The plots of intensity versus wavelength for three black bodies at temperatures $T_1$,$T_2$ and $T_3$ respectively are as shown. Their temperatures are such that

$A$ particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^4 \ K$. The wavelength in nanometers at which its radiation becomes maximum is $(b = 0.0029 \ mK)$.

On investigation of light from three different stars $A, B$ and $C$, it was found that in the spectrum of $A$ the intensity of red colour is maximum, in $B$ the intensity of blue colour is maximum and in $C$ the intensity of yellow colour is maximum. From these observations, it can be concluded that

The wavelength corresponding to the maximum energy emitted by a body at a certain temperature is ${\lambda _0}$. If the temperature of the body is increased such that the new maximum wavelength becomes $\frac{3{\lambda _0}}{4}$, then by what factor does the emissive power increase?

$A$ star which appears blue will be

$A$ black body radiates maximum energy at wavelength $\lambda$ and its emissive power is $E$. Now,due to a change in the temperature of that body,it radiates maximum energy at wavelength $\frac{2 \lambda}{3}$. At that temperature,the emissive power is: