JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    A machine gun fires a bullet of mass 40 g with a velocity \[1200\,\,m{{s}^{-1}}.\] 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]

    A)             One    

    B)             Four

    C)             Two    

    D)             Three

    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 \[F=u\ \left( \frac{dm}{dt} \right)\] \[\therefore \] \[F=u\times {{m}_{B}}\times N\] Þ \[N=\frac{F}{{{m}_{B}}\times u}=\frac{144}{40\times {{10}^{-3}}\times 1200}=3\]


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