If the force on a rocket moving with a velocity of 300 m/s is 345 N, then the rate of combustion of the fuel is
(a) 0.55 kg/s (b) 0.75 kg/s (c) 1.15 kg/s (d) 2.25 kg/s

Options

A. 0.55 kg/s
B. 0.75 kg/s
C. 1.15 kg/s
D. 2.25 kg/s

Answer

To solve the problem, we will use the formula related to the force acting on a rocket due to the combustion of fuel. The force (F) on the rocket is given by the equation:

\[ F = V \frac{dm}{dt} \]

where:
- \( F \) is the force,
- \( V \) is the velocity of the rocket,
- \( \frac{dm}{dt} \) is the rate of combustion of the fuel.

### Step 1: Identify the given values
From the question, we have:
- Force \( F = 345 \, \text{N} \)
- Velocity \( V = 300 \, \text{m/s} \)

### Step 2: Rearrange the formula to find the rate of combustion
We need to find \( \frac{dm}{dt} \), so we rearrange the equation:

\[ \frac{dm}{dt} = \frac{F}{V} \]

### Step 3: Substitute the known values into the equation
Now, substitute the values of force and velocity into the equation:

\[ \frac{dm}{dt} = \frac{345 \, \text{N}}{300 \, \text{m/s}} \]

### Step 4: Perform the calculation
Now, calculate the right-hand side:

\[ \frac{dm}{dt} = \frac{345}{300} \]

Calculating this gives:

\[ \frac{dm}{dt} = 1.15 \, \text{kg/s} \]

### Step 5: Conclusion
Thus, the rate of combustion of the fuel is:

**Answer: \( \frac{dm}{dt} = 1.15 \, \text{kg/s} \)**

The correct option is (c) 1.15 kg/s.

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