Laws Of Motion Question 245
Question: A machine gun fires a bullet of mass 40 g with a velocity $ 1200m{{s}^{-1}}. $ The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most[AIEEE 2004]
Options:
A) One
B) Four
C) Two
D) Three
Show Answer
Answer:
Correct Answer: D
Solution:
u = velocity of bullet $ \frac{dm}{dt}= $
Mass fired per second by the gun $ \frac{dm}{dt} $ = Mass of bullet (mB) × Bullets fired per sec (N)
Maximum force that man can exert $ F=u\ ( \frac{dm}{dt} ) $
$ \therefore $ $ F=u\times m _{B}\times N $
therefore $ N=\frac{F}{m _{B}\times u}=\frac{144}{40\times {{10}^{-3}}\times 1200}=3 $
BETA
