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# Multiple choice question for engineering

## Set 1

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?

a) P1 > P2 > P3
b) P1 < P2 < P3
c) P1 = P2 = P3
d) P1 > P2 < P3

Answer: c [Reason:] 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?

a) 2.5
b) 3.5
c) 4.5
d) 5.5

Answer: b [Reason:] 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?

a) 101.3
b) 101.5
c) 101.7
d) 101.9

Answer: c [Reason:] Total height of the water in the beaker = 2 + 12 * 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?

a) 4.375
b) 4.5
c) 5.5
d) 5.625

Answer: b [Reason:] 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 = 12 * 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 = 12 * 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?

a) 27.33
b) 19.2
c) 10.8
d) 2.67

Answer: b [Reason:] 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 = 13 * 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 = 13 * 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:

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?

a) 0.01
b) 0.05
c) 0.10
d) 0.15

Answer: a [Reason:] 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, 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

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)

a) 4:75% higher
b) 4:75% lower
c) 6:75% higher
d) 6:75% lower

Answer: a [Reason:] 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

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?

a) 0%
b) 5%higher
c) 5%lower
d) 10%higher

Answer: a [Reason:] 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?

a) 588.6
b) 637.65
c) 735.75
d) 833.85

Answer: b [Reason:] 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 [Reason:] Pressure at the base initially = Sw * h3 * g; Pressure at the base after pouring the second liquid = Sw * h3 * g + Sl * 2h3 * g, where Sw and Sl are the specific gravities of water and the second liquid.

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?

a) 23.5
b) 24.5
c) 26.5
d) 27.5

Answer: b [Reason:] 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?

a) 30o
b) 50o
c) 60o
d) 70o

Answer: d [Reason:] 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?

a) 12o
b) 15o
c) 17o
d) 20o

Answer: b [Reason:] 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 = 12 * 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:

a) 766
b) 776
c) 786
d) 796

Answer: c [Reason:] 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?

a) 36
b) 40
c) 76
d) 116

Answer: b [Reason:] Pgas = Patm – ρgH Taking gauge pressure in terms of cm of Hg, -40 = 0 – H; H = 40.

## Set 2

1. When is orifice called ‘large orifice’?
a) If the head of liquid is less than 5 times the depth of orifice
b) If the head of liquid is less than 2.5 times the depth of orifice
c) If the head of liquid is less Hence, 4 times the depth of orifice
d) If the head of liquid is less than 1.5 times the depth of orifice

Answer: a [Reason:] It is the correct parametric definition for ‘large orifice’.

2. In case of any orifice, velocity always remains constant and hence discharge can be calculated.
a) True
b) False

Answer: b [Reason:] In case of large orifice, velocity always remains variable and hence discharge cannot be calculated.

3. Find the discharge through a rectangular orifice 2.2 m wide and 1.3 m deep fitted to a easier tank. The water level in a team is 2.5 m above the top edge of orifice.
a) 13.9 m3/s
b) 11.5 m3/s
c) 16.9 m3/s
d) 8.7 m3/s

Answer: a [Reason:] Q = 2/3 Cd *b*√2g* (H21.5 – H11.5) Here, H1 = 3.8 H2 = 2.5 b = 2.2 Hence, Q = 13.9 m3/s.

4. Find the discharge through a rectangular orifice 3.2 m wide and 1.7 m deep fitted to a easier tank. The water level in a team is 3.3 m above the top edge of orifice. Take Cd = 0.6
a) 29.4 m3/s
b) 58.5 m3/s
c) 67.9 m3/s
d) 78.7 m3/s

Answer: a [Reason:] Q = 2/3 Cd *b*√2g* (H21.5 – H11.5) Here, H1 = 5 H2 = 3.3 b = 3.2 Hence, Q = 29.4 m3/s.

5. Find the discharge through totally drowned orifice of width 2.3 m if the difference of water levels on both side of the orifice be 40 cm. The height of water from to and bottom of the orifice are 2.6 m and 2.75 m respectively.
a) .56 m3/s
b) .64 m3/s
c) .75 m3/s
d) .55 m3/s

Answer: a [Reason:] Q = Cd * b * (H2 – H1) √2gH Here, b = 2.3 H2 = 2.75 H1 = 2.6 H = 40 Q = .56 m3/s.

6. Find the discharge through totally drowned orifice of width 3.3 m if the difference of water levels on both side of the orifice be 50 cm. The height of water from to and bottom of the orifice are 2.25 m and 2.67 m respectively.
a) 2.8 m3/s
b) 2.7 m3/s
c) 2.6 m3/s
d) 2.5 m3/s

Answer: a [Reason:] Q = Cd * b * (H2 – H1) √2gH Here, b = 3.3 H2 = 2.67 H1 = 2.25 H = 50 Q = 2.6 m3/s.

7. A rectangular orifice of 2 m width and 1.2 m deep is fitted in one side of large tank. The easier level on one side of the orifice is 3m above the top edge of the orifice while on the other side of the orifice the water level is 0.5 m below it’s top edge. Calculate discharge if Cd = .64
a) 4.95 m3/s
b) 5.67 m3/s
c) 3.56 m3/s
d) 6.75 m3/s

Answer: a [Reason:] Explanation: Q = Cd * b * (H2 – H) √2gH Here, b = 2 H2 = 4.2

H = 3.5 Q = 4.94 m3/s.

8. The time taken to empty the tank is independent of Cd but depends only on the height and acceleration due to gravity.
a) True
b) False

Answer: b [Reason:] The time taken to empty the tank is dependent on Cd as well as depends only on the height and acceleration due to gravity.

9. The discharge rate is independent of the height difference and dependent only on the height.
a) True
b) False

Answer: b [Reason:] The discharge rate is dependent of the height difference and dependent only on the height.

10. In case of submerged orifice the discharge is substantially dependent on temperature of fluid
a) True
b) False

Answer: b [Reason:] Discharge is dependent on temperature but minimally.

## Set 3

1. 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 [Reason:] 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.

2. 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 [Reason:] The specific gravity of a liquid is the ratio of two similar quantities (densities) which makes it unitless.

3. The specific volume of a liquid is the reciprocal of
a) weight density
b) mass density
c) specific weight
d) specific volume

Answer: b [Reason:] 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.

4. 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 [Reason:] Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e., γ = w / v Thus, unit of is N = m3.

5. Which one of the following is the dimension of mass density?
a) [M1 L-3 T0].
b) [M1 L3 T0].
c) [M0 L-3 T0].
d) [M0 L3 T0].

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

6. Which one of the following is the dimension of specific gravity of a liquid?
a) [M1 L-3 T0].
b) [M1 L0 T0].
c) [M0 L-3 T0].
d) [M0 L0 T0].

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

7. Which one of the following is the dimension of specific volume of a liquid?
a) [M1 L-3 T0].
b) [M-1 L3 T0].
c) [M-1 L-3 T0].
d) [M0 L3 T0].

Answer: b [Reason:] Specific volume(v) is defined as the volume(V ) per unit mass(m). Thus, [v] = [V]/[m] = [L3]/[M] = [M-1L3].

8. 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].
d) [ML2 T-2].

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

9. 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 [Reason:] 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.

10. 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 [Reason:] 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.

11. 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 [Reason:] 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.

12. 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 [Reason:] Specific volume(v) is defined as the volume(V ) per unit mass(m). Thus,

## Set 4

1. Two horizontal plates placed 250mm have an oil of viscosity 20 poises. Calculate the shear stress in oil if upper plate is moved with velocity of 1250mm/s.
a) 20 N/m2
b) 2 N/m2
c) 10 N/m2
d) None of the mentioned

Answer:c [Reason:] Shear Stress = Viscosity * Velocity Gradient = 20/10* 1.25/0.25 = 10 N/m2.

2. The kinematic viscosity of oil of specific gravity .8 is .0005 .This oil is used for lubrication of shaft of diameter .4 m and rotates at 190 rpm. Calculate the power lost in the bearing for a sleeve length of 90mm. The thickness of the oil film is 1.5mm.
a) 477.65 Watts
b) 955.31 Watts
c) 238.83 Watts
d) None of the mentioned

Answer: a [Reason:] Power lost= torque * angular velocity = force* radius* angular velocity = shear stress * area* radius* angular velocity Shear Stress = viscosity* velocity gradient Power lost= 0.0005*0.8*1000* 2*3.142*190/60*0.2*3.142*0.23 * 190/60 = 477.65 Watts.

3. Find the kinematic viscosity of oil having density 1962 g/m3. the force experienced for area of 20 m2 is 4.904 kN and velocity of gradient at that point is 0.2/s.
a) 0.625
b) 1.25
c) 2.5
d) None of the mentioned

Answer: a [Reason:] kinematic viscosity = dynamic viscosity / density = (shear stress*density)/velocity gradient = (4904* 1962)/(20*0.2) = .625.

4. The velocity distribution for fluid flow over a flat plate is given by u=2y-6y2 in which u is the velocity in metre per second at a distance of y metre above the plate. Determine the shear stress at y=0.15m.Take dynamic viscosity of fluid as 8.6 poise.
a) 0.172 N/m2
b) 0.344 N/m2
c) 0.086 N/m2
d) None of the mentioned

Answer:a [Reason:] for y = 0.15m, velocity gradient = 0.2 viscosity= shear stress/velocity gradient shear stress = 0.86*0.2 = 0.172N/m2.

5. In which types of fluids it is observed that momentum transfer dominates cohesive forces with increase in temperature and hence viscosity increases
a) Gases
b) Liquids
c) Solids
d) None of the mentioned

Answer:a [Reason:] It is the characteristic property of gases which show increase in viscosity with increase in temperature.

6. What is the characteristic variation shown by the thixotropic fluids in their shear stress vs. rate of shear strain graph?
a) shear stress increases with increase in rate of shear strain
b) shear stress decreases with increase in rate of shear strain
c) shear stress shows variation only after a definite shear stress is reached
d) shear stress has decreasing constant and then variation relationship with rate of shear strain

Answer: c [Reason:] Thixotropic fluid show a Non-Newtonian variation for shear stress vs. rate of shear strain graph after a characteristic limiting value of shear stress is reached.

7. What happens to viscosity in the case of incompressible fluids as temperature is increased?
a) It remains constant
b) It increases
c) It decreases
d) None of the mentioned

Answer: c [Reason:] In case of incompressible fluids, cohesive forces govern the viscosity. As temperature increases the cohesive forces between fluid molecules decreases due to increase in molecular agitation. Hence, as a result, viscosity decreases.

8. If a fluid, which has a constant specific gravity, is taken to a planet where acceleration due to gravity is 3 times compared to its value on earth, what will happen to its kinematic viscosity.
a) It increases
b) It decreases
c) It remains constant
d) None of the above

Answer: c [Reason:] Kinematic viscosity depends on density and dynamic viscosity. Both, density and dynamic viscosity, are independent of acceleration due to gravity. Therefore, kinematic viscosity is independent of acceleration due to gravity.

9. In liquids in order to measure the viscosity of fluid experimentally we consider the variation of shear stress with respect to what property?
a) strain
b) shear strain
c) rate of shear strain
d) none of the mentioned

Answer: c [Reason:] By definition, viscosity is shear stress per unit ‘rate of shear strain’.

10. For a compressible fluid the kinematic viscosity is affected by temperature and pressure variation.
a) True
b) False

Answer: a [Reason:] Viscosity shows variation for change in temperature and pressure for compressible fluids. Hence, kinematic viscosity is affected by temperature and pressure variation.

## Set 5

1. Find the total pressure on a rectangular plate of dimensions 2×3 m immersed in a fluid of specific gravity 0.65 at a depth of 6 m from the surface.
a) 22.9 N/cm2
b) 45.8 N/cm2
c) 11.5 N/cm2
d) None of the mentioned

Answer: a [Reason:] Total pressure, F=w*a*y =9.81*650*6*2*3 =22.9 N/cm2.

2. Find the total pressure on a circular plate of diameter 3 m immersed in a fluid of specific gravity 0.75 at a depth of 5 m from the surface on a planet having acceleration dueto gravity 7 m/s2.
a) 18.5 N/cm2
b) 37 N/cm2
c) 9.25 N/cm2
d) None of the mentioned

Answer: a [Reason:] Total pressure, F=w*a*y =7*750*π*1.5*1.5*5 =18.5 N/cm2.

3. A rectangular plate surface 3m wide and 5m deep lies in fluid of specific gravity .9 such that its plane makes an angle of 45⁰ with water surface, upper edge 3m below free water surface. Determine the total pressure.
a) 86.5 N/cm2
b) 173 N/cm2
c) 43.25 N/cm2
d) None of the mentioned

Answer: a [Reason:] Total pressure, F=w*a*ŷ=w*A*(y+dsinθ)=9.81*900*15*(3+5sin45) =86.5 N/cm2.

4. A rectangular plan surface 5 m wide and 7 m deep lies in water in such a way that its plane makes an angle 60⁰ with the free surface of water. Determine the total pressure force when the upper edge is 3 m below the free surface.
a) 311.15 N/cm2
b) 622.3 N/cm2
c) 155.5 N/cm2
d) None of the mentioned

Answer: a [Reason:] Total pressure, F=w*a*ŷ=w*A*(y+dsinθ)=9.81*1000*35*(3+7sin60)=311.15 N/cm2.

5. A circular plate 5.0 m diameter is immersed in such a way that its greatest and least depth below the free surface are 3 m and 1 m respectively. determine the position of the centre of pressure.
a) 2.5 m
b) 5 m
c) 4.5 m
d) 6 m

Answer: a [Reason:] centre of pressure, ŷ=I*sin²θ/Aĥ + ĥ …….ĥ=(y+dsinθ) =3.142*2.54*sin²23.58/3.142*2.52*(1+2.5sin23.58)+*(1+2.5sin23.58) =2.5 m.

6. For an inclined plate the pressure intensity at every point differs.
a) True
b) False

Answer: a [Reason:] Due to inclination the depth of every point is different from the free liquid surface. Hence, the pressure intensity varies with depth.

7. The pressure intensity for a horizontal plate is maximum on the surface of the earth and decreases as we move further away from the surface of the earth either downward or upward.
a) True
b) False

Answer: a [Reason:] As we move away from the surface of the earth either downward or upward g decreases, w decreases. Hence pressure intensity decreases.

8. For an inclined plane for which position, maximum total pressure acts on it.
a) Horizontal
b) Vertical
c) Inclined
d) None of the mentioned

Answer: b [Reason:] Total pressure, F=w*a*ŷ=w*A*(y+dsinθ) For vertical plate, θ=90⁰ Hence, total pressure is maximum.

9. The total pressure or pressure intensity is zero for any point on inclined surface in space.
a) True
b) False