300+ REAL TIME Strength of Materials Objective Questions & Answers 2023

250+ TOP MCQs on Power of Shaft and Answers

This set of Strength of Materials Multiple Choice Questions on “Power of Shaft”.

1. Calculate the power transmitted in the shaft at 150 rpm. Take torque as 9000Nm.
a) 140 kW
b) 150 kW
c) 160 kW
d) 175 kW
Answer: a
Clarification: To find power transmitted (P) is P = 2 π N T / 60 watts.
P = 2 π × 150 × 9000 / 60
P = 140 kW.

2. Which of the following is not a cross drainage work?
a) Aqueduct
b) Head regulator
c) Super passage
d) Level crossing
Answer: b
Clarification: The head regulator is hydraulic structure constructed at the head of a canal system where it takes off from a reservoir behind a weir or a dam. It is used as a measuring device.

3. Stucco is a type of _________
a) Varnishing
b) Distempering
c) Plastering
d) Whitewashing
Answer: c
Clarification: Stucco is the name given to a decorative type of plaster, which provides an excellent finish like that with marble’s lining.

4. The thickness of cement plaster should not be more than _______
a) 15 mm
b) 12 mm
c) 16 mm
d) 20 mm
Answer: b
Clarification: The cement plaster is applied in one or two coats. The surface is polished with the trowel or iron float. The thickness of the coat should not be more than 12 mm.

5. __________ mm thick plastering is done for stone masonry.
a) 10 mm
b) 15 mm
c) 18 mm
d) 20 mm
Answer: d
Clarification: Normally 12 mm thick plastering is done for brick masonry and 20 mm thick plastering is done for the stone masonry. The plastered surface is then cured by sprinkling water over the surface for one or two weeks.

6. The thickness of lime plaster varies from _______ to ________ mm.
a) 15 – 20 mm
b) 12 – 15 mm
c) 18 – 25 mm
d) 20 – 25 mm
Answer: d
Clarification: The proportioning of the ingredients of lime plaster is adapted according to a number of coats to be applied. The thickness of lime plaster varies from 20 to 25 mm.

7. Which of the following plastering is widely adopted in rural areas?
a) Stucco Plastering
b) Mud plastering
c) Lime plastering
d) Asphalt plastering
Answer: b
Clarification: Mud plastering is done on the walls of temporary Sheds and widely adopted in rural areas. The Plaster is evenly dashed against the wall with a wooden float. After 24 hours the surface is tapped.

8. Which of the following blasters contains pulverized alum?
a) Water proof plaster
b) Plaster on lathe
c) C plaster
d) Marble plaster
Answer: a
Clarification: Waterproof plaster is made by mixing 1 part of cement, 2 parts of sand and pulverized alum at the rate of 120 Newton per metre and in the water to be used.

9. Which of the following is known as” laying trowel”?
a) Float
b) Gauge trowel
c) Floating Rule
d) Skimming float
Answer: a
Clarification: The tool which is used to spread the mortar on the surface is known as float. It is also known as laying trowel. It is made of thin tempered Steel.

10. _________ is used to check the level of plastered surface.
a) Gauging trowel
b) Plumb bob
c) Floating Rule
d) Float
Answer: c
Clarification: Floating rule is the tool which is used to check the level of plastered surface between the successive screeds.

11. Skimming float is ____________
a) Wooden float
b) Metalled float
c) Tempered steel float
d) Asbestos cement sheet
Answer: a
Clarification: The wooden floor is known as the skimming float and it is used for final and finishing coat of plaster. The Plaster is evenly spread against the wall surface with a wooden float.

12. Which of the following is a defect in plastering?
a) Flaking
b) Scrap
c) Rust
d) Staining
Answer: a
Clarification: Flaking is a defect in plastering. It is a formation of a very loose mass of plastered surface due to poor bond between successive coats. This is obtained due to poor workmanship.

13. ________ is a process of mixing various constituents of plaster.
a) Grazing
b) Blistering
c) Gauging
d) Hacking
Answer: c
Clarification: Gauging is defined as a process of mixing various constituents of plaster. It is to be done after the brick work had carried out to the best workmanship. Efflorescence can be removed to some extent of dry brushing.

14. The small projections of plaster are known as ________
a) Back
b) Dado
c) Dot
d) Hack
Answer: c
Clarification: The small projections of plaster laid on the background are known as dots. These are laid for fixing of screeds. The size of the dots may be 15×15 cm.

15. ______ openings or indentations of corrugations in plaster.
a) Helms
b) Grains
c) Keys
d) Flake
Answer: c
Clarification: Keys are the openings or indentations of corrugations on the background or surface of undercoat, to which plaster will form mechanical bond.

250+ TOP MCQs on Normal & Shear Stress and Answers

Strength of Materials Multiple Choice Questions on “Normal & Shear Stress”.

1. In the given figure a stepped column carries loads. What will be the maximum normal stress in the column at B in the larger diameter column if the ratio of P/A here is unity?
a) 1/1.5
b) 1
c) 2/1.5
d) 2
Answer: c
Clarification: Normal stress at B = Total load acting at B / Area of a cross-section at B
= (P + P) / 1.5 A = 2P/ 1.5A = 2/1.5.

2. The stress which acts in a direction perpendicular to the area is called ____________
a) Shear stress
b) Normal stress
c) Thermal stress
d) None of the mentioned
Answer: b
Clarification: Normal stress acts in a direction perpendicular to the area. Normal stress is of two types tensile and compressive stress.

3. Which of these are types of normal stresses?
a) Tensile and compressive stresses
b) Tensile and thermal stresses
c) Shear and bending
d) Compressive and plane stresses
Answer: a
Clarification: The normal stress is divided into tensile stress and compressive stress.

4. In a body loaded under plane stress conditions, what is the number of independent stress components?
a) 1
b) 2
c) 3
d) 6
Answer: c
Clarification: In a body loaded under plane stress conditions, the number of independent stress components is 3 I.e. two normal components and one shear component.

5. If a bar of large length when held vertically and subjected to a load at its lower end, its won-weight produces additional stress. The maximum stress will be ____________
a) At the lower cross-section
b) At the built-in upper cross-section
c) At the central cross-section
d) At every point of the bar
Answer: b
Clarification: The stress is the load per unit area. After the addition of weight in the bar due to its loading on the lower end the force will increase in the upper cross-section resulting in the maximum stress at the built-in upper cross-section.

6. Which type of stress does in a reinforcement bar is taken by the concrete?
a) Tensile stress
b) Compressive stress
c) Shear stress
d) Bending stress
Answer: b
Clarification: Concrete has the property of taking a good amount of compressive stress. So, In the reinforcement bar, the compressive stress is taken by the concrete.

7. A material has a Poisson’s ratio of 0.5. If uniform pressure of 300GPa is applied to that material, What will be the volumetric strain of it?
a) 0.50
b) 0.20
c) 0.25
d) Zero
Answer: d
Clarification: As volumetric strain = (1-2μ)σ/E
Here the value of μ is 0.5 so 1 – 2 * 0.5 becomes zero.
Therefore whatever be the stress the value of volumetric strain will be zero.

8. A diagram which shows the variations of the axial load for all sections of the pan of a beam is called ____________
a) Bending moment diagram
b) Shear force diagram
c) Thrust diagram
d) Stress diagram
Answer: d
Clarification: The stress diagram shows the variation of the axial load for all sections of the pan. The bending moment diagram shows the variation of moment in a beam. The shear force diagram shows the variation in the shear force due to loading in the beam.

9. The stress induced in a body, when subjected to two equal and opposite forces which are acting tangentially across the resisting section resulting the shearing of the body across its section is called ____________
a) Bending stress
b) Compressive stress
c) Shear strain
d) Shear stress
Answer: d
Clarification: Shear stress makes the body to shear off across the section. It is tangential to the area over which it acts. The corresponding strain is the shear strain.

10. What is the formula for shear stress?
a) Shear resistance/shear area
b) Force/unit area
c) Bending strain/area
d) Shear stress/length
Answer: a
Clarification: When force is applied, the twisting divides the body. The resistance is known as shear resistance and shear resistance per unit area is known as shear stress.

11. Which of the following stresses are associated with the tightening of a nut on a bolt?
P. Crushing and shear stress in threads
Q. Bending stress due to the bending of bolt
R. Torsional shear stress due to frictional resistance between the nut and the bolt
Select the correct answer using the codes given below:
a) P and Q
b) P and R
c) Only P
d) Only R
Answer: a
Clarification: Bending stress comes when there is some kind of eccentric load. Torsional stress will come when the nut is rotating. Shear stress will come in tightening of a nut on bolt.

12. The transverse shear stress acting in a beam of rectangular cross-section, subjected to a transverse shear load, is ____________
a) variable with maximum at the bottom of the beam
b) Variable with maximum at the top of the beam
c) Uniform
d) Variable with maximum on the neutral axis
Answer: d
Clarification: Maximum value of shear stress at neutral axis is τ = 3/2 τ_mean
So, transverse shear stress is variable with a maximum in the neutral axis.

13. A block 100mm x 100mm base and 10mm height. What will the direct shear stress in the element when a tangential force of 10kN is applied to the upper edge to a displacement 1mm relative to lower face?
a) 1Pa
b) 1MPa
c) 10MPa
d) 100Pa
Answer: b
Clarification: Shear stress = 10kN / 100mmx100mm = 1 N/mm² = 1MPa.

250+ TOP MCQs on Center of Gravity of Section and Answers

This set of Strength of Materials Assessment Questions and Answers on “Center of Gravity of Section”.

1. The center of gravity of the rod shown in figure will be _____________

a) 5cm
b) 10cm
c) 15cm
d) 20cm
Answer: b
Clarification: The center of gravity of a rod will be on its center. Here it will be at 10cm.

2. The center of gravity of a circle of radius 10 cm will be _____________
a) At its center of the diameter
b) At the center of the radius
c) Anywhere on the circumference
d) Anywhere in its area
Answer: a
Clarification: The whole weight of a circle can be assumed to act through its center. So the center of gravity of a circle is at its center. Whatever may be the radius of the circle the center of gravity will be on its center.

3. A rectangle has dimension of 10cm x 20cm. where will be its center of gravity?
a) (10,10)
b) (20,5)
c) (10,5)
d) (5,10)
Answer: c
Clarification: The centre of gravity of this rectangular area will be half of 10cm from x-axis and half of 20cm from y-axis. therefore the center of gravity will be at (10,5).

4. Where will be the centre of gravity of the T section shown in the figure?

a) 8
b) 8.5
c) 10.5
d) 11.5
Answer: d
Clarification: The center of gravity is given by, y = (a₁y₁ + a₂y₂) / (a₁ + a₂) = (100×17.5 + 150×7.5) / (100+150) = 11.5cm.

5. Where will be the center of gravity of the L-section shown in figure?

a) (4.33, 2.33)
b) (4, 6)
c) (2.33, 4.33)
d) (1, 5)
Answer: c
Clarification: The center of gravity is given by, y = (a₁y₁ + a₂y₂) / (a₁ + a₂) = (20×7 + 16×1) / (20+16) = 4.33cm.
This will on for the y-axis.
For the x-axis, The center of gravity is given by, x = (a₁x₁ + a₂x₂) / (a₁ + a₂) = (20×1 + 16×4) / (20+16) = 2.33cm.
So the center of gravity will be at (2.33, 4.33).

6. Where will be the center of gravity of the figure shown?

a) (3.45,4.52)
b) (3.59,4.52)
c) (3.66,5.17)
d) (4.01,5.15)
Answer: b
Clarification: Area of triangle = 20, area of rectangle = 50
The center of gravity is given by, y = (a₁y₁ + a₂y₂) / (a₁ + a₂) = (20×10/3 + 50×5) / (20+50) = 4.52cm.
This will on for the y-axis.
For the x-axis, The center of gravity is given by, x = (a₁x₁ + a₂x₂) / (a₁ + a₂) = (20×6.33 + 50×2.5) / (20+50) = 3.59cm.
So the center of gravity will be at (2.33, 4.33).

7. Where will be the center of gravity of the shown figure?

a) (4.66,6.332)
b) (4.34,3.24)
c) (4.25,6.45)
d) (4.87,6.41)
Answer: a
Clarification: Area of triangle = 25, area of rectangle = 100
The center of gravity is given by, y = (a₁y₁ + a₂y₂) / (a₁ + a₂) = (25×10/3 + 100×5) / (25+100) = 4.66cm.
This will on for the y-axis.
For the x-axis, The center of gravity is given by, x = (a₁x₁ + a₂x₂) / (a₁ + a₂) = (25×11.66 + 100×5) / (25+100) = 6.332cm.
So the center of gravity will be at (2.33, 4.33).

8. Where will be the center of gravity of an I section will be if the dimension of web is 2x20cm and that of flange is 2x15cm If the y-axis will pass through the center of the section?
a) 8.5cm
b) 9.5cm
c) 10.5cm
d) 11.5cm
Answer: b
Clarification: The center of gravity is given by, y = (a₁y₁ + a₂y₂ + a₃y₃) / (a₁ + a₂ + a₃) = (40×18 + 30×9.5 +40×1 / (40 +30+40) = 9.5cm.

9. Where will be the center of gravity of an T section will be if the dimension of web is 2x20cm and that of flange is 2x15cm If the y-axis will pass through the center of the section?
a) 10.5cm
b) 11.45cm
c) 12.35cm
d) 12.85cm
Answer: b
Clarification: The center of gravity is given by, y = (a₁y₁ + a₂y₂) / (a₁ + a₂) = (40×16 + 30×7.5)/ (30+40) = 12.35cm.

10. Where will be the center of gravity of the following section?
a) 7.33cm
b) 8.33cm
c) 9.33cm
d) 10.33
Answer: b
Clarification: Area of triangle = 50, area of rectangle = 50
The center of gravity is given by, y = (a₁y₁ + a₂y₂) / (a₁ + a₂) = (50×11.66 + 50×5)/(50+50) = 8.33cm.

11. Where will be the centre of gravity of the following L-section?
a) (18.31,30.81)
b) (19.45, 29.87)
c) (20,30)
d) (19.62,29.62)
Answer: a
Clarification: The center of gravity is given by, y = (a₁y₁ + a₂y₂) / (a₁ + a₂) = (600×50 + 414×3) / (600+414) = 18.31cm.
This will on for the y-axis.
For the x-axis, The center of gravity is given by, x = (a₁x₁ + a₂x₂) / (a₁ + a₂) = (600×3 + 414×40.5) / (600+414) = 30.81cm.
So the center of gravity will be at (2.33, 4.33).

12. Where will be the center of gravity of an I section will be if the dimension of upper web is 2x8cm, lower web is 2×16 and that of flange is 2x12cm If the y-axis will pass through the center of the section?
a) 7.611cm
b) 7.44cm
c) 6.53cm
d) 6.44cm
Answer: d
Clarification: Area of upper web a1 = 16cm, area of flange a2 = 24, area of lower web a3 = 32.
The center of gravity is given by, y = (a₁y₁ + a₂y₂ + a₃y₃) / (a₁ + a₂ + a₃) = (16×15 + 24×8 +32×1 / (16 +24+32)) = 6.44cm.

Strength of Materials Assessment Questions,

250+ TOP MCQs on Introduction to Shear Stress and Answers

This set of Strength of Materials Multiple Choice Questions on “Introduction to Shear Stress”.

1. At ________ the shearing stress in a beam are maximum.
a) Extreme fibres
b) Modulus of section
c) Neutral axis
d) Along the cross-sectional area
Answer: c
Clarification: Shearing stress in a beam is maximum at the neutral axis. Shearing stress is defined as the resistance offered by the internal stress to the shear force.

2. Determine the shear stress at the level of neutral axis, if a beam has a triangle cross section having base “b” and altitude “h”. Let the shear force be subjected is F.
a) 3F/8bh
b) 4F/3bh
c) 8F/3bh
d) 3F/6bh
Answer: c
Clarification: For a triangular section subjected to shear force the shear stress in neutral axis is
Shear stress at NA = 4/3 [Average shear stress].
= 4/3 [F/0.5 bh] = 8F / 3bh.

3. The maximum shear stress is ______ times the average shear stress [For rectangular beams].
a) 2.5
b) 3
c) 1.2
d) 1.5
Answer: d
Clarification: The maximum shear stress occurs at neutral axis. Then y = 0.
Max shear stress = 3F/2bd = 3/2 [F/bd].
= 1.5 Average shear stress.

4. Shear stress in a beam is zero at ______
a) Neutral axis
b) Extreme fibres
c) Cross section
d) Junctions
Answer: b
Clarification: The resistance offered by the internal stress to shear is known as shearing stress. Shearing stress is zero at extreme fibres of the beam. The bending stresses are maximum at extreme fibres of the beam cross section.

5. Shear stress distribution over rectangular section will be _________
a) parabolic
b) elliptical
c) triangular
d) trapezoidal
Answer: a
Clarification:

Maximum shear stress is 1.5 times that of average shear stress.
The shear stress distribution is parabolic.

6. A round Steel rod of 100 mm diameter is bent into an arc of radius 100m. What is the maximum stress in the rod? Take E = 2×10⁵ N/mm².
a) 150 N/mm²
b) 200 N/mm²
c) 100 N/mm²
d) 300 N/mm²
Answer: c
Clarification: D = 100m
y = 50 mm
R = 10 × 10³mm
By equation of flexure; E/R = f/y
f=E/R ×y
= 2×10⁵/ 100 × 10³× 50
= 100 N/mm².

7. For circular section, the maximum shear stress is equal to ____________ times of average shear stress.
a) 2/3
b) 3/2
c) 4/3
d) 3/4
Answer: c
Clarification: Maximum shear stress occurs at neutral axis & y = 0.
Max. Shear stress = 4/3 [ F/A ].
F/A is average shear stress.
The maximum shear stress distribution is 33% more than average shear stress.

8. A steel beam is 200 mm wide and 300 mm deep. The beam is simply supported and carries a concentrated load w. If the maximum stress are 2 N/mm². What will be the corresponding load?
a) 50 kN
b) 80 kN
c) 40 kN
d) 85 kN
Answer: b
Clarification: For a rectangular cross section
Max. Shear stress = 3/2 [F/A ]
2 = 3/2 [w/200 × 300].
w = 80 kN.

9. Maximum shear stress in thin cylindrical shell be ___________
a) pr/2t
b) pr/3t
c) pr/4t
d) pr/ 5t
Answer: c
Clarification: Hoop stress P(h) = pr/t
Longitudinal stress P(l) = pr/2t
Thus, hoop stress is twice the longitudinal stress
Max. Shear stress = P(h) – P(l) / 2
= pr/4t.

10. Circumferential stress is same as of _________
a) Hoop stress
b) Longitudinal stress
c) Transverse stress
d) Phreatic stress
Answer: a
Clarification: In a thin cylindrical shell of internal radius r thickness t when subjected to internal fluid pressure P, the stress developed in the internal walls can be termed as circumferential stress or hoop stress.
P(h) = pr/t.

250+ TOP MCQs on Deflection of Cantilever and Answers

This set of Strength of Materials Multiple Choice Questions on “Deflection of Cantilever”.

1. The ratio of maximum deflection of a beam to its ___________ is called stiffness of the beam.
a) Load
b) Slope
c) Span
d) Reaction at the support
Answer: c
Clarification: The stiffness of a beam is a measure of it’s resistance against deflection. The ratio of the maximum deflection of a beam to its span can be termed as stiffness of the beam.

2. Stiffness of the beam is inversely proportional to the _____ of the beam.
a) Slope
b) Support reaction
c) Deflection
d) Load
Answer: c
Clarification: Stiffness of a beam is inversely proportional to the deflection. Smaller the deflection in a beam due to given external load, greater is its stiffness.

3. The maximum ____ should not exceed the permissible limit to the span of the beam.
a) Slope
b) Deflection
c) Load
dl Bending moment
Answer: b
Clarification: The maximum deflection of a loaded beam should not exceed the permissible limit in relation to the span of a beam. While designing the beam the designer should be keep in mind that both strength and stiffness criteria.

4. In cantilever beam the deflection occurs at ______
a) Free end
b) Point of loading
c) Through out
d) Fixed end
Answer: a
Clarification: Deflection can be defined as the perpendicular displacement of a point on straight access to the curved axis. In cantilever beams, the maximum deflection occurs at free end.

5. The maximum deflection in cantilever beam of span “l”m and loading at free end is “W” kN.

a) Wl³/2EI
b) Wl³/3EI
c) Wl³/4EI
d) Wl²/2EI
Answer: b
Clarification: Maximum deflection occurs at free end distance between centre of gravity of bending moment diagram and free end is x = 2l/3.
As deflection is equal to the slope × “x”. The slope = Wl2/2EI radians
Maximum deflection (y) = Ax/EI = Wl³/3EI.

6. In an ideal fluid, the ____________ stresses are pretend to be absent.
a) Bending
b) Shearing
c) Tensile
d) Compressive
Answer: b
Clarification: An ideal fluid is a fluid where there is no resistance to the deformation. Ideal Fluids are those Fluids which have no viscosity surface tension. The shear stress is also absent. This fluid is also called as perfect fluid.

7. Air and water are the examples of ___________
a) Non Newtonian fluids
b) Vortex fluids
c) Real fluids
d) Ideal fluids
Answer: d
Clarification: The ideal Fluids are imaginary fluids in nature, they are incompressible. These fluids possess low viscosity. Air and water are considered as ideal fluids.

8. _______ fluids are practical fluids
a) Ideal
b) Real
c) Vortex
d) Newtonian
Answer: b
Clarification: These fluids possess properties such as viscosity, surface tension. They are compressible in nature. The certain amount of resistance is always offered by the fluids, they also possess shear stress. They are also known as practical fluids.

9. Specific weight of water at 4°C is ____________ N/m³.
a) 9810
b) 9760
c) 9950
d) 9865
Answer: a
Clarification: The specific weight (weight density) of a fluid is weight per unit volume. It is represented by symbol w & it is expressed in Newton per metre cube (N/m³). The specific weight of water at 4 degree centigrade is 9810 N/m3or 9.81 kN/m³.

10. The inverse of specific weight of a fluid is __________
a) Specific gravity
b) Specific Volume
c) Compressibility
d) Viscosity
Answer: b
Clarification: Specific volume is the volume of the fluid by Unit Weight it is the reciprocal of specific weight is denoted by “v”. SI units are m³/N.
v= 1/specific weight.

11. Calculate the specific gravity of mercury.
a) 12.5
b) 14.7
c) 13.6
d) 11.8
Answer: c
Clarification: The specific gravity of any fluid is the ratio of the specific weight of fluid by specific weight of water. For mercury, the specific weight is 133416 N/m³. For water, w = 9810 N/m³.
S = 133416/9810
S= 13.6.

12. Specific gravity of water is __________
a) 0.8
b) 1
c) 1.2
d) 1.5
Answer: b
Clarification: The specific gravity is also called as relative density. It is dimensionless quantity and it has no units. The specific gravity of water is the ratio of specific weight of fluid to specific weight of water, as both the numerator and denominator are same. The value is 1.

13. Compute the maximum deflection at free end of a cantilever beam subjected to udl for entire span of l metres.
a) wl⁴/8EI
b) wl⁴/4EI
c) wl³/8EI
d) wl²/6EI
Answer: a
Clarification: The slope at free end = A/EI = wl³/6EI
Maximum deflection at free end is Ax/EI; [x= ¾ l]
y= wl³/6EI × ¾ l = wl⁴/8EI.

14. Calculate the maximum deflection of a cantilever beam with udl on entire span of 3m the intensity of you udl be 25 kN/m. Take EI as 4000 kN/m².
a) 0.052m
b) 0.063m
c) 0.076m
d) 0.09m
Answer: b
Clarification: For cantilever beams with udl on entire span, the maximum deflection = wl⁴/8EI
y = wl⁴/8EI = 25 × 3⁴/ 8 × 4000 = 0.063m.

15. Which of the following is not an example of Malleability?
a) Wrought Iron
b) Ornamental silver
c) Torsteel
d) Ornamental gold
Answer: c
Clarification: Torsteel is an example of mechanical property ductility. The ductility is a property of a material by which material can be fractured into thin wires after undergoing a considerable deformation without any rupture.

250+ TOP MCQs on Composite Shaft and Answers

This set of Strength of Materials Multiple Choice Questions on “Composite Shaft”.

1. Calculate that torque, if the diameter of the shaft is 50 mm and revolutions @ 130 rpm. The maximum shear stress is 62.5 N/mm².
a) 1564 Nm
b) 1478 Nm
c) 1534 Nm
d) 1494 Nm
Answer: c
Clarification: Diameter of shaft = 50 mm
Revolutions of shaft = 130 rpm.
Maximum shear stress = f= 62.5 N/mm².
T = f π D³/ 16
T = 62.5 ×50³× π / 16.
T = 1534 Nm.

2. What is the example for a centrifugal pump?
a) Reciprocating pump
b) Suction pump
c) Rotodynamic pump
d) Delivery pump
Answer: c
Clarification: Rotodynamic pumps have a rotating element through which as the liquid passes its angular momentum changes, due to which the pressure energy of the liquid is increased. The centrifugal pump is a rotodynamic pump.

3. Reciprocating pump is an example of ___________
a) Positive displacement pump
b) Delivery pump
c) Suction pump
d) Rotodynamic pump
Answer: a
Clarification: Positive displacement pumps are those pumps in which the liquid is sucked and then it is pushed to the thrust exerted on it by a piston. The most common example of the positive displacement pump is the reciprocating pump.

4. ____ is the difference between theoretical discharge and the actual discharge of the pump.
a) Crank
b) Hook
c) Slip
d) Centile
Answer: c
Clarification: c
Clarification: Slip of a pump is defined as the difference between the theoretical discharge and actual discharge after pump.

5. _____ is a phenomenon by which the study and continuous flow of liquid are obstructed.
a) Slip
b) Separation
c) Air vessels
d) Knockage
Answer: b
Clarification: Separation of reciprocating pump is that phenomenon by which the steady and continuous flow of liquid is affected by the presence of air and dissolved gases.

6. Negative slip occurs when the______ is more than theoretical discharge.
a) Virtual discharge
b) Actual discharge
c) Mean discharge
d) Mode discharge
Answer: b
Clarification: When the delivery valve opens before the suction stroke is completed, the actual discharge is more than the theoretical discharge. In such cases, the slip of the pump is known as a negative slip.

7. _____ slip occurs, when the delivery pipe is short and the suction pipe is long.
a) Positive
b) Critical
c) Negative
d) Zero
Answer: b
Clarification: The slip occurs when the delivery pipe is short and the suction pipe is long. The pump is running at high speeds as the delivery valve open before a suction stroke is completed, the slip of the pump is known as negative slip.

8. ________ reduces the possibility of separation.
a) Air vessels
b) Casing
c) Impeller
d) Vortex
Answer: a
Clarification: An air vessel may be a closed chamber having the compressed air in a top portion and the water at the bottom. It reduces the possibility of separation and it ensures the pump to run at high speed.

9. If the absolute pressure falls below ___________ m, the pump prone to separation.
a) 3 m
b) 2 m
c) 1.5 m
d) 2.5 m
Answer: d
Clarification: If the absolute pressure falls below 2.5 metres of water, the dissolved gases will be appearing in a liquid and continuous flow will be chocked. This phenomenon can be termed as separation.

10. The phenomenon of separation can also be known as ___________
a) Cavitation
b) Priming
c) Positive head
d) Pulsate
Answer: a
Clarification: Separation is a phenomenon of obstructing the flow by the presence of dissolved gases when the absolute pressure falls below 2.5 metres of water. This phenomenon of separation can also be known as knocking (or) cavitation in the reciprocating pump.

11. The work done against friction is reduced due to _____________
a) Impeller
b) Priming
c) Air vessel
d) Vortex
Answer: c
Clarification: An air vessel is fitted to the suction and delivery pipes at a point close to the cylinder of a single acting reciprocating. The pump increases the length of the suction pipe and reduces the work done against friction.

12. Volute is a type of _____________
a) Delivery pipe
b) Casing
c) Impeller
d) Suction pipe
Answer: b
Clarification: Casing is an airtight chamber covering the impeller. The different types of casing
i.Volute casing
ii.Vortex casing
iii.Casing with guide blades.

13. ______ pumps, the torque is uniform.
a) Reciprocating pump
b) Suction pump
c) Delivery pump
d) Centrifugal pump
Answer: d
Clarification: Centrifugal pump is used for lifting highly viscous liquids such as oils, muddy and sewage water, paper pulp etc. In centrifugal pump, torque is uniform and no air vessels are required.

14. What is the practical maximum suction lift in a reciprocating pump?
a) 3.5 m
b) 4.5 m
c) 5 m
d) 6.5 m
Answer: d
Clarification: Reciprocating pump can handle only pure water or less viscous liquids free from impurities. It can be operated at low speeds only. The practical maximum section lift is 6.5 metres.

15. _____ pumps give a larger discharge.
a) Suction
b) Reciprocating
c) Centrifugal
d) Positive displacement
Answer: c
Clarification: Centrifugal pump are an example of rotodynamic pump the basic principle of centrifugal pump is that “when a certain mass of liquid is rotated by an external force, then the centrifugal head is impressed which enables it to rise to a higher level”. A centrifugal pump discharges a larger quantity when compared to other pumps.