Category: Fluid Mechanics Objective Questions

250+ TOP MCQs on Ideal Indicator Diagram and Answers

Fluid Mechanics Multiple Choice Questions on “Ideal Indicator Diagram”.

1. Which among the following can be assessed based on the indicator diagram?
a) Overall performance of engine
b) Area of the reservoir
c) Load the engine can take
d) Speed of the Engine
Answer: a
Clarification: The main purpose of the indicator diagram is to asses the overall performance of the engine. It is done by assessing the performance of each unit of the main engine. It plays an important role in the manufacture of ship engine.

2. How are the indicator diagrams drawn for the ship?
a) By providing the ship’s mass
b) By providing the ship’s volume
c) By matching ships sea trial diagrams
d) By differentiating the diagrams
Answer: c
Clarification: Indicator diagrams are drawn by taking the diagrams at regular intervals. This time taken at regular interval is matched with the ship’s sea trial diagrams. The comparison is to check if there is any significant difference in the performance.

3. Which among the following is not a type of indicator diagram?
a) Power card
b) Draw card
c) Compression diagram
d) Power house
Answer: d
Clarification: Power house is not a type of indicator diagram. It is also called as the power station. The main purpose of the power house is to supply electrical power to any system for an efficient working of the system.

4. Which among the following cannot be determined by indicator diagram?
a) The actual power generated
b) Cogeneration
c) Compression inside the cylinder
d) Exhausting and scavenging process
Answer: b
Clarification: Heat generated from clean heat and power is called as cogeneration. Cogeneration can be defined as combined heat and power that used to heat engine or the power station to generate electricity. Therefore, cogeneration cannot be determined by an indicator diagram.

5. High loading in engines will lead to_________
a) Speed rise
b) Temperature rise
c) Bearing damage
d) Volume expansion
Answer: c
Clarification: High loading in engines will lead to bearing damage. It also leads to several other problems like cracking, breaking etc. So, it is very important in assessing and reading these diagrams correctly.

6. What type of indicator device is used nowadays?
a) Digital pressure indicator
b) Mechanical indicator
c) Electrical indicator
d) Ionization Gauge
Answer: a
Clarification: In earlier days, the indicator diagram was assessed with the help of mechanical indicator. But now, we use a digital pressure indicator. It has got a lot of advantages over the mechanical one. It has a compact hand-held unit with a computer display system.

7. How can we asses the indicator diagram by looking at the card diagram?
a) Comparing the maximum firing pressure and compression pressure
b) Dead weight method
c) Conveyor method
d) Ionization method
Answer: a
Clarification: Using a card diagram, we can asses the indicator diagram. This process is performed using multiple parameters. One of them is the maximum firing pressure and the other is the comparison pressure.

8. Which among the following is not an effect of deficiency in the indicator diagram?
a) Bad quality of fuel
b) Fuel pump leakage
c) Low fuel pressure
d) Fuel temperature rise
Answer: d
Clarification: Deficiencies in an indicator diagram are classified into many types. Deficiencies can be determined using the card diagram. Card diagram contains maximum firing pressure and compression pressure which does not depend on the temperature rise of fluid.

9. Deficiency type 3 is not caused due to_________
a) Leaking exhaust valve
b) Resistance to flow of air
c) Low scavenging pressure
d) High linear wear
Answer: b
Clarification: Deficiencies in an indicator diagram are classified into many types. Deficiencies can be determined using the card diagram. Card diagram contains maximum firing pressure and compression pressure. In deficiency 3, the compression pressure is low and peak pressure is also low.

10. Deficiency type 4 is caused due to_________
a) Resistance to flow of air
b) Leaking exhaust valve
c) Low scavenging pressure
d) Overload of engine
Answer: d
Explanation Deficiencies in an indicator diagram are classified into many types. Deficiencies can be determined using the card diagram. Card diagram contains maximum firing pressure and compression pressure. In deficiency 4, compression pressure is high and peak pressure is also high leading to the overload of an engine.

250+ TOP MCQs on Velocity of Sound or Pressure Wave in a Fluid – 2 and Answers

Fluid Mechanics Puzzles on “Velocity of Sound or Pressure Wave in a Fluid – 2”.

1. A wave that has propagating disturbance is called ________
a) Shock wave
b) Compression wave
c) Compressed wave
d) Longitudinal wave
Answer: a
Clarification: Shock wave is defined as the wave that has a propagating disturbance. When the propagating wave move faster than the local speed of fluid, it is called as a shock wave. It can propagate through any medium.

2. A shock wave carries _______
a) Heat
b) Pressure
c) Energy
d) Temperature
Answer: c
Clarification: The shock wave carries energy and can propagate through a medium. It can be characterized as abrupt. It is due to the discontinuous changes in pressure, density and temperature of the medium.

3. Shock waves can be normal, oblique and bow.
a) True
b) False
Answer: a
Clarification: Shock waves can be at normal angles, oblique angles and bow angles. Normal angle is perpendicular to the shock medium’s flow direction. Oblique angle is at the direction of flow. Bow occurs at the front of the object.

4. Bow occurs at what Mach number?
a) Mach number = 0
b) Mach number is negative
c) Mach number = 1
d) Mach number greater than 1
Answer: d
Clarification: Bow occurs when the Mach number is greater than one. It occurs at the upstream of the front of a blunt object. It happens when the velocity exceeds Mach 1.

5. A boundary over which physical conditions undergo changes is called _______
a) Shear front
b) Shock front
c) Contact front
d) Cap front
Answer: b
Clarification: Shock front is defined as the boundary over which the physical conditions of a fluid undergo abrupt changes. This happens due to the shock wave. Thus, the correct choice is option Shock front.

6. A shock wave caused by driver gas is called _______
a) Shear front
b) Shock front
c) Contact front
d) Cap front
Answer: c
Clarification: Contact front is defined as a shock wave caused by a driver gas. It is the boundary layer between the driver gases and the driven gases. The main function of the contact front is to trails the shock front.

7. When the fluid flow is discontinuous, what is established?
a) Heat
b) Pressure
c) Control volume
d) Temperature
Answer: c
Clarification: When the fluid flow is discontinuous, a control volume is established. It is established around the shock wave. It is treated as a discontinuity where the entropy increases to over a infinitesimally large region.

8. Shock waves that deviate from the arbitrary angle are called_______
a) Oblique shock
b) Bow shock
c) Normal shock
d) Detonation
Answer: a
Clarification: Shock waves in a flow field which are attached to a body start deviating at some arbitrary angle from the flow direction of the fluid. This shock so developed due to angle deviation is called as oblique shock.

9. When a shock wave forms continuous pattern, it is called ________
a) Oblique shock
b) Bow shock
c) Normal shock
d) Detonation
Answer: b
Clarification: Bow occurs when the Mach number is greater than one. It occurs at the upstream of the front of a blunt object. It happens when the velocity exceeds Mach 1. When a shock wave forms continuous pattern, it is called a bow shock.

10. Shocks generated due to an interaction of two bodies are called ______
a) Oblique shock
b) Bow shock
c) Moving shock
d) Detonation
Answer: c
Clarification: Shocks that are generated by the interaction of two bodies of gas. This interaction takes place at different pressure, with the shock wave propagating into a lower pressure gas and an expansion wave propagates into a high-pressure gas.

11. Shocks generated due to trailing exothermic reaction is called ______
a) Oblique shock
b) Bow shock
c) Moving shock
d) Detonation
Answer: d
Clarification: Shocks that are generated due to trailing exothermic reaction is called as detonation. It is a wave traveling through a medium of highly combustible and chemically unstable medium.

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250+ TOP MCQs on Pressure Distribution in a Fluid – 2 and Answers

Fluid Mechanics Questions and Answers for Experienced people on “Pressure Distribution in a Fluid – 2”.

1. Three beakers 1, 2 and 3 of different shapes are kept on a horizontal table and filled with water up to a height h. If the pressure at the base of the beakers are P1, P2 and P3 respectively, which one of the following will be the relation connecting the three?
fluid-mechanics-questions-answers-experienced-q1
a) P1 > P2 > P3
b) P1 < P2 < P3
c) P1 = P2 = P3
d) P1 > P2 < P3
Answer: c
Clarification: The pressure on the surface of the liquid in the beakers is the same. Pressure varies in the downward direction according to the formula P = ρgh, where ρ is the density of the liquid and h is the height of the liquid column from the top.
P1 = ρgh
P2 = ρgh
P3 = ρgh
Since all the beakers contain water up to to the same height, P1 = P2 = P3.

2. A beaker is filled with a liquid of specific gravity S = 1:2 as shown. What will be the pressure difference (in kN/m2) between the two points A and B, 30 cm below and 10 cm to the right of point A?
fluid-mechanics-questions-answers-experienced-q2
a) 2.5
b) 3.5
c) 4.5
d) 5.5
Answer: b
Clarification: Pressure increases in the vertically downward direction but remains constant in the horizontal direction. Thus,
PB = PA + ρgh
where PB = Pressure at B, PA = Pressure at A, ρ = density of the liquid, g = acceleration due to gravity and h = vertical distance separating the two points.
PB – PA = 1:2 * 103 * 9.81 * 0.3 N/m2 = 3.53 kN/m2

3. The arm of a teapot is 10 cm long and inclined at an angle of 60o to the vertical. The center of the arm base is 2 cm above the base of the beaker. Water is poured into the beaker such that half the arm is filled with it. What will be the pressure at the base of the beaker if the atmospheric pressure is 101.3 kPa?
fluid-mechanics-questions-answers-experienced-q3
a) 101.3
b) 101.5
c) 101.7
d) 101.9
Answer: c
Clarification: Total height of the water in the beaker = 2 + 1⁄2 * 10 cos 60o cm = 4:5 cm. Pressure at the base of the beaker = 101.3 + 103 * 9.81 * 0.045 Pa = 101.3 + 0.44 kPa = 101.74 kPa.

4. A beaker of height 10 cm is half-filled with water (Sw = 1) and half-filled with oil (So = 1). At what distance (in cm) from the base will the pressure be half the pressure at the base of the beaker?
fluid-mechanics-questions-answers-experienced-q4
a) 4.375
b) 4.5
c) 5.5
d) 5.625
Answer: b
Clarification: Gauge pressure at the base of the beaker = So * 103 * 0.05 * g + Sw * 103 * 0.05 * g = 882.9Pa. Let the required height be h m from the base.
If 0.05 ≤ h < 0.1,
800(0.1 – h)g = 1⁄2 * 882.9
Thus, h = 0.04375 (out of the range considered).
If 0 < h ≤ 0:05,
800 * 0.05 * g + 103 * (0.05 – h) * g = 1⁄2 * 882.9
Thus, h = 0.045 (in the range considered). Hence, the correct answer will be 45 cm.

5. A beaker of height 30 cm is filled with water (Sw = 1) up to a height of 10 cm. Now oil (So = 0:9) is poured into the beaker till it is completely filled. At what distance (in cm) from the base will the pressure be one-third the pressure at the base of the beaker?
fluid-mechanics-questions-answers-experienced-q5
a) 27.33
b) 19.2
c) 10.8
d) 2.67
Answer: b
Clarification: Gauge pressure at the base of the beaker = So * 103 * 0.2 * g + Sw * 103 * 0.1 * g = 2550.6Pa. Let the required height be h m from the base.
If 0.1 ≤ h < 0.3,
800(0.3 – h)g = 1⁄3 * 2550.6
Thus, h = 0.192 (in the range considered).
Even if there’s no need to check for the other range, it’s shown here for demonstration purpose.If
0 < h ≤ 0.1,
800 * 0.2 * g + 103 * (0.2 – h) * g = 1⁄3 * 2550.6
Thus, h = 0.2733 (out of the range considered). Hence, the correct answer will be 19.2 cm.

6. An oil tank of height 6 m is half-filled with oil and the air above it exerts a pressure of 200 kPa on the upper surface. The density of oil varies according to the given relation:
fluid-mechanics-questions-answers-experienced-q6
What will be the percentage error in the calculation of the pressure at the base of the tank if the density is taken to be a constant equal to 800?
fluid-mechanics-questions-answers-experienced-q6a
a) 0.01
b) 0.05
c) 0.10
d) 0.15
Answer: a
Clarification: The change of pressure with the vertical direction y is given by
dP/dy = – ρg
dP = -ρg dy
If Pa and Pb be the pressures at the top and bottom surfaces of the tank,
fluid-mechanics-questions-answers-experienced-q6c
Thus, Pb = 223.5746kPa. If the density is assumed to be constant,
Pb = 200 + 800 * 9.81 * 3 * 103 = 223.544 kPa. Hence, precentage error fluid-mechanics-questions-answers-experienced-q6e

7. If a gas X be confined inside a bulb as shown, by what percent will the pressure of the gas be higher or lower than the atmospheric pressure? (Take the atmospheric pressure equal to 101.3 kPa)
fluid-mechanics-questions-answers-experienced-q7
a) 4:75% higher
b) 4:75% lower
c) 6:75% higher
d) 6:75% lower
Answer: a
Clarification: Pa = Patm = 101.3
Pb = Pa + 0.9 * 9.81 * 0.03 = 101.56
Pc = Pb + 13.6 * 9.81 * 0.04 = 106.9
Pd = Pc – 1 * 9.81 * 0.05 = 106.41
Pe = Pd – 0.9 * 9.81 * 0.04 = 106.1
PX = Pe = 106.1
Since, PX > Patm, the percentage by which the pressure of the gas is higher than the atmospheric pressure will be fluid-mechanics-questions-answers-experienced-q7a

8. A tank of height 3 m is completely filled with water. Now two-third of the liquid is taken out and an equal amount of two other immiscible liquids of specific gravities 0.8 and 1.2 are poured into the tank. By what percent will the pressure at the base of the tank change?
fluid-mechanics-questions-answers-experienced-q8
a) 0%
b) 5%higher
c) 5%lower
d) 10%higher
Answer: a
Clarification: Pressure at the base initially = 1 * 9.81 * 3 = 29.43 kPa; Pressure at the base after adding the other two liquids= 0.8 * 9.81 * 1 + 1 * 9.81 * 1 + 1.2 * 9.81 * 1 kPa; Thus the pressure at the base remains the same.

9. A beaker of height 15 cm is completely filled with water. Now two-third of the liquid is taken out and an equal amount of two other immiscible liquids of specific gravities 0.8 and 1.2 are poured into the tank. What will be the pressure (in kPa) at a point situated at a height, half the height of the beaker?
fluid-mechanics-questions-answers-experienced-q9
a) 588.6
b) 637.65
c) 735.75
d) 833.85
Answer: b
Clarification: PA = 0.8 * 103 * 9.81 * 0.05 + 1 * 103 * 9.81 * 0.025 = 637.65 kPa.

10. A beaker of height h is completely filled with water. Now two-third of the liquid is replaced by another liquid. If the pressure at the base of the beaker doubled, what is the specific gravity of the liquid poured?
a) 0.5
b) 1
c) 2
d) 2.5
Answer: d
Clarification: Pressure at the base initially = Sw * h⁄3 * g; Pressure at the base after pouring the second liquid = Sw * h⁄3 * g + Sl * 2h⁄3 * g, where Sw and Sl are the specific gravities of water and the second liquid.
fluid-mechanics-questions-answers-experienced-q10

11. A beaker, partially filled with a liquid is rotated by an angle 30o as shown. If the pressure at point B becomes 12 bar, what will be the height (in cm) of the beaker?
fluid-mechanics-questions-answers-experienced-q11
a) 23.5
b) 24.5
c) 26.5
d) 27.5
Answer: b
Clarification: If the height of the beaker is h, the pressure at point B = 103 * g * h * cos 30o = 12 * 103kPa; h = 24.5 cm.

12. A beaker of height 15 cm is partially filled with a liquid and is rotated by an angle θ as shown.
If the pressure at point B becomes 5 bar, what will be the value of θ?
fluid-mechanics-questions-answers-experienced-q12
a) 30o
b) 50o
c) 60o
d) 70o
Answer: d
Clarification: If the angle of inclination is taken to be θ, the pressure at point B = 103 * g * 0.15 * cos θ = 5 * 103 kPa; θ = 70.12o.

13. A beaker of height 30 cm is partially filled with a liquid and is rotated by an angle θ as shown.
At this point, the pressure at point B is found to be 5 bar. By what angle should θ be increased such that the pressure at B gets halved?
fluid-mechanics-questions-answers-experienced-q13
a) 12o
b) 15o
c) 17o
d) 20o
Answer: b
Clarification: Let θ1 and θ2 be the angles at which the beaker is inclided for the two cases mentioned.
103 * 9.81 * 0.15 * cos θ1 = 5 * 100; θ1 = 70.12o

103 * 9.81 * 0.3 * cos θ2 = 1⁄2 * 5 * 100; θ1 = 85.12o

θ2 – θ1 = 15o

14. A closed tank (each side of 5 m) is partially filled with fluid as shown. If the pressure of the air above the fluid is 2 bar, find the pressure at the bottom of the tank. Assume the density ρ of the fluid to vary according to the given relation:
fluid-mechanics-questions-answers-experienced-q14
a) 766
b) 776
c) 786
d) 796
Answer: c
Clarification:
PA = Patm = 760
PB = PA + 30
PC = PB – 50 / 13.6 = 786.32
PX = PC = 786.3.

15. For what height of the mercury column will the pressure inside the gas be 40 cm Hg vacuum?
fluid-mechanics-questions-answers-experienced-q15
a) 36
b) 40
c) 76
d) 116
Answer: b
Clarification:
Pgas = Patm – ρgH
Taking gauge pressure in terms of cm of Hg,
-40 = 0 – H; H = 40.

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250+ TOP MCQs on Continuity Equation in Two and Three Dimensions and Answers

Fluid Mechanics Multiple Choice Questions on “Continuity Equation in Two and Three Dimensions”.

1. The continuity equation is based on the premise of-
a) Law of conservation of energy
b) Law of conservation of mass
c) Law of conservation of momentum
d) None of the mentioned
Answer: b
Clarification: Continuity equation is based on the the principle of conservation of mass.

2. The continuity equation is only applicable to incompressible fluid.
a) True
b) False
Answer: b
Clarification: The continuity equation is only applicable to incompressible as well as compressible fluid.

3. For incompressible fluid flow, if area reduces then what is the effect on the velocity.
a) increases
b) decreases
c) first increases then decreases
d) first decreases then increases
Answer: a
Clarification: According to continuity equation,
Area × velocity = constant
Hence, as area decreases velocity increases.

4. For compressible fluid flow in a pipe, having decrease in specific gravity what will be the effect of decrease in diameter?
a) It will cause increase in velocity
b) It will cause decrease in velocity
c) It remains constant
d) None of the mentioned
Answer: a
Clarification: According to continuity equation,
ρ*A*v = constant
Hence, as density and area decreases velocity is bound to increase.

5. What is the most common assumption while dealing with fluid flow problems using continuity equation?
a) Flow is assumed to be compressible
b) Flow is assumed to be unsteady
c) Flow is assumed to be steady
d) Flow is assumed to be turbulent
Answer: c
Clarification: In majority of the fluid flow problems, flow is assumed to be steady.

6. The diameters of a pipe at the sections 1 and 2 are 8 cm and 13 cm respectively. Find the discharge through pipe if the velocity of water flowing through the pipe at section 1 is 6 m/s. Determine also the velocity at section 2.
a) 2.27 m/s
b) 4.54 m/s
c) 1.13 m/s
d) 3.25 m/s
Answer: a
Clarification: According to continuity equation,
Area × velocity = constant
Area1*Velocity1 = Area2*Velocity2
Velocity2=(Area1*Velocity1)/Area2
= (82 * 6) / 132=2.27 m/s.

7. The continuity equation can only be used for analysis of conserved quantity.
a) True
b) False
Answer: a
Clarification: Continuity equation is defined on a control volume and hence, is applicable only to Conserved quantities.

8. The diameter of a pipe at the section 1 is 9 cm. If the velocity of water flowing through the pipe at section 1 is 4.8 m/s and section 2 is 9 m/s, Determine the area at section 2.
a) 33.93 m2
b) 67.86 m2
c) 16.96 m2
d) 38.66 m2
Answer: a
Clarification: According to continuity equation,
Area × velocity = constant
Area1*Velocity1 = Area2*Velocity2
(Area1*Velocity1)/Velocity2=Area2
Area 2= 33.93 m2.

9. For a flow to be physically possible it must primarily satisfy which equation?
a) Equation of conservation of energy
b) Equation of conservation of mass or continuity equation
c) Equation of conservation of momentum
d) None of the mentioned
Answer: a
Clarification: Fluid flow must satisfy equation of conservation of mass or continuity equation, for itto be physically possible.

10. Continuity equation can also be derived for polar coordinate system
a) True
b) False
Answer: a
Clarification: Continuity equation in polar coordinate is also used for analysis.

250+ TOP MCQs on Discharge Over different type of weirs and Answers

Fluid Mechanics Assessment Questions and Answers on “Discharge Over Broad-Crested Weir, Narrow Crested Weir, Ogee Weir, Submerged or Drowned Weir”.

1. In discharge of water over narrow crested weir, head of water is directly proportional to Discharge Coefficient.
a) True
b) False
Answer: b
Clarification: In discharge of water over narrow crested weir, head of water is inversely proportional to Discharge Coefficient.

2. In discharge of water over narrow crested weir, discharge is directly proportional to the cube root of acceleration due to gravity.
a) True
b) False
Answer: b
Clarification: In discharge of water over narrow crested weir, discharge is directly proportional to the square root of acceleration due to gravity.

3. In discharge of water over ogee weir, discharge is directly proportional to the second power of length.
a) True
b) False
Answer: b
Clarification: In discharge of water over ogee weir, discharge is directly proportional to the first power of length.

4. For discharge over ogee weir discharge is directly proportional to length but for discharge over narrow crested weir it is inversely proportional to length.
a) True
b) False
Answer: b
Clarification: For both weirs it’s same.

5. An Ogee weir 5 m long had a head of 40 cm of water. If CD = 0.61, find the discharge over the weir.
a) 2.9 m3/s
b) 2.3 m3/s
c) 3.1 m3/s
d) 3.3 m3/s
Answer: a
Clarification: Q = 0.67 * Cd * L * √2g * H1.5
Q = 2.3 m3/s.

6. The height of water on upstream and downstream side of a submerged weir of 4 m length are 24 cm and 13 cm. If Cd for free and drowned portions are .62 and .78 respectively, find the discharge over the weir.
a) .85 m3/s
b) 1.35 m3/s
c) 3.2 m3/s
d) .55 m3/s
Answer: a
Clarification: Q = 0.67 * Cd1 * L * √2g * H1.5 +. Cd2 * L * h * √2g(H – h)
= .67 * .6 * 3 * √2g * (.24 – .13)1.5 + .8 * 3 * .13 √2g(.24 – .13)
= .85 m3/s.

7. An Ogee weir 3.4 m long had a head of 40 cm of water. If CD = 0.63 find the discharge over the weir.
a) 1.61 m3/s
b) 2.5 m3/s
c) 3.1 m3/s
d) 3.3 m3/s
Answer: a
Clarification: Q = 0.67 * Cd * L * √2g * H1.5
Q = 1.61 m3/s.

8. The height of water on upstream and downstream side of a submerged weir of 4 m length are 23.5 cm and 14 cm. If Cd for free and drowned portions are .61 and .75 respectively, find the discharge over the weir.
a) m3/s
b) 1.35 m3/s
c) 3.2 m3/s
d) .55 m3/s
Answer: a
Clarification: Q = 0.67 * Cd1 * L * √2g * H1.5 +. Cd2 * L * h * √2g(H – h)
= .67 * .6 * 3 * √2g * (.24 – .13)1.5 + .8 * 3 * .13 √2g(.24 – .13)
= .85 m3/s.

9. In discharge of water over narrow crested weir, discharge is directly proportional to the second power of height.
a) True
b) False
Answer: b
Examples: In discharge of water over narrow crested weir, discharge is directly proportional to the one and half power of height.

10. In discharge of water over Ogee weir, discharge is directly proportional to the first power of length.
a) True
b) False
Answer: a
Clarification: This is as per empirical relation.

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250+ TOP MCQs on Effect of Acceleration in Suction and Delivery Pipes on Indicator Diagrams and Answers

Fluid Mechanics Multiple Choice Questions on “Effect of Acceleration in Suction and Delivery Pipes on Indicator Diagrams”.

1. What is the full form of NPSH?
a) Net pipe suction head
b) Net positive suction head
c) Non-positive suction head
d) Non-polar suction head
Answer: b
Clarification: The full form of NPSH is net positive suction head. It is defined as the margin of pressure over vapour pressure. It happens at the pump suction nozzle.

2. What is the formula for NPSH?
a) Ps-Pvap
b) Vs-Vvap
c) Ts-Tvap
d) Ms-Mvap
Answer: a
Clarification: NPSH is defined as the margin of pressure over vapour pressure. It happens at the pump suction nozzle. It is the difference between the suction pressure and vapour pressure. Thus, the correct option is Ps-Pvap.

3. What is absolute pressure?
a) Momentum pressure to thermal pressure
b) Thermal pressure to momentum pressure
c) Atmospheric pressure plus gauge pressure
d) Atmospheric pressure by gauge pressure
Answer: c
Clarification: Absolute pressure is defined as the summation of Atmospheric pressure plus gauge pressure. This equation provides the answer in units of pressure. The obtained psi can be converted to units of head.

4. What is the unit of NPSH?
a) kPa
b) kgm
c) kg/m
d) kg
Answer: a
Clarification: Unit of NPSH can be expressed as units of specific energies. All the units in NPSH are psi units. The psi units are kilopascal, bar, pounds per square inch etc. So, kPa is the correct option.

5. What does ‘g’ in psig mean?
a) Pressure is measured below absolute pressure
b) Pressure is measured above absolute pressure
c) Pressure is measured below absolute zero
d) Pressure is measured above absolute zero
Answer: b
Clarification: All the units in NPSH are psi units. The psi units are kilopascal, bar, pounds per square inch etc. The ‘g’ in psig indicates that the pressure is measured above the absolute pressure.

6. What does ‘a’ in psia mean?
a) Pressure is measured below absolute pressure
b) Pressure is measured above absolute pressure
c) Pressure is measured below absolute zero
d) Pressure is measured above absolute zero
Answer: d
Clarification: All the units in NPSH are psi units. The psi units are kilopascal, bar, pounds per square inch etc. The ‘a’ in psia indicates that the pressure is measured above absolute zero which is a perfect vacuum.

7. Which among the following happens at the first half of NPSH?
a) Vapour pressure
b) Atmospheric pressure
c) Suction pressure
d) Discharge pressure
Answer: c
Clarification: During the first half of the NPSH, the suction of air takes place. For the suction of an air, a strong suction pressure is required. Sometimes, the elevation of gauge must also be added. But, most of the times it is considered negligible.

8. Which among the following happens in the second half of NPSH?
a) Vapour pressure
b) Atmospheric pressure
c) Suction pressure
d) Discharge pressure
Answer: a
Clarification: During the second half of the NPSH, vapour pressure is more difficult to determine. It is a measure of the desire of liquids. Cool water has a low vapour pressure. Thus, the option is Vapour pressure.

9. Vapour pressure depends on_______
a) Pressure
b) Temperature
c) Density
d) Viscosity
Answer: b
Clarification: Vapour pressure depends on temperature only. With the increase in temperature, the vapour pressure increases. It increases until it reaches the critical temperature. At the critical temperature, the vapour pressure tends to slowly vanish.

10. Which among the following is not a type of pipe?
a) Stainless steel pipes
b) PVC piping
c) Brass pipes
d) Wooden pipes
Answer: d
Clarification: Wooden pipes for transportation and other important facilities is not feasible. Wood has various properties against water. Water flowing through the wooden pipe will slowly reduce the strength of the wood which may lead to cracks.