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`

Options

A. `Q_A` will be maximum
B. `Q_B` will be maximum
C. `Q_C` will be maximum
D. `Q_A=Q_B=Q_C`

Answer

B

Explanation

We know that `lambda_(m)T=clambda_(A) lt lambda_(B) lt lambda_(C)`
So, `T_(A) gt T_(B) gt T_(C)`
`{ :. T_(A)=(C)/(3xx10^(-7)),T_(B)=(C)/(4xx10^(-7)),T_(C)=(C)/(5xx10^(-7))}`
`Q=esigmaAT^(4)`,(`e=1` black body)
`:. Q=sigmaAT^(4)`
`:. Q_(A)=sigmapi(2xx10^(-2))^(2)xx(C^(4))/(27xx10^(-28))`
and `Q_(B)=sigma(4xx10^(-2))xx(C^(2))/(64xx10^(-28))`
`Q_(C)=sigmapi(6xx10^(-2))^(2)xx(C^(2))/(625xx10^(-28))`
From comparison, `Q_(B)` is maximum.