Radiated energy at $T$ temperature is $E$ for a body of diameter $d$. If the temperature becomes $2T$ and the diameter becomes $d/4$,then the radiated energy will be:

Star $A$ has radius $r$ and surface temperature $T$,while star $B$ has radius $4r$ and surface temperature $T/2$. The ratio of the power radiated by the two stars,$P_A : P_B$,is:

$A$ rectangular black body of temperature $127^{\circ} C$ has surface area $4 \ cm \times 2 \ cm$ and rate of radiation is $E$. If its temperature is increased by $400^{\circ} C$ and surface area is reduced to half of the initial value,then the rate of radiation is:

$A$ sphere of surface area $4 \ m^2$ at temperature $400 \ K$ and having emissivity $0.5$ is located in an environment of temperature $200 \ K$. The net rate of energy exchange of the sphere is (Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8} \ W \ m^{-2} \ K^{-4}$) (in $W$)

The 'Kangri' is an earthen pot used to stay warm in Kashmir during the winter months. Assume that the 'Kangri' is spherical and of surface area $7 \times 10^{-2} \,m^{2}$. It contains $300 \,g$ of a mixture of coal,wood,and leaves with a calorific value of $30 \,kJ/g$ (and provides heat with $10 \%$ efficiency). The surface temperature of the 'Kangri' is $60^{\circ}C$ and the room temperature is $0^{\circ}C$. Then,a reasonable estimate for the duration $t$ (in $h$) that the 'Kangri' heat will last is (take the 'Kangri' to be a black body).

The spectrum of a black body at two temperatures $27\,^oC$ and $327\,^oC$ is shown in the figure. Let $A_1$ and $A_2$ be the areas under the two curves respectively. Find the value of $\frac {A_2}{A_1}$.