Three rods of identical cross-section and lengths 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 \(100^\circ\mathrm{C}\) and the other at \(0^\circ\mathrm{C}\) (see figure). If the joints of the rod are at \(70^\circ\mathrm{C}\) and \(20^\circ\mathrm{C}\) 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 :
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In steady state the same heat current flows through every rod, so \(K\,\Delta T\) is the same for all three (equal length and area).
Temperature drops: rod 1: \(100-70=30^\circ\mathrm{C}\); rod 2: \(70-20=50^\circ\mathrm{C}\); rod 3: \(20-0=20^\circ\mathrm{C}\). Steady state, no side loss: heat current \(H=\dfrac{KA\,\Delta T}{L}\) is the same in all rods, with the same \(A\) and \(L\), so \(K_1(30)=K_2(50)=K_3(20)\). Hence \(K_1:K_2:K_3=\dfrac{1}{30}:\dfrac{1}{50}:\dfrac{1}{20}=10:6:15\). \(K_1:K_3=10:15=2:3\) and \(K_2:K_3=6:15=2:5\). Check other options: \(K_1:K_2=10:6=5:3\) (not 5:2), so (3) is wrong; \(K_2\lt K_3\) but \(K_1\gt K_2\), so (2) and (4) are wrong. Answer: (1) \(K_1 : K_3 = 2 : 3,\ K_2 : K_3 = 2 : 5\)
