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Three thermally radiating carbon black coated discs A, B and C with radii 2 m, 4 m and 6 m have maximum intensity wavelengths 300 nm, 400 nm and 500 nm, respectively. If the powers radiated by them are Qa, Qb and Qc, respectively, then
  • a)
    Qa is maximum
  • b)
    Qb is maximum
  • c)
    Qc is maximum
  • d)
    Qa = Qb = Qc
Correct answer is option 'B'. Can you explain this answer?
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Explanation:
The power radiated by a black body is given by the Stefan-Boltzmann law, which states that the power radiated per unit area is proportional to the fourth power of the absolute temperature of the body:

P = σAT⁴

Where P is the power radiated, σ is the Stefan-Boltzmann constant, A is the surface area, and T is the absolute temperature of the body.

In this case, we have three carbon black coated discs with different radii and maximum intensity wavelengths. The maximum intensity wavelength corresponds to the peak of the black body radiation curve, which is inversely proportional to the temperature of the body. Therefore, we can say that the temperature of disc A is higher than that of disc B, which is higher than that of disc C.

Now, let's compare the powers radiated by the three discs.

Disc A:
Radius = 2 m
Maximum intensity wavelength = 300 nm

Disc B:
Radius = 4 m
Maximum intensity wavelength = 400 nm

Disc C:
Radius = 6 m
Maximum intensity wavelength = 500 nm

Since the temperature of disc A is the highest among the three discs, it will radiate the most power. Therefore, the power radiated by disc A (Qa) is maximum.

So, the correct answer is option B - Qb is maximum.
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Three thermally radiating carbon black coated discs A, B and C with radii 2 m, 4 m and 6 m have maximum intensity wavelengths 300 nm, 400 nm and 500 nm, respectively. If the powers radiated by them are Qa, Qb and Qc, respectively, thena)Qa is maximumb)Qb is maximumc)Qc is maximumd)Qa = Qb = QcCorrect answer is option 'B'. Can you explain this answer?
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