For a Rocket propulsion velocity of exhaust gases relative to rocket is 2km/s. If mass of rocket system is 1000 kg, then the rate of fuel consumption for a rockt to rise up with acceleration `4.9 m//s^(2)` will be:-

Options

A. 12.25 kg/s
B. 17.5 kg/s
C. 7.35 kg/s
D. 5.2 kg/s

Explanation

To solve the problem step by step, we will use the principles of rocket propulsion and Newton's second law of motion.

### Step 1: Understand the Forces Acting on the Rocket
When the rocket is rising, two main forces act on it:
1. The thrust force (upward) due to the exhaust gases.
2. The gravitational force (downward) acting on the rocket.

### Step 2: Write the Equation of Motion
According to Newton's second law, the net force acting on the rocket can be expressed as:
\[ F_{\text{net}} = F_{\text{thrust}} - F_{\text{gravity}} = ma \]
Where:
- \( F_{\text{thrust}} \) is the thrust force generated by the exhaust gases.
- \( F_{\text{gravity}} = mg \) is the weight of the rocket.
- \( m \) is the mass of the rocket.
- \( a \) is the acceleration of the rocket.

### Step 3: Express Thrust Force
The thrust force can be expressed in terms of the rate of fuel consumption (\( \frac{dm}{dt} \)) and the velocity of the exhaust gases (\( V_r \)):
\[ F_{\text{thrust}} = V_r \cdot \frac{dm}{dt} \]
Where:
- \( V_r = 2 \, \text{km/s} = 2000 \, \text{m/s} \)

### Step 4: Substitute Known Values
Now, substituting the expressions for thrust and gravity into the equation of motion:
\[ V_r \cdot \frac{dm}{dt} - mg = ma \]
Substituting \( V_r = 2000 \, \text{m/s} \), \( m = 1000 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( a = 4.9 \, \text{m/s}^2 \):
\[ 2000 \cdot \frac{dm}{dt} - 1000 \cdot 9.8 = 1000 \cdot 4.9 \]

### Step 5: Simplify the Equation
Now simplify the equation:
\[ 2000 \cdot \frac{dm}{dt} - 9800 = 4900 \]
Adding \( 9800 \) to both sides:
\[ 2000 \cdot \frac{dm}{dt} = 4900 + 9800 \]
\[ 2000 \cdot \frac{dm}{dt} = 14700 \]

### Step 6: Solve for Rate of Fuel Consumption
Now, divide both sides by \( 2000 \):
\[ \frac{dm}{dt} = \frac{14700}{2000} \]
\[ \frac{dm}{dt} = 7.35 \, \text{kg/s} \]

### Final Answer
The rate of fuel consumption for the rocket to rise with an acceleration of \( 4.9 \, \text{m/s}^2 \) is:
\[ \frac{dm}{dt} = 7.35 \, \text{kg/s} \]

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