QJEEM23Phc220: JEE Main 2023 Physics - Class 11 JEEM23Phc220 Solution: An average force of 125   N is applied on a machine gun…

To find the number of bullets fired per second, we can use the concept of momentum. When the machine gun fires bullets, it experiences a backward force due to the conservation of momentum. The force applied on the machine gun is used to balance this backward force.

First, let's find the momentum of each bullet:

Momentum = Mass × Velocity

Bullet mass = $$10\mathrm{g}=0.01\mathrm{kg}$$ Bullet velocity = $$250\mathrm{m}/\mathrm{s}$$

Momentum per bullet = $$0.01\mathrm{kg}\cdot 250\mathrm{m}/\mathrm{s}=2.5\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$$

Now let's find the momentum per second that needs to be balanced by the applied force:

Force = $$125\mathrm{N}$$

Momentum per second = Force × Time

Since we are considering a time interval of 1 second, the momentum per second is equal to the applied force:

Momentum per second = $$125\mathrm{N}$$

Now, let's find the number of bullets fired per second:

Number of bullets = Momentum per second / Momentum per bullet

Number of bullets = $$\frac{125\mathrm{N}}{2.5\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}}$$

Number of bullets = 50

Therefore, the machine gun fires 50 bullets per second.