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 will use Stefan-Boltzmann's law, which states that the rate of heat radiation emitted by a body is proportional to the fourth power of its absolute temperature and its surface area.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are given that the temperature of a body is doubled (from \(T_1\) to \(T_2 = 2T_1\)) and the rate of heat radiation remains the same due to a change in its surface area.

2. **Applying Stefan-Boltzmann's Law**:
According to Stefan-Boltzmann's law, the rate of heat radiation \(E\) can be expressed as:
\[ E = \sigma A T^4 \]
where \(E\) is the rate of radiation, \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area, and \(T\) is the absolute temperature.

3. **Setting Up the Initial Conditions**:
Let the initial area of the body be \(A_1\) and the initial temperature be \(T_1\). The rate of radiation in this case can be expressed as:
\[ E_1 = \sigma A_1 T_1^4 \]

4. **Setting Up the Final Conditions**:
After the temperature is doubled, the new temperature is \(T_2 = 2T_1\). Let the new surface area be \(A_2\). The rate of radiation in this case can be expressed as:
\[ E_2 = \sigma A_2 (2T_1)^4 \]
Simplifying this, we get:
\[ E_2 = \sigma A_2 \cdot 16 T_1^4 \]

5. **Equating the Two Rates of Radiation**:
Since it is given that the rate of heat radiation remains the same, we can set \(E_1\) equal to \(E_2\):
\[ \sigma A_1 T_1^4 = \sigma A_2 \cdot 16 T_1^4 \]

6. **Canceling Common Terms**:
We can cancel \(\sigma\) and \(T_1^4\) from both sides (assuming they are non-zero):
\[ A_1 = 16 A_2 \]

7. **Finding the Ratio of Areas**:
Rearranging the equation gives us:
\[ \frac{A_1}{A_2} = 16 \]

### Conclusion:
The ratio of the initial area to the final area is:
\[ \frac{A_1}{A_2} = 16:1 \]