Highly selective Backlog Qs for JEE MainPhysicsThermal Properties of Matter
Two metal slabs of same cross-sectional area have thicknesses d₁ and ; and thermal conductivities K₁ and K₂ respectively, are connected in series. The free ends of the two slabs are kept at temperatures T₁ and T₂ (T₁>T₂ ) . The temperature T of their common junction is
Options
- AK₁ T₁ d₂+K₂ T₂ d₁ K₁ d₂+K₂ d₁
- BK₁ T₁+K₂ T₂ K₁+K₂
- CK₁ T₁+K₂ T₂ T₁+T₂
- DK₁ T₁ d₁+K₂ T₂ d₂ K₁ d₂+K₂ d₁
Correct answer
A. K₁ T₁ d₂+K₂ T₂ d₁ K₁ d₂+K₂ d₁
Step-by-step solution
Heat current, Q ₁= K₁ (T₁-T ) A d₁ For second slab, Heat current, Q ₂= K₂ (T-T₂ ) A d₂ As slabs are in series same heat current flows through them Q ₁= Q ₂ K₁ (T₁-T ) A d₁ = K₂ (T-T₂ ) A d₂ Therefore, temperature T of their common junction is T= K₁ T₁ d₂+K₂ T₂ d₁ K₂ d₁+K₁ d₂