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(B) In a steady state,the rate of heat flow $(H)$ through both rods connected in series is the same.$H_A = H_B$$\frac{K_A A \Delta T_A}{L_A} = \frac{K

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$0.5$

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(B) In a steady state,the rate of heat flow $(H)$ through both rods connected in series is the same.$H_A = H_B$$\frac{K_A A \Delta T_A}{L_A} = \frac{K_B A \Delta T_B}{L_B}$Given: $L_A = 30 	ext{ cm}$,$L_B = 70 	ext{ cm}$,$\Delta T_A = 100 - 70 = 30 	ext{ } ^\circ	ext{C}$,$\Delta T_B = 70 - 35 = 35 	ext{ } ^\circ	ext{C}$.Assuming the cross-sectional area $A$ is the same for both rods:$\frac{K_A \cdot 30}{30} = \frac{K_B \cdot 35}{70}$$K_A = K_B \cdot 0.5$$\frac{K_A}{K_B} = 0.5$

Two different rods $A$ and $B$ are kept as shown in the figure. The ratio of thermal conductivities of $A$ and $B$ is

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