A fluid is in streamline flow across a horizontal pipe of variable area of cross section. For this which of the following statements is correct?

Options

A. The velocity is minimum at the narrowest part of thepipe and the pressure is minimum at the widest part of the pipe
B. The veloeity is maximum at the narrowest part of the pipe and pressure is maximum at the widest part of the pipe
C. Velocity and pressure both are maximum at the narrowest part of the pipe
D. Velocity and pressure both are maximum at the widest part of the pipe

Explanation

To solve the problem regarding the behavior of fluid in streamline flow across a horizontal pipe of variable cross-section, we can follow these steps:

### Step 1: Understand the Continuity Equation
The continuity equation states that for an incompressible fluid flowing through a pipe, the mass flow rate must remain constant. This can be expressed as:
\[ A_1 V_1 = A_2 V_2 \]
where \( A \) is the cross-sectional area and \( V \) is the fluid velocity at two different points in the pipe.

### Step 2: Analyze the Flow in a Variable Cross-Section Pipe
In a horizontal pipe with varying cross-section, if the area decreases (narrowest part), the velocity of the fluid must increase to maintain the constant mass flow rate. Conversely, if the area increases (widest part), the velocity must decrease.

### Step 3: Apply Bernoulli's Theorem
Bernoulli's theorem states that for an incompressible, non-viscous fluid flowing in a streamline flow, the total mechanical energy along a streamline is constant. This can be expressed as:
\[ P + \frac{1}{2} \rho V^2 + \rho gh = \text{constant} \]
where \( P \) is the pressure, \( \rho \) is the fluid density, \( V \) is the fluid velocity, and \( h \) is the height above a reference level. In a horizontal pipe, the height (\( h \)) remains constant, so we can simplify this to:
\[ P + \frac{1}{2} \rho V^2 = \text{constant} \]

### Step 4: Determine Pressure and Velocity Relationships
From the continuity equation, we established that:
- At the narrowest part of the pipe (where area \( A \) is minimum), the velocity \( V \) is maximum.
- At the widest part of the pipe (where area \( A \) is maximum), the velocity \( V \) is minimum.

Using Bernoulli's theorem:
- Since the velocity is maximum at the narrowest part, the kinetic energy term (\( \frac{1}{2} \rho V^2 \)) is also maximum, which implies that the pressure \( P \) at this point must be minimum to keep the total energy constant.
- Conversely, at the widest part of the pipe, the velocity is minimum, so the pressure \( P \) must be maximum.

### Conclusion
Thus, the correct statements are:
- The velocity is maximum at the narrowest part of the pipe.
- The pressure is minimum at the narrowest part of the pipe and maximum at the widest part of the pipe.

### Final Answer
The correct statement is: **Velocity is maximum at the narrowest part of the pipe and pressure is minimum at the narrowest part of the pipe.**

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