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

According to Wien's displacement law:

The wavelength of maximum intensity of radiation emitted by a star is $289.8 \, nm$. The radiation intensity for the star is: (Stefan's constant $\sigma = 5.67 \times 10^{-8} \, W m^{-2} K^{-4}$,Wien's constant $b = 2898 \, \mu m \cdot K$)

For a perfectly black body,a graph is plotted between the frequency of radiation with maximum intensity $(v_m)$ and the absolute temperature $T$. Which of the following graphs is correct?

$A$ black body radiates maximum energy at wavelength $\lambda$ and its emissive power is $E$. Now,due to a change in the temperature of that body,it radiates maximum energy at wavelength $\frac{2 \lambda}{3}$. At that temperature,the emissive power is:

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: