The energy spectrum of a black body has maximum energy at a wavelength of $\lambda_0$. Now, the temperature of the black body is increased such that the maximum energy is obtained at a wavelength of $3\lambda_0/4$. By what factor will the power radiated by the black body increase?

Two spheres $S_{1}$ and $S_{2}$ have radii $R$ and $3R$ and temperatures $T$ and $T/3$ respectively. If they are coated with a material of the same emissivity, and the rate of radiation of $S_{1}$ is $E$, then the rate of radiation of $S_{2}$ is:

$A$ black rectangular surface of area $A$ emits energy $E$ per second at $27^{\circ} C$. If length and breadth are reduced to $1/3$ of their initial values and the temperature is raised to $327^{\circ} C$,then the energy emitted per second becomes:

Two spheres of the same material have radii $1\ m$ and $4\ m$ and temperatures $4000\ K$ and $2000\ K$ respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

Three large identical plates are kept parallel to each other. The outer two plates are maintained at temperatures $T$ and $2T$, respectively. The temperature of the middle plate in steady state will be close to ........... $T$.

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