Fluid Mechanics Solutions Top 1000 Basic Level Multiple Choice Questions OF Fluid Mechanics

44. Which one is in a state of failure?

a) Solid

b) Liquid

c) Gas

d) Fluid                 

Answer: d
                  Explanation: A fluid is a Tresca material with zero cohesion. In simple words, fluid is in a state of failure.

 

45. A small shear force is applied on an element and then removed. If the element regains it’s original position, what kind of an element can it be?

a) Solid

b) Liquid

c) Fluid

d) Gaseous

Answer: a
Explanation: Fluids (liquids and gases) cannot resist even a small shear force and gets permanently deformed. Hence, the element must be a solid element.     

46. In which type of matter, one won’t find a free surface?

a) Solid

b) Liquid

c) Gas

d) Fluid

Answer: c
Explanation: Solid molecules have a definite shape due to large inter-molecular forces. In liquids, molecules are free to move inside the whole mass but rarely escape from itself. Thus, liquids can form free surfaces under the effect of gravity. But, in case of gases, molecules tend to escape due to low forces of attraction. Thus, gases won’t form any free surface.

47. If a person studies about a fluid which is at rest, what will you call his domain of study?

a) Fluid Mechanics

b) Fluid Statics

c) Fluid Kinematics

d) Fluid Dynamics

Answer: b
Explanation: Fluid Mechanics deals with the study of fluid at rest or in motion with or without the consideration of forces, Fluid Statics is the study of fluid at rest, Fluid Kinematics is the study of fluid in motion without consideration of forces and Fluid Dynamics is the study of fluid in motion considering the application forces.

48. The value of the compressibility of an ideal fluid is

a) zero

b) unity

c) infinity

d) more than that of a real fluid

Answer: a
Explanation: Ideal fluids are incompressible which means they will have zero compressibility.

49. The value of the Bulk Modulus of an ideal fluid is

a) zero

b) unity

c) infinity

d) less than that of a real fluid

Answer: c
Explanation: Bulk modulus k is the reciprocal of compressibility fi.
k = ¹⁄_fi
Ideal fluids are incompressible which means fi = 0. Thus, k will be infinity.

50. The value of the viscosity of an ideal fluid is

a) zero

b) unity

c) infinity

d) more than that of a real fluid

Answer: a
Explanation: Ideal fluids are non-viscous which means they will have zero viscosity.

51. The value of the surface tension of an ideal fluid is

a) zero

b) unity

c) infinity

d) more than that of a real fluid.

Answer: a
Explanation: Ideal fluids haze zero surface tension but real fluids have some finite value of surface tension.

52. Which one of the following is the unit of mass density?

a) kg = m3

b) kg = m2

c) kg = m

d) kg = ms

Answer: a
Explanation: Mass Density(p) is defined as the mass(m) per unit volume(V ), i.e., p = m ⁄v
Thus, the unit of p is kg = m³.

53. The specific gravity of a liquid has

a) the same unit as that of mass density

b) the same unit as that of weight density

c) the same unit as that of specific volume

d) no unit

Answer: d
Explanation: The specific gravity of a liquid is the ratio of two similar quantities (densities) which makes it unitless.

54. The specific volume of a liquid is the reciprocal of

a) weight density

b) mass density

c) specific weight

d) specific volume

Answer: b
Explanation: Specific volume(v) is defined as the volume(V ) per unit mass(m).
v = v⁄m = 1 / m⁄v = 1⁄p
where p is the mass density.

55. Which one of the following is the unit of specific weight?

a) N = m3

b) N = m2

c) N = m

d) N = ms

Answer: a
Explanation: Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,
γ = w / v
Thus, unit of is N = m³.

56. Which one of the following is the unit of specific weight?

a) N = m3

b) N = m2

c) N = m

d) N = ms

Answer: a
Explanation: Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,
γ = w / v
Thus, unit of is N = m³.

57. Which one of the following is the dimension of specific gravity of a liquid?

a) [M1 L -3 T 0 ].

b) [M1 L 0 T 0 ].

c) [M0 L -3 T 0 ].

d) [M0 L 0 T 0 ].

Answer: a
Explanation: Mass Density(p) is defined as the mass(m) per unit volume(V ), i.e.,
[p] = [m]/[v] = [m] /[L³] = [ML^-3].

58. Which one of the following is the dimension of specific volume of a liquid?

a) [M1 L -3 T 0 ].

b) [M-1 L 3 T 0 ].

c) [M-1 L -3 T 0 ].

d) [M0 L 3 T 0 ].

Answer: d
Explanation: The specific gravity of a liquid is the ratio of two similar quantities (densities) which makes it dimensionless.

59. Which one of the following is the dimension of specific weight of a liquid?

a) [ML-3 T -2 ].

b) [ML3 T -2 ].

c) [ML-2 T -2 ].

Answer: c
Explanation: Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,

60. Two fluids 1 and 2 have mass densities of p1 and p2 respectively. If p1 > p2, which one of the following expressions will represent the relation between their specific volumes v1 and v2?

a) v1 > v2

b) v1 < v2

c) v1 = v2

d) Cannot be determined due to insufficient information.

Answer: b
Explanation: Specific volume(v) is defined as the volume(V ) per unit mass(m).
v = v⁄m = 1 / m⁄v = 1⁄p
where p is the mass density. Thus, if p1 > p2, the relation between the specific volumes v1 and v2
will be represented by v1 < v2.

61. A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific weight of the liquid will be

a) 6:5 kN = m3

b) 6:6 kN = m3

c) 6:7 kN = m3

d) 6:8 kN = m3

Answer: a
Explanation: Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,
γ = w⁄V
Thus, γ = 6:5 ⁄10^-3 N ⁄ m³ = 6:5 kN/m³.

62. A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific gravity of the liquid will be

a) 0.65

b) 0.66

c) 0.67

d) 0.68

Answer: b
Explanation: Specific gravity(S) of a liquid is defined as the ratio of the density of the liquid(pl) to that of water(pw).

Thus, S = 0:66.

63. A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific volume of the liquid will be

a) 1 l =kg

b) 1:5 l =kg

c) 2 l =kg

d) 2:5 l =kg

Answer: b
Explanation: Specific volume(v) is defined as the volume(V ) per unit mass(m). Thus,

64. Calculate the specific weight and weight of 20dm3 of petrol of specific gravity 0.6.

a) 5886,117.2

b) 5886,234.2

c) 11772,117.2

d) None of the mentioned

Answer: a
Explanation: Specific weight = density*acceleration due to gravity
=.6*1000*9.81=5886N/m³
Weight=volume*specific weight
=5886*0.02=117.2N.

65. If 200m3 of fluid has a weight of 1060N measured on the planet having acceleration due to gravity 6.625m/s2, what will be it’s specific volume?

a) 0.8

b) 0.7

c) 0.9

d) 0.5

Answer: a
Explanation: Specific weight=Weight/volume
= (Mass*acceleration due to gravity)/volume
=density*acceleration due to gravity
=1/(specific volume *acceleration due to gravity)
Specific volume=1060/(200*6.625).

66. For an incompressible fluid does density vary with temperature and pressure?

a) It varies for all temperature and pressure range

b) It remains constant

c) It varies only for lower values of temperature and pressure

d) It varies only for higher values of temperature and pressure

Answer: b
Explanation: For an incompressible fluid, the change in density is negligible. Thus it does not change with temperature and pressure.

67. Specific gravity is what kind of property?

a) Intensive

b) Extensive

c) None of the mentioned

d) It depends on external conditions

Answer: a
Explanation: It is independent of quantity of matter present.

68. If there is bucket full of oil and bucket full of water and you are asked to lift them, which one of the two will require more effort given that volume of buckets remains same?

a) Oil bucket

b) Water bucket

c) Equal effort will be required to lift both of them

d) None of the mentioned

Answer: a
Explanation: It is independent of quantity of matter present.

69. If the fluid has specific weight of 10N/m3 for a volume of 100dm3 on a planet which is having acceleration due to gravity 20m/s2 , what will be its specific weight on a planet having acceleration due to gravity 4m/s2?

a) 5 N/m3

b) 50 N/m3

c) 2 N/m3

d) 10 N/m3

Answer: c
Explanation: For same volume, specific weight is directly proportional to acceleration due to gravity
Specific weight=4*10/20=2.

70. Should Specific Wieght of incompressible fluid only be taken at STP?

a) Yes, as specific weight may show large variation with temperature and pressure

b) No, it can be taken for any temperature and pressure

c) It should be taken at standard temperature but pressure may be any value

d) It should be taken at standard pressure but temperature may be any value

Answer: b
Explanation: Specific weight is inversely proportional to volume. For incompressible fluid , variation of volume with temperature and pressure is negligible for practical consideration. Therefore, specific weight remains constant.

71. An instrument with air as fluid was involved in some experiment( specific volume was the characteristic property utilized) which was conducted during day in desert. Due to some reason experiment couldn’t be conducted during day and had to be conducted during night. However there were considerable errors in obtained values. What might be the reason of these errors?

a) It was human error

b) It was instrumental error

c) Error was due to the fact that experiment was conducted at night

d) None of the mentioned

Answer: c
Explanation: In Desert areas, temperature at night is considerably lower than at day. Due to this air contracts at night. Hence, it’s specific volume changes. As specific volume was characteristic property utilized, results obtained showed error due to change in specific volume.

72. A stone weighed 177 N on earth. It was dropped in to oil of specific gravity 0.8 on a planet whose acceleration due to gravity is 5m/s2. It displaced oil having weight of 100N. What was the volume of oil displaced by the stone?

a) 25 Litres

b) 15 Litres

c) 25 m3

d) None of the mentioned

Answer: a
Explanation: Volume displaced=oil displaced/(specific gravity*water density* acceleration due to gravity )=100/ (0.8*1000*5).

73. An compressible fluid’s specific gravity was measured on earth, on a planet having acceleration due to gravity 5.5 times that of earth, and in space at STP. Where will it be having highest value?

a) on the earth

b) on the planet

c) in the space

d) it will be constant everywhere

Answer: d
Explanation: Specific gravity is characteristic property of fluid and is independent of external conditions.

74. Water flows between two plates of which the upper one is stationary and the lower one is moving with a velocity V. What will be the velocity of the fluid in contact with the upper plate?

a) V

b) N ⁄ 2

c) 2V

d) 0

Answer: d
Explanation: According to the No-Slip condition, the relative velocity between the plate and the fluid in contact with it must be zero. Thus, the velocity of the fluid in contact with the upper plate is 0 and that with the lower plate is V.

75. The viscous force the relative motion between the adjacent layers of a fluid in motion. Which one of the flowing fits best in the sentence?

a) opposes

b) never affects

c) facilitates

d) may effect under certain conditions

Answer: a
Explanation: Viscosity is the internal friction of a fluid in motion. It is the property by the virtue of which the relative motion between two adjacent fluid layers is opposed.

76. The viscosity of a fluid in motion is 1 Poise. What will be it’s viscosity (in Poise) when the fluid is at rest?

a) 0

b) 0.5

c) 1

d) 2

Answer: c
Explanation: Viscosity is the property of a fluid and is constant for a given fluid under given conditions, irrespective of the fact whether the fluid is at rest or in motion.

77. Which of the following correctly states how the viscosities of a liquid and a gas will change with temperature?

a) Viscosity increases with the increase in temperature of a liquid and decreases with the increase in temperature of a gas

b) Viscosity increases with the increase in temperature of a liquid and increases with the increase in temperature of a gas

c) Viscosity decreases with the increase in temperature of a liquid and decreases with the increase in temperature of a gas

d) Viscosity decreases with the increase in temperature of a liquid and increases with the increase in temperature of a gas

Answer: a
Explanation: Viscosity of a liquid is due to the cohesion between it’s molecules. With the increase in temperature of a liquid, cohesion increases, leading to the rise in viscosity. Viscosity of a gas is due to the momentum transfer between it’s molecules. With the increase in the temperature of a liquid, molecular motion increases, leading to the fall in viscosity.

78. Which one of the following is not a unit of dynamic viscosity?

a) Pa-s

b) N-s/m2

c) Poise

d) Stokes

Answer: d
Explanation: 

where F= viscous force, A= area, du ⁄ dx = velocity gradient, μ = co-effcient of viscosity. Therefore,

SI unit of μ is N-s/m² = Pa-s and CGS unit of μ is dyne-s/cm². 1 Poise= 1 dyne-s/cm² and 1 Stokes= 1 cm²/s. Thus, Stokes is not an unit of μ, rather it is a unit of kinematic viscosity υ.

79. Which of the following is a unit of dynamic viscosity?

 a) [M1 L 1 T -1 ].

b) [M1 L -1 T -1 ].

c) [M1 L -2 T -2 ].

d) [M1 L -2 T -2 ].

Answer: b
Explanation: 

where F= viscous force, A= area, du ⁄ dx = velocity gradient, μ = co-effcient of viscosity. Therefore,

 

80. Which one of the following is the CGS unit of dynamic viscosity?

a) Stokes

b) Pa-s

c) m2 /s

d) Poise

Answer: d
Explanation:

 

where F= viscous force, A= area, du ⁄ dx = velocity gradient, μ = co-effcient of viscosity. Therefore,

CGS unit of μ is = dyne-s/cm². 1 Poise= 1 dyne-s/cm² and 1 Stokes= 1 cm²/s. Thus, the CGS unit of μ will be Poise. Stokes is the CGS unit of kinematic viscosity.

81. The dynamic viscosity of a fluid is 1 Poise. What should one multiply to it to get the answer in N-s/m2 ?

a) 0.1

b) 1

c) 10

d) 100

Answer: a
Explanation:
1 Poise = 1 dyne-s/cm²

                  

82. Which of the following is a unit of kinematic viscosity?

a) Stokes

b) Pa-s

c) m2=s

d) Poise

Answer: a
Explanation: ν = μ/ρ, where ν = kinematic viscosity, μ = dynamic viscosity and ρ = density of the fluid. Unit of μ is dyne-s/cm² and that of ρ is kg/cm³.
Thus, the unit of ν is cm²/s = Stokes Poise is the unit of dynamic viscosity.
1 Poise = 1 dyne-s/cm²

                     

83. Which of the following is the dimension of kinematic viscosity?

 a) [L1 T -1 ].

b) [L1 T -2 ].

c) [L2 T -1 ].

d) [L2 T -2 ].

Answer: c
Explanation: ν = μ/ρ, where ν = kinematic viscosity, μ = dynamic viscosity and ρ = density of the fluid.

84. The kinematic viscosity of a fluid is 0.1 Stokes. What will be the value is m2 /s?

a) 10-2

b) 10-3

c) 10-4

d) 10-5

Answer: d
Explanation: 1Stokes = 1cm²/s = 10^-4m²/s Therefore, 0.1Stokes = 10^-1cm²/s = 10^-5m²/s

85. The shear stress at a point in a liquid is found to be 0.03 N/m2 . The velocity gradient at the point is 0.15 s-1 . What will be it’s viscosity (in Poise)?

a) 20

 b) 2

c) 0.2

d) 0.5

Answer: b
Explanation: 

where F= viscous force, A= area, du ⁄ dx = velocity gradient, μ = co-effcient of viscosity. Therefore,

86. The space between two plates (20cm*20cm*1cm), 1 cm apart, is filled with a liquid of viscosity 1 Poise. The upper plate is dragged to the right with a force of 5N keeping the lower plate stationary.

 

What will be the velocity in m/s of flow at a point 0.5 cm below the lower surface of the upper plate if linear velocity profile is assumed for the flow?

a) 1.25                  

b) 2.5                     

c) 12.5                   

d) 0.25

Answer: c
Explanation: 

where F_ν = viscous force, A = area, du ⁄ dx = velocity gradient, μ = co-effcient of viscosity. If linear velocity profile is assumed, du⁄dx = U/x, where U = velocity of the upper plate and x = distance between the two plates. Now, the viscous force Fv = -F= -5N. Substituting all the values in the equation, U becomes 12.5 m/s.