Three discs, A, B and C having radii 2 m, 4 m and 6 m respectively are coated with carbon black on t

Q: Why did we take the area as proportional to $r^2$ instead of $4\pi r^2$?

The problem states they are 'discs', so their surface area is $2\pi r^2$ (both sides) or $\pi r^2$ (one side). In either case, the area is directly proportional to $r^2$, and the constant factors cancel out when taking ratios.

Q: Why can we use 3, 4, and 5 for wavelengths instead of 300, 400, and 500?

Since we are only interested in the ratio of the powers, any common multiplying factor (like $100\text{ nm}$) will be raised to the same power and cancel out across all terms.

Q: What does 'coated with carbon black' mean in this context?

It implies that the discs behave as ideal black bodies. Therefore, their emissivity $e$ is considered to be 1, allowing us to use the standard Stefan-Boltzmann equation without worrying about different emissivities.

Q: Why did we substitute $T$ instead of finding its exact value?

We don't have the value of Wien's constant $b$ provided in the problem, and calculating the exact temperature would be tedious and unnecessary. Proportionality is a much faster and elegant way to compare quantities.

Q: Is it possible to solve this without converting to decimals?

Yes, you could find a common denominator for 81, 16, and 625, but since these numbers are large and relatively prime, converting to approximate decimals is much quicker for comparison.

Question

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

Options:

(A) $Q_A$ is maximum

(B) $Q_B$ is maximum

(C) $Q_C$ is maximum

(D) $Q_A = Q_B = Q_C$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine three discs, A, B, and C, coated with carbon black. This coating ensures they behave as perfect black bodies, meaning their emissivity $e \approx 1$. We are given their radii: $r_A = 2	ext{ m}$, $r_B = 4	ext{ m}$, and $r_C = 6	ext{ m}$. We are also given the wavelengths at which they emit maximum intensity: $\lambda_A = 300	ext{ nm}$, $\lambda_B = 400	ext{ nm}$, and $\lambda_C = 500	ext{ nm}$.

Our goal is to determine which of these three discs radiates the maximum power. To do this, we need to connect the physical dimensions and the emission spectra to the total radiated power.

### The Master Equation

To find the radiated power, we first recall the **Stefan-Boltzmann Law**, which states that the power $Q$ radiated by a black body is proportional to its surface area $A$ and the fourth power of its absolute temperature $T$:

$$Q = \sigma A T^4$$

Since the area of a disc is proportional to the square of its radius ($A \propto r^2$), we can write:

$$Q \propto r^2 T^4$$

However, we don't have the temperatures of the discs. This is where **Wien's Displacement Law** comes to our rescue. It tells us that the wavelength of maximum emission $\lambda_m$ is inversely proportional to the absolute temperature $T$:

$$\lambda_m T = b \implies T \propto \frac{1}{\lambda_m}$$

By substituting this temperature proportionality into our power equation, we derive our master relation:

$$Q \propto \frac{r^2}{\lambda_m^4}$$

### Final Calculation

Now, we can set up a ratio to compare the radiated powers of the three discs:

$$Q_A : Q_B : Q_C = \frac{r_A^2}{\lambda_A^4} : \frac{r_B^2}{\lambda_B^4} : \frac{r_C^2}{\lambda_C^4}$$

To simplify the calculation, we can use the proportional values for the wavelengths (3, 4, and 5 instead of 300, 400, and 500). Substituting the given values:

$$Q_A : Q_B : Q_C = \frac{2^2}{3^4} : \frac{4^2}{4^4} : \frac{6^2}{5^4}$$

Let's compute these fractions:

$$Q_A : Q_B : Q_C = \frac{4}{81} : \frac{16}{256} : \frac{36}{625}$$

Simplifying the middle term gives us $\frac{1}{16}$. Now, to easily compare these fractions, we convert them into decimals:

$$\frac{4}{81} \approx 0.049$$

$$\frac{1}{16} = 0.0625$$

$$\frac{36}{625} = 0.0576$$

Comparing these values, it is clear that $0.0625$ is the largest. Therefore, **disc B radiates the maximum power**.

Select Answer:

Visualized Solution

Visualizing the Setup

Stefan-Boltzmann Law

Wien's Displacement Law

Master Relation

Setting up the Ratio

Substituting Values

Simplifying Fractions

Decimal Conversion

Final Conclusion

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

A black body is at a temperature of $2880 K$ . The energy of radiation emitted by this body with wavelength between $499 nm$ and $500 nm$ is $U_{1}$ , between $999 nm$ and $1000 nm$ is $U_{2}$ and between $1499 nm$ and $1500 nm$ is $U_{3}$ . The Wien constant, $b = 2.88 \times 1 0^{6} nm-K$ . Then,

A body with area $A$ and temperature $T$ and emissivity $e = 0.6$ is kept inside a spherical black body. What will be the maximum energy radiated?

A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the north star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperature of the sun and the north star is

Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K, respectively. The ratio of the energy radiated per second by the first sphere to that by the second is