Two plates of equal area are placed in contact with each other. The thickness of the plates are 2.0 cm and 3.0 cm respectively. The outer face of first plate is at `-25^(@) C` and that of second plate is at `+25^(@) C`. The conductivities of the plates are in the ratio `2:3`. Calculate the temperature of the common surface of the plates.

Explanation

To solve the problem, we need to find the temperature of the common surface of two plates in contact, given their thicknesses, outer temperatures, and thermal conductivities. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
- Thickness of Plate 1, \( l_1 = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Thickness of Plate 2, \( l_2 = 3 \, \text{cm} = 0.03 \, \text{m} \)
- Temperature of the outer face of Plate 1, \( \theta_1 = -25^\circ C \)
- Temperature of the outer face of Plate 2, \( \theta_2 = +25^\circ C \)
- Ratio of thermal conductivities, \( k_1 : k_2 = 2 : 3 \)

### Step 2: Express the Thermal Conductivities
From the ratio of thermal conductivities, we can express:
\[ k_1 = \frac{2}{3} k_2 \]

### Step 3: Set Up the Heat Transfer Equation
For steady-state heat transfer, the heat flow through both plates must be equal:
\[ \frac{k_1 A (\theta_1 - \theta)}{l_1} = \frac{k_2 A (\theta - \theta_2)}{l_2} \]
Where \( A \) is the area of the plates and \( \theta \) is the temperature at the common surface.

### Step 4: Substitute Known Values
Substituting \( k_1 \) in terms of \( k_2 \) into the equation:
\[ \frac{\frac{2}{3} k_2 A (-25 - \theta)}{0.02} = \frac{k_2 A (\theta - 25)}{0.03} \]

### Step 5: Cancel Out Common Terms
Since \( A \) and \( k_2 \) are common on both sides, we can cancel them out:
\[ \frac{\frac{2}{3} (-25 - \theta)}{0.02} = \frac{(\theta - 25)}{0.03} \]

### Step 6: Simplify the Equation
Cross-multiplying gives:
\[ \frac{2}{3} (-25 - \theta) \cdot 0.03 = (\theta - 25) \cdot 0.02 \]
This simplifies to:
\[ \frac{2 \cdot 0.03}{3} (-25 - \theta) = 0.02 (\theta - 25) \]
Calculating \( \frac{2 \cdot 0.03}{3} = 0.02 \):
\[ 0.02 (-25 - \theta) = 0.02 (\theta - 25) \]

### Step 7: Eliminate the Common Factor
Dividing both sides by \( 0.02 \):
\[ -25 - \theta = \theta - 25 \]

### Step 8: Solve for \( \theta \)
Rearranging gives:
\[ -25 + 25 = 2\theta \]
\[ 0 = 2\theta \]
Thus,
\[ \theta = 0^\circ C \]

### Conclusion
The temperature of the common surface of the plates is \( 0^\circ C \).