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# Multiple choice question for engineering

## Set 1

1. A cuboidal beaker is half filled with water. By what percent will the hydrostatic force on one of the vertical sides of the beaker increase if it is completely filled?
a) 100
b) 200
c) 300
d) 400

Answer: c [Reason:] Hydrostatic force per unit width on a vertical side of a beaker = 12 * ρgh2, where ρ = density of the liquid and h= height of liquid column. The hydrostatic force when the beaker is completely filled = 12 ρg(2h)2 = 2ρgh2. Thus, percentage increase in hydrostatic force = = 300%.

2. By what factor will the hydrostatic force on one of the vertical sides of a beaker decrease if the height of the liquid column is halved?
a) 12
b) 13
c) 14
d) 23

Answer: c [Reason:] Hydrostatic force per unit width on a vertical side of a beaker = 12 * ρgh2, where ρ = density of the liquid and h= height of liquid column. Thus, if the liquid column is halved, the hydrostatic force on the vertical face will become one-fourth.

3. Equal volume of two liquids of densities ρ1 and ρ2 are poured into two identical cuboidal beakers. The hydrostatic forces on the respective vertical face of the beakers are F1 and F2 respectively. If ρ1 > ρ2, which one will be the correct relation between F1 and F2?
a) F1 > F2
b) F1 ≥ F2
c) F1 < F2
d) F1 ≤ F2

Answer: a [Reason:] Hydrostatic force per unit width on a vertical side of a beaker = 12 * ρgh2, where ρ = density of the liquid and h= height of liquid column. Thus if ρ1 > ρ2, F1 > F2 and F1 ≠ F2, when the h is constant.

4. Which of the following is the correct relation between centroid (G) and the centre of pressure (P) of a plane submerged in a liquid?
a) G is always below P
b) P is always below G
c) G is either at P or below it.
d) P is either at G or below it.

Answer: d [Reason:] The depth of the centroid y and the centre of pressure yCP are related by: where I = the moment of inertia and A= area. None of the quantities I, A and y can be negative. Thus, YCP > y. For horizontal planes, I = 0, hence YCP = y

5. A beaker contains water up to a height of h. What will be the location of the centre of pressure?
a) h3 from the surface
b) h2 from the surface
c) 2h3 from the surface
d) h6 from the surface

Answer: c [Reason:] The depth of the centroid y and the centre of pressure yCP are related by: where I = the moment of inertia and A = area. If y = h2; I = bh3/12 ;A = bh, then

6. A cubic tank is completely filled with water. What will be the ratio of the hydrostatic force exerted on the base and on any one of the vertical sides?
a) 1:1
b) 2:1
c) 1:2
d) 3:2

Answer: b [Reason:] Hydrostatic force per unit width on a vertical side of a beaker Fv = 12 * ρgh2, where ρ = density of the liquid and h= height of the liquid column. Hydrostatic force per unit width on the base of the beaker = Fb = ρgh * h = ρgh2. Thus, Fb : Fv = 2 : 1.

7. A rectangular lamina of width b and depth d is submerged vertically in water, such that the upper edge of the lamina is at a depth h from the free surface. What will be the expression for the depth of the centroid (G)?
a) h
b) h + d
c) h + d2
d) h + d / 2

Answer: c [Reason:] The centroid of the lamina will be located at it’s centre. ( d2). Thus, the depth of the centre of pressure will be h + d2.

8. A rectangular lamina of width b and depth d is submerged vertically in water, such that the
upper edge of the lamina is at a depth h from the free surface. What will be the expression for the depth of the centre of pressure?

Answer: c [Reason:] The depth of the centroid y and the centre of pressure yCP are related by: where I = the moment of inertia and A = area. If y = h + d2; I = bh3/12 ;A = bd. thus,

9. A square lamina (each side equal to 2m) is submerged vertically in water such that the upper edge of the lamina is at a depth of 0.5 m from the free surface. What will be the total water pressure (in kN) on the lamina?

a) 19.62
b) 39.24
c) 58.86
d) 78.48

Answer: c [Reason:] Total liquid pressure on the lamina = F = γyA, where γ = specific weight of the liquid, y = depth of centroid of the lamina from the free surface, A= area of the centroid. Now, γ = 9:81 * 103 N / m3; y = 0.5 + 12 * 2m = 1.5 m, A = 2 * 2 m2 = 4 m2. Hence, F = 58.86 kN.

10. A square lamina (each side equal to 2m) with a central hole of diameter 1m is submerged vertically in water such that the upper edge of the lamina is at a depth of 0.5 m from the free surface. What will be the total water pressure (in kN) on the lamina?

a) 15.77
b) 31.54
c) 47.31
d) 63.08

Answer: c [Reason:] Total liquid pressure on the lamina = F = γyA, where γ = specific weight of the liquid, y = depth of centroid of the lamina from the free surface, A= area of the centroid. Now, γ = 9:81 * 103 N / m3; y = 0.5 + 12 * 2m = 1.5 m, A = 2 * 2 – π4 * 12 m2 = 3.215 m2 Hence, F = 47.31 kN.

11. A square lamina (each side equal to 2m) is submerged vertically in water such that the upper edge of the lamina is at a depth of 0.5 m from the free surface. What will be the depth (in m) of the centre of pressure?

a) 1.32
b) 1.42
c) 1.52
d) 1.72

Answer: d [Reason:] The depth of the centroid y and the centre of pressure yCP are related by: where I = the moment of inertia and A = area. Now,

12. What will be the total pressure (in kN) on a vertical square lamina submerged in a tank of oil (S=0.9) as shown in the figure?

a) 26.5
b) 35.3
c) 44.1
d) 61.7

Answer: c [Reason:] Total liquid pressure on the lamina = F = γyA, where γ = specific weight of the liquid, y = depth of centroid of the lamina from the free surface, A= area of the centroid. Now, γ = 9.81 * 0.9 * 103 N / m3; y = 2.5m, A = 12 * 22 = 2 m2. Hence, F = 44.1 kN.

13. The upper and lower edges of a square lamina of length 4 m are at a depths of 1 m and 3 m respectively in water. What will be the depth (in m) of the centre of pressure?

a) 1.33
b) 1.57
c) 2.17
d) 2.33

Answer: c [Reason:] The depth of the centroid y and the centre of pressure yCP are related by: where I= the moment of inertia and A = area and θ = the angle of inclination of the lamina to the horizontal. Now, = 2:17m.

14. The upper and lower edges of a square lamina of length 4 m are at a depths of 1 m and 3 m respectively in water. What will be the total pressure (in kN) on the lamina?

a) 156.96
b) 235.44
c) 313.92
d) 392.4

Answer: c Total liquid pressure on the lamina = F = γyA, where γ = specific weight of the liquid, y = depth of centroid of the lamina from the free surface, A= area of the centroid. Now, γ = 9:81 * 0.9 * 103 N / m3; y = 1 + 3-1 / 2 = 2m, A = 4 * 4 = 16 m2. Hence, F = 313.92 kN.

15. What will be the depth (in m) of the centre of pressure for a vertical square lamina submerged in a tank of oil (S=0.8) as shown in the figure?

a) 1.45
b) 1.65
c) 1.75
d) 1.95

Answer: c [Reason:] The depth of the centroid y and the centre of pressure yCP are related by: where I = the moment of inertia and A = area. Each side of the lamina = 3/&sqrt;2 Now, y = 1 + 32 = 1.5,

## Set 2

1. Which forces are neglected to obtain Euler’s equation of motion from Newton’s second law of motion?
a) Viscous force, Turbulence force, Compressible force
b) Gravity force, Turbulence force, Compressible force
c) Body force, Gravity force, Turbulence force
d) Viscous force, Turbulence force, Body force

Answer: a [Reason:] Viscous force, Turbulence force, Compressible forces are neglected to obtain Euler’s equation of motion from Newton’s second law of motion.

2. Navier-Stoke’s equation can be obtained from Reynolds’s equation by not considering which type of force?
a) Turbulence force
b) Gravity force
c) Compressible force
d) Viscous force

Answer: a [Reason:] Navier-Stoke’s equation can be obtained from Reynolds’s equation by not considering Turbulence force.

3. In order to apply Bernoulli’s equation across two sections, we have to obtain it from Euler’s equation. What is the operation that needs to be carried out in order to obtain it from Euler’s equation?
a) Partial Differentiation
b) Differentiation
c) Integration
d) None of the mentioned

Answer: c [Reason:] We have to integrate the Euler’s equation to obtain Bernoulli’s equation from it.

4. Which of the following assumption is incorrect in the derivation of Bernoulli’s equation?
a) The fluid is ideal
c) The flow is incompressible
d) The flow is rotational

Answer: d [Reason:] The assumption made is that the flow is irrotational.

5. From mathematical perspective if for a fluid curl of velocity vector is some constant value, it is ideal, steady and incompressible, then we can apply Bernoulli’s equation.
a) True
b) False

Answer: b [Reason:] If curl is some constant value, the fluid is irrotational and hence, Bernoui can’t be applied.

6. If in a fluid, while applying Newton’s second law of motion, compressibility force is neglected then what equation is obtained?
a) Navier Stoke’s Equation
b) Reynold’s equation of motion
c) Euler’s Equation of motion
d) Continuity Equation for fluid flow

Answer: b [Reason:] If in a fluid, while applying Newton’s second law of motion, compressibility force is neglected then Reynold’s equation of motion is obtained.

7. According to statement of Bernoulli’s theorem if Kinetic or Potential Energy is changing then overall energy changes and hence energy is not constant at all the places.
a) True
b) False

Answer: b [Reason:] Energy remains constant at all point as per statement of Bernoulli’s theorem even if Kinetic or Potential Energy is changing.

8. In derivation for Euler’s equation, the weight of fluid element is considered as negligible and ignored in calculation.
a) True
b) False

Answer: b [Reason:] The weight of fluid element even though negligible is consideration in calculation.

9. Water is flowing through a pipe of 3.8 cm diameter under a pressure of 20 N/cm2 (gauge) and with mean velocity of 3.7 m/s. Find the total head or total energy per unit weight of the water at a cross section, which is 7 m above the datum line.
a) 56.14 m
b) 28.07 m
c) 84.18 m
d) 10.52 m

Answer: b [Reason:] Total head = kinetic energy head + potential energy head + Datum head = p/ρg + v2/ 2g + z = 20.38 + 0.69 + 7 = 28.07 m.

10. A fluid with specific gravity 0.85 is flowing through a diameter 250 mm and 150 mm at the bottom and upper ends respectively. Determine the difference in datum head if the rate of flow through pipe is 0.04 m3/s. Take pressure at top and bottom as 27 N/cm2 and 10 N/cm2.
a) 17.1 m
b) 34.2 m
c) 10.5 m
d) None of the mentioned

## Set 3

1. Which one is in a state of failure?
a) Solid
b) Liquid
c) Gas
d) Fluid

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

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

3. In which type of matter, one won’t find a free surface?
a) Solid
b) Liquid
c) Gas
d) Fluid

Answer: c [Reason:] 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.

4. 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 [Reason:] 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.

5. 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 [Reason:] Ideal fluids are incompressible which means they will have zero compressibility.

6. 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 [Reason:] Bulk modulus k is the reciprocal of compressibility fi. k = 1fi Ideal fluids are incompressible which means fi = 0. Thus, k will be infinity.

7. 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 [Reason:] Ideal fluids are non-viscous which means they will have zero viscosity.

8. 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 [Reason:] Ideal fluids haze zero surface tension but real fluids have some finite value of surface tension.

## Set 4

1. Which one of the following is a major loss?
a) frictional loss
b) shock loss
c) entry loss
d) exit loss

Answer: a [Reason:] The major loss for the flflow through the pipes is due to the frictional resistance between adjacent fluid layers sliding over each other. All other losses are considered to be minor losses.

2. Which property of the fluid accounts for the major losses in pipes?
a) density
b) specific gravity
c) viscosity
d) compressibility

Answer: c [Reason:] The major loss for the flow through the pipes is due to the frictional resistance between adjacent fluid layers sliding over each other. This resistance arises due to the presence of viscous property of the fluid.

3. The frictional resistance for fluids in motion is
a) proportional to the velocity in laminar flow and to the square of the velocity in turbulent flow
b) proportional to the square of the velocity in laminar flow and to the velocity in turbulent flow
c) proportional to the velocity in both laminar flow and turbulent flow
d) proportional to the square of the velocity in both laminar flow and turbulent flow

Answer: a [Reason:] According to the laws of fluid friction, rf / v (for steady streamline flow) and rf / v2(for turbulent flow), where rf is the frictional resistance and v is the velocity of flow.

4. The frictional resistance for fluids in motion is
a) dependent on the pressure for both laminar and turbulent flows
b) independent of the pressure for both laminar and turbulent flows
c) dependent on the pressure for laminar flow and independent of the pressure for turbulent flow
d) independent of the pressure for laminar flow and dependent on the pressure for turbulent flow

Answer: b [Reason:] According to the laws of fluid friction, the frictional resistance is independent of the pressure for both laminar and turbulent flows.

5. The frictional resistance for fluids in motion is
a) inversely proportional to the square of the surface area of contact
b) inversely proportional to the surface area of contact
c) proportional to the square of the surface area of contact
d) proportional to the surface area of contact

Answer: d [Reason:] According to the laws of fluid friction, the frictional resistance is proportional to the surface area of contact for both laminar and turbulent flows.

6. The frictional resistance for fluids in motion varies
a) slightly with temperature for both laminar and turbulent flows
b) considerably with temperature for both laminar and turbulent flows
c) slightly with temperature for laminar flow and considerably with temperature for turbulent flow
d) considerably with temperature for laminar flow and slightly with temperature for turbulent flow

Answer: d [Reason:] According to the laws of fluid friction, the frictional resistance for fluids in motion varies considerably with temperature for laminar flow and slightly with temperature for turbulent flow.

7. Which one of the follflowing is correct?
a) the frictional resistance depends on the nature of the surface area of contact
b) the frictional resistance is independent of the nature of the surface area of contact
c) the frictional resistance depends on the nature of the surface area of contact for laminar flows but is independent of the nature of the surface area of contact for turbulent flows
d) the frictional resistance is independent of the nature of the surface area of contact for laminar flows but depends on the nature of the surface area of contact for turbulent flows

Answer: d [Reason:] According to the laws of fluid friction, the frictional resistance is independent of the nature of the surface area of contact for laminar flows but depends on the nature of the surface area of contact for turbulent flows.

8. Which one of the follflowing is correct?
a) the frictional resistance is always dependent on the nature of the surface area of contact
b) the frictional resistance is always independent of the nature of the surface area of contact
c) the frictional resistance is dependent on the nature of the surface area of contact when the liquid flows at a velocity less than the critical velocity
d) the frictional resistance is independent of the nature of the surface area of contact when the liquid flows at a velocity less than the critical velocity

Answer: d [Reason:] Frictional resistance is dependent on the nature of the surface area of contact. But, when the liquid flows at a velocity less than the critical velocity, a thin stationary film of the liquid is formed on the supporting surface. Hence, the frictional resistance becomes independent of the nature of the surface of contact.

9. Which one of the follflowing is correct?
a) Darcy-Weisbach’s formula is generally used for head loss in flow through both pipes and open channels
b) Chezy’s formula is generally used for head loss in flow through both pipes and open channels
c) Darcy-Weisbach’s formula is generally used for head loss in flow through both pipes and Chezy’s formula for open channels
d) Chezy’s formula is generally used for head loss in flow through both pipes and Darcy-Weisbach’s formula for open channels

Answer: c [Reason:] Darcy-Weisbach’s formula is generally used for head loss in flow through both pipes as it takes into consideration the flow velocity whereas Chezy’s formula is used for open channels as it considers the pressure difference.

10. A liquid flows through pipes 1 and 2 with the same flow velocity. If the ratio of their pipe diameters d1 : d2 be 3:2, what will be the ratio of the head loss in the two pipes?
a) 3:2
b) 9:4
c) 2:3
d) 4:9

Answer: c [Reason:] According to Darcy-Weisbach’s formula, wherehf is the head loss in the pipe, f is the co-efficient of friction, L is the length, D is the diameter and V is the flow velocity. Thus, hf1 : hf2 = D2 : D1 = 2 : 3.

11. A liquid flowss through two similar pipes 1 and 2. If the ratio of their flow velocities v1 : v2 be 2:3, what will be the ratio of the head loss in the two pipes?
a) 3:2
b) 9:4
c) 2:3
d) 4:9

Answer: d [Reason:] According to Darcy-Weisbach’s formula, where hf is the head loss in the pipe, f is the co-efficient of friction, L is the length, D is the diameter and V is the flow velocity. Thus, hf1 : hf2 = v1 : v2 = 4 : 9.

12. A liquid flows with the same velocity through two pipes 1 and 2 having the same diameter. If the length of the second pipe be twice that of the first pipe, what should be the ratio of the head loss in the two pipes?
a) 1:2
b) 2:1
c) 1:4
d) 4:1

Answer: a [Reason:] According to Darcy-Weisbach’s formula, where hf is the head loss in the pipe, f is the co-efficient of friction, L is the length, D is the diameter and V is the flow velocity. Thus, hf1 : hf2 = L1 : L2 = 1 : 2.

13. The head loss at the entrance of the pipe is that at it’s exit
a) equal to
b) half
c) twice
d) four times

Answer: b [Reason:] According to Darcy-Weisbach’s formula, hi = o.5v2 / 2g and ho = v2 / 2g, where hi is the head loss at pipe entrance, ho is the head loss at pipe exit and v is the flow velocity. Thus hi = 0.5ho.

14. On which of the factors does the co-efficent of bend in a pipe depend?
a) angle of bend and radius of curvature of the bend
b) angle of bend and radius of the pipe
c) radius of curvature of the bend and pipe
d) radius of curvature of the bend and pipe and angle of bend

Answer: d [Reason:] The co-efficent of bend in a pipe depends on all the three parameters – radius of curvature of the bend, diameter (radius) of the pipe and angle of bend.

## Set 5

1. A rectangular tank is moving horizontally in the direction of its length with a constant acceleration of 3.6 m/s2. If tank is open at the top then calculate the angle of water surface to the horizontal.
a) 20.15
b) 69.84
c) 40.30
d) None of the mentioned

Answer: a [Reason:] tanθ=a/g tanθ=3.6/9.8 θ=20.15⁰.

2. A rectangular tank is moving horizontally in the direction of its length with a constant acceleration of 4.8 m/s2. The length of tank is 7 m and depth is 1.5 m. If tank is open at the top then calculate the maximum pressure intensity at the bottom.
a) 6.3 N/cm2
b) 3.15 N/cm2
c) 12.6 N/cm2
d) 1.6 N/cm2

Answer: b [Reason:] tanθ=a/g tanθ=4.8/9.8 θ=26.07⁰ h= d+(L/2)tanθ = 1.5+3.5tan26.07 = 3.21 m p=ρ*g*h =3.15 N/cm2.

3. A rectangular tank is moving horizontally in the direction of its length with a constant acceleration of 5.5 m/s2. The length of tank is 5.5 m and depth is 2 m. If tank is open at the top then calculate the minimum pressure intensity at the bottom.
a) 3.8 N/cm2
b) 1.9 N/cm2
c) 5.7 N/cm2
d) 2.6 N/cm2

Answer: b [Reason:] tanθ=a/g tanθ=5.5/9.8 θ=29.28⁰ h= d-(L/2)tanθ = 2-2.75 tan29.28 = 3.21 m

p=ρ*g*h =1.9 N/cm2.

4. A rectangular tank is moving horizontally in the direction of its length with a constant acceleration of 4.5 m/s2.The length, width and depth of tank are 7 m, 3m, 2.5m respectively. If tank is open at the top then calculate the total force due to water acting on higher pressure end of the tank.
a) 1.07 MN
b) 2.14 MN
c) 4.28 MN
d) 4.35 MN

Answer: a [Reason:] tanθ=a/g tanθ=4.5/9.8 θ=24.64⁰ h= d+(L/2)tanθ = 2.5+3.5tan24.64 = 4.1 m

F=wAĥ =9810*2.68*4.1 = 1.07 MN.

5. A tank containing water upto a depth of 500 mm is moving vertically upward with a constant acceleration of 2.45 m/s2. Find the force exerted by fluid of specific gravity .65 on the side of tank,width of tank is 1m.
a) 996.1 N
b) 1992.2 N
c) 498.06 N
d) 124.5 N

Answer: a [Reason:] p=ρ*g*h*(1+a/g) =650*9.81*0.5*(1+2.45/9.81) =3984.5 N/m2 F=wAĥ = 650*9.81*0.5*0.5*3984.5 = 996.1N.

6. A tank containing water upto a depth of 750 mm is moving vertically downward with a constant acceleration of 3.45 m/s2. Find the force exerted by fluid of specific gravity .85 on the side of tank,width of tank is 2m
a) 2682.75 N
b) 5365.5 N
c) 1341.25 N
d) 4024.5 N

Answer: a [Reason:] p=ρ*g*h*(1-a/g) =750*9.81*0.75*(1-3.45/9.81) =3577 N/m2 F=wAĥ = 0.5*0.75*3984.5*2 = 2682.75 N.

7. A tank containing water upto a depth of 650 mm is stationary. Find the force exerted by fluid of specific gravity .55 on the side of tank,width of tank is 1.5m
a) 1709.9 N
b) 3419.4N
c) 6838.8 N
d) 1367.75 N

Answer: a [Reason:] p=ρ*g*h =550*9.81*0.65 =3507 N/m2 F=wAĥ = 0.5*0.65*3507*1.5 = 1709.7N.

8. The pressure intensity at the bottom remains same, even if the tank moves with constant horizontal acceleration.
a) True
b) False

Answer: b [Reason:] The pressure intensity at the bottom differs due to variation in height as tank moves with constant acceleration.

9. There will be development of shear stress due to the dynamic motion of tank or container.
a) True
b) False

Answer: b [Reason:] The water in tank is at rest even if the tank is moving.

10. If the tank is moving vertically, which of its component is subjected to maximum total pressure?
a) Lower part of vertical walls
b) Higher part of vertical walls
c) Base
d) None of the mentioned