If a black wire of platinum is heated, its colour first appears red, then yellow, and finally white. This can be understood on the basis of:

The plots of intensity $(I)$ versus wavelength $(\lambda)$ for three black bodies at temperatures $T_1, T_2$ and $T_3$ respectively are as shown. Their temperatures are such that:

The radiation energy density per unit wavelength at a temperature $T$ has a maximum at a wavelength $\lambda_0$. At temperature $2T$,it will have a maximum at a wavelength:

If $\lambda$ denotes the wavelength at which the radiative emission from a black body at a temperature $T$ is maximum,then

$A$ piece of iron is heated in a flame. It first becomes dull red, then becomes reddish yellow, and finally turns to white hot. The correct explanation for the above observation is possible by using

Two bodies $A$ and $B$ of equal surface area have thermal emissivities of $0.01$ and $0.81$ respectively. The two bodies are radiating energy at the same rate. Maximum energy is radiated from the two bodies $A$ and $B$ at wavelengths $\lambda_A$ and $\lambda_B$ respectively. The difference in these two wavelengths is $1 \mu m$. If the temperature of body $A$ is $5802 \ K$,then the value of $\lambda_B$ is: