Correct Answer - Option 4 : Dependent on the friction factor

__Concept:__

Common mean velocity distribution for turbulent flow,

\(\frac{{u - \bar u}}{{{V^*}}} = 5.75{\log _{10}}\frac{y}{R} + 3.75\)

Where, V* = shear velocity, u = Flow velocity, u̅ = avg. velocity

\({V^*} = \bar u \times \sqrt {\frac{f}{8}} \)

f = friction factor

At y = R, u = u_{max}

\(\frac{{{u_{max}} - \bar u}}{{\bar u \times \sqrt {\frac{f}{8}} }} = 3.78\)

\(\frac{{{u_{max}}}}{{\bar u}} - 1 = \frac{{3.75}}{{\sqrt 8 }} \times \sqrt f \)

\(\frac{{{u_{max}}}}{{\bar u}} = 1 + 1.33\;\sqrt f \)

\(\frac{{{u_{max}}}}{{\bar u}} = 1 + 1.43\;\sqrt f \)

So, from the above equations, we can conclude that, Ratio of maximum velocity to mean velocity is dependent on friction factor.