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Step-by-Step Solution
Step 1: Identify the Physical Situation
We have two identical metal wires with different thermal conductivities, $K_1$ and $K_2$. Since the wires are identical, both have the same length (say $L$) and cross-sectional area (say $A$). They are connected in series for heat conduction.
Step 2: Write the Formula for Heat Flow in Series
When heat flows in series through two rods or wires of identical dimensions, the total temperature drop is distributed across each conductor. Let the heat current (rate of heat flow) be $Q$ (in J/s). For a rod of thermal conductivity $K$, length $L$, and cross-sectional area $A$, subjected to a temperature difference $ \Delta T$, the rate of heat flow is given by:
$ Q = \frac{KA\,\Delta T}{L} \,.$
In the series arrangement, the same heat current $Q$ flows through both wires.
Step 3: Express the Equivalent Thermal Conductivity
Let $K_\text{eff}$ be the effective thermal conductivity of the combination. Since the two wires are joined end to end in series, their effective length becomes $2L$ (because each wire has length $L$). The total area remains $A$. Hence the overall rate of heat flow by conduction for the series combination is:
$ Q = \frac{K_\text{eff}\,A\,\Delta T}{2L} \,.$
Step 4: Equate the Heat Flow for Each Segment
Each individual wire carries the same current $Q$. For the first wire with conductivity $K_1$:
$ Q = \frac{K_1 \, A \, \Delta T_1}{L} \,,
$
and for the second wire with conductivity $K_2$:
$ Q = \frac{K_2 \, A \, \Delta T_2}{L} \,,
$
where $\Delta T_1$ and $\Delta T_2$ are the temperature differences across the first and second wires respectively, and $\Delta T = \Delta T_1 + \Delta T_2$ is the total temperature difference across both.
Step 5: Derive the Final Relation
Using the fact that $Q$ is the same through both wires and combining the equations leads to a relationship for $K_\text{eff}$. After algebraic simplification (since the wires are identical in length and area), we obtain:
$ K_\text{eff} = \frac{2 K_1 K_2}{K_1 + K_2} \,.
$
This matches the given correct answer.
Final Answer
$ \displaystyle K_\text{eff} = \frac{2 K_1 K_2}{K_1 + K_2}\,.
$
