Back to Directory
NEET PHYSICSEasy

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).
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started