The rectangular surface of area `8 cm xx 4 cm` of a black body at temperature `127^(@)C` emits energy `E` per section if length and breadth are reduced to half of the initial value and the temperature is raised to `327^(@)C`, the ratio of emission of energy becomes

Options

A. `(3)/(8)E`
B. `(81)/(16)E`
C. `(9)/(16)E`
D. `(81)/(64)E`

Explanation

To solve the problem, we need to find the ratio of energy emitted by a black body when its area and temperature change. Let's break it down step by step.

### Step 1: Determine the initial area and temperature
The initial dimensions of the rectangular surface are given as:
- Length (L) = 8 cm
- Breadth (B) = 4 cm

The area \( A_1 \) can be calculated as:
\[ A_1 = L \times B = 8 \, \text{cm} \times 4 \, \text{cm} = 32 \, \text{cm}^2 \]
To convert this to square meters:
\[ A_1 = 32 \, \text{cm}^2 = 32 \times 10^{-4} \, \text{m}^2 = 3.2 \times 10^{-2} \, \text{m}^2 \]

The initial temperature \( T_1 \) in Kelvin is:
\[ T_1 = 127^\circ C + 273 = 400 \, \text{K} \]

### Step 2: Determine the new area and temperature
The new dimensions after reducing length and breadth to half are:
- New Length = \( \frac{8}{2} = 4 \, \text{cm} \)
- New Breadth = \( \frac{4}{2} = 2 \, \text{cm} \)

The new area \( A_2 \) is:
\[ A_2 = \frac{L}{2} \times \frac{B}{2} = 4 \, \text{cm} \times 2 \, \text{cm} = 8 \, \text{cm}^2 \]
In square meters:
\[ A_2 = 8 \, \text{cm}^2 = 8 \times 10^{-4} \, \text{m}^2 = 8 \times 10^{-4} \, \text{m}^2 \]

The new temperature \( T_2 \) in Kelvin is:
\[ T_2 = 327^\circ C + 273 = 600 \, \text{K} \]

### Step 3: Calculate the energy emitted
The energy emitted per second \( E \) by a black body is given by the formula:
\[ E = \sigma \cdot A \cdot T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant.

For the initial state:
\[ E_1 = \sigma \cdot A_1 \cdot T_1^4 = \sigma \cdot (3.2 \times 10^{-2}) \cdot (400^4) \]

For the new state:
\[ E_2 = \sigma \cdot A_2 \cdot T_2^4 = \sigma \cdot (8 \times 10^{-4}) \cdot (600^4) \]

### Step 4: Find the ratio of emissions
To find the ratio \( \frac{E_1}{E_2} \):
\[ \frac{E_1}{E_2} = \frac{\sigma \cdot (3.2 \times 10^{-2}) \cdot (400^4)}{\sigma \cdot (8 \times 10^{-4}) \cdot (600^4)} \]
The \( \sigma \) cancels out:
\[ = \frac{(3.2 \times 10^{-2}) \cdot (400^4)}{(8 \times 10^{-4}) \cdot (600^4)} \]

### Step 5: Simplify the expression
\[ = \frac{3.2 \times 10^{-2}}{8 \times 10^{-4}} \cdot \frac{400^4}{600^4} \]
\[ = \frac{3.2 \times 10^{2}}{8} \cdot \left(\frac{400}{600}\right)^4 \]
\[ = 40 \cdot \left(\frac{2}{3}\right)^4 \]
\[ = 40 \cdot \frac{16}{81} \]
\[ = \frac{640}{81} \]

### Final Ratio
Thus, the ratio of the energy emissions is:
\[ \frac{E_1}{E_2} = \frac{640}{81} \]