Ordinary bodies $A$ and $B$ radiate maximum energy at wavelengths differing by $4 \mu m$. The absolute temperature of body $A$ is $3$ times that of body $B$. The wavelength at which body $B$ radiates maximum energy is: (in $\mu m$)

$A$ black body radiates power $P$ and maximum energy is radiated by it at a wavelength $\lambda_0$. The temperature of the black body is now changed such that it radiates maximum energy at the wavelength $\frac{\lambda_0}{4}$. The power radiated by it at the new temperature is (in $P$)

The maximum wavelength of radiation emitted at $2000\;K$ is $4\;\mu m$. What will be the maximum wavelength of radiation emitted at $2400\;K$?

If a graph is plotted by taking spectral emissive power along $y$-axis and wavelength along $x$-axis,then the area under the graph above the wavelength axis is ...........

The wavelength of maximum emitted energy of a body at $700 \ K$ is $4.08 \ \mu m$. If the temperature of the body is raised to $1400 \ K$,the wavelength of maximum emitted energy will be ........ $\mu m$.

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