The ratio of the energy of emitted radiation of a black body at $27^{\circ}C$ and $927^{\circ}C$ is:

Two electric lamps $A$ and $B$ radiate the same power. Their filaments have the same dimensions and have emissivities $e_A$ and $e_B$. Their surface temperatures are $T_A$ and $T_B$. The ratio $\frac{T_A}{T_B}$ will be equal to :-

The rate of radiation by a black body is $R$ at temperature $T$. Another body has the same area but an emissivity of $0.2$ and a temperature of $3T$. Its rate of radiation is: (in $R$)

$A$ black body of mass $34.38 \ g$ and surface area $19.2 \ cm^2$ is at an initial temperature of $400 \ K$. It is allowed to cool inside an evacuated enclosure kept at a constant temperature of $300 \ K$. The rate of cooling is $0.04 \ ^{\circ}C/s$. The specific heat of the body in $J \ kg^{-1} \ K^{-1}$ is (Stefan's constant $\sigma = 5.73 \times 10^{-8} \ W \ m^{-2} \ K^{-4}$)

The original temperature of a black body is $727^{\circ}C$. The temperature at which this black body must be raised so as to double the total radiant energy is ....... $K$.

Two spherical black bodies of radii $R_1$ and $R_2$ and with surface temperatures $T_1$ and $T_2$ respectively radiate the same power. The ratio of $R_1$ to $R_2$ will be