Two slabs A and B of different materials but of the same thicknesss are joined end to end to form a composite slab. The thermal conductivities of A and B are `K_(1)` and `K_(2)` respectively. A steady temperature difference of `12^(@)`C is maintained across the composite slab. If `K_(1) = K_(2)/2`, the temperature difference across slabs A is

Options

A. `4^(@)`C
B. `6^(@)`C
C. `8^(@)`C
D. `10^(@)`C

Explanation

To solve the problem, we will use the concept of heat conduction through the composite slab and the relationship between thermal conductivity, temperature difference, and thickness.

### Step-by-Step Solution:

1. **Understand the Setup**: We have two slabs, A and B, with thermal conductivities \( K_1 \) and \( K_2 \) respectively. The total temperature difference across the composite slab is \( 12^\circ C \).

2. **Given Relation**: We know that \( K_1 = \frac{K_2}{2} \).

3. **Heat Flow Equation**: The rate of heat flow through both slabs must be equal since they are in series. We can express this as:
\[ \frac{12 - T_A}{K_1} = \frac{T_A - 0}{K_2} \]
where \( T_A \) is the temperature at the interface between slabs A and B.

4. **Substituting \( K_1 \)**: Substitute \( K_1 \) in terms of \( K_2 \):
\[ \frac{12 - T_A}{\frac{K_2}{2}} = \frac{T_A - 0}{K_2} \]

5. **Simplify the Equation**: This simplifies to:
\[ \frac{2(12 - T_A)}{K_2} = \frac{T_A}{K_2} \]
By multiplying both sides by \( K_2 \) (assuming \( K_2 \neq 0 \)):
\[ 2(12 - T_A) = T_A \]

6. **Rearranging the Equation**: Rearranging gives:
\[ 24 - 2T_A = T_A \]
\[ 24 = 3T_A \]
\[ T_A = 8^\circ C \]

7. **Finding Temperature Difference Across Slab A**: The temperature difference across slab A is:
\[ \Delta T_A = 12 - T_A = 12 - 8 = 4^\circ C \]

### Final Answer:
The temperature difference across slab A is \( 4^\circ C \).

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