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

Step 1: Identify the given parameters.

Convert the cross-sectional areas to standard SI units.

$$A_1 = 10^{-3}\,\text{m}^2$$ $$v_1 = 1\,\text{m s}^{-1}$$ $$P_1 = 2000\,\text{Pa}$$ $$A_2 = 5 \times 10^{-4}\,\text{m}^2$$ $$\rho = 10^{3}\,\text{kg m}^{-3}$$

Step 2: Apply the equation of continuity to find the velocity at the second point.

The volume flow rate remains constant throughout the pipeline.

$$A_1 v_1 = A_2 v_2$$ $$v_2 = \frac{A_1 v_1}{A_2}$$ $$v_2 = \frac{10^{-3} \times 1}{5 \times 10^{-4}}$$ $$v_2 = 2\,\text{m s}^{-1}$$

Step 3: Apply Bernoulli's equation for a horizontal pipe.

Since the pipe is horizontal, the gravitational potential energy terms are equal and cancel out.

$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$

Step 4: Calculate the dynamic pressure terms.

Compute the kinetic energy per unit volume at both points.

$$\frac{1}{2}\rho v_1^2 = \frac{1}{2} \times 10^{3} \times 1^2 = 500\,\text{Pa}$$ $$\frac{1}{2}\rho v_2^2 = \frac{1}{2} \times 10^{3} \times 2^2 = 2000\,\text{Pa}$$

Step 5: Solve for the pressure at the second point.

Rearrange Bernoulli's equation to find $P_2$.

$$P_2 = P_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2$$ $$P_2 = 2000 + 500 - 2000$$ $$P_2 = 500\,\text{Pa}$$

Answer: 500