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
(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 heat transfer through the inner cylinder and the outer cylindrical shell step by step.

### Step 1: Identify the Components
We have:
- An inner cylinder of radius \( R \) and thermal conductivity \( K_1 \).
- An outer cylindrical shell with inner radius \( R \) and outer radius \( 2R \), made of a material with thermal conductivity \( K_2 \).

### Step 2: Write the Heat Flow Equations
In steady state, the heat flow through both the inner cylinder and the outer shell must be equal.

The heat flow \( \Delta Q \) through a cylindrical surface can be expressed as:
\[ \Delta Q = \frac{K \cdot A \cdot \Delta T}{L} \]
where:
- \( K \) is the thermal conductivity,
- \( A \) is the cross-sectional area,
- \( \Delta T \) is the temperature difference,
- \( L \) is the length of the cylinder.

### Step 3: Calculate the Heat Flow for the Inner Cylinder
For the inner cylinder:
- Area \( A_1 = \pi R^2 \)
- Heat flow through the inner cylinder:
\[ \Delta Q_1 = \frac{K_1 \cdot \pi R^2 \cdot \Delta T}{L} \]

### Step 4: Calculate the Heat Flow for the Outer Cylinder
For the outer cylindrical shell:
- The area \( A_2 \) is the difference between the outer and inner surfaces:
\[ A_2 = \pi ( (2R)^2 - R^2 ) = \pi (4R^2 - R^2) = \pi (3R^2) \]
- Heat flow through the outer cylinder:
\[ \Delta Q_2 = \frac{K_2 \cdot \pi (3R^2) \cdot \Delta T}{L} \]

### Step 5: Set Up the Equation for Steady State
Since the heat flow through both components is equal:
\[ \Delta Q_1 = \Delta Q_2 \]
Substituting the expressions we derived:
\[ \frac{K_1 \cdot \pi R^2 \cdot \Delta T}{L} = \frac{K_2 \cdot \pi (3R^2) \cdot \Delta T}{L} \]

### Step 6: Simplify the Equation
Cancel out common terms (\( \pi, R^2, \Delta T, L \)):
\[ K_1 = 3K_2 \]

### Step 7: Find the Effective Thermal Conductivity
Now, we need to express the effective thermal conductivity \( K \) of the entire system. The effective thermal conductivity can be found by combining the contributions from both cylinders:
\[ K = \frac{K_1 + 3K_2}{4} \]

### Conclusion
Thus, the effective thermal conductivity of the system is:
\[ \boxed{\frac{K_1 + 3K_2}{4}} \]