Two sheets of thickness d and 3d, are touching each other. The temperature just outside the thinner sheet is `T_1` and on the side of the thicker sheet is `T_3`. The interface temperature is `T_2. T_1, T_2 and T_3` are in arithmetic progression. The ratio of thermal conductivity of thinner sheet to thicker sheet is .

Options

A. `1:3`
B. `3:1`
C. `2:3`
D. `1:9`

Answer

To solve the problem, we need to find the ratio of the thermal conductivity of the thinner sheet to that of the thicker sheet given that the temperatures \( T_1, T_2, T_3 \) are in arithmetic progression and the thicknesses of the sheets are \( d \) and \( 3d \).

### Step-by-step Solution:

1. **Understanding the Problem**:
- We have two sheets: one with thickness \( d \) (thinner sheet) and another with thickness \( 3d \) (thicker sheet).
- The temperatures at the outer surfaces are \( T_1 \) (thinner sheet), \( T_2 \) (interface), and \( T_3 \) (thicker sheet).
- The temperatures \( T_1, T_2, T_3 \) are in arithmetic progression.

2. **Using the Arithmetic Progression**:
- Since \( T_1, T_2, T_3 \) are in arithmetic progression, we can express this as:
\[ T_1 - T_2 = T_2 - T_3 \]
- Rearranging gives:
\[ T_1 + T_3 = 2T_2 \tag{1} \]

3. **Applying the Heat Flow Equation**:
- The rate of heat flow \( \dot{Q} \) through the sheets can be expressed as:
\[ \dot{Q} = \frac{T_1 - T_2}{R_1} = \frac{T_2 - T_3}{R_2} \]
- Where \( R_1 \) and \( R_2 \) are the thermal resistances of the thinner and thicker sheets, respectively.

4. **Calculating Thermal Resistance**:
- The thermal resistance \( R \) for a sheet is given by:
\[ R = \frac{L}{KA} \]
- For the thinner sheet (thickness \( d \)):
\[ R_1 = \frac{d}{K_1 A} \]
- For the thicker sheet (thickness \( 3d \)):
\[ R_2 = \frac{3d}{K_2 A} \]

5. **Setting Up the Heat Flow Equations**:
- Substituting the resistances into the heat flow equations gives:
\[ \dot{Q} = \frac{T_1 - T_2}{\frac{d}{K_1 A}} = \frac{K_1 A (T_1 - T_2)}{d} \]
\[ \dot{Q} = \frac{T_2 - T_3}{\frac{3d}{K_2 A}} = \frac{K_2 A (T_2 - T_3)}{3d} \]

6. **Equating the Two Expressions for Heat Flow**:
- Since \( \dot{Q} \) is constant, we can set the two expressions equal to each other:
\[ \frac{K_1 (T_1 - T_2)}{d} = \frac{K_2 (T_2 - T_3)}{3d} \]
- Canceling \( d \) from both sides gives:
\[ K_1 (T_1 - T_2) = \frac{K_2 (T_2 - T_3)}{3} \]

7. **Substituting from Equation (1)**:
- From equation (1), we know \( T_1 - T_2 = T_2 - T_3 \). Let \( T_1 - T_2 = x \), then \( T_2 - T_3 = x \).
- Substitute \( x \) into the equation:
\[ K_1 x = \frac{K_2 x}{3} \]
- Dividing both sides by \( x \) (assuming \( x \neq 0 \)):
\[ K_1 = \frac{K_2}{3} \]

8. **Finding the Ratio**:
- Rearranging gives:
\[ \frac{K_1}{K_2} = \frac{1}{3} \]
- Thus, the ratio of thermal conductivity of the thinner sheet to the thicker sheet is:
\[ K_1 : K_2 = 1 : 3 \]

### Final Answer:
The ratio of thermal conductivity of the thinner sheet to the thicker sheet is \( 1 : 3 \).