Water is flowing at a speed of `1.5 ms ^(-1)` through horizontal tube of cross-sectional area `10 ^(-2) m ^(2)` and you are trying to stop the flow by your palm. Assuming that the water stops immediately after hitting the palm, the minimum force that you must every should be
(density of water `= 10 ^(3) kgm ^(-3))`

Options

A. 15 N
B. `22. 5 N`
C. `33.7N`
D. `45N`

Explanation

To solve the problem, we need to calculate the minimum force required to stop the flow of water using the concept of momentum change. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given values
- Speed of water, \( v = 1.5 \, \text{m/s} \)
- Cross-sectional area of the tube, \( A = 10^{-2} \, \text{m}^2 \)
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)

### Step 2: Calculate the volume of water flowing in 1 second
The volume \( V \) of water flowing through the tube in 1 second can be calculated using the formula:
\[ V = A \times v \]
Substituting the values:
\[ V = 10^{-2} \, \text{m}^2 \times 1.5 \, \text{m/s} = 0.015 \, \text{m}^3 \]

### Step 3: Calculate the mass of water flowing in 1 second
Using the density of water, we can find the mass \( m \) of the water:
\[ m = \rho \times V \]
Substituting the values:
\[ m = 10^3 \, \text{kg/m}^3 \times 0.015 \, \text{m}^3 = 15 \, \text{kg} \]

### Step 4: Calculate the change in momentum
The momentum \( p \) of the water is given by:
\[ p = m \times v \]
Substituting the values:
\[ p = 15 \, \text{kg} \times 1.5 \, \text{m/s} = 22.5 \, \text{kg m/s} \]

### Step 5: Calculate the force required to stop the water
The force \( F \) required to stop the water can be calculated using the formula for the rate of change of momentum:
\[ F = \frac{\Delta p}{\Delta t} \]
Assuming the time \( \Delta t \) is 1 second:
\[ F = \frac{22.5 \, \text{kg m/s}}{1 \, \text{s}} = 22.5 \, \text{N} \]

### Conclusion
The minimum force that you must exert to stop the flow of water is \( 22.5 \, \text{N} \).