250+ TOP MCQs on Classification of Fluids and Answers

Bioprocess Engineering Multiple Choice Questions on “Classification of Fluids”.

1. “The density of liquids is practically dependent of pressure”?
A. True
B. False

Answer: B
Explanation: The density of liquids is practically independent of pressure; liquids are incompressible fluids. If the density of a fluid changes with pressure, the fluid is compressible. Gases are generally classed as compressible fluids.

2. What is the property of an ideal or perfect fluid?
A. Compressible and zero viscosity
B. Compressible and zero density
C. Incompressible and zero viscosity
D. Incompressible and zero density

Answer: C
Explanation: Viscosity is the property of fluids responsible for internal friction during flow. An ideal or perfect fluid is a hypothetical liquid or gas which is incompressible and has zero viscosity. The term inviscid applies to fluids with zero viscosity.

3. Fluids which undergo strain rates proportional to the applied shear stress are termed as?
A. Inviscid fluid
B. Newtonian fluid
C. Non- Newtonian fluid
D. Viscous fluid

Answer: B
Explanation: Newtonian fluids undergo strain rates proportional to the applied shear stress. It is defined to be a fluid whose shear stress is linearly proportional to the velocity gradient in the direction perpendicular to the plane of shear. This definition means regardless of the forces acting on a fluid, it continues to flow.

4. Which of the following is not an example of a Newtonian fluid?
A. Water
B. Glycerine
C. Non- Drip Paints
D. Alcohol

Answer: C
Explanation: Stirring a non-Newtonian fluid can cause the viscosity to decrease, so the fluid appears “thinner” (this is seen in non-drip paints). There are many types of non-Newtonian fluids, as they are defined to be something that fails to obey a particular property – for example, most fluids with long molecular chains can react in a non-Newtonian manner.

5. Which of the following is not an example of a Non- Newtonian fluids?
A. Gels
B. Water
C. Suspensions
D. Pudding

Answer: B
Explanation: Water is a Newtonian fluid, because it continues to display fluid properties no matter how much it is stirred or mixed. A slightly less rigorous definition is that the drag of a small object being moved slowly through the fluid is proportional to the force applied to the object. (Compare friction). Important fluids, like water as well as most gases, behave – to good approximation – as a Newtonian fluid under normal conditions on Earth.

6. What do you mean by the term “Rheology”?
A. Study of materials with both solid and fluid characteristics
B. Study of materials with only solid characteristics
C. Study of materials with only fluid characteristics
D. Study of material with both fluid and gas characteristics

Answer: A
Explanation: It is the study of the flow of matter, primarily in a liquid state, but also as “soft solids” or solids under conditions in which they respond with plastic flow rather than deforming elastically in response to an applied force. It is a branch of physics which deals with the deformation and flow of materials, both solids and liquids.

7. What is the unit of viscosity of fluids in C.G.S?
A. ml/s
B. l/s
C. Poise
D. Newton

Answer: C
Explanation: The unit of viscosity (.in C.G.S) is poise.
One poise = 0.1 N.s / m²,
µ_water = 10^-3 Ns / m²
µ_air = 1.81 x 10^-5 Ns / m².

8. If 5 m³ of certain oil weighs 45 kN calculate the specific weight of the oil.
A. 10 kN/m³
B. 9 kN/m³
C. 5 kN/m³
D. 2 kN/m³

9. Refer to Q8 and Calculate specific gravity of the oil. (Specific weight of water = 9.807 kN/m³)
A. 0.991
B. 0.918
C. 0.917
D. 0.992

10. A liquid has a mass density of 1550 kg/m³. Calculate its specific weight.
A. 1.50×102 N/m³
B. 1.52×104 N/m³
C. 1.54×104 N/m³
D. 1.50×104 N/m³

Answer: B
Explanation: Given data: Mass density = 1550 kg/m³
Specific weight = mass density × Acceleration due to gravity
= 1550 kg/m³ × 9.8 m/s²
= 1.52×10⁴ N/m³.

11. Refer to Q10 and calculate specific gravity.
A. 1.55
B. 1.56
C. 1.50
D. 1.54

12. Refer to Q 10 and Q11, and calculate specific volume.
A. 6.51 × 10^-5 m³/N
B. 6.78 × 10^-5 m³/N
C. 6.45 × 10^-5 m³/N
D. 6.57 × 10^-5 m³/N

Answer: D
Explanation: Specific volume = 1/(Specific weight)
= 1/(1.52×104 N/m³)
= 6.57 × 10^-5 m³/N.

13. If the equation of a velocity profile over a plate is v = 5y² + y (where v is the velocity in m/s) determine the shear stress at y = 0. Given the viscosity of the liquid is 8.35 poise.
A. 0.830
B. 0.832
C. 0.835
D. 0.834

14. Refer to Q13 and determine the shear stress at y =7.5cm.
A. 1.46 N/m³
B. 1.45 N/m³
C. 1.40 N/m³
D. 1.43 N/m³

15. “Mach is the Dimensionless quantity”.
A. True
B. False