A cylinder of radius R made of a material of thermal conductivity `K_1` is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity `K_2`. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is

Options

A. `K_1 + K_2`
B. `(K_1K_2)/(K_1+K_2)`
C. `(K_1 + 3K_2)/4`
D. `(3K_1 + K_2)/4`

Explanation

To find the effective thermal conductivity of the given system, we will analyze the problem step by step.

### Step 1: Understand the Configuration
We have two cylindrical components:
1. A central cylinder with radius \( R \) and thermal conductivity \( K_1 \).
2. A cylindrical shell surrounding the central cylinder with inner radius \( R \) and outer radius \( 2R \) and thermal conductivity \( K_2 \).

### Step 2: Identify the Heat Transfer Mechanism
Since the system is in a steady state and there is no heat loss across the cylindrical surfaces, we can treat the heat transfer through the two cylindrical components as being in parallel.

### Step 3: Calculate the Resistance of Each Component
The thermal resistance \( R \) for a cylindrical object can be calculated using the formula:
\[ R = \frac{L}{K \cdot A} \]
Where:
- \( L \) is the length of the cylinder,
- \( K \) is the thermal conductivity,
- \( A \) is the cross-sectional area.

#### For the inner cylinder (radius \( R \)):
- The cross-sectional area \( A_1 = \pi R^2 \).
- Therefore, the thermal resistance \( R_1 \) is given by:
\[ R_1 = \frac{L}{K_1 \cdot \pi R^2} \]

#### For the outer cylindrical shell (inner radius \( R \) and outer radius \( 2R \)):
- The cross-sectional area \( A_2 \) is the difference between the outer and inner areas:
\[ A_2 = \pi (2R)^2 - \pi R^2 = \pi (4R^2 - R^2) = 3\pi R^2 \]
- Therefore, the thermal resistance \( R_2 \) is given by:
\[ R_2 = \frac{L}{K_2 \cdot 3\pi R^2} \]

### Step 4: Calculate the Total Resistance in Parallel
Since the resistances are in parallel, the total thermal resistance \( R_{total} \) can be calculated using:
\[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting the values of \( R_1 \) and \( R_2 \):
\[ \frac{1}{R_{total}} = \frac{K_1 \cdot \pi R^2}{L} + \frac{K_2 \cdot 3\pi R^2}{L} \]
Factoring out common terms:
\[ \frac{1}{R_{total}} = \frac{\pi R^2}{L} \left( K_1 + 3K_2 \right) \]

### Step 5: Find the Effective Thermal Conductivity
The effective thermal conductivity \( K_{eff} \) can be related to the total resistance:
\[ R_{total} = \frac{L}{K_{eff} \cdot A_{total}} \]
Where \( A_{total} = \pi (2R)^2 = 4\pi R^2 \).

Thus:
\[ \frac{1}{R_{total}} = \frac{K_{eff} \cdot 4\pi R^2}{L} \]

Equating the two expressions for \( \frac{1}{R_{total}} \):
\[ \frac{\pi R^2}{L} (K_1 + 3K_2) = \frac{K_{eff} \cdot 4\pi R^2}{L} \]

Cancelling \( \frac{\pi R^2}{L} \) from both sides:
\[ K_1 + 3K_2 = 4K_{eff} \]

### Step 6: Solve for \( K_{eff} \)
Rearranging gives:
\[ K_{eff} = \frac{K_1 + 3K_2}{4} \]

### Final Answer
The effective thermal conductivity of the system is:
\[ K_{eff} = \frac{K_1 + 3K_2}{4} \]

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