Two uniform solid spheres `A` and `B` of same material, painted completely black and placed in free space separately. Their radii are `R` and `2R` respectively and the domainating wavelengths in their spectrum are observed to be in the ratio `1:2`, which of the following is not correct.

Options

A. Ratio of their rates of heat loss is `4:1`
B. Ratio of their rates of cooling is `32:1`
C. Ratio of their temperatures is `2:1`
D. Ratio of their emissive powers is `4:1`

Explanation

To solve the problem, we need to analyze the given information about the two solid spheres A and B, their properties, and the implications of their sizes and temperatures based on the laws of thermodynamics and radiation.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two spheres, A with radius \( R \) and B with radius \( 2R \).
- Both spheres are made of the same material and are painted black, which means they are perfect black bodies.
- The dominating wavelengths in their spectrum are in the ratio \( 1:2 \).

2. **Using Wien's Displacement Law**:
- Wien's Displacement Law states that the wavelength \( \lambda \) at which the emission of a black body spectrum is maximum is inversely proportional to its absolute temperature \( T \):
\[ \lambda \propto \frac{1}{T} \]
- If the ratio of the wavelengths is \( \lambda_A : \lambda_B = 1 : 2 \), then we can express this in terms of temperature:
\[ \frac{\lambda_A}{\lambda_B} = \frac{T_B}{T_A} = \frac{1}{2} \]
- This implies:
\[ T_B = 2T_A \]

3. **Calculating Heat Loss**:
- The rate of heat loss \( Q \) for a black body is given by:
\[ Q = \sigma A T^4 \]
- Where \( A \) is the surface area and \( \sigma \) is the Stefan-Boltzmann constant.
- For sphere A:
\[ A_A = 4\pi R^2 \quad \text{and} \quad Q_A = \sigma (4\pi R^2) T_A^4 \]
- For sphere B:
\[ A_B = 4\pi (2R)^2 = 16\pi R^2 \quad \text{and} \quad Q_B = \sigma (16\pi R^2) T_B^4 \]

4. **Finding the Ratio of Heat Loss**:
- The ratio of heat loss \( \frac{Q_A}{Q_B} \):
\[ \frac{Q_A}{Q_B} = \frac{\sigma (4\pi R^2) T_A^4}{\sigma (16\pi R^2) T_B^4} = \frac{4 T_A^4}{16 T_B^4} = \frac{1}{4} \left(\frac{T_A}{T_B}\right)^4 \]
- Substituting \( T_B = 2T_A \):
\[ \frac{Q_A}{Q_B} = \frac{1}{4} \left(\frac{T_A}{2T_A}\right)^4 = \frac{1}{4} \left(\frac{1}{2}\right)^4 = \frac{1}{4} \cdot \frac{1}{16} = \frac{1}{64} \]

5. **Analyzing the Options**:
- The options likely involve the ratios of heat loss, rates of cooling, temperatures, and emissive powers.
- We have established that:
- The ratio of heat loss \( Q_A : Q_B = 1 : 64 \)
- The ratio of temperatures \( T_A : T_B = 1 : 2 \)
- The emissive power is proportional to \( T^4 \), leading to a ratio of \( 1 : 16 \).

6. **Conclusion**:
- The statement that is not correct will be the one that contradicts our findings. For instance, if any option states that the ratio of heat loss is \( 1:4 \) or \( 1:16 \), it would be incorrect based on our calculations.