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**Hint:**To solve this problem,we will apply Reynold's number formula for the laminar flow. Reynold’s number is dimensionless number as it is the ratio of Inertial force to viscous force. Reynold’s number is denoted by \[{N_{\operatorname{Re} }}\]. Mathematically, the Reynold’s number is calculated as,

\[{N_{\operatorname{Re} }} = \dfrac{{\rho vd}}{\mu }\] ……(1)

Where, \[{N_{\operatorname{Re} }}\] is Reynolds’s number and its numerical value for the laminar flow is 2000.

\[\rho \] Is the density of the fluid

\[v\] is the average flow speed or average velocity of the flow

\[d\] is the diameter of the pipe

\[\mu \] is the coefficient of viscosity or dynamic viscosity

**Complete step by step solution:**

Given that:

Diameter of the pipe is, \[d = 5cm\]

The coefficient of viscosity is, \[\mu = {10^{ - 3}}Pa\sec .\]

The density of water is, \[\rho = 1000\dfrac{{kg}}{{m{}^3}}\]

Now put all the given values in equation (1)

${N_{\operatorname{Re} }} = \dfrac{{\rho vd}}{\mu } \\

v = \dfrac{{{N_{\operatorname{Re} }}\mu }}{{\rho d}} \\

v = \dfrac{{2000 \times {{10}^{ - 3}}}}{{1000 \times 0.05}} \\

v = 4\dfrac{{\text{m}}}{{\text{s}}} \\

$

As we know, for laminar flow the maximum flow speed is twice the average flow speed.

Thus the maximum flow speed of water for laminar flow in a pipe is calculated as,

\[

{v_{\max }} = 2 \times v \\

= 2 \times 4 \\

= 8\dfrac{{\text{m}}}{{\text{s}}} \\

\]

**Therefore, the maximum flow speed of water for laminar flow in a pipe is \[{v_{\max }} = 8\dfrac{{\text{m}}}{{\text{s}}}\].**

**Note:**Laminar flow is also known as stream line flow because the fluid particles travel in a regular, smooth and in straight line path. On the other side, if the fluid particles travel in irregular or zig-zag ways then the flow is known as turbulent flow.

If the value of Reynolds’s number is equal to or less than 2000 then the flow is considered as laminar flow. If the value of Reynolds’s number is more than 4000 then the flow is considered as turbulent flow.

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