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← JEE Advanced 2018 · Paper 1
PhysicsNumerical+3Question 14 of 54

JEE Advanced 2018 · Paper 1, Physics question 14

Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T1=300T_1 = 300 K and T2=100T_2 = 100 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K1K_1 and K2K_2 respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K, then K1/K2K_1/K_2 = ______.

Two cylinders of equal length L joined end to end between heat baths T₁ (left) and T₂ (right): a thinner one of conductivity K₁ next to T₁ and a thicker one of conductivity K₂ next to T₂, surrounded by insulating material.
Answer and worked solution

Official answer: 4

Equal heat current in series:
K1A1(300−200)L=K2A2(200−100)L  ⟹  K1A1=K2A2.\frac{K_1A_1(300-200)}L=\frac{K_2A_2(200-100)}L\implies K_1A_1=K_2A_2.

Areas. The larger radius is twice the smaller, so A2=4A1A_2=4A_1:
K1K2=A2A1=4.\frac{K_1}{K_2}=\frac{A_2}{A_1}=4.

Answer: 4

Topics:Conduction