The sun radiates energy in all directions. The average radiations received on the earth surface from the sun is `1.4 "kilowatt"//m^2`. The average earth-sun distance is `1.5 xx 10^11` meters. The mass lost by the sun per day is.

Options

A. `4.4xx10^(9)kg`
B. `7.6xx10^(14)kg`
C. `3.8xx10^(12)kg`
D. `3.8xx10^(14)kg`

Explanation

To find the mass lost by the Sun per day, we can follow these steps:

### Step 1: Understand the Energy Radiated
The average radiation received on the Earth's surface from the Sun is given as \(1.4 \, \text{kW/m}^2\). We need to convert this into energy per day.

1. Convert kilowatts to joules:
\[ 1 \, \text{kW} = 1000 \, \text{J/s} \]
Therefore,
\[ 1.4 \, \text{kW/m}^2 = 1.4 \times 1000 \, \text{J/s/m}^2 = 1400 \, \text{J/s/m}^2 \]

### Step 2: Calculate the Total Energy Received by Earth in One Day
The total energy received by the Earth in one day can be calculated using the formula:
\[ E = \text{Power} \times \text{Time} \]
Where:
- Power = \(1400 \, \text{J/s/m}^2\)
- Time = \(86400 \, \text{s}\) (1 day)

Thus, the total energy received by the Earth per square meter in one day is:
\[ E = 1400 \, \text{J/s/m}^2 \times 86400 \, \text{s} = 120960000 \, \text{J/m}^2 \]

### Step 3: Calculate the Total Surface Area of a Sphere with Radius Equal to Earth-Sun Distance
The average Earth-Sun distance is given as \(1.5 \times 10^{11} \, \text{m}\). The surface area \(A\) of a sphere is given by:
\[ A = 4 \pi r^2 \]
Substituting \(r = 1.5 \times 10^{11} \, \text{m}\):
\[ A = 4 \pi (1.5 \times 10^{11})^2 \]

Calculating this:
\[ A = 4 \pi (2.25 \times 10^{22}) \approx 4 \times 3.14 \times 2.25 \times 10^{22} \approx 28.26 \times 10^{22} \, \text{m}^2 \]

### Step 4: Calculate the Total Energy Radiated by the Sun
Now, we can find the total energy radiated by the Sun to the entire surface area:
\[ E_{\text{total}} = E \times A \]
Substituting the values:
\[ E_{\text{total}} = 120960000 \, \text{J/m}^2 \times 28.26 \times 10^{22} \, \text{m}^2 \]

Calculating this:
\[ E_{\text{total}} \approx 3.42 \times 10^{30} \, \text{J} \]

### Step 5: Relate Energy to Mass Using \(E = mc^2\)
According to Einstein's mass-energy equivalence principle:
\[ E = mc^2 \]
Rearranging gives:
\[ m = \frac{E}{c^2} \]
Where \(c = 3 \times 10^8 \, \text{m/s}\).

Substituting the values:
\[ m = \frac{3.42 \times 10^{30} \, \text{J}}{(3 \times 10^8 \, \text{m/s})^2} \]
Calculating \(c^2\):
\[ c^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \]
Thus,
\[ m = \frac{3.42 \times 10^{30}}{9 \times 10^{16}} \approx 3.8 \times 10^{13} \, \text{kg} \]

### Final Answer
The mass lost by the Sun per day is approximately \(3.8 \times 10^{13} \, \text{kg}\).

---