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)

Let us analyze the problem step-by-step to understand why this is the correct answer.

1. Understanding the Arrangement:
We have two rods of identical dimensions. Each rod has:
- Length = L
- Diameter = d (which means they have the same cross-sectional area, A)

These two rods are joined end-to-end (in series) to form a composite cylindrical rod.
- The first rod has thermal conductivity $k_1$.
- The second rod has thermal conductivity $k_2$.
- The total length of the composite rod is $2L$, and its cross-sectional area remains $A$.
- Heat transfer in the radial direction and thermal contact resistance at the joint are negligible, meaning heat flows axially through both rods in series.

2. Concept of Thermal Resistance:
For axial conduction, the thermal resistance $R$ of a conductor is given by the formula:

$R=\frac{\mathrm{Length}}{k\cdot A}$

Since the two rods are connected in series, their individual thermal resistances ($R_1$ and $R_2$) add up to give the total equivalent thermal resistance ($R_{\mathrm{eq}}$) of the composite rod:

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

3. Substituting the Expressions:
Let $k_{\mathrm{eq}}$ be the effective thermal conductivity of the composite rod of length $2L$.
The individual thermal resistances are:

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

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

The total thermal resistance of the composite rod is:

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

4. Solving for Effective Thermal Conductivity ($k_{\mathrm{eq}}$):
Equating the resistances:

$\frac{2L}{k_{\mathrm{eq}}A}=\frac{L}{k_{1}A}+\frac{L}{k_{2}A}$

We can cancel the common terms $L$ and $A$ from both sides:

$\frac{2}{k_{\mathrm{eq}}}=\frac{1}{k_1}+\frac{1}{k_2}$

Combine the fractions on the right-hand side using a common denominator:

$\frac{2}{k_{\mathrm{eq}}}=\frac{k_1+k_2}{k_1{k}_2}$

Taking the reciprocal to solve for $k_{\mathrm{eq}}$:

$k_{\mathrm{eq}}=\frac{2k_1{k}_2}{k_1+k_2}$

Thus, the effective thermal conductivity of the composite rod is indeed the harmonic mean of the individual conductivities, which matches the correct answer: 2k₁k₂/(k₁+k₂).