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# Multiple choice question for engineering

## Set 1

1. The right limb of a simple U-tube manometer containing mercury is open to the atmosphere while the lift limb is connected to a pipe in which a fluid of specific gravity 0.85 is flowing. The centre of the pipe is 14 cm below the level of mercury in the right limb.Evaluate the pressure of fluid flowing in the pipe if the difference of mercury level in the two limbs is 22 cm.
a) 2.86 N/cm2
b) 5.73 N/cm2
c) 1.43 N/cm2
d) None of the mentioned

Answer: a [Reason:] Pressure at centre of pipe + Pressure at depth 8 cm in left limb = Pressure at depth 22 cm in right limb P = 13600×9.81×0.22 – 850×9.81×.08 = 2.86 N/cm2.

2. A single coloumn manometer is connected to a pipe containing a liquid of specific gravity 0.75. Find the pressure in the pipe if the area of reservoir is 250 times the area of tube for the manometer reading. The difference in mercury level is 40 cm. On the left limb the fluid is upto the height of 20 cm.
a) 10.42 N/cm2
b) 5.21 N/cm2
c) 2.60 N/cm2
d) None of the mentioned

Answer: b [Reason:] Pressure = a/A height × (density of mercury × 9.81-density of fluid × 9.81)+ height in right limb × density of mercury × 9.81 – height in left limb × density of fluid × 9.81 = 5.21 N/cm2 { Here a/A = 1/ 250}.

3. A Differential manometer is connected at the points A and Bat the centre of two pipes. The pipe A(left limb) contains a liquid of specific gravity = 1.5 while pipe B (right limb)contains a liquid of specific gravity 0.85. The pressure at A and B are .5 kgf/cm2 and 1.2 kgf/cm2 respectively. Find the difference in level of mercuru in the differential manometer. A is 2.5m above B and 5 m above the mercury in its own limb. B is 2.5 m above the mercury level in limb A.
a) 12.7 cm
b) 25.5 cm
c) 6.28 cm
d) 10.85 cm

Answer: a [Reason:] Total pressure at the datum line in limb A = Total pressure at the datum line in limb B 0.5*9.81*10000 + 5*9.81*1500 + h*9.81*13600 = 1.2*9.81*10000 + (h+2)*9.81*850 After solving, h=12.7 cm.

4. An inverted differential manometer is connected to two pipes A and B which covey water. The fluid in manometer is oil of specific gravity 0.75. For the manometer readings, find the pressure difference between A and B. Datum in left limb is 40 cm above point A. Point B is 60 cm below datum line. Difference in level of fluid is 20 cm.
a) 1471 N/m2
b) 2943 N/m2
c) 735.75 N/m2
d) None of the mentioned

Answer: a [Reason:] Total pressure at the datum line in limb A = Total pressure at the datum line in limb B Pressure difference between A and B = -0.4*9.81*100 + 0.2*9.81*750 + 0.4*9.81*1000 = 1471 N/m2.

5. In the inverted U-tube Differential manometer, how is the specific gravity of manometric fluid used relative to the fluid flowing in the pipes
a) Specific gravity is more than that of fluid flowing in pipes
b) Specific gravity is less than that of fluid flowing in pipes
c) Specific gravity is equal to that of fluid flowing in pipes
d) None of the mentioned

Answer: b [Reason:] In the inverted U-tube Differential manometer, specific gravity of manometric fluid used is less than relative to the fluid flowing in the pipes as the manonmetric fluid is at the top.

6. Why is large reservoir used in single column manometer?
a) In order to enhance the change in level of liquid in reservoir
b) In order to negate the effects of change in level due to pressure variation
c) In order to reduce the effect due to dynamic pressure variation due to motion
d) None of the mentioned

Answer: b [Reason:] Single column manometer directly gives the pressure by measuring the height in the other limb and due to large cross sectional area of the reservoir, for any variation in pressure, the change can be neglected.

7. Manometers are the pressure measuring devices which use the principle of dynamic pressure to measure the pressure difference.
a) True
b) False

Answer: b [Reason:] Manometers are the pressure measuring devices which use the principle of pressure due to static fluid (i.e the column height) to measure the pressure difference.

8. The distance moved by liquid will be more in which type of manometer?
a) Inclined Single coloumn manometer
b) Vertical Single coloumn manometer
c) Horizontal Single coloumn manometer
d) None of the mentioned

Answer: a [Reason:] The distance moved by liquid will be more in Inclined Single column manometer due to its inclination.

9. Differential manometer gives the pressure reading with respect to atmospheric pressure.
a) True
b) False

Answer: b [Reason:] Differential manometer gives the pressure difference between the fluid flowing in two pipes with respect to each other.

10. Which device is popularly used for measuring difference of low pressure?
a) Inverted U-tube Differential Manometer
b) U-tube Differential Manometer
c) Inclined Single column manometer
d) Vertical Single column manometer

Answer: a [Reason:] Inverted U-tube Differential Manometer has lighter manometric fluid, Hence it is used for measuring the low pressure difference.

## Set 2

1. A rectangular pontoon is 5 m long, 3 m wide and 1.40 m high. The depth of immersion of the pontoon is 0.60 m in seawater. If the centre of gravity is 0.7 m above the bottom of the pontoon, determine the metacentric height. The density for seawater = 1045 kg/m3.
a) 0.135
b) 0.271
c) 0.543
d) 0.068

Answer: a [Reason:] BG=Centre of pontoon – Centre of immersed portion=0.7-0.3=0.4 Metacentric height=I/∀ -BG I=bd³/12 = 5*3³/12 ∀=5*3*1.4 Metacentric height=0.135 m.

2. A uniform body of size 4 m long * 2.5 m wide * 1.5 m deep floats in water. What is the weight of the body if depth of immersion is 1 m ?
a) 147.1 kN
b) 294.3 kN
c) 73.5 kN
d) 588.6 kN

Answer: a [Reason:] Weight of Body = Weight of water displaced = ρ*g*Volume of displaced water=9.81*1000*4*2.5*1.5=147.1kN.

3. A block of material of specific gravity 0.45 floats in water. Determine the meta-centric height of the block if its size is 3 m * 2 m* 0.8 m.
a) 0.506 m
b) 0.376 m
c) 1.012 m
d) 0.127 m

Answer: b [Reason:] BG= Centre of pontoon – Centre of immersed portion=0.4 – 0.55*0.8=0.04 Metacentric height=I/∀ -BG I=bd³/12 = 3*2³/12 ∀=3*2*0.8 Metacentric height=0.376 m.

4. A solid cylinder of diameter 4.5 has a height of 2.5 metres. Find the meta-centric height of the cylinder when it is floating in water with its axis vertical. The sp. gr. of the cylinder=0.45.
a) 1.9 m
b) 3.8 m
c) 5.7 m
d) .95 m

Answer: a [Reason:] BG= Centre of pontoon – Centre of immersed portion=1.25-0.45*2.5=0.125 Metacentric height=I/∀ -BG I=π*r⁴ ∀= π*r*r*h Metacentric height=1.9 m.

5. In case of spherically shaped bodies of uniform mass distribution and completely immersed in fluid and floating, the centre of buoyancy coincides with centre of gravity.
a) True
b) False

Answer: a [Reason:] The volume of fluid displaced by the body is equal to the actual volume of body in air. Hence, In case of spherically shaped bodies of uniform mass distribution and completely immersed in fluid and floating, the centre of buoyancy coincides with centre of gravity.

6. Proper explanation for metacentre is:
a) Point at which line of action of force meets the normal axis of body when it is given angular displacement
b) Intersection of line passing through new centre of buoyancy and centre of gravity.
c) point about which body starts oscillating when it is given small angular displacement
d) All of the mentioned

Answer: d [Reason:] All of the above explanation are apt.

7. The metacentric height is affected by the change in density.
a) True
b) False

Answer: True [Reason:] Metacentre does depend on the density. Hence, the metacentric height is affected by the change in density.

8.For a completely immersed body, the metacentric height is always zero.
a) True
b) False

Answer: b [Reason:] The metacentric height may or may not be zero as metacentre will not always coincide with centre of gravity.

9. Meta centre always lies below the centre of gravity
a) True
b) False

Answer: b [Reason:] It depends on the stability of floating body.

10. The principle of floatation of bodies is based on the premise of
a) Metacentre
b) Newtons first law
c) Newtons law of viscosity
d) None of the mentioned

Answer: a [Reason:] The principle of floatation of bodies is based on the premise of Metacentre.

## Set 3

1. When a uniform flow is flowing through a doublet, resultant flow obtained is
a) Flow past a Rankine oval of equal axes
b) Flow past a circular cylinder
c) All of the mentioned
d) None of the mentioned

Answer: c [Reason:] They both mean the same thing. They are different ways of interpreting the same phenomenon.

2. How many stagnation points are present in a source and sink pair in a uniform flow
a) Two
b) Three
c) One
d) None

Answer: a [Reason:] There are two stagnation points are present in a source and sink pair in a uniform flow.

3. Which of the following is not a type of superimposed flow?
a) Source and sink pair in uniform flow.
b) Double flow
c) A source and sink pair in turbulent flow.
d) A plane source in uniform flow

Answer: c [Reason:] There is no such kind of superimposed flow.

4. In the equation for steam function due to source steam function is inversely proportional to magnitude at discharge
a) True
b) False

Answer: a [Reason:] Steam function is directly proportional to magnitude of discharge.

5. Streamlines of doublet flow are family of circles tangent to a common axis.
a) True
b) False

Answer: a [Reason:] This is special characteristic of doublet flow.

6. What is the characteristic of stagnation point
a) Velocity is zero
b) Acceleration is uniform
c) Velocity is zero
d) Acceleration is zero

Answer: a [Reason:] At stagnation point velocity is zero, as the fluid comes at rest. This characteristic at stagnation point.

7. Potential lines for the source-sink pair will be eccentric non intersecting circles with their centers on the axis.
a) True
b) False

Answer: b [Reason:] The potential line for the source sink pair will be eccentric non intersecting circles with their centers on the axis.

8. What is the term used for a case where source and sink( both of them are of equal magnitude) approach each other such, distance between them reduces and product of distance and discharge magnitude remains constant.
a) A plain pair
b) A doublet source in uniform
c) Hagen Poiseuille flow
d) Coutte flow

Answer: b [Reason:] This is a special case of superimposed flow called double flow.

9. What type of flow is obtained by superimposing two definite flow types{( considering ideal condition)
a) Ideal
b) Potential
c) Both
d) None of the mentioned

Answer: c [Reason:] Any linear combination of two different types of flows will result in a potential and ideal flow.

10. The nature of streamlines in a flow net obtained by the combination of source and sink is-
a) Linear
b) Curvilinear
c) Circular
d) Random

Answer: c [Reason:] The combination of the source and sink would result in a flow net where streamlines will be circular axes.

## Set 4

1. The characteristic of Ideal fluid are
a) Incompressible
b) Inviscid
c) Fluid velocity is uniform
d) Shear stress has a constant, non zero value

Answer : c [Reason:] : As ideal fluid is inviscid, shear stress is zero.

2. Which of the following is not a case of ideal fluid flow?
a) Forced vortex Flow
b) Uniform Flow
c) Sink Flow
d) Superimposed flow

Answer: a [Reason:] Forced vortex Flow does not satisfy the characteristic of ideal fluid flow.

3.What is a special characteristic of uniform flow parallel to X axis?
a) Velocity is constant
b) Acceleration is constant
c) X- component of velocity is constant
d) None of the mentioned

Answer: a [Reason:] Velocity is constant in uniform flow.

4. The source flow is flow coming from a point and moving out in a circular manner.
a) True
b) False

Answer: a [Reason:] The source flow is flow coming from a point and moving out in a radial manner.

5. The sink flow is flow in which fluid moves radially inwards towards a point where it disappears at a variable rate.
a) True
b) False

Answer: b [Reason:] The sink flow is flow in which fluid moves radially inwards towards a point where it disappears at a constant rate.

6. The pattern for streamlines and equipotential lines is different for source and sink flow.
a) True
b) False

Answer: b [Reason:] The pattern for streamlines and equipotential lines is different for source and sink flow.

7. In free vortex flow, the flow is linear in nature.
a) True
b) False

Answer : b [Reason:] : In free vortex flow, the flow is circular in nature.

8. What is the nature of streamlines of free vortex flow?
a) Concentric
b) Non-concentric
c) Linear
d) None of the mentioned

Answer: a [Reason:] The nature of streamlines of free vortex flow is concentric.

9. For source flow, the radial velocity increases as we move radially outward.
a) True
b) False

Answer : b [Reason:] There is an inverse relation between velocity and radial distance for source flow.

10. When is air assumed to be incompressible?
a) At low speed
b) At high speed
c) Independent of its speed
d) None of the mentioned

Answer : a [Reason:] Air is assumed to be incompressible at low speed.

## Set 5

1. Which one of the following is the unit of pressure?
a) N
b) N/m
c) N/m2
d) N/m3

Answer: c [Reason:] Pressure is defined as the force per unit area acting normal to a surface. The SI unit of force is N and area is m2. Thus, the unit of pressure will be N = m2.

2. Which one of the following is the dimension of pressure?
a) [MLT2].
b) [MLT-2].
c) [ML-1T2].
d) [ML-1T-2].

Answer: d [Reason:] Pressure (p) is defined as the force (F) per unit area (A) acting normal to a surface. Thus, 3. Which one of the following statements is true regarding pressure?
a) Pressure is a scalar quantity
b) Pressure is a vector quantity
c) Pressure is a scalar quantity only when the area is infinitesimally small
d) Pressure is a vector quantity only when the area is infinitesimally small

Answer: a [Reason:] Pressure is defined as the force per unit area acting normal to a surface. Both force and area are vectors, but the division of one by the other leads to a scalar quantity.

4. A beaker half-filled with water is exposed to the atmosphere. If the pressure at points A, B and C as shown are Pa, Pb and Pc respectively, which one of the following will be the relation connecting the three? a) Pa > Pb = Pc
b) Pa > Pb > Pc
c) Pa < Pb < Pc
d) Pa < Pb = Pc

Answer: d [Reason:] Since the beaker is exposed to the atmosphere, the pressure at point A will be atmospheric, Pa = 0. Pressure increases in the vertically downward direction, Pa < Pb and Pa < Pc. Pressure remains constant in the horizontal direction, Pb = Pc. Therefore, Pa < Pb = Pc.

5. A beaker is filled with a liquid up to a height h. If A and B are two points, one on the free surface and one at the base as shown, such that the minimum distance between the two is l, what will be the pressure at point B?  Answer: b [Reason:] For a constant density liquid, pressure varies linearly in the vertically downward 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. Since A is at the free surface, PA = 0, PB = ρgh.

6. A beaker of height h is filled with a liquid of density ρ up to a certain limit. The beaker is rotated by an angle θ such that further increase in the angle will result in over flow of the liquid. If the liquid surface is exposed to the atmosphere, what will be the gauge pressure at point B? a) ρgh
b) ρgh sin θ
c) ρgh cos θ
d) ρgh=2

Answer: c [Reason:] Vertical distance below the free surface at which the point B is located will be h cos θ. Since the pressure at the free surface is atmospheric, the gauge pressure at B will be = 0 + ρgh cos θ.

7. An arm of a teapot is completely filled with tea (density=ρ) If the arm has a length of l and is inclined at 30o to the horizontal, what will be the pressure difference between the two points, C at the mouth and D at the base of the arm? a) ρgl
b) ρgl/2
c) √2ρgl
d) 2ρgl

Answer: b [Reason:] Vertical distance difference between the two points, C at the mouth and D at the base of the arm will be l sin θ = l sin 30o = l=2. Thus, pressure difference between C and D is = ρgl/2.

8. A beaker is filled with a liquid of density ρ1 up to a certain height. The pressure at the base of the beaker id Pb. If the liquid is replaced by an equal volume of another liquid of density ρ2, what will be the pressure at the base of the beaker now?  Answer: d [Reason:] PB = ρ1gh, where h=height up to which the liquid is filled. Since equal volume of the second liquid is poured, it’ll also rise to a height of h. Thus, the pressure at the base will become 9. A beaker is filled with a liquid of density ρ1 up to a certain height. A is a point, h m downwards from the free surface of the liquid as shown. The liquid is replaced by equal volume of another liquid of density ρ2. If ρ1 > ρ2, how will the pressure at point A change? a) remain same
b) increase
c) decrease
d) become zero

Answer: c [Reason:] P1= ρ1gh and P2 = ρ2gh, where P1 and P2 are the pressures at point A when liquids of density ρ1 and ρ2 are poured. If ρ1 > ρ2, P1 > P2. Thus the pressure at point A will decrease.

10. A beaker is filled with a liquid of density ρ1 up to a certain height. A is a point, h m downwards from the free surface such that the pressure at A is P. If the liquid is replaced by equal volume of another liquid of density ρ2, at what distance from the free surface will the pressure be P now? Answer: c [Reason:] P = ρ1gh. Let the point inside the liquid where the pressure is P be at a distance of hx from the surface. Thus, P = ρ2ghx. Hence, ρ1 * h = ρ2 * hx, ie, hx = ρ1/ρ2 h.

11. If the pressure at a point is 1m of water, what will be it’s value in terms of m of oil? (Take, the specific gravity of oil to be 0.8)
a) 0.8
b) 1
c) 1.25
d) 2.5

Answer: c [Reason:] Pressure at a point P is equal to ρgh, where ρ is the density and h is the height of the liquid column. Therefore, ρwater * 1 * g = ρoil * h * g, where h is the pressure in terms of m of oil. Thus, h = ρwater / ρoil = 1/0.8 = 1.25.

12. A beaker is filled with a liquid of density ρ up to a height h. If half the liquid is replaced by equal volume of another liquid of twice the density, what will be the change in the base pressure? a) increased by ρgh
b) decreased by ρgh
c) increased by ρgh=2
d) decreased by ρgh=2

Answer: c [Reason:] Base pressure when the beaker is filled with a liquid of density ρ up to a height h = ρgh Base pressure when half the liquid is replaced by equal volume of another liquid of twice the density = ρg h2 + 2ρg h2 = 32 ρgh Thus the change in base pressure is = ρgh / 2. Since, P2 > P1, there will be an increase in pressure.

13. A cuboidal container (each side of 30 cm0) is completely filled with water. A is a point, 25 cm above the base such that the pressure at point A is P. At what height (in cm) from the base will the pressure be 2P? a) 20
b) 15
c) 12.5
d) 10

Answer: a [Reason:] Pressure at a point P is equal to ρgh, where ρ is the density and h is the height of the liquid column from the top. Thus, ρ * g *(30 – h) = 2 * ρ * g *( 30-25), where h from the base where the pressure will be 2P. Thus, h = 30 – 2(30 – 25) = 20.

14. A closed tank (of height 5 m) is PArtially filled with a liquid 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 of the liquid to vary according to the following relation: where y is the height from the base a) 2.12
b) 2.15
c) 2.18
d) 2.5

Answer: c [Reason:] The change of pressure P with vertical direction y is given by  