Laws Of Motion Question 202

Question: If force on a rocket having exhaust velocity of 300 m/sec is 210 N, then rate of combustion of the fuel is [CBSE PMT 1999; MH CET 2003; Pb. PMT 2004]

Options:

A)0.7 kg/s

B) 1.4 kg/s

C) 0.07 kg/s

D) 10.7 kg/s

Show Answer

Answer:

Correct Answer: A

Solution:

$ F=u\ ( \frac{dm}{dt} ) $

therefore $ \frac{dm}{dt}=\frac{F}{u}=\frac{210}{300}=0.7\ kg/s $



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