
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
162.3k+ views
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.
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.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
