁ ₁+K₂ ₂ K₁+K₂

Correct answer

C. K₁ ₁ d₂+K₂ ₂ d₁ K₁ d₂+K₂ d₁

Step-by-step solution

For first slab, Heat current, H₁= K₁ ( ₁- ) A d₁ For second slab, Heat current, H₂= K₂ ( - ₂ ) A d₂ As slabs are in series, H₁=H₂ array ll & K₁ ( ₁- ) A d₁ = K₂ ( - ₂ ) A d₂ & = K₁ ₁ d₂+K₂ ₂ d₁ K₂ d₁+K₁ d₂ array

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