The temperature of a body is doubled but the rate fo theat radiaton remains same due to change in its surface area. The ratio of initial area to final area is

Options

A. 2
B. 4
C. 8
D. 16

Answer

To solve the problem, we need to find the ratio of the initial area (A1) to the final area (A2) when the temperature of a body is doubled, but the rate of heat radiation remains the same due to a change in its surface area.

### Step-by-Step Solution:

1. **Define Initial and Final Temperatures:**
- Let the initial temperature be \( T_1 = T \).
- The final temperature is \( T_2 = 2T \).

2. **Use Stefan-Boltzmann Law:**
- According to Stefan-Boltzmann Law, the rate of heat radiation \( E \) is given by:
\[ E = \sigma A T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the temperature.

3. **Write Expressions for Initial and Final Heat Radiation:**
- For the initial state (with area \( A_1 \)):
\[ E_1 = \sigma A_1 T_1^4 = \sigma A_1 T^4 \]
- For the final state (with area \( A_2 \)):
\[ E_2 = \sigma A_2 T_2^4 = \sigma A_2 (2T)^4 = \sigma A_2 \cdot 16T^4 \]

4. **Set the Heat Radiations Equal:**
- Since the rate of heat radiation remains the same:
\[ E_1 = E_2 \]
- Therefore:
\[ \sigma A_1 T^4 = \sigma A_2 \cdot 16T^4 \]

5. **Cancel Common Terms:**
- Cancel \( \sigma \) and \( T^4 \) from both sides:
\[ A_1 = 16 A_2 \]

6. **Find the Ratio of Areas:**
- Rearranging gives:
\[ \frac{A_1}{A_2} = 16 \]

### Final Answer:
The ratio of the initial area to the final area is:
\[ \frac{A_1}{A_2} = 16 \]