A bullet travelling horizontally looses `1//20^(th)` of its velocity while piercing a wooden plank. Then the number of such planks required to stop the bullet is

Options

A. 6
B. 9
C. 11
D. 13

Explanation

To solve the problem of how many wooden planks are required to stop a bullet that loses \( \frac{1}{20} \) of its velocity while piercing through each plank, we can follow these steps:

### Step-by-Step Solution:

1. **Initial Velocity and Velocity Loss**:
- Let the initial velocity of the bullet be \( v \).
- The bullet loses \( \frac{1}{20} \) of its velocity while passing through one plank.
- Therefore, the velocity after passing through one plank is:
\[ v' = v - \frac{1}{20}v = \frac{19}{20}v \]

2. **Determine the Change in Velocity**:
- The change in velocity when the bullet passes through one plank is:
\[ \Delta v = v - v' = v - \frac{19}{20}v = \frac{1}{20}v \]

3. **Acceleration Due to Plank**:
- The bullet experiences a constant retardation (negative acceleration) while passing through the plank. Let this retardation be \( a \).
- We can use the kinematic equation:
\[ v'^2 = v^2 + 2(-a)d \]
- Substituting \( v' = \frac{19}{20}v \):
\[ \left(\frac{19}{20}v\right)^2 = v^2 - 2ad \]

4. **Simplifying the Equation**:
- Expanding the left side:
\[ \frac{361}{400}v^2 = v^2 - 2ad \]
- Rearranging gives:
\[ 2ad = v^2 - \frac{361}{400}v^2 = \frac{39}{400}v^2 \]
- Therefore:
\[ ad = \frac{39}{800}v^2 \]

5. **Total Distance to Stop the Bullet**:
- Let \( n \) be the number of planks required to stop the bullet.
- The total distance the bullet travels through \( n \) planks is \( n \cdot d \).
- Using the same kinematic equation for stopping the bullet:
\[ 0 = v^2 + 2(-a)(nd) \]
- Rearranging gives:
\[ v^2 = 2and \]

6. **Substituting \( ad \)**:
- From the previous step, we know \( ad = \frac{39}{800}v^2 \).
- Substituting this into the equation:
\[ v^2 = 2n\left(\frac{39}{800}v^2\right) \]
- Simplifying gives:
\[ v^2 = \frac{78n}{800}v^2 \]

7. **Solving for \( n \)**:
- Dividing both sides by \( v^2 \) (assuming \( v \neq 0 \)):
\[ 1 = \frac{78n}{800} \]
- Rearranging gives:
\[ n = \frac{800}{78} \approx 10.26 \]

8. **Final Number of Planks**:
- Since \( n \) must be a whole number, we round up to the nearest whole number:
\[ n = 11 \]

### Conclusion:
The number of wooden planks required to stop the bullet is **11**.