Two stars emit maximum radiation at wavelengths $3600 \mathring A$ and $4800 \mathring A$ respectively. The ratio of their temperatures is:

$A$ black body at $1227^{\circ}C$ emits radiations with maximum intensity at a wavelength of $5000\;\mathring{A}$. If the temperature of the body is increased by $1000^{\circ}C$,the maximum intensity will be observed at ...... $\mathring{A}$.

Two bodies $A$ and $B$ radiate maximum energy with a wavelength difference of $4 \mu m$. The absolute temperature of body $A$ is $3$ times that of body $B$. The wavelength at which body $B$ radiates maximum energy is: (in $\mu m$)

The graph of relative intensity versus wavelength for three perfectly black bodies at temperatures $T_1, T_2,$ and $T_3$ is shown below. The relationship between their temperatures is:

Two spherical stars $A$ and $B$ emit blackbody radiation. The radius of $A$ is $400$ times that of $B$ and $A$ emits $10^4$ times the power emitted from $B$. The ratio $(\lambda_A / \lambda_B)$ of their wavelengths $\lambda_A$ and $\lambda_B$ at which the peaks occur in their respective radiation curves is

$A$ 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: