Three discs $A, B$ and $C$ having radii $2 m, 4 m$ , and $6 m$ respectively are coated with carbon black on their outer surfaces. The wavelength corresponding to maximum intensity are $300 n m, 400 n m$ and $500 n m$ respectively. The power radiated by them are $Q_{A}, Q_{B}$ and $Q_{C}$ respectively.

Q_{A}

is maximum

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Q_{C}

is maximum

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Q_{B}

is maximum

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Q_{A} = Q_{B} = Q_{C}

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Solution

The correct option is D $Q_{B}$ is maximum
As we know that from wein's displacement law $λ_{m a x} T = c o n s t a n t$
Also, $Q = σ A T^{4}$ and $T \propto \frac{1}{λ}$
So, $Q \propto \frac{A}{λ_{m a x}^{4}}$ where $A = {\frac{π}{r}}^{2}$
Now, $Q_{A} : Q_{B} : Q_{C} = \frac{2^{2}}{3^{4}} : \frac{4^{2}}{4^{4}} : \frac{6^{2}}{5^{4}}$
So, $\frac{4}{81} : \frac{1}{16} : \frac{36}{625}$ $\Rightarrow Q_{B}$ is maximum.

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Similar questions

Q. Three discs

A, B, C

having radii

2 m

4 m

and

6 m

respectively are coated with carbon black on their outer surface. The wavelengths corresponding to the maximum intensity for each of them are

300 n m, 400 n m, 500 n m

respectively. The power radiated by the discs are

Q_{A}, Q_{B}

and

Q_{C}

respectively. Then,

Three discs A, B and C having radii 2 m, 4m and 6m respectively are coated with carbon black on their outer
surfaces. The wavelengths corresponding to the maximum intensity are 300 nm, 400 nm, 500 nm respectively.
The power radiated by the are $Q_{A}, Q_{B}$ and $Q_{C}$ respectively. Then