A horizontal jet of water coming out of a pipe of area of cross-section `20cm^(2)` hits a vertical wall with a velocity of `10ms^(-1)` and rebounds with the same speed. The force exerted by water on the wall is .

Options

A. `0.2 N`
B. `10 N`
C. `400N`
D. `200N`

Explanation

To solve the problem, we need to determine the force exerted by the water jet on the wall when it hits and rebounds. We can use the principle of conservation of momentum and the definition of force as the rate of change of momentum.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Area of cross-section of the pipe, \( A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 = 0.002 \, \text{m}^2 \)
- Velocity of water, \( v = 10 \, \text{m/s} \)

2. **Calculate the Volume Flow Rate**:
The volume flow rate \( Q \) can be calculated using the formula:
\[ Q = A \cdot v \]
Substituting the values:
\[ Q = 0.002 \, \text{m}^2 \cdot 10 \, \text{m/s} = 0.02 \, \text{m}^3/\text{s} \]

3. **Calculate the Mass Flow Rate**:
The mass flow rate \( \dot{m} \) can be found using the density of water \( \rho \) (approximately \( 1000 \, \text{kg/m}^3 \)):
\[ \dot{m} = \rho \cdot Q \]
Substituting the values:
\[ \dot{m} = 1000 \, \text{kg/m}^3 \cdot 0.02 \, \text{m}^3/\text{s} = 20 \, \text{kg/s} \]

4. **Calculate Change in Momentum**:
The change in momentum \( \Delta p \) when the water hits the wall and rebounds is given by:
\[ \Delta p = \dot{m} \cdot (v_{\text{final}} - v_{\text{initial}}) \]
Here, \( v_{\text{initial}} = 10 \, \text{m/s} \) (towards the wall) and \( v_{\text{final}} = -10 \, \text{m/s} \) (away from the wall).
Therefore:
\[ \Delta p = 20 \, \text{kg/s} \cdot (-10 \, \text{m/s} - 10 \, \text{m/s}) = 20 \, \text{kg/s} \cdot (-20 \, \text{m/s}) = -400 \, \text{kg m/s} \]
The negative sign indicates the direction of momentum change.

5. **Calculate the Average Force**:
The average force \( F \) exerted by the water on the wall can be calculated using:
\[ F = \frac{\Delta p}{\Delta t} \]
Since we are considering the change in momentum per second, we have \( \Delta t = 1 \, \text{s} \):
\[ F = \frac{-400 \, \text{kg m/s}}{1 \, \text{s}} = 400 \, \text{N} \]

### Final Answer:
The force exerted by the water on the wall is \( 400 \, \text{N} \).

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