Two plates of same area are placed in contact. Their thickness as well as thermal conductivities are in the ratio `2:3`. The outer surface of one plate is maintained at `10^(@)`C and that of the other at `0^(@)`C. What is the temperature at the common surface?

Options

A. `0^(@)`C
B. `2.5^(@)`C
C. `5^(@)`C
D. `6.5^(@)`C

Explanation

To solve the problem, we will use the concept of thermal conduction and the principle of heat transfer through two slabs in contact. We can set up the equations based on the given information.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Let the thickness of the first plate be \( L_1 \) and the second plate be \( L_2 \).
- The thermal conductivities are \( K_1 \) and \( K_2 \).
- The ratio of thicknesses and thermal conductivities is given as \( \frac{L_1}{L_2} = \frac{2}{3} \) and \( \frac{K_1}{K_2} = \frac{2}{3} \).
- The temperatures at the outer surfaces are \( T_1 = 10^\circ C \) and \( T_2 = 0^\circ C \).
- Let the temperature at the common surface be \( T \).

2. **Set Up the Heat Transfer Equations:**
- According to Fourier's law of heat conduction, the heat flow through each plate can be expressed as:
\[ Q_1 = \frac{K_1 A (T_1 - T)}{L_1} \]
\[ Q_2 = \frac{K_2 A (T - T_2)}{L_2} \]
- Since the heat flow through both plates is the same, we can set \( Q_1 = Q_2 \):
\[ \frac{K_1 A (T_1 - T)}{L_1} = \frac{K_2 A (T - T_2)}{L_2} \]

3. **Cancel Common Terms:**
- The area \( A \) cancels out from both sides:
\[ \frac{K_1 (T_1 - T)}{L_1} = \frac{K_2 (T - T_2)}{L_2} \]

4. **Substitute Ratios:**
- Substitute \( K_1 = 2x \), \( K_2 = 3x \), \( L_1 = 2y \), and \( L_2 = 3y \) where \( x \) and \( y \) are constants:
\[ \frac{2x (10 - T)}{2y} = \frac{3x (T - 0)}{3y} \]
- Simplifying gives:
\[ \frac{10 - T}{y} = \frac{T}{y} \]

5. **Eliminate \( y \):**
- Since \( y \) is common, we can eliminate it:
\[ 10 - T = T \]

6. **Solve for \( T \):**
- Rearranging gives:
\[ 10 = 2T \implies T = 5^\circ C \]

### Final Answer:
The temperature at the common surface is \( T = 5^\circ C \).