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. (a) `1:3`
B. (b) `3:1`
C. (c) `2:3`
D. (d) `3:9`

Explanation

To solve the problem, we need to analyze the thermal conduction through the two sheets and use the information that the temperatures \( T_1 \), \( T_2 \), and \( T_3 \) are in arithmetic progression (AP).

### Step 1: Understand the relationship between temperatures
Since \( T_1 \), \( T_2 \), and \( T_3 \) are in arithmetic progression, we can express this relationship mathematically:
\[ T_2 - T_1 = T_3 - T_2 \]
This implies:
\[ 2T_2 = T_1 + T_3 \quad \Rightarrow \quad T_2 = \frac{T_1 + T_3}{2} \]

### Step 2: Write the heat transfer equations
The heat transfer through the thinner sheet (thickness \( d \)) can be expressed as:
\[ H = \frac{T_1 - T_2}{R_1} \]
where \( R_1 = \frac{d}{k_1 A} \) (thermal resistance of the thinner sheet).

For the thicker sheet (thickness \( 3d \)):
\[ H = \frac{T_2 - T_3}{R_2} \]
where \( R_2 = \frac{3d}{k_2 A} \) (thermal resistance of the thicker sheet).

### Step 3: Set the heat transfer equations equal
Since the heat transfer \( H \) through both sheets is the same, we can equate the two expressions:
\[ \frac{T_1 - T_2}{\frac{d}{k_1 A}} = \frac{T_2 - T_3}{\frac{3d}{k_2 A}} \]

### Step 4: Simplify the equation
Cancelling \( A \) and \( d \) from both sides gives:
\[ \frac{T_1 - T_2}{\frac{1}{k_1}} = \frac{T_2 - T_3}{\frac{3}{k_2}} \]
Rearranging this, we have:
\[ (T_1 - T_2) \cdot k_2 = 3(T_2 - T_3) \cdot k_1 \]

### Step 5: Substitute \( T_2 \) in terms of \( T_1 \) and \( T_3 \)
Using the relationship from Step 1, substitute \( T_2 \):
\[ T_1 - \frac{T_1 + T_3}{2} = \frac{T_1 - T_3}{2} \]
Thus, we can rewrite the equation:
\[ \frac{T_1 - T_3}{2} \cdot k_2 = 3 \left( \frac{T_1 + T_3}{2} - T_3 \right) \cdot k_1 \]
This simplifies to:
\[ \frac{T_1 - T_3}{2} \cdot k_2 = 3 \left( \frac{T_1 - T_3}{2} \right) \cdot k_1 \]

### Step 6: Solve for the ratio of thermal conductivities
Dividing both sides by \( \frac{T_1 - T_3}{2} \) (assuming \( T_1 \neq T_3 \)):
\[ k_2 = 3k_1 \]
Thus, the ratio of thermal conductivities is:
\[ \frac{k_1}{k_2} = \frac{1}{3} \]
This means:
\[ \frac{k_1}{k_2} = \frac{1}{3} \quad \Rightarrow \quad \text{Ratio of } k_1 : k_2 = 1 : 3 \]

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

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