Two identical plates of different metals are joined to form a single plate whose thickness is double the thickness of each plate. If the coefficients of conductivity of each plate are 2 and 3 respectively, Then the conductivity of composite plate will be

Answer

To find the conductivity of the composite plate formed by two identical plates of different metals, we can follow these steps:

### Step 1: Understand the Setup
We have two identical plates made of different metals, with thermal conductivities \( k_1 = 2 \) and \( k_2 = 3 \). The thickness of each plate is \( l \), so the total thickness of the composite plate is \( 2l \).

### Step 2: Apply the Concept of Thermal Conductivity
The thermal current \( I \) through a material can be expressed using Fourier's law of heat conduction:
\[ I = \frac{k \cdot A \cdot (T_h - T_c)}{L} \]
where:
- \( I \) is the thermal current,
- \( k \) is the thermal conductivity,
- \( A \) is the cross-sectional area,
- \( T_h \) and \( T_c \) are the temperatures on either side of the plate,
- \( L \) is the thickness of the plate.

### Step 3: Calculate Thermal Current for Each Plate
For the first plate (with conductivity \( k_1 = 2 \)):
\[ I_1 = \frac{2 \cdot A \cdot (T_h - T_c)}{l} \]

For the second plate (with conductivity \( k_2 = 3 \)):
\[ I_2 = \frac{3 \cdot A \cdot (T_h - T_c)}{l} \]

### Step 4: Total Thermal Current
Since the plates are in series, the total thermal current \( I \) through the composite plate is the sum of the currents through each plate:
\[ I = I_1 + I_2 = \frac{2 \cdot A \cdot (T_h - T_c)}{l} + \frac{3 \cdot A \cdot (T_h - T_c)}{l} \]
\[ I = \frac{(2 + 3) \cdot A \cdot (T_h - T_c)}{l} = \frac{5 \cdot A \cdot (T_h - T_c)}{l} \]

### Step 5: Equivalent Conductivity of the Composite Plate
For the composite plate, which has an area of \( 2A \) and thickness \( 2l \), we can express the thermal current as:
\[ I = \frac{k_{eq} \cdot 2A \cdot (T_h - T_c)}{2l} \]

### Step 6: Set the Two Expressions for Thermal Current Equal
Equating the two expressions for \( I \):
\[ \frac{5 \cdot A \cdot (T_h - T_c)}{l} = \frac{k_{eq} \cdot 2A \cdot (T_h - T_c)}{2l} \]

### Step 7: Simplify and Solve for \( k_{eq} \)
Cancelling \( A \) and \( (T_h - T_c) \) from both sides (assuming they are not zero):
\[ 5 = \frac{k_{eq} \cdot 2}{2} \]
\[ 5 = k_{eq} \]

### Final Result
Thus, the equivalent conductivity \( k_{eq} \) of the composite plate is:
\[ k_{eq} = 5 \]