A horizontal pipeline carries water in a streamline flow. At a point along the tube where the cross sectional area is `10^(-2)(m^2)`, the water velcity is `2m//s` and the pressure is 8000Pa. The pressure of water at another point where cross sectional area is `0.5xx(10)^(-2)(m^2)` is

Options

A. `4000Pa`
B. `1000Pa`
C. `2000Pa`
D. `3000Pa`

Explanation

To solve the problem, we will use Bernoulli's equation and the principle of continuity. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
We have two points in a horizontal pipeline carrying water:
- **Point 1**:
- Cross-sectional area \( A_1 = 10^{-2} \, \text{m}^2 \)
- Velocity \( V_1 = 2 \, \text{m/s} \)
- Pressure \( P_1 = 8000 \, \text{Pa} \)

- **Point 2**:
- Cross-sectional area \( A_2 = 0.5 \times 10^{-2} \, \text{m}^2 \)
- Velocity \( V_2 = ? \)
- Pressure \( P_2 = ? \)

### Step 2: Apply the Continuity Equation
According to the continuity equation, the product of the cross-sectional area and velocity at two points in a streamline flow is constant:
\[ A_1 V_1 = A_2 V_2 \]
Substituting the known values:
\[ 10^{-2} \times 2 = 0.5 \times 10^{-2} \times V_2 \]
This simplifies to:
\[ 0.02 = 0.005 \times V_2 \]
Now, solve for \( V_2 \):
\[ V_2 = \frac{0.02}{0.005} = 4 \, \text{m/s} \]

### Step 3: Apply Bernoulli's Equation
Bernoulli's equation states:
\[ P_1 + \frac{1}{2} \rho V_1^2 + \rho g H_1 = P_2 + \frac{1}{2} \rho V_2^2 + \rho g H_2 \]
Since the pipeline is horizontal, \( H_1 = H_2 \) and the terms \( \rho g H_1 \) and \( \rho g H_2 \) cancel out. Thus, we can simplify the equation to:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]
Rearranging gives:
\[ P_2 = P_1 + \frac{1}{2} \rho V_1^2 - \frac{1}{2} \rho V_2^2 \]

### Step 4: Substitute Known Values
Assuming the density of water \( \rho = 1000 \, \text{kg/m}^3 \):
\[ P_2 = 8000 + \frac{1}{2} \times 1000 \times (2^2) - \frac{1}{2} \times 1000 \times (4^2) \]
Calculating the kinetic energy terms:
\[ P_2 = 8000 + \frac{1}{2} \times 1000 \times 4 - \frac{1}{2} \times 1000 \times 16 \]
\[ P_2 = 8000 + 2000 - 8000 \]
\[ P_2 = 2000 \, \text{Pa} \]

### Final Answer
The pressure of water at the second point is \( P_2 = 2000 \, \text{Pa} \).

---