If the force on a rocket having an exhaust velocity of $300 ext{ m/s}$ is $210 ext{ N}$, then the rate of

Question

If the force on a rocket having an exhaust velocity of $300 	ext{ m/s}$ is $210 	ext{ N}$, then the rate of combustion of the fuel is:

Accepted Answer

1. **Identify the Principle:** The force (thrust) exerted on a rocket is equal to the rate of change of momentum of the ejected fuel. For a variable mass system, the thrust force ($F$) is defined as the product of the exhaust velocity ($v$) and the rate of mass ejection (combustion rate, $\frac{dm}{dt}$).
   $$ F = v \frac{dm}{dt} $$
2. **Substitute Values:**
   Given:
   - Thrust Force, $F = 210 	ext{ N}$
   - Exhaust Velocity, $v = 300 	ext{ m/s}$
   Substitute these into the equation:
   $$ 210 = 300 	imes \frac{dm}{dt} $$
3. **Calculate Rate:**
   $$ \frac{dm}{dt} = \frac{210}{300} = \frac{2.1}{3} = 0.7 	ext{ kg/s} $$
   (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion; see Exercise 5.9 for similar rocket propulsion problems).

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