The temperatures of two bodies A and B are `727^(@)C` and `127^(@)C` . The ratio of rate of emission of radiations will be

Options

A. `727//127`
B. `625//16`
C. `1000//400`
D. `100//16`

Answer

To find the ratio of the rate of emission of radiation from two bodies A and B with temperatures \(727^\circ C\) and \(127^\circ C\), we can use the Stefan-Boltzmann Law, which states that the power emitted per unit area of a black body is proportional to the fourth power of its absolute temperature.

### Step-by-Step Solution:

1. **Convert the temperatures from Celsius to Kelvin:**
- The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]
- For body A:
\[ T_A = 727 + 273 = 1000 \, K \]
- For body B:
\[ T_B = 127 + 273 = 400 \, K \]

2. **Apply the Stefan-Boltzmann Law:**
- The rate of emission of radiation \(U\) for each body can be expressed as:
\[ U_A = \sigma A T_A^4 \]
\[ U_B = \sigma A T_B^4 \]
- Here, \(\sigma\) is the Stefan-Boltzmann constant and \(A\) is the area of the bodies. Since we are looking for the ratio, we can ignore \(\sigma\) and \(A\).

3. **Calculate the ratio of the rates of emission:**
- The ratio of the rates of emission \( \frac{U_A}{U_B} \) is given by:
\[ \frac{U_A}{U_B} = \frac{T_A^4}{T_B^4} \]
- Substituting the temperatures:
\[ \frac{U_A}{U_B} = \frac{(1000)^4}{(400)^4} \]

4. **Simplify the ratio:**
- This can be simplified as:
\[ \frac{U_A}{U_B} = \left(\frac{1000}{400}\right)^4 = \left(\frac{10}{4}\right)^4 = \left(\frac{5}{2}\right)^4 \]
- Now calculate \(\left(\frac{5}{2}\right)^4\):
\[ \left(\frac{5}{2}\right)^4 = \frac{5^4}{2^4} = \frac{625}{16} \]

5. **Final Result:**
- Therefore, the ratio of the rate of emission of radiation from bodies A and B is:
\[ \frac{U_A}{U_B} = \frac{625}{16} \]