A stream of a liquid of density p flowin horizontally with speed v rushes out of a tube of radius r and hits a verticla wall nearly normally. Assumi g that the liquid does not rebound from the wall, the force exerted o the wall by the impact of the liquid is given by

Options

A. `pirrhov`
B. `pirrhov^(2)`
C. `pir^(2)rhov`
D. `pir^(2)rhov^(2)`

Answer

To solve the problem of finding the force exerted on the wall by the impact of a liquid stream, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Parameters**:
- Density of the liquid, \( \rho \)
- Speed of the liquid, \( v \)
- Radius of the tube, \( r \)

2. **Understand the Flow of Liquid**:
- The liquid flows out of the tube horizontally and hits a vertical wall normally (perpendicularly).
- We assume that the liquid does not rebound from the wall.

3. **Calculate the Volume Flow Rate**:
- The volume flow rate \( Q \) of the liquid can be expressed as:
\[ Q = A \cdot v \]
- Where \( A \) is the cross-sectional area of the tube. For a circular tube, the area \( A \) is given by:
\[ A = \pi r^2 \]
- Therefore, the volume flow rate becomes:
\[ Q = \pi r^2 \cdot v \]

4. **Calculate the Mass Flow Rate**:
- The mass flow rate \( \dot{m} \) can be calculated using the density \( \rho \):
\[ \dot{m} = \rho \cdot Q = \rho \cdot (\pi r^2 \cdot v) \]

5. **Determine the Change in Momentum**:
- When the liquid hits the wall, it comes to a stop. The change in momentum \( \Delta p \) for the mass flow rate is:
\[ \Delta p = \dot{m} \cdot v = \rho \cdot (\pi r^2 \cdot v) \cdot v = \rho \cdot \pi r^2 \cdot v^2 \]

6. **Calculate the Force Exerted on the Wall**:
- The force \( F \) exerted on the wall by the impact of the liquid is equal to the rate of change of momentum:
\[ F = \Delta p = \rho \cdot \pi r^2 \cdot v^2 \]

### Final Answer:
The force exerted on the wall by the impact of the liquid is given by:
\[ F = \pi r^2 \rho v^2 \]