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 for fluid flow.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Cross-sectional area at point A, \( A_1 = 10^{-2} \, m^2 \)
- Velocity at point A, \( V_1 = 2 \, m/s \)
- Pressure at point A, \( P_1 = 8000 \, Pa \)
- Cross-sectional area at point B, \( A_2 = 0.5 \times 10^{-2} \, m^2 = 5 \times 10^{-3} \, m^2 \)

2. **Use the Continuity Equation**:
The continuity equation states that \( A_1 V_1 = A_2 V_2 \).
\[ A_1 V_1 = A_2 V_2 \implies (10^{-2})(2) = (5 \times 10^{-3}) V_2 \]
\[ 0.02 = 5 \times 10^{-3} V_2 \]
\[ V_2 = \frac{0.02}{5 \times 10^{-3}} = 4 \, m/s \]

3. **Apply Bernoulli's Equation**:
Bernoulli's equation for horizontal flow states:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]
Here, we can assume the density of water \( \rho = 1000 \, kg/m^3 \).

Substitute the known values into the equation:
\[ 8000 + \frac{1}{2} (1000)(2^2) = P_2 + \frac{1}{2} (1000)(4^2) \]
\[ 8000 + \frac{1}{2} (1000)(4) = P_2 + \frac{1}{2} (1000)(16) \]
\[ 8000 + 2000 = P_2 + 8000 \]

4. **Solve for \( P_2 \)**:
\[ 10000 = P_2 + 8000 \]
\[ P_2 = 10000 - 8000 = 2000 \, Pa \]

5. **Final Result**:
The pressure of water at point B is \( P_2 = 2000 \, Pa \).