In the figure,the distribution of energy density of the radiation emitted by a black body at a given temperature is shown. The possible temperature of the black body is ....... $K$.

Two stars emit maximum radiation of wavelength $3600 \ \mathring{A}$ and $4800 \ \mathring{A}$ respectively. The ratio of their temperatures is

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

Three stars $A, B, C$ have surface temperatures $T_{A}, T_{B}, T_{C}$ respectively. Star $A$ appears bluish,star $B$ appears reddish,and star $C$ appears yellowish. Hence,

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:

If the temperature of a black body is doubled,the frequency at which the spectral intensity becomes maximum will be