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:

When the temperature of a black body is increased, the value of $\lambda_m$ changes from $0.26 \mu m$ to $0.13 \mu m$. What is the ratio of its emissive power corresponding to these temperatures?

The rectangular surface of area $8 \text{ cm} \times 4 \text{ cm}$ of a black body at a temperature of $127^{\circ}\text{C}$ emits energy at the rate of $E$ per second. If the length and breadth of the surface are each reduced to half of the initial value and the temperature is raised to $327^{\circ}\text{C}$,the rate of emission of energy will become

$A$ thin square steel plate with each side equal to $10$ cm is heated by a blacksmith. The rate of radiated energy by the heated plate is $1134$ $W$. The temperature of the hot steel plate is ....... $K$ (Stefan's constant $\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}$,emissivity of the plate = $1$).

Two spheres made of the same material have radii in the ratio $1:2$. Both are at the same temperature. The ratio of heat radiation energy emitted per second by them is:

The rate of radiation of a black body at $0^{\circ} C$ is $E \text{ J}s^{-1}$. The rate of radiation of the black body at $273^{\circ} C$ will be