Three conductors of same length having thermal conductivity k₁, k₂ and k₃ are connected as shown in figure.
Area of cross sections of 1ˢᵗ and 2ⁿᵈ conductor are same and for 3ʳᵈ conductor it is double of the 1ˢᵗ conductor. The temperatures are given in the figure. In steady state condition, the value of θ is _________ ° C. (Given : k₁=60 Js⁻¹ m⁻¹ K⁻¹, k₂=120 Js⁻¹ m⁻¹ K⁻¹, k₃=135 Js⁻¹ m⁻¹ K⁻¹ )

Solution

Physics sir · 4 steps

In steady state, the heat entering the junction through rods 1 and 2 must all leave through rod 3.

1. Read the circuit. Rods 1 and 2 both run from 100°C to θ, so they are in parallel. Rod 3 runs from θ to 0°C, in series with that pair. Each rod has the same length L. Rods 1 and 2 have area A, and rod 3 has area 2A.
2. Heat current balance. Heat current through a rod is H = (kA Δ T)/L. Nothing collects at the junction in steady state, so the current in equals the current out.

   (k₁ + k₂)A(100 - θ)/L = (k₃ (2A)(θ - 0))/L

3. Put in numbers. k₁ + k₂ = 180 and 2k₃ = 270, so 180(100 - θ) = 270 θ. That gives 18000 = 450 θ.
4. So. θ = 18000/450 = 40 °C.

The common slip: 57.1. Forgetting that rod 3 has double the area gives 180(100 - θ) = 135 θ, so θ ≈ 57.1 instead of 40.

The idea to keep. Treat conduction like current in a circuit. Rods between the same two temperatures are in parallel, and at any junction in steady state the heat current in equals the heat current out.

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