A wall has two layers A and B made of two different materials thermal conductivities `K_A` and `K_B` (`K_A = 3K_B`). The thickness of both the layers is same. The temperature across the wall is `20^@C` in thermal equilibrium. Then

Options

A. temperature difference across A is `12^@`C
B. rate of heat transfer across A is more than that across B
C. rate of heat transfer is same for both
D. temperature difference across A is `15^@C `

Answer

To solve the problem, we need to analyze the heat transfer through the two layers of the wall. Here are the steps to arrive at the solution:

### Step 1: Understand the Setup
We have two layers A and B with thermal conductivities \( K_A \) and \( K_B \) respectively. We know that \( K_A = 3K_B \). The thickness of both layers is the same, and the temperature difference across the wall is \( 20^\circ C \).

### Step 2: Define the Temperatures
Let:
- \( T_1 \) be the temperature on one side of the wall.
- \( T_2 \) be the temperature on the other side of the wall.
- \( T \) be the temperature at the interface between the two layers.

Given that \( T_1 - T_2 = 20^\circ C \), we can express \( T_2 \) as:
\[ T_2 = T_1 - 20 \]

### Step 3: Write the Heat Transfer Equations
Since the layers are in series, the heat transfer (or heat current) through both layers must be equal. The heat transfer rate \( H \) can be expressed as:
\[ H = \frac{T_1 - T}{R_A} = \frac{T - T_2}{R_B} \]
where \( R_A \) and \( R_B \) are the thermal resistances of layers A and B respectively.

### Step 4: Calculate the Thermal Resistances
The thermal resistance for each layer can be defined as:
\[ R_A = \frac{L}{K_A \cdot A} \]
\[ R_B = \frac{L}{K_B \cdot A} \]
where \( L \) is the thickness of the layers and \( A \) is the cross-sectional area (which we can assume to be the same for both layers).

### Step 5: Substitute the Resistances into the Heat Transfer Equation
Substituting the resistances into the heat transfer equations gives:
\[ H = \frac{T_1 - T}{\frac{L}{K_A \cdot A}} = \frac{K_A \cdot A (T_1 - T)}{L} \]
\[ H = \frac{T - T_2}{\frac{L}{K_B \cdot A}} = \frac{K_B \cdot A (T - T_2)}{L} \]

### Step 6: Set the Heat Transfer Equations Equal
Setting the two equations for \( H \) equal to each other:
\[ K_A (T_1 - T) = K_B (T - T_2) \]

### Step 7: Substitute \( K_A \) and \( T_2 \)
Substituting \( K_A = 3K_B \) and \( T_2 = T_1 - 20 \):
\[ 3K_B (T_1 - T) = K_B (T - (T_1 - 20)) \]
Canceling \( K_B \) (assuming \( K_B \neq 0 \)):
\[ 3(T_1 - T) = T - T_1 + 20 \]

### Step 8: Rearranging the Equation
Rearranging gives:
\[ 3T_1 - 3T = T - T_1 + 20 \]
\[ 4T_1 - 4T = 20 \]
\[ T_1 - T = 5 \]

### Step 9: Conclusion
Thus, the temperature difference across layer A is:
\[ T_1 - T = 5^\circ C \]