A sphere at temperature `600 K` is placed in an enviroment to temperature is `200 K`. Its cooling rate is `R`. If its temperature reduced to `400 K` then cooling rate in same enviorment will become

Options

A. `(3)/(16)R`
B. `(16)/(3)R`
C. `(9)/(27)R`
D. None

Explanation

To solve the problem, we will use Stefan-Boltzmann's law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Initial temperature of the sphere, \( T_1 = 600 \, K \)
- Temperature of the environment, \( T = 200 \, K \)

2. **Calculate Initial Cooling Rate (R)**:
According to Stefan-Boltzmann's law, the cooling rate \( R \) can be expressed as:
\[ R = \sigma (T_1^4 - T^4) \]
where \( \sigma \) is the Stefan-Boltzmann constant.

3. **Substitute Values for Initial Cooling Rate**:
Plugging in the values:
\[ R = \sigma (600^4 - 200^4) \]

4. **Identify New Conditions**:
- New temperature of the sphere, \( T_2 = 400 \, K \)

5. **Calculate New Cooling Rate (R1)**:
The new cooling rate \( R_1 \) is given by:
\[ R_1 = \sigma (T_2^4 - T^4) \]
Substituting the new temperature:
\[ R_1 = \sigma (400^4 - 200^4) \]

6. **Set Up the Ratio of Cooling Rates**:
We can find the ratio of the new cooling rate to the initial cooling rate:
\[ \frac{R_1}{R} = \frac{T_2^4 - T^4}{T_1^4 - T^4} \]

7. **Substitute the Values into the Ratio**:
\[ \frac{R_1}{R} = \frac{400^4 - 200^4}{600^4 - 200^4} \]

8. **Calculate the Fourth Powers**:
- \( 400^4 = 256 \times 10^4 \)
- \( 200^4 = 16 \times 10^4 \)
- \( 600^4 = 1296 \times 10^4 \)

9. **Substitute the Fourth Powers into the Ratio**:
\[ \frac{R_1}{R} = \frac{(256 - 16) \times 10^4}{(1296 - 16) \times 10^4} = \frac{240 \times 10^4}{1280 \times 10^4} \]

10. **Simplify the Ratio**:
\[ \frac{R_1}{R} = \frac{240}{1280} = \frac{3}{16} \]

11. **Calculate New Cooling Rate**:
Therefore, the new cooling rate \( R_1 \) is:
\[ R_1 = \frac{3}{16} R \]

### Final Answer:
The cooling rate when the temperature of the sphere is reduced to \( 400 \, K \) will be \( \frac{3}{16} R \).