JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
An average force of \(125\,N\) is applied on a machine gun firing bullets each of mass \(10\,g\) at the speed of \(250\,m / s\) to keep it in position. The number of bullets fired per second by the machine gun is:
- A \(5\)
- B \(50\)
- C \(100\)
- D \(25\)
Answer & Solution
Correct Answer
(B) \(50\)
Step-by-step Solution
Detailed explanation
\(F = n mv\) where \(n =\) number of bullets fired per second \(n =\frac{ f }{ mv }=\frac{125}{10 \times 10^{-3} \times 250}=50\)
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