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 calculating the force exerted on the wall by the water from the hosepipe, we can follow these steps:

### Step 1: Calculate the Cross-Sectional Area of the Hosepipe
The radius of the hosepipe is given as \( r = 5 \, \text{cm} = 0.05 \, \text{m} \). The area \( A \) of the cross-section of the hosepipe can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (0.05)^2 = \pi (0.0025) \approx 0.00785 \, \text{m}^2 \]

### Step 2: Calculate the Volume Flow Rate
The volume flow rate \( \frac{dV}{dt} \) can be calculated by multiplying the cross-sectional area \( A \) by the velocity \( v \):
\[ \frac{dV}{dt} = A \cdot v \]
Given that the speed of water \( v = 5 \, \text{m/s} \):
\[ \frac{dV}{dt} = 0.00785 \cdot 5 \approx 0.03925 \, \text{m}^3/\text{s} \]

### Step 3: Calculate the Mass Flow Rate
The mass flow rate \( \frac{dm}{dt} \) can be calculated using the volume flow rate and the density of water \( \rho \):
\[ \frac{dm}{dt} = \frac{dV}{dt} \cdot \rho \]
Using the density of water \( \rho \approx 1000 \, \text{kg/m}^3 \):
\[ \frac{dm}{dt} = 0.03925 \cdot 1000 \approx 39.25 \, \text{kg/s} \]

### Step 4: Calculate the Force Exerted on the Wall
The force \( F \) exerted on the wall can be calculated using the formula:
\[ F = \frac{dm}{dt} \cdot v \]
Substituting the values:
\[ F = 39.25 \cdot 5 \approx 196.25 \, \text{N} \]

### Final Result
The force exerted on the wall by the water is approximately \( 196.25 \, \text{N} \).

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