Heat Conduction

As explained in NCERT Class 11, Chapter 11 (Thermal Properties of Matter), heat conduction in the steady state follows $\frac{Q}{t}=\frac{kA\Delta T}{L}$, where the rate of heat flow is the same through both joined rods in series. Because the same heat current passes through identical area and length, the temperature drop across each rod satisfies $\Delta T=\frac{(Q/t)L}{kA}$, so $\Delta T\propto 1/k$. The rod with the smaller thermal conductivity $k$ therefore sustains the larger temperature difference. The option naming the higher-conductivity rod reverses this inverse relation. The option of equal drops would only hold if both conductivities were equal, which they are not. The option depending on which rod is heated first confuses transient behaviour with the steady state, where the heat flow has already become uniform and time-independent. This situation is exactly analogous to two resistors in series carrying the same electric current, where the larger voltage drop appears across the larger resistance. A plausibility check: a poor conductor resists heat flow strongly, so a bigger temperature gradient is needed to drive the same heat current through it, just as a thick blanket sustains a large temperature difference across itself, confirming that the larger drop occurs across the rod of lower thermal conductivity.