Laws of Motion

JEE Main level, single correct, medium

65 of 210

Question 65 of 210

A machine gun fires a bullet of mass 40 g40 \mathrm{~g} with a velocity 1200 ms−11200 \mathrm{~ms}^{-1}. The man holding it can exert a maximum force of 144 N144 \mathrm{~N} on the gun. How many bullets can he fire per second at the most?

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Answer: (D) three

Change in momentum for each bullet fired is
=401000×1200=48 N=\frac{40}{1000} \times 1200=48 \mathrm{~N}
If a bullet fired exerts a force of 48 N48 \mathrm{~N} on man's hand so ρ\rho man can exert maximum force of 144 N144 \mathrm{~N}, number of bullets that can be fired =144/48=3=144 / 48=3 bullets.

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