The diameter of the top of a firebrigade pump is 5 cm. Water is thrown by this pump at a horizontal speed of 18 m/s on a wall. If water rebounds back from the wall, then the force exerted by water on wall will be :-

Options

A. `2.35xx10^(5)` dyne
B. `5.76xx10^(6)` dyne
C. `6.36xx10^(7)` dyne
D. `12.72xx10^(7)` dyne

Answer

To solve the problem of finding the force exerted by water on the wall when it rebounds, we can follow these steps:

### Step 1: Calculate the area of the fire brigade pump
The diameter of the pump is given as 5 cm, so the radius \( r \) is:
\[ r = \frac{5 \text{ cm}}{2} = 2.5 \text{ cm} = 0.025 \text{ m} \]
The area \( A \) of the top of the pump can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 = \pi (0.025)^2 = \pi (0.000625) \approx 0.0019635 \text{ m}^2 \]

### Step 2: Calculate the volume flow rate
The volume flow rate \( Q \) can be calculated using the formula:
\[ Q = A \cdot v \]
where \( v \) is the velocity of the water, given as 18 m/s. Thus,
\[ Q = 0.0019635 \text{ m}^2 \cdot 18 \text{ m/s} \approx 0.0353 \text{ m}^3/\text{s} \]

### Step 3: Calculate the mass flow rate
The mass flow rate \( \dot{m} \) can be calculated using the density of water \( \rho \) (approximately 1000 kg/m³):
\[ \dot{m} = \rho \cdot Q = 1000 \text{ kg/m}^3 \cdot 0.0353 \text{ m}^3/\text{s} \approx 35.3 \text{ kg/s} \]

### Step 4: Calculate the change in momentum
When the water hits the wall, it has an initial momentum \( p_i = \dot{m} \cdot v \) and after rebounding, it has a final momentum \( p_f = -\dot{m} \cdot v \). Therefore, the change in momentum \( \Delta p \) is:
\[ \Delta p = p_f - p_i = -\dot{m} \cdot v - \dot{m} \cdot v = -2 \dot{m} \cdot v \]
Substituting the values:
\[ \Delta p = -2 \cdot 35.3 \text{ kg/s} \cdot 18 \text{ m/s} \approx -1272 \text{ kg m/s} \]

### Step 5: Calculate the force exerted on the wall
The force \( F \) exerted by the water on the wall is given by the rate of change of momentum:
\[ F = \frac{\Delta p}{\Delta t} = \Delta p \text{ (since we are considering per second)} \]
Thus,
\[ F = 1272 \text{ N} \]

### Final Result
The force exerted by the water on the wall is approximately:
\[ F \approx 1272 \text{ N} \]