Two cylindrical rods A and B are joined in series. Rod A has length L and radius r , while rod B has length 2L and radius 2r . The free end of rod A is maintained at 100^ C and the free end of rod B is maintained at 0^ C . The rods are thermally insulated from the outside. In the steady state, the temperature of the junction is found to be 80^ C . The ratio of the thermal conductivities of the two rods, K_A / K_B , i

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In steady state, the rate of heat flow through both rods is equal. H_A = H_B K_A A_A T_A L_A = K_B A_B T_B L_B Substituting the given values: A_A = r^2 , A_B = (2r)^2 = 4 r^2 T_A = 100^ C - 80^ C = 20^ C T_B = 80^ C - 0^ C = 80^ C K_A ( r^2) (20) L = K_B (4 r^2) (80) 2L 20 K_A = 160 K_B K_A K_B = 160 20 = 8 If the radius is not squared, the ratio obtained is 4 . If the lengths are swapped, the ratio obtained is 32 . Answer: 8