The exhaust velocity of gases with respect to a small rocket of mass 25 kg. is `28xx10^(2)m//s`. At what rate the fuel must burn so that it may rise up with an acceleration of `9.8 m//s^(2)`?

Options

A. 175 Kg
B. 1.75 Kg/s
C. 0.175 Kg/s
D. Zero

Answer

To solve the problem, we need to determine the rate at which fuel must burn (denoted as \( \frac{dm}{dt} \)) for the rocket to achieve an upward acceleration of \( 9.8 \, \text{m/s}^2 \). We will use the rocket equation and Newton's second law.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the rocket, \( m = 25 \, \text{kg} \)
- Exhaust velocity of gases, \( v = 28 \times 10^2 \, \text{m/s} = 2800 \, \text{m/s} \)
- Acceleration, \( a = 9.8 \, \text{m/s}^2 \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

2. **Calculate the Total Force Required:**
The total force \( F \) needed for the rocket to rise with an acceleration of \( 9.8 \, \text{m/s}^2 \) can be calculated using Newton's second law:
\[ F = m(g + a) \]
Substituting the values:
\[ F = 25 \, \text{kg} \times (9.8 \, \text{m/s}^2 + 9.8 \, \text{m/s}^2) = 25 \, \text{kg} \times 19.6 \, \text{m/s}^2 = 490 \, \text{N} \]

3. **Apply the Rocket Equation:**
According to the rocket equation, the thrust \( F \) can also be expressed as:
\[ F = v \frac{dm}{dt} \]
Where \( \frac{dm}{dt} \) is the rate of fuel consumption.

4. **Set the Forces Equal:**
Now we can set the two expressions for force equal to each other:
\[ v \frac{dm}{dt} = 490 \, \text{N} \]

5. **Solve for \( \frac{dm}{dt} \):**
Rearranging the equation to find \( \frac{dm}{dt} \):
\[ \frac{dm}{dt} = \frac{490 \, \text{N}}{v} \]
Substituting the value of \( v \):
\[ \frac{dm}{dt} = \frac{490 \, \text{N}}{2800 \, \text{m/s}} = \frac{490}{2800} = 0.175 \, \text{kg/s} \]

6. **Final Result:**
The rate at which the fuel must burn is:
\[ \frac{dm}{dt} = 0.175 \, \text{kg/s} \]

### Conclusion:
The required rate of fuel consumption for the rocket to rise with an acceleration of \( 9.8 \, \text{m/s}^2 \) is \( 0.175 \, \text{kg/s} \).