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. `3 : 9`

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two sheets:
- Thinner sheet with thickness \(d\) and thermal conductivity \(k_1\).
- Thicker sheet with thickness \(3d\) and thermal conductivity \(k_2\).
- The temperatures are given as follows:
- Temperature just outside the thinner sheet: \(T_1\)
- Interface temperature: \(T_2\)
- Temperature just outside the thicker sheet: \(T_3\)
- It is given that \(T_1\), \(T_2\), and \(T_3\) are in arithmetic progression.

2. **Using the Concept of Heat Current**:
- The heat current \(I_H\) through both sheets must be equal since they are in contact:
\[ I_H = \frac{T_2 - T_1}{R_H1} = \frac{T_3 - T_2}{R_H2} \]
- Where \(R_H1\) and \(R_H2\) are the thermal resistances of the thinner and thicker sheets, respectively.

3. **Calculating Thermal Resistance**:
- The thermal resistance \(R_H\) is given by:
\[ R_H = \frac{\Delta x}{kA} \]
- For the thinner sheet:
\[ R_H1 = \frac{d}{k_1 A} \]
- For the thicker sheet:
\[ R_H2 = \frac{3d}{k_2 A} \]

4. **Setting Up the Equation**:
- Substitute the resistances into the heat current equation:
\[ \frac{T_2 - T_1}{\frac{d}{k_1 A}} = \frac{T_3 - T_2}{\frac{3d}{k_2 A}} \]
- Simplifying this gives:
\[ (T_2 - T_1) \cdot \frac{k_1}{d} = (T_3 - T_2) \cdot \frac{k_2}{3d} \]
- Canceling \(d\) and \(A\) from both sides, we have:
\[ k_1 (T_2 - T_1) = \frac{k_2}{3} (T_3 - T_2) \]

5. **Expressing the Temperatures**:
- Since \(T_1\), \(T_2\), and \(T_3\) are in arithmetic progression, we can express:
\[ T_2 = \frac{T_1 + T_3}{2} \]
- Therefore, we can find:
\[ T_3 - T_2 = T_3 - \frac{T_1 + T_3}{2} = \frac{T_3 - T_1}{2} \]
\[ T_2 - T_1 = \frac{T_1 + T_3}{2} - T_1 = \frac{T_3 - T_1}{2} \]

6. **Substituting Back**:
- Substitute these into the equation:
\[ k_1 \cdot \frac{T_3 - T_1}{2} = \frac{k_2}{3} \cdot \frac{T_3 - T_1}{2} \]
- Canceling \(\frac{T_3 - T_1}{2}\) (assuming \(T_3 \neq T_1\)):
\[ k_1 = \frac{k_2}{3} \]

7. **Finding the Ratio**:
- Rearranging gives:
\[ \frac{k_1}{k_2} = \frac{1}{3} \]

### Final Answer:
The ratio of the thermal conductivity of the thinner sheet to the thicker sheet is:
\[ \frac{k_1}{k_2} = \frac{1}{3} \]