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)` is maximum
B. `Q_(B)` is maximum
C. `Q_(C)` is maximum
D. `Q_(A)=Q_(B)=Q_(C)`

Answer

B

Explanation

According to Wien's displacement law
`lambda T =b` wien's constant
`:. lamda_(A) T_(A) =b or T_(A) = (b)/(3xx10^(-7))`
`T_(A) = (b xx 10^(7))/(3) = (z)/(3)` where `z = (b xx 10^(7))`
Similarly `T_(B) = (b xx 10^(7))/(4) = (z)/(4)` and `T_(C) = (bxx10^(7))/(5) = (z)/(5)`
Again according to Stefan's law
`Q` = Power radiated by black body `= AsigmaT^(4)`
where `A` = area of disc `=piR^(2)`
`Q_(A) = (piR_(A)^(2)) xx sigma xx (T_(A))^(4) or Q_(A) = pi(2 xx 10^(-2))^(2) xx sigma ((z)/(3))^(4)`
`Q_(4) = pisigma xx 10^(-4) xx z^(4) xx (2^(2))/(3^(4))`
or `Q_(A) = (pi sigma xx 10^(-4) xx z^(4)) xx (4)/(81)`
Put `pi sigma xx 10^(-4) xx z^(4) = k = constant`
or `Q_(A) = (4k)/(81) = 0.049k`
Similary `Q_(B) = (kxx (4)^(2))/((4)^(2))=(K)/(16)=0.062k`
`Q_(C) = (k xx (6)^(2))/((5)^(4)) = (36K)/(625) =0.037k` Hence `Q_(B)` is maximum .