Question

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

Given,
$R_A=2\text{m};R_B=4\text{m};R_C=6\text{m}$

${\lambda }_{mA}=300\text{nm}{\lambda }_{mB}=400\text{nm}{\lambda }_{mC}=500\text{nm}$

Using Wein's displacement law,

${\lambda }_m.T=b$, where $b$ is constant.

Power radiated by a body at temperature $T$ having Area $A$ is given by Stefan's law

$Q=e\sigma A{T}^4$

For a thin disc of radius $R$ emitting from both surfaces,
$A=2\pi R^2$

$\Rightarrow Q=e\sigma \left(2\pi R^2\right){\left(\frac{b}{{\lambda }_m}\right)}^4$

Since $e,\sigma$ and $b$ are same for all three disc.

$\Rightarrow Q\propto \frac{R^2}{({\lambda }_m{)}^4}$

$\Rightarrow Q_A:Q_B:Q_C=\frac{(R_A{)}^2}{({\lambda }_{mA}{)}^4}:\frac{(r_B{)}^2}{({\lambda }_{mB}{)}^4}:\frac{(r_C{)}^2}{({\lambda }_{mC}{)}^4}$

$\Rightarrow Q_A:Q_B:Q_C=\frac{(2{)}^2}{(300{)}^4}:\frac{(4{)}^2}{(400{)}^4}:\frac{(6{)}^2}{(500{)}^4}$

$\Rightarrow Q_A:Q_B:Q_C=0.049:0.0625:0.057$

$\therefore Q_B$ is maximum.

Hence, option $\left(B\right)$ is correct.