Three very large plates of same area are kept parrallel and close to each other. They are considered as ideal black surfaces and have high thermal conductivity. The first and third plates are maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e., second) plate under steady state condition is

Options

A. `(65/2)^(1//4) T`
B. `(97/4)^(1//4)T`
C. `(97/2)^(1//4) T`
D. `(97)^(1//4) T`

Explanation

To find the temperature of the middle plate (the second plate) under steady-state conditions, we can follow these steps:

### Step 1: Understand the setup
We have three large parallel plates:
- Plate 1 is at temperature \( 2T \)
- Plate 2 (the middle plate) is at an unknown temperature \( T' \)
- Plate 3 is at temperature \( 3T \)

### Step 2: Write the power absorbed by the middle plate
The power absorbed by the middle plate (Plate 2) comes from both Plate 1 and Plate 3. The power absorbed from each plate can be calculated using the Stefan-Boltzmann law:

\[ P_{\text{absorbed}} = P_1 + P_3 = \sigma A (2T)^4 + \sigma A (3T)^4 \]

### Step 3: Write the power radiated by the middle plate
The power radiated by the middle plate (Plate 2) is given by:

\[ P_{\text{radiated}} = 2 \sigma A (T')^4 \]

### Step 4: Set up the equation for steady-state
At steady state, the power absorbed by Plate 2 equals the power radiated by Plate 2:

\[ \sigma A (2T)^4 + \sigma A (3T)^4 = 2 \sigma A (T')^4 \]

### Step 5: Simplify the equation
We can cancel \( \sigma A \) from both sides of the equation:

\[ (2T)^4 + (3T)^4 = 2 (T')^4 \]

### Step 6: Calculate \( (2T)^4 \) and \( (3T)^4 \)
Calculating the powers:

\[ (2T)^4 = 16T^4 \]
\[ (3T)^4 = 81T^4 \]

### Step 7: Substitute back into the equation
Substituting these values back into the equation gives:

\[ 16T^4 + 81T^4 = 2 (T')^4 \]

### Step 8: Combine terms
Combining the terms on the left side:

\[ 97T^4 = 2 (T')^4 \]

### Step 9: Solve for \( (T')^4 \)
Dividing both sides by 2:

\[ (T')^4 = \frac{97}{2} T^4 \]

### Step 10: Take the fourth root
Taking the fourth root of both sides gives:

\[ T' = \left(\frac{97}{2}\right)^{1/4} T \]

### Final Answer
Thus, the temperature of the middle plate (Plate 2) under steady-state conditions is:

\[ T' = \frac{97^{1/4}}{2^{1/4}} T \]