Water from a hosepipe of radius 5 cm strikes a wall at a speed of `5ms^(-1)` normally and stops. The force exerted on the wall in newton is

Options

A. `13.5pi`
B. `6.25pi`
C. `62.5pi`
D. `27pi`

Answer

To solve the problem of finding the force exerted on the wall by water from a hosepipe, we will follow these steps:

### Step 1: Convert the radius to meters
Given the radius of the hosepipe is 5 cm, we need to convert this to meters:
\[ r = 5 \text{ cm} = 5 \times 10^{-2} \text{ m} \]

### Step 2: Calculate the cross-sectional area of the hosepipe
The cross-sectional area \( A \) of the circular hosepipe can be calculated using the formula:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (5 \times 10^{-2})^2 \]
\[ A = \pi (25 \times 10^{-4}) \]
\[ A = 25\pi \times 10^{-4} \text{ m}^2 \]

### Step 3: Use the velocity of water
The speed \( v \) of the water is given as:
\[ v = 5 \text{ m/s} \]

### Step 4: Use the density of water
The density \( \rho \) of water is approximately:
\[ \rho = 1000 \text{ kg/m}^3 \]

### Step 5: Calculate the force exerted on the wall
The force \( F \) exerted on the wall can be calculated using the formula:
\[ F = A \cdot v^2 \cdot \rho \]
Substituting the values we have:
\[ F = (25\pi \times 10^{-4}) \cdot (5^2) \cdot (1000) \]
\[ F = (25\pi \times 10^{-4}) \cdot 25 \cdot 1000 \]
\[ F = 625\pi \times 10^{-4} \cdot 1000 \]
\[ F = 625\pi \times 10^{-1} \]
\[ F = 62.5\pi \text{ N} \]

### Step 6: Calculate the numerical value of the force
Using \( \pi \approx 3.14 \):
\[ F \approx 62.5 \times 3.14 \]
\[ F \approx 196.25 \text{ N} \]

### Final Answer
The force exerted on the wall is approximately:
\[ F \approx 196.25 \text{ N} \]