The rate of cooling at `600 K`. If surrounding temperature is `300 K` is `H`. The rate of cooling at `900 K` is

Options

A. `16/3 H`
B. `2 H`
C. `3H`
D. `2/3 H`

Explanation

To find the rate of cooling at `900 K` given that the rate of cooling at `600 K` is `H` and the surrounding temperature is `300 K`, we can use the Stefan-Boltzmann law, which states that the rate of cooling is proportional to the difference in the fourth power of the absolute temperature of the object and the surrounding temperature.

### Step-by-Step Solution:

1. **Understand the relationship**: The rate of cooling \( H \) is directly proportional to \( T^4 - T_0^4 \), where \( T \) is the temperature of the object and \( T_0 \) is the surrounding temperature.

\[ H \propto T^4 - T_0^4 \]

2. **Set up the equation for the given temperatures**:
- For \( T = 600 K \):
\[ H = k (600^4 - 300^4) \]
- For \( T = 900 K \):
\[ H' = k (900^4 - 300^4) \]

Here, \( k \) is the proportionality constant.

3. **Form the ratio of the two rates of cooling**:
\[ \frac{H'}{H} = \frac{900^4 - 300^4}{600^4 - 300^4} \]

4. **Simplify the expression**:
- Calculate \( 900^4 - 300^4 \) and \( 600^4 - 300^4 \):
\[ 900^4 - 300^4 = (3 \cdot 300)^4 - 300^4 = 3^4 \cdot 300^4 - 300^4 = 81 \cdot 300^4 - 300^4 = (81 - 1) \cdot 300^4 = 80 \cdot 300^4 \]
\[ 600^4 - 300^4 = (2 \cdot 300)^4 - 300^4 = 2^4 \cdot 300^4 - 300^4 = 16 \cdot 300^4 - 300^4 = (16 - 1) \cdot 300^4 = 15 \cdot 300^4 \]

5. **Substitute back into the ratio**:
\[ \frac{H'}{H} = \frac{80 \cdot 300^4}{15 \cdot 300^4} = \frac{80}{15} = \frac{16}{3} \]

6. **Calculate \( H' \)**:
\[ H' = \frac{16}{3} H \]

### Final Answer:
The rate of cooling at `900 K` is \( \frac{16}{3} H \).