A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

Options

A. 225
B. 450
C. 1000
D. 1800

Explanation

To solve the problem, we will use the Stefan-Boltzmann Law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature and the surface area of the body. The formula for the power radiated by a black body is given by:

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

where:
- \( P \) is the power radiated,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the body,
- \( T \) is the absolute temperature.

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- Initial radius \( R_1 = 12 \, \text{cm} = 0.12 \, \text{m} \)
- Initial temperature \( T_1 = 500 \, \text{K} \)
- Initial power \( P_1 = 450 \, \text{W} \)

2. **Calculate the initial surface area**:
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi R^2 \]
For the initial radius:
\[ A_1 = 4 \pi (0.12)^2 \]

3. **Determine the new conditions**:
- New radius \( R_2 = \frac{R_1}{2} = \frac{0.12}{2} = 0.06 \, \text{m} \)
- New temperature \( T_2 = 2 \times T_1 = 2 \times 500 = 1000 \, \text{K} \)

4. **Calculate the new surface area**:
For the new radius:
\[ A_2 = 4 \pi (0.06)^2 \]

5. **Use the Stefan-Boltzmann Law to find the new power**:
The ratio of the new power \( P_2 \) to the initial power \( P_1 \) can be expressed as:
\[ \frac{P_2}{P_1} = \frac{A_2 T_2^4}{A_1 T_1^4} \]

6. **Express the areas in terms of the radii**:
Since \( A_1 = 4 \pi R_1^2 \) and \( A_2 = 4 \pi R_2^2 \), we can simplify:
\[ \frac{P_2}{P_1} = \frac{R_2^2 T_2^4}{R_1^2 T_1^4} \]

7. **Substituting the values**:
Substitute \( R_2 = \frac{R_1}{2} \) and \( T_2 = 2T_1 \):
\[ \frac{P_2}{P_1} = \frac{\left(\frac{R_1}{2}\right)^2 (2T_1)^4}{R_1^2 T_1^4} \]

8. **Simplifying the equation**:
\[ \frac{P_2}{P_1} = \frac{\frac{R_1^2}{4} \cdot 16 T_1^4}{R_1^2 T_1^4} = \frac{16}{4} = 4 \]

9. **Calculate \( P_2 \)**:
\[ P_2 = 4 P_1 = 4 \times 450 \, \text{W} = 1800 \, \text{W} \]

### Final Answer:
The power radiated when the radius is halved and the temperature is doubled is \( P_2 = 1800 \, \text{W} \).