Three discs $X, Y$ and $Z$ having radii $2 \mathrm{~m}, 2 \mathrm{~m}$ and $6 \mathrm{~m}$, respectively are…
Three discs $X, Y$ and $Z$ having radii $2 \mathrm{~m}, 2 \mathrm{~m}$ and $6 \mathrm{~m}$, respectively are coated with carbon black on outer surfaces. The wavelength corresponding to maximum intensity are 300 nm, 400 nm and 500 nm respectively. The power radiated by them is Px,Py and Pz. Then,
$P_y$ is maximum
$P_{\mathrm{z}}$ is maximum
$P_x=P_y=P_z$
$P_x$ is maximum
Solution
As we know that from Wien's displacement law $T \propto \frac{1}{\lambda_{\max }}$
From Stefan's Law, the emitted power is proportional to temperature raised to power four and directly poroportional to the surface area of the blackbody:
$P \propto A T^4$
So, $P \propto \frac{A}{\lambda_{\max }^4}$ where $A=4 \pi r^2$ is the surface area of the body with spherical shape of radius $r$.
Therefore,
$P_x: P_y: P_z=\frac{2^2}{3^4}: \frac{2^2}{4^4}: \frac{6^2}{5^4}=\frac{4}{81}: \frac{4}{16}: \frac{36}{625}=\frac{32}{648}: \frac{156}{624}: \frac{36}{625}$
$\Rightarrow P_y>P_x>P_z$