Question

A rectangular black body of temperature \(127^{\circ} \mathrm{C}\) has surface area \(4 \mathrm{~cm} \times 2 \mathrm{~cm}\) and rate of radiation is E . If its temperature is increased by \(400^{\circ} \mathrm{C}\) and surface area is reduced to half of the initial value then the rate of radiation is

• A. 4 E
• B. E
• C. 2 E
• D. 8 E

Hint:

The rate of radiation from a black body is proportional to area times the fourth power of the absolute temperature.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a black body the rate of radiation is \(E\propto A\,T^{4}\), where \(T\) is in kelvin.

Step 2: Convert temperatures
Initial \(T_1=127+273=400\ \text{K}\).
Final temperature is \(127+400=527^{\circ}\text{C}\), so \(T_2=800\ \text{K}\).

Step 3: Calculation
\(\dfrac{E_2}{E_1}=\dfrac{A_2}{A_1}\left(\dfrac{T_2}{T_1}\right)^{4}=\dfrac{1}{2}\times 2^{4}=8\).

Step 4: Compare the options
The new rate is \(8E\), option (D).

Final Answer:
The new rate of radiation is 8 times the original, 8E. \[ \boxed{\text{(D) }8\,\text{E}} \]