Question

Two black spheres P & Q have radii in the ratio 4 : 3. The wavelength of maximum intensity of radiation are in the ratio 4 : 5 respectively. The ratio of radiated power by P to Q is ______.

• A. $\frac{625}{144}$
• B. $\frac{125}{81}$
• C. $\frac{25}{9}$
• D. $\frac{5}{3}$

Hint:

Combine Wien's Law and Stefan's Law into one powerful super-ratio: $\frac{E_1}{E_2} = \left( \frac{R_1}{R_2} \right)^2 \left( \frac{\lambda_2}{\lambda_1} \right)^4$. This skips the intermediate temperature step entirely!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We combine Wien's Displacement Law ($\lambda_m T = b$) and Stefan-Boltzmann Law ($P = \sigma A T^4$).

Step 2: Formula Application:
$T \propto \frac{1}{\lambda_m}$ and $A \propto R^2$. Therefore, Power $P \propto R^2 \left( \frac{1}{\lambda_m} \right)^4 \propto \frac{R^2}{\lambda_m^4}$.

Step 3: Explanation:
$\frac{P_P}{P_Q} = \left( \frac{R_P}{R_Q} \right)^2 \times \left( \frac{\lambda_{mQ}}{\lambda_{mP}} \right)^4$. Given $\frac{R_P}{R_Q} = \frac{4}{3}$ and $\frac{\lambda_{mP}}{\lambda_{mQ}} = \frac{4}{5} \implies \frac{\lambda_{mQ}}{\lambda_{mP}} = \frac{5}{4}$. Ratio $= \left( \frac{4}{3} \right)^2 \times \left( \frac{5}{4} \right)^4 = \frac{16}{9} \times \frac{625}{256} = \frac{625}{9 \times 16} = \frac{625}{144}$.

Step 4: Final Answer:
The ratio of radiated power is 625 : 144.