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 dimensions and temperature change. We will use the Stefan-Boltzmann law for this calculation, which states that the energy emitted per unit time (E) by a black body is given by:

\[ E = \sigma A T^4 \]

where:
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the area of the surface,
- \( T \) is the absolute temperature in Kelvin.

### Step 1: Calculate the initial area and temperature
The initial dimensions of the rectangular surface are:
- Length = 8 cm
- Breadth = 4 cm

Thus, the initial area \( A \) is:
\[ A = 8 \, \text{cm} \times 4 \, \text{cm} = 32 \, \text{cm}^2 \]

The initial temperature \( T_1 \) is given as:
\[ T_1 = 127^\circ C = 127 + 273 = 400 \, K \]

### Step 2: Calculate the initial energy emitted
Using the Stefan-Boltzmann law, the initial energy emitted \( E_1 \) is:
\[ E_1 = \sigma A T_1^4 \]
\[ E_1 = \sigma (32 \, \text{cm}^2) (400 \, K)^4 \]

### Step 3: Calculate the new area and temperature
When the length and breadth are reduced to half, the new dimensions are:
- New Length = \( \frac{8}{2} = 4 \, \text{cm} \)
- New Breadth = \( \frac{4}{2} = 2 \, \text{cm} \)

Thus, the new area \( A' \) is:
\[ A' = 4 \, \text{cm} \times 2 \, \text{cm} = 8 \, \text{cm}^2 \]

The new temperature \( T_2 \) is given as:
\[ T_2 = 327^\circ C = 327 + 273 = 600 \, K \]

### Step 4: Calculate the new energy emitted
Using the Stefan-Boltzmann law, the new energy emitted \( E_2 \) is:
\[ E_2 = \sigma A' T_2^4 \]
\[ E_2 = \sigma (8 \, \text{cm}^2) (600 \, K)^4 \]

### Step 5: Calculate the ratio of emissions
Now, we can find the ratio of the new energy emitted to the initial energy emitted:
\[ \text{Ratio} = \frac{E_2}{E_1} = \frac{\sigma (8 \, \text{cm}^2) (600 \, K)^4}{\sigma (32 \, \text{cm}^2) (400 \, K)^4} \]

The \( \sigma \) cancels out:
\[ \text{Ratio} = \frac{8 \times (600)^4}{32 \times (400)^4} \]

### Step 6: Simplify the ratio
This can be simplified further:
\[ \text{Ratio} = \frac{8}{32} \times \frac{(600)^4}{(400)^4} = \frac{1}{4} \times \left(\frac{600}{400}\right)^4 \]
\[ \text{Ratio} = \frac{1}{4} \times \left(\frac{3}{2}\right)^4 = \frac{1}{4} \times \frac{81}{16} = \frac{81}{64} \]

### Final Answer
The ratio of emission of energy becomes:
\[ \frac{81}{64} \]