A stream of water flowing horizontally with a speed of 15 m/s gushes out of a tube of cross sectional area `10^(-2)m^(2)` and hits at a vertical wall and after collision falls dead. The force exerted on the wall by the impact of water is

Options

A. 225 N
B. 2250 N
C. 520 N
D. none of these

Answer

To solve the problem of calculating the force exerted on the wall by the impact of the water stream, we can follow these steps:

### Step 1: Understand the Problem
We have a stream of water flowing horizontally with a speed of 15 m/s, gushing out of a tube with a cross-sectional area of \(10^{-2} \, m^2\). The water hits a vertical wall and comes to a stop, meaning its final velocity after the collision is 0 m/s.

### Step 2: Calculate the Mass Flow Rate
The mass flow rate (\( \frac{dm}{dt} \)) can be calculated using the formula:
\[ \frac{dm}{dt} = \rho \cdot A \cdot v \]
where:
- \( \rho \) is the density of water (approximately \(1000 \, kg/m^3\)),
- \( A \) is the cross-sectional area of the tube (\(10^{-2} \, m^2\)),
- \( v \) is the velocity of the water (\(15 \, m/s\)).

Substituting the values:
\[ \frac{dm}{dt} = 1000 \, kg/m^3 \cdot 10^{-2} \, m^2 \cdot 15 \, m/s \]
\[ \frac{dm}{dt} = 1000 \cdot 0.01 \cdot 15 = 150 \, kg/s \]

### Step 3: Calculate the Change in Momentum
The change in momentum (\( \Delta p \)) per unit time is given by:
\[ F = \frac{dp}{dt} = \frac{d(mv)}{dt} \]
Since the final velocity after hitting the wall is 0, the change in momentum is simply the initial momentum:
\[ F = \frac{dm}{dt} \cdot v \]
Substituting the values:
\[ F = 150 \, kg/s \cdot 15 \, m/s \]
\[ F = 2250 \, N \]

### Step 4: Conclusion
The force exerted on the wall by the impact of the water is \(2250 \, N\).