Two cylindrical rods A and B are joined end-to-end. Their ratios are L A / L B L_A/L_B L A / L B = 1/2, r A / r B r_A/r_B r A / r B = 2 and K A / K B K_A/K_B K A / K B = 1/2. The free ends are maintained at 400 K and 200 K respectively. Find the steady interface temperature in kelvin.
Physics · Thermodyanmics · heat transfer in solid and liquid
JEE Mains 2025 7 April Evening Shift

Question

Two cylindrical rods A and B are joined end-to-end. Their ratios are L A / L B L_A/L_B L A / L B = 1/2, r A / r B r_A/r_B r A / r B = 2 and K A / K B K_A/K_B K A / K B = 1/2. The free ends are maintained at 400 K and 200 K respectively. Find the steady interface temperature in kelvin.

Physics · Thermodyanmics · heat transfer in solid and liquid

JEE Mains 2025 7 April Evening Shift

G1 Physics · Accepted Answer

Thermal resistance of a rod is R t h R_th R t ​ h = L/(KA), where A = πr². At steady state, heat currents through A and B are equal: (400-T)/ R A R_A R A ​ = (T-200)/ R B . R_B. R B ​ . Now, R A / R B R_A/R_B R A ​ / R B ​ = ( L A / L B ) ( K B / K A ) ( A B / A A ) . L_A/L_B)(K_B/K_A)(A_B/A_A). L A ​ / L B ​ ) ( K B ​ / K A ​ ) ( A B ​ / A A ​ ) . Given, L A / L B L_A/L_B L A ​ / L B ​ = 1/2, K B / K A K_B/K_A K B ​ / K A ​ = 2, and A B / A A A_B/A_A A B ​ / A A ​ = ( r B / r A ) 2 r_B/r_A)² r…