Question

Three discs $A,B,C$ having radii $2m$, $4m$ and $6m$ respectively are coated with carbon black on their outer surface. The wavelengths corresponding to the maximum intensity for each of them are $300nm,400nm,500nm$ respectively. The power radiated by the discs are $Q_A,Q_B$ and $Q_C$ respectively. Then,

• A. $Q_A$ is maximum
• B. $Q_B$ is maximum
• C. $Q_C$ is maximum
• D. $Q_A=Q_B=Q_C$

Answer 1

The correct option is B $Q_B$ is maximum

According to Stefan boltzmann law
$Q=\sigma A{T}^4$, Here $A=\pi r^2$

According to wien's law
${\lambda }_{m}T=b$
or, $T=\left(\frac{b}{{\lambda }_m}\right)\therefore Q=\sigma \left(\pi r^2\right){\left(\frac{b}{{\lambda }_m}\right)}^4$
$Q\propto \frac{r^2}{{\lambda }_m^4}$

$Q_A\propto \frac{2^2}{{300}^4}$

$Q_B\propto \frac{4^2}{{400}^4}$

$Q_C\propto \frac{6^2}{{500}^4}$

$\therefore Q_B>Q_C>Q_A$