The tungsten filament of an electric lamp has a surface area A and a power rating P. If the emissivity of the filament is ϵ and σ is Stefan's constant, the steady temperature of the filament will be
T=(PAϵσ)14
The energy radiated per second per unit area at temperature T is σϵT4.
Thus, the energy radiated per second (or power radiated) from the filament of area A will be given by P=AσϵT4⇒T=(PAσϵ)14
Hence, the correct choice is (d).