Correct Answer - Option 3 : 62500

__Concept__

__Thickness of Laminar Boundary layer:__

\(δ = \frac{{5x}}{{\sqrt {R{e_x}} }}\)

Where,

x = distance from the leading edge

Rex = local Reynold's Number = \(R{e_x} = \frac{{ρ Vx}}{μ } = \frac{{Vx}}{ν }\)

where, ρ = density of fluid in kg/m3, V = average velocity in m/s

μ = dynamic viscosity in N-s/m2 and ν = kinematic viscosity in m2/s.

**Calculation:**

Given, δ = 4 mm, x = 20 cm = 200 mm

then

\(δ = \frac{{5x}}{{\sqrt {R{e_x}} }}\)

\(R{e_x} = {\left( {\frac{{5x}}{\delta }} \right)^2} = {\left( {\frac{{5 \times 200}}{4}} \right)^2} = 62500\)