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Air flows past a cricket ball of 40 mm diameter. It is observed that the boundary layer becomes turbulent at a Reynolds number of 200000. If the kinematic viscosity of air is 1.5 × 10-5 m2/s, then the speed of air when the flow becomes turbulent is:

1. 25 m/s
2. 75 m/s
3. 100 m/s
4. 50 m/s

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Correct Answer - Option 2 : 75 m/s

Concept:

Reynolds number signifies the ratio of inertia force to viscous force in velocity boundary layer. i.e. Re = \(\frac{F_i}{F_v}\)

Reynolds number also represents where the boundary layer changes from laminar to turbulent.

It is given in terms of viscosity:

\(Re=\frac{{\rho V L}}{\mu}=\frac{{ V L}}{ν}\)

where, ρ = density of fluid, V = velocity of fluid, L = characteristics length, μ = dynamic viscosity of fluid and ν = kinematic viscosity.

Characteristic length = \(\frac{4A}{P}=D\)

Calculation:

Given:

d = 40 mm = 0.04 m, Re = 200000 = 2 × 105, v = 1.5 × 10-5 m2/s.

\(Re=\frac{{\rho V D}}{\mu}=\frac{{ V D}}{ν}\)

\(2\times 10^5=\frac{{ V \times 0.04}}{1.5\times 10^{-5}}=75\;m/s\)

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