A rocket of mass 1000 kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity `100m//s` with respect to the rocket. (What is the minimum rate of burning fuel, so as to just lift the rocket upwards against the gravitational attraction?) (Take `g=10m//s^(2)`)

Options

A. 50 kg/s
B. 100kg/s
C. 200kg/s
D. 400kg/s

Explanation

To solve the problem of determining the minimum rate of burning fuel for the rocket to lift off against gravitational force, we can follow these steps:

### Step 1: Identify the forces acting on the rocket
The primary force acting on the rocket is the gravitational force, which can be calculated using the formula:
\[ F_g = m \cdot g \]
where:
- \( m \) is the mass of the rocket (1000 kg),
- \( g \) is the acceleration due to gravity (10 m/s²).

### Step 2: Calculate the gravitational force
Substituting the values into the formula:
\[ F_g = 1000 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10,000 \, \text{N} \]

### Step 3: Establish the condition for lift-off
For the rocket to just lift off, the thrust produced by the rocket must be equal to the gravitational force acting on it. The thrust \( F \) can be expressed in terms of the rate of mass burning \( \frac{dm}{dt} \) and the velocity of the exhaust gases \( v \):
\[ F = \frac{dm}{dt} \cdot v \]

### Step 4: Set up the equation for thrust
To achieve lift-off:
\[ \frac{dm}{dt} \cdot v = F_g \]
Substituting the known values:
\[ \frac{dm}{dt} \cdot 100 \, \text{m/s} = 10,000 \, \text{N} \]

### Step 5: Solve for the rate of burning fuel
Rearranging the equation to find \( \frac{dm}{dt} \):
\[ \frac{dm}{dt} = \frac{10,000 \, \text{N}}{100 \, \text{m/s}} \]
\[ \frac{dm}{dt} = 100 \, \text{kg/s} \]

### Conclusion
The minimum rate of burning fuel required for the rocket to just lift off is:
\[ \frac{dm}{dt} = 100 \, \text{kg/s} \]

### Final Answer
The answer is \( 100 \, \text{kg/s} \).

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