Three discs $$A, B$$  and $$C$$ having radii $$2, 4$$  and $$6\,cm$$  respectively are coated with carbon black. Wavelength for maximum intensity for the three discs are $$300, 400$$   and $$500\,nm$$  respectively. If $${Q_A},$$ $${Q_B}$$ and $${Q_C}$$ are power emitted by $$A, B$$  and $$D$$ respectively, then

Solution :
We know that
$$\eqalign{ & {\lambda _m}T = {\text{Constant}} \cr & {\lambda _A} < {\lambda _B} < {\lambda _C} \cr & {\text{So, }}{T_A} > {T_B} > {T_C} \cr & \left\{ {\because \,\,{T_A} = \frac{C}{{3 \times {{10}^{ - 7}}}},{T_B} = \frac{C}{{4 \times {{10}^{ - 7}}}},{T_C} = \frac{C}{{5 \times {{10}^{ - 7}}}}} \right\} \cr & Q = e\sigma A{T^4} \cr & e = 1\,\,{\text{black }}{\text{body}} \cr & \therefore \,\,Q = \sigma A{T^4} \cr & \therefore \,\,{Q_A} = \sigma .\pi {\left( {2 \times {{10}^{ - 2}}} \right)^2} \times \frac{{{C^4}}}{{27 \times {{10}^{ - 28}}}} \cr & {Q_B} = \sigma .\pi {\left( {4 \times {{10}^{ - 2}}} \right)^2} \times \frac{{{C^2}}}{{64 \times {{10}^{ - 28}}}} \cr & {\text{and, }}{Q_C} = \sigma .\pi {\left( {6 \times {{10}^{ - 2}}} \right)^2} \times \frac{{{C^2}}}{{625 \times {{10}^{ - 28}}}} \cr} $$
From comparison $${Q_B}$$ is maximum.