Three discs, A, B and C having radii 2m, 4m and6m respectively are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensity are `300nm, 400nm` and `500 nm`, respectively. The power radiated by them are `Q_A, Q_B and Q_C` respectively
(a) `Q_A` is maximum (b) `Q_B` is maximum (c) `Q_C` is maximum (d) `Q_A = Q_B = Q_C`

Explanation

To solve the problem, we need to determine the power radiated by the three discs A, B, and C, which are treated as black bodies. The power radiated by a black body is given by the Stefan-Boltzmann law, which states that the power \( P \) radiated by a black body is proportional to the fourth power of its temperature \( T \) and its surface area \( A \). The relevant formula for power can be expressed as:

\[ P \propto A T^4 \]

Since the discs are coated with carbon black, we can assume they behave like ideal black bodies. The area \( A \) of a disc is given by:

\[ A = \pi r^2 \]

Where \( r \) is the radius of the disc. The wavelength corresponding to maximum intensity \( \lambda_m \) is given for each disc. According to Wien's displacement law, the temperature \( T \) of a black body is inversely proportional to the wavelength of maximum intensity:

\[ T \propto \frac{1}{\lambda_m} \]

Therefore, we can express the power radiated by each disc in terms of its radius and the wavelength corresponding to maximum intensity.

### Step 1: Calculate the area of each disc
- For disc A (radius = 2m):
\[ A_A = \pi (2^2) = 4\pi \, \text{m}^2 \]

- For disc B (radius = 4m):
\[ A_B = \pi (4^2) = 16\pi \, \text{m}^2 \]

- For disc C (radius = 6m):
\[ A_C = \pi (6^2) = 36\pi \, \text{m}^2 \]

### Step 2: Calculate the temperature for each disc using Wien's law
- For disc A (\(\lambda_m = 300 \, \text{nm} = 300 \times 10^{-9} \, \text{m}\)):
\[ T_A \propto \frac{1}{300 \times 10^{-9}} \implies T_A = k \cdot \frac{1}{300 \times 10^{-9}} \]

- For disc B (\(\lambda_m = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}\)):
\[ T_B \propto \frac{1}{400 \times 10^{-9}} \implies T_B = k \cdot \frac{1}{400 \times 10^{-9}} \]

- For disc C (\(\lambda_m = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}\)):
\[ T_C \propto \frac{1}{500 \times 10^{-9}} \implies T_C = k \cdot \frac{1}{500 \times 10^{-9}} \]

### Step 3: Express the power for each disc
Using the formula \( P \propto A T^4 \), we can express the power for each disc:
- For disc A:
\[ P_A \propto A_A T_A^4 = (4\pi) \left(k \cdot \frac{1}{300 \times 10^{-9}}\right)^4 \]

- For disc B:
\[ P_B \propto A_B T_B^4 = (16\pi) \left(k \cdot \frac{1}{400 \times 10^{-9}}\right)^4 \]

- For disc C:
\[ P_C \propto A_C T_C^4 = (36\pi) \left(k \cdot \frac{1}{500 \times 10^{-9}}\right)^4 \]

### Step 4: Compare the powers
Now we can compare the ratios of the powers:
\[ \frac{P_A}{P_B} = \frac{4\pi \left(\frac{1}{300 \times 10^{-9}}\right)^4}{16\pi \left(\frac{1}{400 \times 10^{-9}}\right)^4} \]
\[ \frac{P_B}{P_C} = \frac{16\pi \left(\frac{1}{400 \times 10^{-9}}\right)^4}{36\pi \left(\frac{1}{500 \times 10^{-9}}\right)^4} \]

Calculating these ratios will show which disc radiates the most power.

### Conclusion
After evaluating the ratios, we find that disc A radiates the maximum power due to its smaller wavelength corresponding to maximum intensity, which results in a higher temperature and thus more power radiated.

### Final Answer
The correct option is (a) \( Q_A \) is maximum.