Two solid cylindrical rods 1 and 2 of different materials are joined end-to-end in series. The free end of rod 1 is maintained at 100^ C and the free end of rod 2 is kept at 0^ C . In the steady state, the temperature of the interface between the two rods is 80^ C . If the ratio of their lengths is L₁ L₂ = 2 and the ratio of their radii is r₁ r₂ = 2 , then the ratio of their thermal conductivities K₁ K₂ is ________.

Step-by-step solution

Let the thermal resistances of rod 1 and rod 2 be R₁ and R₂ respectively. In steady state, the rate of heat flow through both rods is the same: 100 - 80 R₁ = 80 - 0 R₂ 20 R₁ = 80 R₂ R₂ R₁ = 4 The thermal resistance of a cylindrical rod is given by R = L K r^2 . Therefore, the ratio of their resistances is: R₂ R₁ = ( L₂ L₁ ) ( K₁ K₂ ) ( r₁ r₂ )^2 Given L₁ L₂ = 2 L₂ L₁ = 1 2 and r₁ r₂ = 2 . Substituting these values: 4 = ( 1 2 ) ( K₁ K₂ ) (2)^2 4 = ( 1 2 ) ( K₁ K₂ ) (4) 4 = 2 ( K₁ K₂ ) K₁ K₂ = 2 Answer: 2