Aslender rod of length L, diameter d (L>>d) and thermal conductivity k₁ is joined with another rod of identical dimensions, but of thermal conductivity k₂, to from a composite cylindrical rod of length 2L. The heat transfer in radial direction and contact resistance are negligible. The effective thermal conductivity of the composite rod is

The correct option is 2k₁k₂/(k₁ + k₂).

To find the effective thermal conductivity of the composite rod, we can analyze the arrangement of the two rods as a series thermal circuit.

Let the cross-sectional area of each rod be $A$ (since they have identical dimensions with diameter $d$, where $A=\frac{\pi d^2}{4}$).

The thermal resistance of a conductor is given by the formula:

$R=\frac{L}{kA}$

where:
• $L$ is the length of the conductor,
• $k$ is the thermal conductivity,
• $A$ is the cross-sectional area.

For the first rod of thermal conductivity $k_1$ and length $L$, the thermal resistance $R_1$ is:

$R_1=\frac{L}{k_{1}A}$

For the second rod of thermal conductivity $k_2$ and length $L$, the thermal resistance $R_2$ is:

$R_2=\frac{L}{k_{2}A}$

Since the two rods are joined end-to-end (in series) to form a composite rod of total length $2L$, and radial heat transfer is negligible, the equivalent thermal resistance $R_{\text{eq}}$ is the sum of their individual thermal resistances:

$R_{\text{eq}}=R_1+R_2$

Substitute the values of $R_1$ and $R_2$ into the equation:

$R_{\text{eq}}=\frac{L}{k_{1}A}+\frac{L}{k_{2}A}=\frac{L}{A}(\frac{1}{k_1}+\frac{1}{k_2})=\frac{L}{A}\left(\frac{k_1+k_2}{k_1{k}_2}\right)$

Now, let $k_{\text{eff}}$ be the effective thermal conductivity of the composite rod of length $2L$ and cross-sectional area $A$. The equivalent resistance can also be expressed as:

$R_{\text{eq}}=\frac{2L}{k_{\text{eff}}A}$

Equating the two expressions for $R_{\text{eq}}$:

$\frac{2L}{k_{\text{eff}}A}=\frac{L}{A}\left(\frac{k_1+k_2}{k_1{k}_2}\right)$

Cancel $\frac{L}{A}$ from both sides of the equation:

$\frac{2}{k_{\text{eff}}}=\frac{k_1+k_2}{k_1{k}_2}$

Rearranging to solve for $k_{\text{eff}}$ gives:

$k_{\text{eff}}=\frac{2k_1{k}_2}{k_1+k_2}$