n bullet strikes per second elastically on wall and rebound then what will be the force exerted on the wall by bullets if mass of each bullet is m:-

Options

A. mnv
B. 5mnv
C. 2mnv
D. `(mnv)/(2)`

Explanation

To solve the problem of finding the force exerted on the wall by bullets striking it elastically, we can follow these steps:

### Step 1: Understand the Problem
We have \( n \) bullets striking a wall per second, and each bullet has a mass \( m \). The bullets strike the wall elastically, meaning they rebound with the same speed but in the opposite direction.

### Step 2: Calculate the Change in Momentum for One Bullet
When a bullet strikes the wall, it has an initial momentum given by:
\[ P_{\text{initial}} = mv \]
where \( v \) is the velocity of the bullet before it hits the wall.

After rebounding, the bullet's momentum becomes:
\[ P_{\text{final}} = -mv \]
The negative sign indicates that the direction of the bullet's velocity has reversed.

The change in momentum (\( \Delta P \)) for one bullet is:
\[ \Delta P = P_{\text{final}} - P_{\text{initial}} = (-mv) - (mv) = -2mv \]

### Step 3: Calculate the Total Change in Momentum for \( n \) Bullets
Since \( n \) bullets strike the wall per second, the total change in momentum for \( n \) bullets is:
\[ \Delta P_{\text{total}} = n \times (-2mv) = -2nmv \]

### Step 4: Calculate the Force Exerted on the Wall
The force (\( F \)) exerted on the wall can be calculated using the formula for force, which is the rate of change of momentum:
\[ F = \frac{\Delta P_{\text{total}}}{\Delta t} \]
Since \( n \) bullets strike per second, \( \Delta t = 1 \) second. Thus:
\[ F = -2nmv \]
The negative sign indicates the force is in the opposite direction of the bullet's initial motion, but we are interested in the magnitude of the force:
\[ F = 2nmv \]

### Conclusion
The force exerted on the wall by the bullets is:
\[ F = 2nmv \]