Two identical bodies have temperatures $277^{\circ} C$ and $67^{\circ} C$. If the surroundings temperature is $27^{\circ} C$, the ratio of loss of heats of the two bodies during the same interval of time is (approximately) (in $ : 1$)

Two bodies $P$ and $Q$ have thermal emissivities of $\varepsilon_P$ and $\varepsilon_Q$ respectively. Surface areas of these bodies are same and the total radiant power emitted is also at the same rate. If the temperature of $P$ is $\theta_P$ Kelvin,then the temperature of $Q$ i.e.,$\theta_Q$ is:

The temperature of a body is increased from $27\,^{\circ}C$ to $127\,^{\circ}C$. The radiation emitted by it increases by a factor of:

The temperature of a body is increased from $-73^{\circ}C$ to $327^{\circ}C$. The ratio of energy emitted per second is:

The temperature of a perfect black body is $727^{\circ}C$ and its surface area is $0.1\, m^{2}$. If the Stefan-Boltzmann constant is $\sigma = 5.67 \times 10^{-8} \, W/m^{2} \cdot K^{4}$,then the heat radiated in $1\, min$ is ........ $cal$.

$A$ wire of length $L$ and diameter $d$ is used in a bulb. The temperature of the wire is $T$ and the power radiated by the wire is $P$. Its emissivity is $e$ ($\sigma$ = Stefan's constant). (Assume that the emissivity of the wire material is the same at all wavelengths).