A stream of water flowing horizontally with a speed of `25ms^-1` gushes out of a tube of cross-sectional area `10^(-3) m^(2)`, and hits at a vertical wall nearby. What is the force exerted on the wall by the impact of water?

Options

A. 125N
B. 625N
C. `-650N`
D. `-1125N`

Explanation

To solve the problem of finding the force exerted on a wall by the impact of water flowing out of a tube, we can follow these steps:

### Step 1: Identify the given data
- Speed of water, \( v = 25 \, \text{m/s} \)
- Cross-sectional area of the tube, \( A = 10^{-3} \, \text{m}^2 \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)

### Step 2: Calculate the volume of water flowing out per second
The volume of water flowing out per second can be calculated using the formula:
\[ \text{Volume flow rate} = A \times v \]
Substituting the values:
\[ \text{Volume flow rate} = 10^{-3} \, \text{m}^2 \times 25 \, \text{m/s} = 25 \times 10^{-3} \, \text{m}^3/s = 0.025 \, \text{m}^3/s \]

### Step 3: Calculate the mass of water flowing out per second
Using the density of water, we can find the mass flow rate:
\[ \text{Mass flow rate} = \text{Volume flow rate} \times \rho \]
Substituting the values:
\[ \text{Mass flow rate} = 0.025 \, \text{m}^3/s \times 1000 \, \text{kg/m}^3 = 25 \, \text{kg/s} \]

### Step 4: Calculate the change in momentum
When the water hits the wall, it comes to rest. The change in momentum per second (which is equal to the force exerted) can be calculated as:
\[ \text{Change in momentum} = \text{mass} \times \text{change in velocity} \]
The initial velocity \( u = 25 \, \text{m/s} \) and the final velocity \( v = 0 \, \text{m/s} \). Thus:
\[ \text{Change in momentum} = 25 \, \text{kg/s} \times (0 - 25) \, \text{m/s} = 25 \, \text{kg/s} \times (-25) \, \text{m/s} = -625 \, \text{kg m/s}^2 \]

### Step 5: Calculate the force exerted on the wall
According to Newton's second law, the force exerted is equal to the rate of change of momentum:
\[ F = \text{Change in momentum} = -625 \, \text{N} \]
The negative sign indicates the direction of the force is opposite to the direction of the water flow.

### Final Answer
The magnitude of the force exerted on the wall by the impact of water is:
\[ \boxed{625 \, \text{N}} \]