If the rate of emission of energy from a star is `2.7 xx 10^(36)` J/ sec, the rate of loss of mass in the star will be

Options

A. `3xx10^(18) kg//sec`
B. `3xx10^(19) kg//sec`
C. `3xx10^(20) kg//sec`
D. `3xx10^(21) kg//sec`

Answer

To find the rate of loss of mass in the star given the rate of emission of energy, we can use Einstein's mass-energy equivalence principle, which is expressed by the equation:

\[ E = mc^2 \]

Where:
- \( E \) is the energy,
- \( m \) is the mass,
- \( c \) is the speed of light in a vacuum (approximately \( 3 \times 10^8 \) m/s).

### Step-by-Step Solution:

1. **Identify the given values:**
- Rate of emission of energy from the star, \( \frac{dE}{dt} = 2.7 \times 10^{36} \) J/s.
- Speed of light, \( c = 3 \times 10^8 \) m/s.

2. **Differentiate the energy-mass relation:**
- From \( E = mc^2 \), if we differentiate both sides with respect to time, we get:
\[ \frac{dE}{dt} = c^2 \frac{dm}{dt} \]
- Rearranging this gives us:
\[ \frac{dm}{dt} = \frac{1}{c^2} \frac{dE}{dt} \]

3. **Substitute the values into the equation:**
- We know \( \frac{dE}{dt} = 2.7 \times 10^{36} \) J/s and \( c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \) m²/s².
- Substitute these values into the equation:
\[ \frac{dm}{dt} = \frac{2.7 \times 10^{36}}{9 \times 10^{16}} \]

4. **Perform the division:**
- Calculate the division:
\[ \frac{dm}{dt} = \frac{2.7}{9} \times 10^{36 - 16} = 0.3 \times 10^{20} = 3 \times 10^{19} \text{ kg/s} \]

5. **Final result:**
- Therefore, the rate of loss of mass in the star is:
\[ \frac{dm}{dt} = 3 \times 10^{19} \text{ kg/s} \]