A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be
(a)225 (b)450
(c) 900 (d)1800

Explanation

To solve the problem, we need to determine the power radiated by a spherical black body when its radius is halved and its temperature is doubled. 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 \( Q_1 = 450 \, \text{W} \)

2. **Calculate the initial surface area:**
The surface area \( A_1 \) of a sphere is given by the formula:
\[ A_1 = 4 \pi r_1^2 \]
Substituting the value of \( r_1 \):
\[ A_1 = 4 \pi (0.12)^2 = 4 \pi (0.0144) \approx 0.18096 \, \text{m}^2 \]

3. **Determine the new parameters:**
- 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:**
The new surface area \( A_2 \) is:
\[ A_2 = 4 \pi r_2^2 = 4 \pi \left( \frac{r_1}{2} \right)^2 = 4 \pi \left( \frac{0.12}{2} \right)^2 = 4 \pi \left(0.06\right)^2 = 4 \pi (0.0036) \approx 0.04524 \, \text{m}^2 \]
This can also be expressed as:
\[ A_2 = \frac{A_1}{4} \]

5. **Apply the Stefan-Boltzmann law:**
The power radiated is given by:
\[ Q \propto A \cdot T^4 \]
Therefore, the new power \( Q_2 \) can be expressed as:
\[ Q_2 = k \cdot A_2 \cdot T_2^4 \]
where \( k \) is a constant.

6. **Taking the ratio of the new power to the old power:**
\[ \frac{Q_2}{Q_1} = \frac{A_2 \cdot T_2^4}{A_1 \cdot T_1^4} \]
Substituting \( A_2 = \frac{A_1}{4} \) and \( T_2 = 2T_1 \):
\[ \frac{Q_2}{Q_1} = \frac{\frac{A_1}{4} \cdot (2T_1)^4}{A_1 \cdot T_1^4} \]
Simplifying this gives:
\[ \frac{Q_2}{Q_1} = \frac{\frac{A_1}{4} \cdot 16T_1^4}{A_1 \cdot T_1^4} = \frac{16}{4} = 4 \]

7. **Calculate the new power:**
\[ Q_2 = 4 \cdot Q_1 = 4 \cdot 450 \, \text{W} = 1800 \, \text{W} \]

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