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JEE Main 202311 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
11 April
Shift
Morning Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2023 · 11 April · Morning Shift · Physics · Center of Mass, Momentum and Collision

An average force of 125N is applied on a machine gun firing bullets...

An average force of 125N125N is applied on a machine gun firing bullets each of mass 10g10g at the speed of 250ms−1250m{s}^{- 1} to keep it in position. The number of bullets fired per second by the machine gun is :

Options

  1. A5050
  2. B2525
  3. C100100
  4. D55

Correct answer

A. 5050

Step-by-step solution

The formula to calculate the net momentum of the gun and bullet system can be written as

F=nmv...(1)F = nmv . . . \left(1\right)

Substitute the values of the known parameters into equation (1) and solve to calculate the required number of bullets.

{125=n×101000×(250)⇒n=50\begin{cases}125 & = & n \times \dfrac{10}{1000} \times ( 250 ) \\ ⇒ & n = 50\end{cases}

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