A horizontal pipeline carries water in a streamline flow. At a point along the pipe, where the cross- sectional area is `10cm^2`, the water velocity is `1ms^-1` and the pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area is `5cm^2`, is........Pa. (Density of water `=10^3kg.m^-3`)

Explanation

To solve the problem, we will use the principles of fluid dynamics, specifically the continuity equation and Bernoulli's equation.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Cross-sectional area at point 1, \( A_1 = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 0.001 \, \text{m}^2 \)
- Velocity at point 1, \( v_1 = 1 \, \text{m/s} \)
- Pressure at point 1, \( p_1 = 2000 \, \text{Pa} \)
- Cross-sectional area at point 2, \( A_2 = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 = 0.0005 \, \text{m}^2 \)
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)

2. **Use the Continuity Equation to Find Velocity at Point 2:**
The continuity equation states that the product of cross-sectional area and velocity is constant along the flow:
\[ A_1 v_1 = A_2 v_2 \]
Rearranging gives:
\[ v_2 = \frac{A_1}{A_2} v_1 \]
Substituting the values:
\[ v_2 = \frac{0.001 \, \text{m}^2}{0.0005 \, \text{m}^2} \times 1 \, \text{m/s} = 2 \, \text{m/s} \]

3. **Apply Bernoulli's Equation:**
Bernoulli's equation for horizontal flow is given by:
\[ p_1 + \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2} \rho v_2^2 \]
Rearranging to find \( p_2 \):
\[ p_2 = p_1 + \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2 \]

4. **Substituting Known Values:**
\[ p_2 = 2000 \, \text{Pa} + \frac{1}{2} \times 1000 \, \text{kg/m}^3 \times (1 \, \text{m/s})^2 - \frac{1}{2} \times 1000 \, \text{kg/m}^3 \times (2 \, \text{m/s})^2 \]
Calculating each term:
- Kinetic energy term at point 1:
\[ \frac{1}{2} \times 1000 \times 1^2 = 500 \, \text{Pa} \]
- Kinetic energy term at point 2:
\[ \frac{1}{2} \times 1000 \times 2^2 = 2000 \, \text{Pa} \]
Now substituting these values back into the equation:
\[ p_2 = 2000 \, \text{Pa} + 500 \, \text{Pa} - 2000 \, \text{Pa} \]
Simplifying:
\[ p_2 = 2000 + 500 - 2000 = 500 \, \text{Pa} \]

5. **Final Answer:**
The pressure of water at the point where the cross-sectional area is \( 5 \, \text{cm}^2 \) is \( 500 \, \text{Pa} \).