A water pipe has an internal diameter of 10 cm. Water flows through it at the rate of 20 m/sec. The water jet strikes normally on a wall and falls dead. Find the force on the wall.

Answer

To solve the problem, we need to find the force exerted on the wall by the water jet. We can break down the solution into several steps.

### Step 1: Calculate the Area of Cross-Section of the Pipe
The internal diameter of the pipe is given as 10 cm. First, we need to convert this into meters:
\[ \text{Diameter} = 10 \text{ cm} = 0.1 \text{ m} \]
The radius \( r \) is half of the diameter:
\[ r = \frac{0.1}{2} = 0.05 \text{ m} \]
Now, we can calculate the area \( A \) of the cross-section of the pipe using the formula for the area of a circle:
\[ A = \pi r^2 = \pi (0.05)^2 = \pi (0.0025) \approx 0.007854 \text{ m}^2 \]

### Step 2: Calculate the Volume Flow Rate
The water flows through the pipe at a velocity \( v = 20 \text{ m/s} \). The volume flow rate \( Q \) can be calculated as:
\[ Q = A \cdot v = 0.007854 \text{ m}^2 \cdot 20 \text{ m/s} \approx 0.15708 \text{ m}^3/\text{s} \]

### Step 3: Calculate the Mass Flow Rate
The density of water \( \rho \) is approximately \( 1000 \text{ kg/m}^3 \). The mass flow rate \( \dot{m} \) can be calculated using the formula:
\[ \dot{m} = Q \cdot \rho = 0.15708 \text{ m}^3/\text{s} \cdot 1000 \text{ kg/m}^3 \approx 157.08 \text{ kg/s} \]

### Step 4: Calculate the Change in Momentum
When the water jet strikes the wall and falls dead, the change in momentum per second (which is equal to the force exerted on the wall) can be calculated as:
\[ F = \dot{m} \cdot v = 157.08 \text{ kg/s} \cdot 20 \text{ m/s} \approx 3141.6 \text{ N} \]

### Step 5: Final Answer
Thus, the force exerted on the wall by the water jet is approximately:
\[ F \approx 3142 \text{ N} \]

### Summary
The force on the wall is approximately **3142 N**.

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