Water is flowing through a horizontal pipe of non-uniform cross-section. At the extreme narrow portion of the pipe, the water will have

Correct Answer: Maximum speed and least pressure

To understand the behavior of water flowing through a horizontal pipe of non-uniform cross-section, we need to apply two fundamental principles of fluid dynamics: the Equation of Continuity and Bernoulli's Principle.

1. Principle of Continuity:

For an incompressible fluid like water flowing through a pipe, the volume flow rate remains constant throughout the pipe. Mathematically, this is expressed as:

$A·v=\text{constant}$

where $A$ is the cross-sectional area of the pipe and $v$ is the speed of the fluid. From this equation, we see that speed is inversely proportional to the cross-sectional area:

$v\propto \frac{1}{A}$

Therefore, at the extreme narrow portion of the pipe where the cross-sectional area $A$ is at its minimum, the speed $v$ of the water will be at its maximum.

2. Bernoulli's Principle:

For horizontal flow (where gravitational potential energy stays constant, i.e., $h=\text{constant}$), Bernoulli's equation is given by:

$P+\frac{1}{2}\alpha v^2=\text{constant}$

where $P$ is the static pressure, $\alpha$ is the fluid density, and $v$ is the fluid speed.

Since the total mechanical energy remains constant, an increase in fluid speed ($v$) leads to a corresponding increase in kinetic energy per unit volume ($\frac{1}{2}\alpha v^2$). To compensate and keep the sum constant, the static pressure ($P$) must decrease.

Conclusion:

At the narrowest region of the pipe:
- The speed of the water is at its maximum due to the smallest area.
- The pressure of the water is at its least due to Bernoulli's principle.

Hence, the water will have maximum speed and least pressure.