A stream of water flowing horizontally with a speed of `15 ms^(-1)` pushes out of a tube of cross sectional area `10^(-2)m^(2)` and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water assuming that it does not rebound? (Density of water `=1000 kg m^(-3)`)

Options

A. `1.25xx10^(3)N`
B. `2.25xx10^(3)N`
C. `3.25xx10^(3)N`
D. `4.25xx10^(3)N`

Explanation

To solve the problem, we need to calculate the force exerted on the wall by the impact of the water stream. We can follow these steps:

### Step 1: Understand the Given Data
- Speed of water, \( V = 15 \, \text{m/s} \)
- Cross-sectional area of the tube, \( A = 10^{-2} \, \text{m}^2 \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)

### Step 2: Calculate the Volume of Water Flowing Per Second
The volume of water flowing out of the tube in one second can be calculated using the formula:
\[ \text{Volume} = \text{Area} \times \text{Velocity} \]
Substituting the values:
\[ \text{Volume} = A \times V = 10^{-2} \, \text{m}^2 \times 15 \, \text{m/s} = 0.15 \, \text{m}^3 \]

### Step 3: Calculate the Mass of Water Flowing Per Second
Using the density of water, we can find the mass of water flowing per second:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
Substituting the values:
\[ \text{Mass} = 1000 \, \text{kg/m}^3 \times 0.15 \, \text{m}^3 = 150 \, \text{kg} \]

### Step 4: Calculate the Rate of Change of Momentum
The force exerted on the wall can be calculated using the rate of change of momentum. The momentum \( p \) is given by:
\[ p = \text{mass} \times \text{velocity} \]
The rate of change of momentum (which is equal to force) can be calculated as:
\[ F = \frac{\Delta p}{\Delta t} = \text{mass} \times \text{velocity} \]
Substituting the values:
\[ F = 150 \, \text{kg} \times 15 \, \text{m/s} = 2250 \, \text{N} \]

### Step 5: Final Result
Thus, the force exerted on the wall by the impact of the water is:
\[ F = 2250 \, \text{N} \quad \text{or} \quad 2.25 \times 10^3 \, \text{N} \]