A bullet looses `((1)/(n))^(th)` of its velocity passing through one plank.The number of such planks that are required to stop the bullet can be:

Options

A. `(n^(2))/(2n-1)`
B. `(2n^(2))/(n-1)`
C. infinite
D. n

Explanation

To solve the problem of how many planks are required to stop a bullet that loses \(\frac{1}{n}\) of its velocity after passing through one plank, we can follow these steps:

### Step 1: Define Initial Conditions
Let the initial velocity of the bullet be \(u\). When the bullet passes through one plank, it loses \(\frac{1}{n}\) of its velocity.

### Step 2: Calculate Velocity After One Plank
After passing through one plank, the velocity \(v\) of the bullet can be calculated as:
\[ v = u - \frac{1}{n}u = u\left(1 - \frac{1}{n}\right) = u\left(\frac{n-1}{n}\right) \]

### Step 3: Calculate Kinetic Energy
The initial kinetic energy \(KE_i\) of the bullet is given by:
\[ KE_i = \frac{1}{2}mu^2 \]
After passing through one plank, the kinetic energy \(KE_f\) becomes:
\[ KE_f = \frac{1}{2}m v^2 = \frac{1}{2}m \left(u\left(\frac{n-1}{n}\right)\right)^2 = \frac{1}{2}m \frac{(n-1)^2}{n^2}u^2 \]

### Step 4: Calculate Work Done by One Plank
The work done by the resistive force \(F\) of one plank can be expressed as:
\[ W = KE_i - KE_f = \frac{1}{2}mu^2 - \frac{1}{2}m \frac{(n-1)^2}{n^2}u^2 \]
Simplifying this gives:
\[ W = \frac{1}{2}mu^2 \left(1 - \frac{(n-1)^2}{n^2}\right) = \frac{1}{2}mu^2 \left(\frac{n^2 - (n-1)^2}{n^2}\right) \]
Calculating \(n^2 - (n-1)^2\):
\[ n^2 - (n-1)^2 = n^2 - (n^2 - 2n + 1) = 2n - 1 \]
Thus, the work done by one plank becomes:
\[ W = \frac{1}{2}mu^2 \cdot \frac{2n - 1}{n^2} \]

### Step 5: Total Work Done by N Planks
If \(N\) planks are used, the total work done \(W_{total}\) is:
\[ W_{total} = N \cdot W = N \cdot \left(\frac{1}{2}mu^2 \cdot \frac{2n - 1}{n^2}\right) \]

### Step 6: Set Total Work Equal to Initial Kinetic Energy
To stop the bullet, the total work done must equal the initial kinetic energy:
\[ N \cdot \left(\frac{1}{2}mu^2 \cdot \frac{2n - 1}{n^2}\right) = \frac{1}{2}mu^2 \]

### Step 7: Solve for N
Cancelling \(\frac{1}{2}mu^2\) from both sides gives:
\[ N \cdot \frac{2n - 1}{n^2} = 1 \]
Thus,
\[ N = \frac{n^2}{2n - 1} \]

### Conclusion
The number of planks required to stop the bullet is:
\[ N = \frac{n^2}{2n - 1} \]