$A$ black rectangular surface of area $A$ emits energy $E$ per second at $127^\circ C$. If length and breadth are reduced to half of their initial values and the temperature is raised to $527^\circ C$, then the energy emitted becomes:

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

$A$ black body radiates maximum energy at wavelength $\lambda$ and its emissive power is $E$. Now,due to a change in temperature of that body,it radiates maximum energy at wavelength $\frac{\lambda}{3}$. At that new temperature,the emissive power is: (in $E$)

$A$ 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 $\sigma = 5.73 \times 10^{-8} \ W \ m^{-2} \ K^{-4}$)

Radiation from a black body at the thermodynamic temperature $T_1$ is measured by a small detector at distance $d_1$ from it. When the temperature is increased to $T_2$ and the distance to $d_2$,the power received by the detector is unchanged. What is the ratio $d_2/d_1$?

$A$ black metal foil is warmed by radiation from a small sphere at temperature $T$ and at a distance $d$. It is found that the power received by the foil is $P$. If both the temperature and distance are doubled,the power received by the foil will be: