The distribution of relative intensity $I(\lambda)$ of blackbody radiation from a solid object versus the wavelength $\lambda$ is shown in the figure. If the Wien displacement law constant is $2.9 \times 10^{-3} \ mK$,what is the approximate temperature of the object in $K$?

In determining the temperature of a distant star,one makes use of

Solar radiation emitted by the sun resembles that emitted by a black body at a temperature of $6000 \ K$. Maximum intensity is emitted at a wavelength of about $4800 \ \mathring{A}$. If the sun were to cool down from $6000 \ K$ to $3000 \ K$,then the peak intensity would occur at a wavelength of ....... $\mathring{A}$.

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 wavelengths of maximum intensity of radiations emitted by the sun and the moon are $0.5 \times 10^{-6} \ m$ and $10^{-4} \ m$ respectively,the ratio of their temperatures is:

$Assertion :$ For higher temperature, the peak emission wavelength of a blackbody shifts to lower wavelengths.
$Reason :$ Peak emission wavelength of a blackbody is proportional to the fourth power of temperature.