A horizontal pipeline carries water in a stream-line flow. At a point along the pipe, where the cross-sectional area is $10 \mathrm{~cm}^2$, the velocity of water is $1 \mathrm{~m} / \mathrm{s}$ and pressure is 2000 Pa . The pressure of water at another point where the cross-sectional area $5 \mathrm{~cm}^2$ is $\left[\text { Given } \rightarrow \text { density of water }=1000 \mathrm{~kg} / \mathrm{m}^3\right]$

The equation of continuity gives $A_1 v_1 = A_2 v_2$ for incompressible flow. With $A_1 = 10 \times 10^{-4} \, \text{m}^2$ and $A_2 = 5 \times 10^{-4} \, \text{m}^2$, while $v_1 = 1 \, \text{m/s}$, it follows that $v_2 = \frac{A_1}{A_2} v_1 = \frac{10 \times 10^{-4}}{5 \times 10^{-4}} \cdot 1 = 2 \, \text{m/s}$.

For a horizontal pipeline, Bernoulli’s equation simplifies to $P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$, where $\rho = 1000 \, \text{kg/m}^3$ and $P_1 = 2000 \, \text{Pa}$. Substituting known values:

$2000 + \frac{1}{2} \cdot 1000 \cdot 1^2 = 2000 + 500 = 2500 \, \text{Pa}$

and

$\frac{1}{2} \cdot 1000 \cdot 2^2 = 2000 \, \text{Pa}$.

Solving yields $P_2 = 2500 - 2000 = 500 \, \text{Pa}$.

The pressure at the constricted point is 500 Pa.