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 $
BETA
