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KCET2010PhysicsThermal Properties of Matter

Two slabs are of the thicknesses d₁ and d₂ . Their thermal conductivities are K ₁ and K ₂ 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. AK ₁ ₁+ K ₂ ₂ ₁+ ₂
  2. BK ₁ ₁ ~d ₁+ K ₂ ₂ ~d ₂ ~K ₁ ~d ₂+ K ₂ ~d ₁
  3. CK ₁ ₁ ~d ₂+ K ₂ ₂ ~d ₁ ~K ₁ ~d ₂+ K ₂ ~d ₁
  4. DK ₁ ₁+ K ₂ ₂ ~K ₁+ K ₂

Correct answer

C. K ₁ ₁ ~d ₂+ K ₂ ₂ ~d ₁ ~K ₁ ~d ₂+ K ₂ ~d ₁

Step-by-step solution

For first slab Heat current, H ₁= K ₁ ( ₁- ) A d ₁ For second slab, Heat current, H ₂= K ₂ ( - ₂ ) A d ₂ As slabs are in series aligned H ₁ &= H ₂ & & K ₁ ( ₁- ) A d ₁ &= K ₂ ( - ₂ ) A d ₂ & &= K ₁ ₁ ~d ₂+ K ₂ ₂ ~d ₁ ~K ₂ ~d ₁+ K ₁ ~d ₂ aligned

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