Sun radiates energy at the rate of `3.6xx10^(26)J//s`. The rate of decrease in mass of sun is `(Kgs^(1))`.

Options

A. `12xx10^(10)`
B. `1.3xx10^(20)`
C. `4xx10^(9)`
D. `3.6xx10^(36)`

Explanation

To solve the problem, we need to find the rate of decrease in mass of the Sun based on the energy it radiates. We will use Einstein's mass-energy equivalence principle, which is given by the equation \(E = mc^2\).

### Step-by-step Solution

1. **Identify Given Values**:
- The rate of energy radiation from the Sun, \( \frac{dE}{dt} = 3.6 \times 10^{26} \, \text{J/s} \).
- The speed of light, \( c = 3 \times 10^8 \, \text{m/s} \).

2. **Use the Mass-Energy Equivalence**:
- According to Einstein's equation, the relationship between energy and mass is given by:
\[ E = mc^2 \]
- If we differentiate this equation with respect to time, we get:
\[ \frac{dE}{dt} = c^2 \frac{dm}{dt} \]

3. **Rearranging the Equation**:
- We can rearrange the equation to solve for the rate of change of mass:
\[ \frac{dm}{dt} = \frac{1}{c^2} \frac{dE}{dt} \]

4. **Substituting the Known Values**:
- Substitute \( \frac{dE}{dt} \) and \( c^2 \):
\[ c^2 = (3 \times 10^8 \, \text{m/s})^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \]
- Now substituting into the equation:
\[ \frac{dm}{dt} = \frac{1}{9 \times 10^{16}} \times 3.6 \times 10^{26} \]

5. **Calculating the Rate of Mass Loss**:
- Perform the calculation:
\[ \frac{dm}{dt} = \frac{3.6}{9} \times 10^{26 - 16} = 0.4 \times 10^{10} = 4 \times 10^{9} \, \text{kg/s} \]

6. **Final Result**:
- The rate of decrease in mass of the Sun is:
\[ \frac{dm}{dt} \approx 4 \times 10^{9} \, \text{kg/s} \]

### Summary
The rate of decrease in mass of the Sun is approximately \(4 \times 10^{9} \, \text{kg/s}\).