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Test: Newton's Law of Viscosity - Civil Engineering (CE) MCQ


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10 Questions MCQ Test - Test: Newton's Law of Viscosity

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Test: Newton's Law of Viscosity - Question 1

The following results are obtained on shear stress (τ) and rate of deformation (du/dy) at constant temperature for a fluid.

The above fluid is classified as

Detailed Solution for Test: Newton's Law of Viscosity - Question 1

The mathematical model for describing the behavior of fluids is the Power-Law model, which is given as:

If
n = 1; ⇒ fluid is Newtonian Fluid.
n ≠ 1; ⇒ fluid is non-Newtonian
n < 1; ⇒ fluid is called Pseudoplastic fluids.  
n > 1; ⇒ fluids called dilatants fluids.  
The mathematical model for describing the ideal Plastic fluid is given as:

From the given table:
If du/dy = 0 then τ =  10 i.e. non – zero ⇒ Fluid can not be Newtonian and non-Newtonian

Further Shear stress is a linear function of du/dy, so the fluid is ideal plastic

Test: Newton's Law of Viscosity - Question 2

A Newtonian fluid is filled in the clearance between a shaft and a concentric sleeve. The sleeve attains a speed of 80 cm/s, when a force of 40 N is applied to the sleeve parallel to the shaft. Determine the speed if a force of 300N is applied.

Detailed Solution for Test: Newton's Law of Viscosity - Question 2

Concept:
Newtonian fluids always follow Newton's law of Viscosity. Mathematically Newton's law of viscosity is represented as

Force equation from the above equation can be obtained as

where, μ = Coefficient of Viscosity, τ = Shear Stress
From the given question we can observe that Area(A), Viscosity(μ) and clearance(dy) is constant, hence Force is directly proportional to velocity.

Calculation:
Given:
F1 = 40 N, F2 = 300 N, V1 = 80 cm/s
40/300 = 80/V2
V2 = 600 cm/s

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Test: Newton's Law of Viscosity - Question 3

Kinematic viscosity also known as stoke is equal to:

Detailed Solution for Test: Newton's Law of Viscosity - Question 3

Kinematic viscosity:
It is defined as the ratio between the dynamic viscosity and density of the fluid.
v = μ/ρ
Units of kinematic viscosity:


∴ the SI unit of kinematic viscosity is m2/s and the C.G.S unit of kinematic viscosity is cm2/s or 'Stoke'.
1 Stoke = 10-4 m2/s.

Test: Newton's Law of Viscosity - Question 4

Which is the fluid whose viscosity does NOT change with the rate of deformation?

Detailed Solution for Test: Newton's Law of Viscosity - Question 4

Newtonian fluids: fluids for which the shear stress is linearly proportional to the shear strain rate and follow Newton's law of viscosity.

  •  where μ is shear viscosity of the fluid
  • Newtonian fluids are analogous to elastic solids (Hooke’s law: stress proportional to strain)
  • Any common fluids, such as air and other gases, water, kerosene, gasoline, and other oil-based liquids, are Newtonian fluids
  • A fluid whose viscosity does not change with the rate of deformation or shear strain is known as a Newtonian fluid.

Non-Newtonian fluid: fluids for which the shear stress is not linearly related to the shear strain rate are called non-Newtonian fluids 

  • In non-Newtonian, the viscosity is dependent on the shear rate (Shear Thinning or Thickening)
  • examples include slurries and colloidal suspensions, polymer solutions, blood, paste, and cake batter 

Ideal fluid: fluids which don’t have viscosity and are incompressible are termed as an ideal fluid

  • The ideal fluid does not offer shear resistance i.e no resistance is encountered as the fluid moves.

Real fluid: fluids which do possess viscosity are termed as real fluids.

  • These fluids always offer shear resistance i.e. Certain amount of resistance is always offered by these fluids as they move.
Test: Newton's Law of Viscosity - Question 5

Consider two flat parallel plates placed in horizontal condition 1.2 cm apart and the space between them is filled with the oil of viscosity 15.0 poise. The upper plate is moved with a velocity of 3.25 m/s, then the shear stress in the oil is

Detailed Solution for Test: Newton's Law of Viscosity - Question 5

Concept:
Shear stress in the oil

where: μ = Dynamic viscosity, τ = Shear stress, V = velocity of plate, y = distance between the plates.

Calculation:
Given:

Thickness = t = 1.2 cm = 0.012 m, μ = 15 poise = 1.5 Ns/m2, ΔV = 3.25 m/s

Test: Newton's Law of Viscosity - Question 6

For a Newtonian fluid

Detailed Solution for Test: Newton's Law of Viscosity - Question 6

For a Newtonian fluid,
shear stress, 
or,  where τ = shear stress,
du/dy = dθ/dt = rate of shear strains,
θ = displacement in fluid layer

Test: Newton's Law of Viscosity - Question 7

The velocity distribution for flow over a flat plate is given  in which u is the velocity in m/s at a distance y meter above the plate. Shear stress at y = 0.15 m is (Take μ for fluid = 8.5 poise)

Detailed Solution for Test: Newton's Law of Viscosity - Question 7

Concept:
Shear stress in flow over a flat plate is given as:

where, μ = dynamic viscosity, y = distance in meter above the plate.
1 P = 0.1 Pa-sec

Calculation:
Given:
μ = 8.5 poise = 8.5 × 10-1 Pa-sec, y = 0.15 m, 


∴ τ = 8.5 × 10-1 × 0.45 
τ = 0.3825 N/m2 or Pa

Test: Newton's Law of Viscosity - Question 8

A cubic wooden block of edge 100 mm and weight 1 kN is sliding down on an inclined plane of inclination 30° with the horizontal. A Newtonian fluid with the viscosity 0.2 Ns/m2 is layered on the inclined plane. If the thickness of the layer is 0.02 mm, then the terminal velocity of the block in m/s.

Detailed Solution for Test: Newton's Law of Viscosity - Question 8

Concept:


For a Newtonian fluid,
share stress (τ)  = dynamic visocity (μ) × rate of share strain (du/dy)
For the above situation, 

where, A = Area in contact of a Newtonian fluid, V = Terminal velocity of the block
Calculation:
Given:
Edge of the block, a = 100 mm = 0.1 m , weight, W = 1 kN, θ = 30°, y = 0.02 mm, 

Shear stress acting at the face of the cube in contact with fluid,

V = 5 m/s

Test: Newton's Law of Viscosity - Question 9

Which of the following is correct?

Detailed Solution for Test: Newton's Law of Viscosity - Question 9

According to Newton's law of viscosity, shear stress is directly proportional to the rate of angular deformation (shear strain) or velocity gradient across the flow.


where, τ = shear stress, μ = absolute or dynamic viscosity, du/dy = velocity gradient ⇒ dα/dt = rate of angular deformation (shear-strain).
Units of Dynamic viscosity:

∴ the unit of dynamic viscosity in the SI unit is Ns/m2 or Pa-s.
The dimension of dynamic viscosity is ML-1T-1.
1 Pa-s = 10 Poise
∴ 1 Poise = 0.1 Pa-s or Ns/m2
Kinematic viscosity: 
It is defined as the ratio between the dynamic viscosity and density of the fluid.
v = μ/ρ
Units of kinematic viscosity:

∴ the SI unit of kinematic viscosity is m2/s and the CGS unit of kinematic viscosity is cm2/s or 'Stoke'.
1 Stoke = 10-4 m2/s = 0.0001 m2/s.
The dimensional formula for Kinematic viscosity (ν) is ν = L2T-1.

Test: Newton's Law of Viscosity - Question 10

A Newtonian fluid fills the clearance between a shaft and a sleeve. When a force of 0.9 kN is applied to the shaft parallel to the sleeve, the shaft attains a speed of 1.25 cm/s. What will be the speed of the shaft if a force of 3 kN is applied?

Detailed Solution for Test: Newton's Law of Viscosity - Question 10

Concept:
From Newton's law of viscosity

We know that 
F = τ × A

Calculation:
Given: F2 = 3 kN, F1 = 0.9 kN, u1 =1.25 cm/s

∴ u2 = 4.166 cm/s

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