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NEET PHYSICSLAWS OF MOTIONEasy

Question

If the force on a rocket having an exhaust velocity of 300 m/s300 \text{ m/s} is 210 N210 \text{ N}, then the rate of combustion of the fuel is:

A

0.7 kg/s

B

1.4 kg/s

C

0.07 kg/s

D

10.7 kg/s

Step-by-Step Solution

  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 (FF) is defined as the product of the exhaust velocity (vv) and the rate of mass ejection (combustion rate, dmdt\frac{dm}{dt}). F=vdmdtF = v \frac{dm}{dt}
  2. Substitute Values: Given:
  • Thrust Force, F=210 NF = 210 \text{ N}
  • Exhaust Velocity, v=300 m/sv = 300 \text{ m/s} Substitute these into the equation: 210=300×dmdt210 = 300 \times \frac{dm}{dt}
  1. Calculate Rate: dmdt=210300=2.13=0.7 kg/s\frac{dm}{dt} = \frac{210}{300} = \frac{2.1}{3} = 0.7 \text{ kg/s} (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion; see Exercise 5.9 for similar rocket propulsion problems).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSLAWS OF MOTIONrockethavingexhaustvelocitycombustion

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