Star A has radius r surface temperature T while star B has radius 4 r and surface temperature T /2. The ratio of the power of two starts, P A : P B is

A. 1 : 16
B. 1 : 16
C. 1 : 1
D. 1 : 4

Options

A. 16 : 1
B. 1 : 16
C. 1 : 1
D. 1 : 4

Answer

To solve the problem of finding the ratio of the power of two stars, A and B, we will use Stefan-Boltzmann Law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature and its surface area.

### Step-by-Step Solution:

1. **Identify the parameters for both stars:**
- For Star A:
- Radius = \( r \)
- Surface Temperature = \( T \)
- For Star B:
- Radius = \( 4r \)
- Surface Temperature = \( \frac{T}{2} \)

2. **Write the formula for power using Stefan-Boltzmann Law:**
The power \( P \) radiated by a star is given by:
\[ P = \sigma A T^4 \]
where \( A \) is the surface area and \( \sigma \) is the Stefan-Boltzmann constant.

3. **Calculate the power for Star A:**
- Surface area \( A_A \) of Star A:
\[ A_A = 4\pi r^2 \]
- Power \( P_A \):
\[ P_A = \sigma (4\pi r^2) T^4 = 4\pi \sigma r^2 T^4 \]

4. **Calculate the power for Star B:**
- Surface area \( A_B \) of Star B:
\[ A_B = 4\pi (4r)^2 = 4\pi (16r^2) = 64\pi r^2 \]
- Power \( P_B \):
\[ P_B = \sigma (64\pi r^2) \left(\frac{T}{2}\right)^4 = 64\pi \sigma r^2 \left(\frac{T^4}{16}\right) = 4\pi \sigma r^2 T^4 \]

5. **Find the ratio of powers \( P_A : P_B \):**
\[ \frac{P_A}{P_B} = \frac{4\pi \sigma r^2 T^4}{4\pi \sigma r^2 T^4} = 1 \]
Thus, the ratio \( P_A : P_B = 1 : 1 \).

### Conclusion:
The ratio of the power of the two stars \( P_A : P_B \) is \( 1 : 1 \).