Three rods of identical cross-section and length are made of three different materials of thermal conductivity $K_1$, $K_2$ and $K_3$, respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at $100\ ^\circ\text{C}$ and the other at $0\ ^\circ\text{C}$. If the joints of the rod are at $70\ ^\circ\text{C}$ and $20\ ^\circ\text{C}$ in steady state and there is no loss of energy from the surface of the rod, the correct relationship between $K_1$, $K_2$ and $K_3$ is:
Answer: (A) $K_1 : K_3 = 2 : 3,\ K_2 : K_3 = 2 : 5$
In steady state the same heat current flows through each rod. With equal lengths and areas, $K\,\Delta T$ is the same for all three:
$$K_1(100 - 70) = K_2(70 - 20) = K_3(20 - 0)$$
$$30K_1 = 50K_2 = 20K_3$$
So $K_1 : K_2 : K_3 = \dfrac1{30} : \dfrac1{50} : \dfrac1{20} = 10 : 6 : 15$.
Hence $K_1 : K_3 = 2 : 3$ and $K_2 : K_3 = 2 : 5$.
Solution by Sreeraj P, M.Sc Physics