The Sun emits electromagnetic energy at a rate of $3.9 \times 10^{25} \ W$. Its radius is $6.96 \times 10^8 \ m$. The intensity of sunlight at the solar surface in $W \ m^{-2}$ is:

$A$ black body at a temperature of $227^{\circ} C$ radiates heat energy at the rate of $5 \ cal/cm^{2}-s$. At a temperature of $727^{\circ} C$,the rate of heat radiated per unit area in $cal/cm^{2}-s$ will be:

The energy spectrum of a black body has maximum energy at a wavelength of $\lambda_0$. Now, the temperature of the black body is increased such that the maximum energy is obtained at a wavelength of $3\lambda_0/4$. By what factor will the power radiated by the black body increase?

The rate of radiation of a black body at $0^{\circ}C$ is $E \ J/s$. The rate of radiation of this black body at $273^{\circ}C$ will be:

If the temperature of the $Sun = 6000 \, K$,the radius of the Sun is $7.2 \times 10^{5} \, km$,the radius of the Earth is $6000 \, km$,and the distance between the Earth and the Sun is $15 \times 10^{7} \, km$,find the intensity of light on Earth.

The temperature of a spherical black body is inversely proportional to its radius. If its radius is doubled,then the power radiating from it will be