$A$ 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:

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:

Two spherical black bodies of radius $r_{1}$ and $r_{2}$ with surface temperatures $T_{1}$ and $T_{2}$ respectively,radiate the same power. Then the ratio $r_{1}: r_{2}$ is:

Two spherical black bodies of radii $r_1$ and $r_2$ at temperatures $T_1$ and $T_2$ respectively radiate power in the ratio $1:2$. Then $r_1:r_2$ is:

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).

$A$ black body radiates energy at a rate of $E \, W/m^2$ at a temperature of $T \, K$. When the temperature is reduced to $T/2 \, K$,the radiated energy will be .....