Step 1: Understanding Bernoulli's principle.
Bernoulli's principle states that for an incompressible, non-viscous fluid in streamline flow: \[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}, \] where: \(P\): Pressure of the fluid, \(\rho\): Density of the fluid, \(v\): Velocity of the fluid, \(h\): Height of the fluid.
Step 2: Application to a horizontal pipe.
For a horizontal pipe (\(h\) is constant), the equation simplifies to: \[ P + \frac{1}{2} \rho v^2 = \text{constant}. \] This implies that when the velocity (\(v\)) of the fluid increases, the pressure (\(P\)) decreases, and vice versa.
Step 3: Narrowest part of the pipe.
At the narrowest part of the pipe: The cross-sectional area is minimum. By the equation of continuity (\(A_1v_1 = A_2v_2\)), the velocity is maximum. By Bernoulli's principle, the pressure is minimum.
Step 4: Conclusion.
At the narrowest part of the pipe, the velocity is maximum, and the pressure is minimum. \[ \therefore \text{The correct answer is: (A)}. \]