A spherical black body with radius 12 cm radiates 640 w power at 500 K. If the radius is halved and the temperature doubled, the power radiated in watts would be

Options

A. 160W
B. 409W
C. 2560W
D. 640W

Answer

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 temperature and the surface area of the body.

### Step-by-step Solution:

1. **Identify the initial parameters:**
- Initial radius \( R_1 = 12 \, \text{cm} = 0.12 \, \text{m} \)
- Initial temperature \( T_1 = 500 \, \text{K} \)
- Initial power \( P_1 = 640 \, \text{W} \)

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

3. **Determine the new parameters:**
- New radius \( R_2 = \frac{R_1}{2} = \frac{12 \, \text{cm}}{2} = 6 \, \text{cm} = 0.06 \, \text{m} \)
- New temperature \( T_2 = 2 \times T_1 = 2 \times 500 \, \text{K} = 1000 \, \text{K} \)

4. **Calculate the new surface area:**
- For the new radius:
\[ A_2 = 4 \pi (R_2)^2 = 4 \pi (0.06)^2 \]

5. **Use the Stefan-Boltzmann law to find the new power:**
- The power radiated is given by:
\[ P \propto A \cdot T^4 \]
- Therefore, we can write the ratio of the powers:
\[ \frac{P_1}{P_2} = \frac{A_1 T_1^4}{A_2 T_2^4} \]

6. **Substituting the values:**
- Substitute \( A_1 \) and \( A_2 \):
\[ P_2 = P_1 \cdot \frac{A_1}{A_2} \cdot \frac{T_2^4}{T_1^4} \]
- Since \( A_2 = 4 \pi (0.06)^2 \) and \( A_1 = 4 \pi (0.12)^2 \):
\[ \frac{A_1}{A_2} = \frac{(0.12)^2}{(0.06)^2} = \frac{0.0144}{0.0036} = 4 \]
- And for the temperature:
\[ \frac{T_2^4}{T_1^4} = \frac{(1000)^4}{(500)^4} = \left(\frac{1000}{500}\right)^4 = 2^4 = 16 \]

7. **Final calculation:**
- Now substituting back:
\[ P_2 = 640 \cdot 4 \cdot 16 = 640 \cdot 64 = 40960 \, \text{W} \]

### Conclusion:
The new power radiated by the spherical black body is \( 40960 \, \text{W} \).