Set up the system

Two identical metal wires are connected in series.

Let each wire have:

• length $L$
• cross-sectional area $A$
• thermal conductivities $K_1$ and $K_2$

Since they are in series, heat flows through one and then the other.

Use thermal resistance

Thermal resistance of a wire is $R = \frac{L}{KA}$

So for the two wires: $R_1 = \frac{L}{K_1 A}, \qquad R_2 = \frac{L}{K_2 A}$

In series, thermal resistances add: $R_{\text{eq}} = R_1 + R_2 = \frac{L}{K_1 A} + \frac{L}{K_2 A}$

$R_{\text{eq}} = \frac{L}{A}\left(\frac{1}{K_1} + \frac{1}{K_2}\right)$

Define equivalent conductivity

The total length of the combination is $2L$.

If the effective thermal conductivity is $K_{\text{eq}}$, then $R_{\text{eq}} = \frac{2L}{K_{\text{eq}} A}$

Equate the two expressions

$\frac{2L}{K_{\text{eq}} A} = \frac{L}{A}\left(\frac{1}{K_1} + \frac{1}{K_2}\right)$

Cancel $\frac{L}{A}$ from both sides: $\frac{2}{K_{\text{eq}}} = \frac{1}{K_1} + \frac{1}{K_2}$

$\frac{2}{K_{\text{eq}}} = \frac{K_1 + K_2}{K_1 K_2}$

Therefore, $K_{\text{eq}} = \frac{2K_1K_2}{K_1 + K_2}$

Match with options

$K_{\text{eq}} = \frac{2K_1K_2}{K_1 + K_2}$

This matches Option A.

Verification with stored answer

Stored correct answer: A

Derived answer: A

Hence, the derived answer agrees with the stored answer.