Question

Three discs $A, B$ and $C$ having radii $2 \mathrm{~m}, 4 \mathrm{~m}$, and $6 \mathrm{~m}$ respectively are coated with carbon black on their outer surfaces. The wavelength corresponding to maximum intensity are $300 \mathrm{~nm}, 400 \mathrm{~nm}$ and $500 \mathrm{~nm}$ respectively. The power radiated by them are $Q_{A^{\prime}} Q_{B}$ and $Q_{C}$ respectively.

A. $Q_{A}$ is maximum
B. $Q_{B}$ is maximum
C. $Q_{C}$ is maximum
D. $Q_{A}=Q_{B}=Q_{C}$

Answer 1

Hint: Here three discs are coated with carbon black on their outer surfaces. Wavelength of corresponding maximum intensity is given. We have to find the relation between power radiated by them. We use Wein's displacement law and Stephan's law to find the relation between the power of three discs.






Formula used:
From Wein's displacement law,
$\lambda_{\max } T=$ constant
Stephan's law can be mathematically represented as:
$Q=\sigma A T^{4}$
Where $\mathrm{A}$ is the area of disc, here $A=\pi \mathrm{r}^{2}$







Complete answer:
There are three discs with carbon black coating on the outer surface. Wavelengths corresponding to maximum intensities are also given. Power radiated by each disc is given as $Q_{A^{\prime}} Q_{B}$ and $Q_{C}$. We have to find the relation between these powers radiated by these discs.
We can use Wein's displacement law and Stephan's law to connect the given details in the equation to find the relation between power radiated by these discs.
Wein's displacement law states that black body radiation has different peaks of temperature at different wavelengths and temperature and wavelength are inversely proportional to each other.
Mathematically, $\lambda_{\max } T=$ constant $-(1)$
Stephan's law states that the radiation emitted by a black body per unit area is directly proportional to the fourth power of the temperature.
Mathematically, $Q=\sigma A T^{4}-(2)$
We can substitute (1) in (2) and adding value of area in (2)
We get power radiated, $Q \propto \dfrac{r^{2}}{\lambda_{\max }^{4}}$ Power radiated by disc A, $Q_{A}=\dfrac{2^{2}}{(300 n m)^{2}}$
Power radiated by disc B, $Q_{B}=\dfrac{4^{2}}{(400 n m)^{2}}$
Power radiated by disc $\mathrm{C}, Q_{c}=\frac{6^{2}}{(500 \mathrm{~nm})^{2}}$
Taking ratios of power, we get:
$Q_{A}: Q_{B}: Q_{C}=\dfrac{2^{2}}{300^{4}}: \dfrac{4^{2}}{400^{4}}: \dfrac{6^{2}}{500^{4}}=\dfrac{4}{3^{4}}: \dfrac{4^{2}}{4^{4}}: \dfrac{36}{5^{4}}=\dfrac{4}{81}: \dfrac{1}{16}: \dfrac{36}{625}=0.049: 0.063: 0.058$
That is, $Q_{B}$ is maximum
Therefore, the answer is option (B)





Note:Here it is not necessary to take the ratios to find relation between power. But if we take ratio, then we can avoid the constant values and it will be easier to find the solution.