Laminar Flow MCQ Quiz - Objective Question with Answer for Laminar Flow - Download Free PDF

Last updated on Apr 29, 2025

Latest Laminar Flow MCQ Objective Questions

Laminar Flow Question 1:

In a laminar flow between two fixed plates held parallel to each other at a distance d, the shear stress is:

1) Maximum at plane \(\frac{d}{2}\) away from each plate and zero at the plate boundaries.

2) Zero throughout the passage.

3) Maximum at the plate boundaries and zero at plane \(\frac{d}{2}\) away from each plate.

Which of the above statements is/are correct?

  1. 1 only
  2. 3 only
  3. 2 only 
  4. 1, 2 and 3 
  5. 2 and 3

Answer (Detailed Solution Below)

Option 2 : 3 only

Laminar Flow Question 1 Detailed Solution

The velocity distribution through laminar flow in fixed parallel plates is given by:

F1 A.M Madhu 11.06.20 D1

 

\(u = \frac{1}{{2\mu }}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)(dy - {y^2})\)

Now

\(\frac{{du}}{{dy}} = \frac{1}{{2\mu }}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left( {d - 2y} \right)\)

Shear stress distribution, τ is given by

\(\tau = \frac{{\mu \;du}}{{dy}}\)

\(\tau = \frac{1}{2}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left( {d - 2y} \right)\)

From above, following can be concluded:

1. Shear Stress distribution, ‘τ’ is linear.

2. At y = d/2 i.e. mid point; τ = 0 i.e. At center shear stress = 0.

3. At y = 0 i.e. at boundary; \(\tau = {\tau _{max}} = \frac{1}{2}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left( d \right)\) i.e. shear stress is maximum at boundary.

∴ Shear stress is maximum at the plate boundaries and zero at plane \(\frac{d}{2}\)  away from each plate

Laminar Flow Question 2:

In a laminar flow between two parallel plates, where one plate is stationary and the other moves at a constant velocity, how does Newton's Law of Viscosity describe the velocity gradient in the fluid?

  1. The velocity gradient is directly proportional to the shear stress and inversely proportional to the fluid's viscosity.
  2.  The velocity gradient is inversely proportional to the shear stress and directly proportional to the fluid's viscosity. 
  3. The velocity gradient remains constant regardless of the fluid's viscosity.
  4. The velocity gradient decreases with an increase in the distance between the plates. 

Answer (Detailed Solution Below)

Option 1 : The velocity gradient is directly proportional to the shear stress and inversely proportional to the fluid's viscosity.

Laminar Flow Question 2 Detailed Solution

Explanation:

Viscosity:

Viscosity is the physical property that characterizes the flow resistance of simple fluids.

Newton’s law of viscosity defines, the shear stress between adjacent fluid layers is proportional to the velocity gradients between the two layers.

The ratio of shear stress to shear rate is a constant, for a given temperature and pressure, and is defined as the viscosity or coefficient of viscosity.

F2 J.K 2.7.20 Pallavi D6

Shear stress between two layers of fluid is directly proportional to the rate of change of velocity with respect to perpendicular distance from the fixed point (Velocity Gradient)

\({\rm{\tau \;\propto \;}}\frac{{{\rm{du}}}}{{{\rm{dy}}}}\)

\({\rm{\tau }} = {\rm{\mu \;}}\frac{{{\rm{du}}}}{{{\rm{dy}}}}\)

\({\rm{\mu }} = \frac{{\rm{\tau }}}{{\frac{{{\rm{du}}}}{{{\rm{dy}}}}}}{\rm{\;}}\)

where, τ = Shear stress and \(\frac{{{\rm{du}}}}{{{\rm{dy}}}}\) = Velocity gradient.

Laminar Flow Question 3:

Which of the following hypothesis satisfies the attraction between two closely parallel moving boats?

  1. Macaulay’s equation
  2. Euler’s equation
  3. Bernoulli’s equation
  4. Momentum equation

Answer (Detailed Solution Below)

Option 3 : Bernoulli’s equation

Laminar Flow Question 3 Detailed Solution

Explanation:

Attraction between Two Closely Parallel Moving Boats

The phenomenon where two closely parallel moving boats experience an attraction towards each other can be explained using Bernoulli's equation. This principle is rooted in fluid dynamics and describes the behavior of fluid flow.

  • Bernoulli's Principle: It states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy.

  • Application to Boats: When two boats move parallel to each other, the water flow between the boats speeds up, leading to a decrease in pressure between them.

  • Resulting Attraction: The lower pressure between the boats compared to the higher pressure on their outer sides creates a net force that pulls the boats towards each other.

Analyzing the Given Options

  1. Macaulay’s equation  (Incorrect)

    • Macaulay’s equation is used in structural analysis for determining deflections and moments in beams, which is unrelated to the fluid dynamics of moving boats.

  2. Euler’s equation  (Incorrect)

    • Euler’s equation describes the motion of a fluid, but it does not specifically explain the pressure changes due to fluid speed in the context of two parallel moving boats.

  3. Bernoulli’s equation  (Correct)

    • Bernoulli’s equation directly relates the speed of the fluid to the pressure, explaining the attraction due to the faster water flow between the boats creating lower pressure.

  4. Momentum equation  (Incorrect)

    • The momentum equation deals with the conservation of momentum in fluid flow but does not specifically describe the pressure variations causing the boats to attract each other.

Laminar Flow Question 4:

In the pipe network shown in the figure, all pipes have the same cross-section and can be assumed to have the same friction factor. The pipes, connecting points W, N, and S with point J have an equal length L. The pipe connecting points J and E has a length 10L. The pressure at the ends N, E, and S are equal. The flow rate in the pipe connecting W and J is Q. Assume that the fluid flow is steady, incompressible, and the pressure losses at the pipe entrance and junction are negligible. Consider the following statements:

I. The flow rate in pipe connecting J and E is Q/21.

II. The pressure difference between J and N is equal to the pressure difference between J and E.

Which one of the following option is CORRECT?

qImage678a5fc79f398901f2a81c0a28-4-2025 IMG-874 -61

  1. I is False and II is True 
  2. I is True and II is False
  3. Both I and II are False
  4. Both I and II are True

Answer (Detailed Solution Below)

Option 4 : Both I and II are True

Laminar Flow Question 4 Detailed Solution

Explanation:

qImage6790ca60c5444defc0898b1028-4-2025 IMG-874 -62

PN - PJ = PE - PJ

⇒ \(\frac{P_{N}-P_{J}}{\rho g}=\frac{P_{E}-P_{J}}{\rho g}\)

⇒ \(\left(h_{f}\right)_{N J}=\left(h_{f}\right)_{E J}\)

⇒ \(\frac{32 \mu v_{1} L}{\rho g d}=\frac{32 \mu v_{2}(10 L)}{\rho g d}\)

⇒ v1 = 10v2

Using continuity equation,

Q = 2Q1 + Q2

⇒ \(\frac{\pi}{4} d^{2} v=\frac{\pi}{4} d^{2}\left[2 v_{1}+v_{2}\right]\) 

⇒ v = 2v1 + v2

Using equation (i) and (ii)

\(v_{2}=\frac{1}{21} v \text { and } v_{1}=10 v_{2}=\frac{10}{21} v\) 

Discharge, \(Q_{2}=\frac{\pi}{4} d^{2} v_{2}=\frac{\pi}{4} d^{2}\left(\frac{1}{21} v\right)=\left(\frac{\pi}{4} d^{2} v\right) \frac{1}{21}=\frac{Q}{21}\)

Similarly, \(Q_{1}=\frac{\pi}{4} d^{2} v_{1}=\frac{\pi}{4} d^{2}\times\left(\frac{10}{21} v\right)=\left(\frac{\pi}{4} d^{2} v\right)\times \frac{10}{21}=\frac{10 Q}{21}\)

Now, clearly \(Q_{2}=\frac{Q}{21}\)

Hence, statement I is correct.

and (Pressure)N = (Pressure)E

⇒ PN - PE

⇒ PJ - PN = PJ - PE

Hence, the pressure difference between J and N is equal to the pressure difference between J and E.

So, statement II is true.

Laminar Flow Question 5:

Which of the following types of flow is also known as viscous flow?

  1. Compressible flow
  2. Turbulent flow
  3. Laminar flow
  4. Ideal fluid flow

Answer (Detailed Solution Below)

Option 3 : Laminar flow

Laminar Flow Question 5 Detailed Solution

Explanation:

Laminar flow is also known as viscous flow because in this type of flow:

  • The fluid particles move in parallel layers without mixing.
  • There is no turbulence, and the flow is smooth and orderly.
  • Viscous forces dominate over inertial forces, meaning that fluid friction (viscosity) plays a major role in determining the motion of the fluid.

There are six types of fluid flow:

Compressible flow

  1. The flow in which the density is not constant which means the density of the fluid changes from point to point.
  2. The flow in which the density is constant which means the density of the fluid does not change from point to point.

Incompressible flow

Uniform flow

  1. The type of flow in which the velocity at any given time does not change with respect to space (i.e. length of direction of the flow).
  2. The type of flow in which the velocity at any given time changes with respect to space (i.e. length of the direction of the flow).

Non - uniform flow

Steady flow and 

  1. Type of flow in which the fluid characteristics like velocity, density, pressure, etc at a point do not change with time.
  2. Type of flow in which the fluid characteristics like velocity, density, pressure, etc at a point change with time.

Unsteady flow

Rotational flow

  1. The type of fluid flow in which the fluid particles while flowing along streamlines and also rotate about their own axis.
  2. The type of fluid flow in which the fluid particles while flowing along streamline and do not rotate about their own axis.

Irrotational flow

Laminar flow

  1. The type of flow in which the fluid particles move along well-defined paths or streamline and all the streamlines are straight and parallel.
  2. The type of flow in which the fluid particles move in a zigzag way, the eddies formation takes place which is responsible for high energy loss.

Turbulent flow

1D flow 

  1. The type of flow in which the flow parameter such as velocity is a function of time and one space coordinate only, say x.
  2.  The type of flow in which velocity is a function of time and two rectangular space coordinates say x, y.
  3. The type of flow in which velocity is a function of time and three mutually perpendicular directions. The function of 3 space coordinates (x, y, z).

2D flow

3D flow

Top Laminar Flow MCQ Objective Questions

In which of the following case flow net cannot be drawn?

  1. Irrotational flow
  2. Steady flow
  3. When flow is governed by gravity
  4. When flow is not governed by gravity

Answer (Detailed Solution Below)

Option 3 : When flow is governed by gravity

Laminar Flow Question 6 Detailed Solution

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A flow net is a grid obtained by drawing a series of equipotential lines and streamlines. Flow net is very much useful in analyzing the two dimensional, irrotational flow problems.

The flow nets can be constructed only in the following situations -

(i) The flow should be steady. This is so because streamline pattern for unsteady flow does not remain constant, it changes from instant to instant.

(ii) The flow should be irrotational, which is possible only when flowing fluid is an ideal fluid (having no viscosity) or it has negligible viscosity.

(iii) The flow should not be governed by the force of gravity, because under the action of gravity, the shape of the free surface changes constantly and hence no fixed flow net pattern can be obtained.

∴ Thus flow net cannot be drawn when the flow is governed by gravity.

For laminar flow, Kinetic energy correction factor is:

  1. 1
  2. 1.33
  3. 2
  4. 2.7

Answer (Detailed Solution Below)

Option 3 : 2

Laminar Flow Question 7 Detailed Solution

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Explanation:

Kinetic energy correction factor(α): 

  • It is defined as the ratio of kinetic energy/second based on actual velocity to the kinetic energy/second based on average velocity.
  • \(\alpha = \frac{1}{A}\mathop \smallint \limits_A^{} {\left( {\frac{u}{V}} \right)^3}dA\)
  • where A = area, V= average velocity, u= local velocity at distance r.
  • In circular pipe laminar flow α = 2

The discharge in m3/s for laminar flow through a pipe of diameter 0.04 m having a centre line velocity of 1.5 m/s.

  1. \(\frac{{3\pi}}{{50}}\)
  2. \(\frac{{3\pi}}{{2500}}\)
  3. \(\frac{{3\pi}}{{5000}}\)
  4. \(\frac{{3\pi}}{{10000}}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{{3\pi}}{{10000}}\)

Laminar Flow Question 8 Detailed Solution

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Concept:

For laminar flow, Maximum velocity occurs at centre.

∴ Umax = Centre–line velocity 

For laminar flow, \({U_{avg}} = \frac{{{U_{max}}}}{2} \)

Discharge (Q) = Area × Average velocity

Calculation:

Given:

Pipe diameter, d = 0.04 m, Umax = Centre–line velocity = 1.5 m/s

\({U_{avg}} = \frac{{{U_{max}}}}{2} =\frac{{1.5}}{{2}}\Rightarrow 0.75\) m/s

Discharge (Q) = Area × Average velocity = \( \frac{\pi }{4} \times {\left( {0.04} \right)^2} \times 0.75 =\frac{{3\pi}}{{10000}}\)  m3/s.

The Reynold's number for the flow of a fluid in a horizontal circular tube of constant diameter is 1200. If the diameter of the tube and the kinematic viscosity of the fluid are doubled and that discharged at the pipe exit is unchanged, then the new Reynold's number for the flow in the tube will be

  1. 4800
  2. 300
  3. 1200
  4. 600

Answer (Detailed Solution Below)

Option 2 : 300

Laminar Flow Question 9 Detailed Solution

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Concept:

Reynolds Number:

  • Reynold's number is defined as the ratio of inertia force to viscous force.

\(Re~=~\frac{F_i}{F_v}\) = \(\frac{ρ× V× D}{μ}\) ; where ρ = density, V = velocity, D = diameter of pipe, μ = Dynamic viscosity of fluid

∵ Kinematic viscosity, = \(\frac{Dynamic~viscosity}{density}~= \frac{\mu}{\rho}\)

∴ \(Re~=~\frac{VD}{ν}\) ; Clearly, Reynold's number is directly proportional to the velocity and diameter, and inversely proportional to the kinematic viscosity.

Calculation:

Given:

Initially Reynold's number, Re1 = 1200, D1 = D, D2 = 2 × D, (kinematic viscosity)2 = 2 × (kinematic viscosity)1 ⇒ \(\nu_2~=~2~\times~\nu_1\)

Also, discharge at the pipe exit is unchanged.

So, \(Q_1~=~Q_2\\A_1~\times~V_1~=~A_2~\times~V_2\\{D_1}^2~\times~V_1~=~{D_2}^2~\times~V_2\)

putting the value of D1 and D2 in the above equation, we get

\(V_2~=~\frac{V_1}{4}\)

∵ \(Re~=~\frac{VD}{ν}\)

\(\frac{Re_1}{Re_2}~=~\frac{D_1~\times~V_1}{\nu_1}~\times~\frac{\nu_2}{D_2~\times~V_2}~=~\frac{D~\times~V_1}{\nu_1}~\times~\frac{2~\times~\nu_1}{2~\times~D~~\times~V_1\over 4~}~\)

\(Re_2~=~\frac{Re_1}{4}~=~300\)

So, New Reynold's number for the flow in the tube will be 300.

The shear stress between two fixed parallel plates with a laminar flow between them

  1. a constant across the gap
  2. varies parabolically as the distance from the mid plane
  3. varies inversely as the distance from the mid plane
  4. varies directly as the distance from the mid plane

Answer (Detailed Solution Below)

Option 4 : varies directly as the distance from the mid plane

Laminar Flow Question 10 Detailed Solution

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Explanation:

For the flow between two parallel spaced plates (as shown in the figure below): 

GATE 2017  set 1 images Q11

The velocity distribution is given by:

\(\rm{v = \frac{1}{{2\mu }}\left( {\frac{{ - \partial p}}{{\partial x}}} \right)\left( {By - {y^2}} \right)}\)

Shear stress is given by:

\(\rm{\begin{array}{l} \rm{\tau = \mu \frac{{du}}{{dy}}}\\ \rm{\tau= \frac{1}{2} \times \left( { - \frac{{\partial p}}{{\partial x}}} \right)\left( {B - 2y} \right)}\\ y = \frac{B}{2},\;\tau = 0 \end{array}}\)

At y = 0, τ = τmax

Hence, shear stress varies directly as the distance from the midplane.

Thus, the stress distribution across a section is depicted below:

GATE 2017  set 1 images Q11a

∴ The shear stress is maximum at the boundaries and zero at the centre.

A plate at a distance of 0.03 mm from a fixed plate moves at 0.8 m/s and requires a force of 1.50 N/m2 area of plate. Determine the dynamic viscosity of liquid between the plates.

  1. 50.25 × 10−6 N-S/m2
  2. 56.25 × 10−6 N-S/m2
  3. 6.25 × 10−6 N-S/m2
  4. 66.25 × 10−6 N-S/m2

Answer (Detailed Solution Below)

Option 2 : 56.25 × 10−6 N-S/m2

Laminar Flow Question 11 Detailed Solution

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Concept:

According to newton's law of viscosity-

This law states that the shear stress (\(\tau\)) on a fluid element layer is directly proportional to the rate of shear strain. The constant of proportionality is called the coefficient of viscosity.

\(\tau=\mu\, .({du\over dy})\)

Where \(\tau=\) shear stress \(={Force\over Area}\)

\(\mu=\) dynamic viscosity

\({du\over dy}=\) velocity gradient

Calculation:

 F5 Vinanti Engineering 21.12.22 D5

Given data:

Distance between plates (y) = 0.03 mm or 3 × 10-5 m

Velocity of moving plate (\(v\)) = 0.8 m/s

Force per unit area on plate (F/A) = 1.5 N/m2

Dynamic viscosity (μ) =?

\(Shear\, stress(\tau)=\mu× ({v\over y})\)

\(1.5=\mu× ({0.8\over 3\times 10^{-5}})\)

\(\mu={1.5\times 3\times 10^{-5}\over 0.8}\)

\(\mu=5.625\times 10^{-5}\, N-s/m^2\)

\(\mu=56.25\times 10^{-6}\, N.s/m^2\)

At the centre line of a pipe flowing under pressure where the velocity gradient is zero, the shear stress will be ________.

  1. Minimum
  2. Maximum
  3. Zero
  4. Could be any value

Answer (Detailed Solution Below)

Option 3 : Zero

Laminar Flow Question 12 Detailed Solution

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Explanation:

Shear stress equation for a laminar flow through the pipe is given by:

\({\rm{\tau }} = - \frac{{\rm{r}}}{2} \times \frac{{\partial {\rm{P}}}}{{\partial {\rm{x}}}}\)

Where,

τ = Shear stress at any distance “r” from the center of the pipe

∂P/∂x = Pressure gradient

r = distance from the center of the pipe,

At r = 0, τ = 0,

Hence, at the center line of a pipe flowing under pressure where the velocity gradient is zero, the shear stress will be zero.

The velocity is zero at the wall of the pipe increasing to a maximum at the center, then symmetrically to the other wall, and the velocity distribution is parabolic.

Shear stress maximum at the wall of the pipe decreases to a minimum (zero) at the center, then symmetrically to the other wall, shear stress increase and shear distribution is linear

For a laminar flow through a circular pipe of diameter 400 mm, the maximum velocity is 2 m/s. What will be the velocity at 8 cm from the wall of the pipe?

  1. 1.1 m/s 
  2. 2.2 m/s
  3. 1.28 m/s 
  4. 1.2 m/s 

Answer (Detailed Solution Below)

Option 3 : 1.28 m/s 

Laminar Flow Question 13 Detailed Solution

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Concept:

When a flow is fully developed steady laminar flow of an incompressible fluid in a pipe (circular).

Then velocity profile at any radius r:

\(U = - \frac{1}{{4\mu }}\left( {\frac{{dP}}{{dx}}} \right)\left( {{R^2} - {r^2}} \right)\)

For r = 0

\(\)\(u\left( r \right) = u_{max}( {1 - \frac{{{r^2}}}{{{R^2}}}} )\)

Calculation:

Given :

r = R/2; u = ? at r = 200 - 80 = 120 mm (radial distance from the central line), umax = 2 m/s

\(u\left( r \right) = 2( {1 - \frac{{{120^2}}}{{{200^2}}}} ) = 1.28 \ m/s\)

Laminar developed flow at an average velocity of 5 m/s occurs in a pipe of 10 cm radius. The velocity at 5 cm radius is

  1. 7.5 m/s
  2. 10 m/s
  3. 2.5 m/s
  4. 5 m/s

Answer (Detailed Solution Below)

Option 1 : 7.5 m/s

Laminar Flow Question 14 Detailed Solution

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Concept:

GATE ME 2009 Images-Q34.1

Distribution of velocity profile:

In a fully developed laminar pipe flow, the average velocity is one half of the maximum velocity i.e. Umax = 2Uavg

\(\begin{array}{l} U (r) = U_{max}\left[ {1 - {{\left( {\frac{r}{R}} \right)}^2}} \right] \end{array}\)

\(\begin{array}{l} U (r) = 2{U_{av}}\left[ {1 - {{\left( {\frac{r}{R}} \right)}^2}} \right] \end{array}\)

Calculation:

Given:

Uavg = 5 m/s, R = 10 cm, r = 5 cm

\(U = 2 \times 5\left[ {1 - {{\left( {\frac{5}{{10}}} \right)}^2}} \right] = 7.5\;m/s\)

Glycerine (μ = 1.50 Pa.s: ρ = 1260 kg/m3) flows at a velocity of 6.0 m/s in 10 cm diameter pipe. Head loss in a length of 7 m pipe will be (g = 10 m/s2). 

  1. 14 m
  2. 16 m
  3. 7 m
  4. 8 m

Answer (Detailed Solution Below)

Option 2 : 16 m

Laminar Flow Question 15 Detailed Solution

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Concept:

Hagen Poiseuille equation

Loss of pressure head  \(= \dfrac{{{p_1} - {p_2}}}{{ρ g}} = \dfrac{{32μ VL}}{{ρ g{D^2}}} = {h_f}\)

Calculation:

Given:

Viscosity μ = 1.50 Pa.s, diameter d = 0.1 m, V = 6 m/s, L = 7 m, ρ = 1260 kg/m3

Loss of pressure head  \(= \dfrac{{{p_1} - {p_2}}}{{ρ g}} = \dfrac{{32 \times 6\times1.50\times 7 }}{{1260\times 10\times 0.01{}}} = 16 ~m\)

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