Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
189 views
in General by (95.2k points)
closed by

The thickness of the boundary layer for a fluid flowing over a flat plate at a point 20 cm from the leading edge is found to be 4 mm. The Reynolds number at the point (adopting 5 as the relevant constant) is:


1. 48400
2. 57600
3. 62500
4. 77600

1 Answer

0 votes
by (95.4k points)
selected by
 
Best answer
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  

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

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

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...