Question

A black sphere has radius R whose rate of radiation is E at temperature T. If radius is made $R/2$ and temperature $3T$, the rate of radiation will be

• A. $3E/2$
• B. $27E/8$
• C. $81E/4$
• D. $9E/4$

Hint:

Radiation power is highly sensitive to temperature changes ($\text{Power} \propto T^4$).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the new rate of radiation for a sphere when its dimensions and temperature change, based on Stefan-Boltzmann law.

Step 2: Key Formula or Approach:
According to Stefan-Boltzmann law, the rate of radiation \( E \) is:
\[ E = \sigma A T^4 \] For a sphere, Surface Area \( A = 4\pi R^2 \), so:
\[ E \propto R^2 T^4 \]

Step 3: Detailed Explanation:
Let initial rate be \( E_1 = k R^2 T^4 \).
Final radius \( R_2 = \frac{R}{2} \).
Final temperature \( T_2 = 3T \).
The new rate of radiation \( E_2 \) is:
\[ E_2 \propto R_2^2 T_2^4 \] \[ E_2 \propto \left( \frac{R}{2} \right)^2 (3T)^4 \] \[ E_2 \propto \frac{R^2}{4} \times 81 T^4 \] \[ E_2 = \frac{81}{4} (k R^2 T^4) \] \[ E_2 = \frac{81E}{4} \]

Step 4: Final Answer:
The new rate of radiation will be \( \frac{81E}{4} \).