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 $



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