JEE Physics: Temperature at the Junction of Two Rods (Different Materials) Explained ⚙️
❓ Question Two cylindrical rods A and B made of different materials are joined end-to-end in a straight line. The ratios are: L A L B = 1 2 , r A r B = 2 , K A K B = 1 2 . The free ends of rods A and B are held at 400 K and 200 K respectively. Find the interface temperature (temperature at the junction) when steady-state equilibrium is established. 🖼️ Question Image ✍️ Short Solution In steady state the heat current I I (rate of heat flow) is the same through both rods. Thermal resistance of a rod of length L L , cross-sectional area A A , and thermal conductivity K K is: R = L K A . For series rods, temperature drops are proportional to their resistances. If T i T_i is the interface temperature then: 400 − T i R A = T i − 200 R B . So compute R A R_A and R B R_B up to a common factor using given ratios. Step 1 — Express areas via radii Area A = π r 2 A=\pi r^2 . Given r A / r B = 2 r_A/r_B=2 → A A / A B = ( r A / r B )...