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# Multiple choice question for engineering

## Set 1

1. If the atmospheric pressure at sea level is 7.5 N/cm2, determine the pressure at a height of 3000m assuming the pressure variation follows isothermal law. The density of air is given as 1.2 km/m3.
a) 4.68 N/cm2
b) 9.37 N/cm2
c) 2.34 N/cm2
d) None of the mentioned

Answer: a [Reason:] pressure at any height Z = p*e-gZ/RT =75000*e -9.81*3000*1.2/75000 = 4.68 N/cm2.

2. The barometric pressure at sea level is 760 mm of Mercury while that on a mountain top is 715 mm. If the density of air is assumed constant at 1.2 kg/m3 , what is the elevation of the mountain top?
a) 510 m
b) 1020 m
c) 255 m
d) 128 m

Answer: a [Reason:] Gauge pressure at any height h = pressure at sea level – pressure at that height h=(9.81*13600*0.76)-9.81*13600*0.715)/1.2*9.81 =510 m.

3. Calculate the pressure at a height of 6500m above the sea level if the atmospheric pressure is 10.145 N/cm2 and temperature is 25℃ assuming air is incompressible. Take density of air as 1.2 kg/m3. Neglect variation of g.
a) 4.98 N/cm2
b) 2.49 N/cm2
c) 1.24 N/cm2
d) None of the mentioned

Answer: b [Reason:] Pressure= p – density of air*g*height =101450-9.81*1.2*6500 = 2.49 N/cm2.

4. Calculate the pressure of air at a height of 3500m from sea level where pressure and temperature of air are 10 N/cm2 and 25℃ respectively. The temperature lapse rate is given as 0.0065 ℃ /m. Take density of air at sea level equal to 1.2 kg/m3.
a) 19.7 N/cm2
b) 9.85 N/cm2
c) 4.93 N/cm2
d) 6.24 N/cm2

Answer: b [Reason:] pressure=p * (1-(k-1/k)*g*h*density/p)k/(k-1) =9.85 N/cm2

Here, Lapse rate= -g/R*(k/k-1).

5. Pressure variation for compressible fluid is maximum for which kind of process?
a) Isothermal
c) Quasi Static
d) None of the mentioned

Answer: a [Reason:] Due to constant temperature, pressure variation for compressible fluid is maximum for isothermal process.

6. Why can’t the density be assumed as constant for compressible fluids?
a) It shows variation with temperature and pressure
b) It remains constant with temperature and pressure
c) It becomes almost constant at very high temperature
d) None of the mentioned

Answer: a [Reason:] Volume and hence density changes with change in temperature and pressure.

7.What is the variation observed in temperature in atmosphere with respect to elevation?
a) It goes on decreasing with height
b) It goes on increasing with height
c) It first increases then decreases
d) It first decreases then increases

Answer: d [Reason:] It goes on decreasing first and shows increase after 32000 m.

8. As we go upwards, at height there is slight decrease in pressure variation.
a) True
b) False

Answer: a [Reason:] There is slight decrease in pressure as value of g (acceleration due to gravity) decreases slightly as we go higher.

9. For dynamic fluid motion in a pipe, the pressure measurement cannot be carried out accurately by manometer.
a) True
b) False

Answer: a [Reason:] For fluid moving with variable velocity, fluctuation in pressure is frequent and more in magnitude. Hence, we cannot use manometer.

10. A simple U tube manometer connected to a pipe in which liquid is flowing with uniform speed will give which kind of pressure?
a) Absolute Pressure
b) Vacuum Pressure
c) Gauge Pressure
d) None of the mentioned

Answer: c [Reason:] A simple U tube manometer will give pressure with respect to atmosphere. Hence, it is gauge pressure.

## Set 2

1. Does total pressure takes into the account force exerted by the fluid when it is in the dynamic motion?
a) Yes
b) No
c) Depends on the conditions
d) Depends on the type of Motion

Answer:b [Reason:] Total pressure is defined only for the static fluid at rest. There is no dynamic component as no motion is involved.

2. Can centre of pressure for a vertical plane submerged surface be ever be above centre of Gravity
a) Yes
b) No
c) It can be above in cases where the surface height is very large
d) None of the mentioned

Answer: b [Reason:] Centre of pressure always lies below the centre of gravity. In certain cases it may coincide but it can never be above the centre of gravity.

3. Which principle is used for calculating the centre of pressure?
a) Principle of momentum
b) Principle of conservation of energy
c) Principle of balancing of momentum
d) None of the mentioned

Answer: c [Reason:] We balance the moment in order to calculate the position of centre of pressure.

4. In a vertically submerged plane surface, pressure at evbery point remains same
a) True
b) False

Answer: b [Reason:] Pressure at every point is different as the depth of different point from is different.

5. The magnitude of total pressure and centre of pressure is independent on the shape of the submerged plane surface.
a) True
b) False

Answer: b [Reason:] For differently shaped surfaces, the area and hence position of centroid will be different. Hence, the magnitude of total pressure and centre of pressure is dependent on the shape of the submerged plane surface.

6. What is the variation of total pressure with depth for any submerged surface if we neglect variation in the density?
a) Linear
b) Parabolic
c) Curvilinear
d) Logarithmic

Answer: a [Reason:] Total pressure is given by, F=w*a*y hence, F ∝ y i.e linear relation.

7. A pipe line which is 6 m in diameter contains a gate valve. The pressure at the centre of the pipe is 25 N/cm2. If the pipe is filled with specific gravity 0.8, find the force exerted by the oil upon the gate.
a) 7.06 MN
b) 14.12 MN
c) 3.53 MN
d) 28.24 MN

Answer: a [Reason:] ĥ=p/ρg =250000/9.81*800 =31.855 m F=wAĥ =9.81*800*9*π*31.855 =7.06 MN.

8. Determine the centre of pressure on an isosceles triangle plate of base 6m and altitude 6m when it is immersed vertically in an oil of specific gravity 0.75. The base of the plate coincides with the free surface of oil.
a) 6 m
b) 3 m
c) 9 m
d) 12 m

Answer: b [Reason:] ŷ=I/Aĥ + ĥ I=bh³/36 ŷ=36*2/18 + 2 =3 m.

9. A tank contains water upto a height of 0.5 m above the base. An immiscible liquid of specific gravity 0.75 is filled on the top of water upto 1.5 m height. Calculate total pressure on side of the tank.
a) 17780.61 N/m2
b) 35561.22 N/m2
c) 71122.44 N/m2
d) 8890.31 N/m2

Answer: a [Reason:] F = F1 + F2 + F3 F=w*A*ŷ F= 8277.18+5518.12+3985.31 = 17780.61 N/m2.

10. A circular opening, 6m diameter, in a vertical side of a tank is closed by a disc of 6m diameter which can rotate about a horizontal diameter. Calculate the force on the disc. The centre of circular opening is at the depth of 5 m.
a) 1.38 MN
b) 2.76 MN
c) 5.54 MN
d) 7.85 MN

Answer: a [Reason:] F=w*A*ŷ =9.81*1000*3.142*32*5 =1.38 MN.

## Set 3

1. Which of the following contribute to the reason behind the origin of surface tension?
a) only cohesive forces
c) neither cohesive forces nor adhesive forces
d) both cohesive forces and adhesive forces

Answer: d [Reason:] The molecules on the surface of a liquid experience cohesive forces due to surrounding liquid molecules acting downward and adhesive forces due to surrounding gaseous molecules acting upwards. Surface tension orginates due to this unbalanced force on the surface molecules.

2. A soap film is trapped between a frame and a wire of length 10 cm as shown.

If the surface tension is given as 0.0049 N/m, what will be the value of m (in mg) such that the wire remains in equilibrium?
a) 0.1
b) 1
c) 10
d) 100

Answer: d [Reason:] For the wire to be in equilibrium, Force exerted by the film on the wire due to surface tension (acting upwards) must be equal to the downward force due to the weight of the wire (acting downwards). If σ=surface tension, l=length of the wire 2σl = mg Substituting all the values, m = 2σl/g = 2 * 0.0049 * 0.01 ⁄ 9.81 = 99.9mg.

3. What will be the diameter (in mm) of a water droplet, the pressure inside which is 0.05 N/cm2 greater than the outside pressure? (Take surface tension as 0.075 N/m)
a) 3
b) 0.3
c) 0.6
d) 6

Answer: c [Reason:] p = 4σ/d where p = pressure difference between the liquid droplet and the surrounding medium, σ = surface tension and d = diameter of the droplet. Substituting all the values,

4. A soap bubble of d mm diameter is observed inside a bucket of water. If the pressure inside the bubble is 0.075 N/cm2, what will be the value of d? (Take surface tension as 0.075 N/m)
a) 0.4
b) 0.8
c) 1.6
d) 4

Answer: b [Reason:] p = 8σ/d where p = pressure difference between the bubble and the surrounding medium, σ = surface tension and d = diameter of the bubble. Substituting all the values,

5. A liquid jet of 5 cm diameter has a pressure difference of N/m2. (Take surface tension as 0.075 N/m)
a) 12
b) 6
c) 3
d) 1.5

Answer: d [Reason:] p = σ/d where p = pressure difference between the bubble and the surrounding medium, σ = surface tension and d = diameter of the bubble. Substituting all the values, p = 0.075 / 5 * 10-2 = 1.5 N/m2.

6. The rise in the level of a liquid in a tube is h. What will be the rise in the level if the same amount of liquid is poured into a tube of half the diameter.
a) 0
b) h/2
c) h
d) 2h

Answer: d [Reason:] where h = rise in liquid height in the tube, S = surface tension, θ = the angle of contact, d = diameter of the tube, ρ = density of liquid and g = acceleration due to gravity. All other factors remaining constant, h α d. Thus, if d is halved, h will be doubled.

7. The ratio of the surface tension S and density ρ of liquid 1 and 2 are 1:2 and 1:4 respectively. Equal amount of the two liquids is poured into two identical tubes. what will be the ratio of the rise in the liquid level in the two tubes? (Assume the angle of contact to be same)
a) 1:2
b) 2:1
c) 8:1
d) 1:8

Answer: b [Reason:] where h = rise in liquid height in the tube, S = surface tension, θ = the angle of contact, d = diameter of the tube, ρ = density of liquid and g = acceleration due to gravity. Given, S1 / ρ1 = 1 : 2 and S2 / ρ2 = 1 : 4.

8. The rise in the level of a liquid in a tube is h. If half the amount is poured outside, what will be the new rise in liquid level?
a) 0
b) h/2
c) h
d) 2h

Answer: c [Reason:] The rise in liquid level for a liquid is independent of the amount of liquid present in the tube. Since, same tube is used and same liquid is considered, the rise in the liquid level will remain the same.

9. If a glass tube of 10 mm diameter is immersed in water, what will be the rise or fall in capillary?
(Take surface tension = 0.075 N/m, g = 10 m/s2 and angle of contact = 0)
a) 0.75
b) 1.5
c) 3
d) 6

Answer: c [Reason:] where h = rise in liquid height in the tube, S = surface tension, θ = the angle of contact, d = diameter of the tube, ρ = density of liquid and g = acceleration due to gravity. Substituting all the values,

10. A water drop of diameter 1 cm breaks into 1000 similar droplets of same diameter. What will be the gain or loss in the surface energy? (Take surface tension as 0.075 N/m)
a) gain of 0.424 mJ
b) gain of 0.212 mJ
c) loss of 0.212 mJ
d) loss of 0.424 mJ

Answer: b [Reason:] According to the Principle of Conservation of mass, M = 1000 * m, where M = mass of the big drop, m = mass of each droplet. Assuming density to be constant, D3 = 1000 * d3, i.e. D = 10d, where D = diameter of big drop, d = diameter of a droplet. Change in surface energy = Surface tension * Change in surface area = 0:075*(1000 * πd2 – πD2) = 0:075 * (10 * πD2 – πD2) = 0:075 * 9π * (10-2)2 = 0:212 mJ Since, the change is positive, there will be a gain in the surface energy.

## Set 4

1. Calculate the magnitude of capillary effect in millimeters in a glass tube of 7mm diameter, when immersed in mercury. The temperature of the liquid is 25℃ and the values of surface tension of mercury at 25℃ is 0.51 N/m. The angle of contact for mercury is 130°.
a) 140
b) 280
c) 170
d) 210

Answer: a [Reason:] Capillarity rise or fall h=4*cosθ*σ/ρ*g*d =4*cos130*0.51/13600*9.81*0.007 =140 mm.

2. Determine the minimum size of glass tube that can be used to measure water level if the capillary rise in the tube is restricted to 5mm. Consider surface tension of water in contact with air as 0.073 N/m
a) 5.95mm
b) 11.9mm
c) 2.97mm
d) 4.46mm

Answer: a [Reason:] d=4*cosθ*σ/ρ*g*h =4*1*0.073/1000*9.81*0.005 =5.95mm.

3. An oil of vicosity 7 poise is used for lubrication between shaft and sleeve. The diameter of shaft is 0.6 m and it rotates is 360 rpm. Calculate the power lost in oil for a sleeve length of 160mm. The thickness of oil film is 1.0mm
a) 25.31 kW
b) 50.62 kW
c) 37.97 kW
d) 12.65 kW

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= 7916.8*3.142*0.3*0.3*0.3*2*3.142*60 = 25.31 kW.

4. Find the capillarity rise or fall if a capillary tube of diameter .03m is immersed in hypothetical fluid with specific gravity 6.5, surface tension 0.25 N/m and angle of contact 147°.
a) 0.44mm fall
b) 0.88mm fall
c) 0.44mm rise
d) 0.88mm rise

Answer: a [Reason:] h=4*cosθ*σ/ρ*g*d =4*cos147*0.25/6.5*1000*9.81*0.03 =-0.44 mm i.e 0.44 mm fall.

5. Will capillary rise occur and if it occurs what will be capillary rise if glass capillarity tube is immersed in water and experiment is carried out by astronauts in space.
a) Capillarity rise will not occur
b) Capillarity rise will occur infinitely and will come out in form of fountain
c) Capillarity rise will occur finitely and will be the whole length of tube
d) None of the mentioned

Answer: c [Reason:] Capillary rise is given by h=4*cosθ*σ/ρ*g*d hence rise is inversely proportional to g In space g is 0 m/s2 Hence, capillarity rise will occur finitely and will be the whole length of tube.

6. The surface tension of fluid in contact with air at 25℃ is 0.51N/m. The pressure inside a droplet is to be 0.05 N/cm2 greater than outside pressure. Determine the diameter of the droplet of water.
a) 4.08mm
b) 8.16mm
c) 2.04mm
d) None of the mentioned

Answer: a [Reason:] P=4*σ/d d= 4*.51/500 =4.08 mm.

7. If a fluid of certain surface tension and diameter is used to create a soap bubble and a liquid jet. Which of the two, bubble or liquid jet, will have greater pressure difference on the inside and outside.
a) Liquid jet
b) Soap bubble
c) Both will have same pressure differrence
d) None of the mentioned

Answer: b [Reason:] For soap bubble, P=8*σ/d For liquid jet, P=2*σ/d Hence, soap bubble will be having more pressure difference.

8. Capillarity fall is reduced if we take the appartus (capillary tube immersed in fluid having acute angle of contact) considerable distance inside the earth( i.e below the earth crust).
a) True
b) False

Answer: a [Reason:] Capillary rise is given by h=4*cosθ*σ/ρ*g*d Inside the earth, g (acceleration due to gravity) decreases. Hence, capillary rise will increase compared to that on the earth’s surface.

9. For liquid fluids will capillarity rise (or fall) increase or decrease with rise in temperature.
a) Increase
b) Decrease
c) Remain constant
d) First decrease then increase

Answer: b [Reason:] Capillary rise is given by h=4*cosθ*σ/ρ*g*d As temperature increases, σ(surface tension) decreases. Therefore, correspondingly rise(or fall) will decrease as their is direct proportional relation between the two.

10. Cavitation is more pronounced in rough pipes than smooth surfaced pipes.
a) True
b) False

Answer: a [Reason:] Rough surfaced pipes have more friction with the fluid and hence possibility of cavitation is more pronounced.

## Set 5

1. The greatest and the least depth of a circular plate of 4 m diameter from the free surface of water are 3m and 1 m respectively as shown. What will be the total pressure in (kN) on the plate?

a) 123
b) 185
c) 246
d) 308

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 = 1 + 3 – 1 / 2 – 2m, A = π4 * 42 = 4π m2. Hence, F = 246.55 kN.

2. The greatest and the least depth of a circular plate of 4 m diameter from the free surface of water are 3m and 1 m respectively as shown. What will be the depth (in m) of it’s centre of pressure?

a) 1.125
b) 1.25
c) 2.125
d) 2.25

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, y = 1 + 3 – 1 / 2 = 2, I = π64 * 42 = 4π, A = π4 * 42 = 4π, sin θ = 12 Thus, yCP = 2.125m.

3. The highest and lowest vertices of a diagonal of a square lamina (each side equal to 4m) are 1 m and 3 m respectively as shown. What will be the water force (in kN) on the lamina?

a) 78
b) 118
c) 157
d) 196

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 = 1 + 3 – 1 / 2 = 2m, each side of the lamina = Hence, F = 156:96 kN.

4. The highest and lowest vertices of a diagonal of a square lamina (each side equal to 4m) are 1 m and 3 m respectively as shown. What will be the depth (in m) of it’s centre of pressure?

a) 1.08
b) 1.58
c) 2.08
d) 2.58

Answer: c [Reason:] where I = the moment of inertia and A= area and θ = the angle of inclination of the lamina to the horizontal. Now, y = 1 + 3 – 1 / 2 = 2m, each side of the lamina = =8; sin θ = 3-1/4 = 12. Thus, yCP = 2.08m.

5. 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 water surface. If the pressure on the surface is 12 bar, what will be the total water pressure (in kN) on the lamina?

a) 39
b) 59
c) 64
d) 71

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; A = 2 * 2 = 4 m2. Hence, F = 63.65 kN.

6. A container is filled with two liquids of densities ρ1 and ρ2 up to heights h1 and h2 respectively. What will be the hydrostatic force (in kN) per unit width of the lower face AB?

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,

7. A container is filled with two liquids of densities ρ and 2ρ up to heights h and eh respectively. What will be the ratio of the total pressure on the lower face AB and on the upper face BC?

a) 1 : 1
b) 3 : 1
c) 2 : 1
d) 3 : 2

Answer: b [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.

8. A container is filled with two liquids of densities ρ and 2ρ up to heights h and eh respectively. What will be the ratio of the depths of the centres of pressure of the upper face BC and the lower face AB?

a) 1 : 2
b) 3 : 4
c) 2 : 3
d) 3 : 2

9. A gate of length 5 m is hinged at A as shown to support a water column of height 2.5 m. What should be the minimum mass per unit width of the gate to keep it closed?

a) 3608
b) 4811
c) 7217
d) 9622

Answer: d [Reason:] To keep the gate closed, moment due to weight of the gate should be balanced by the moment due to the hydrostatic force. where m = mass of the plate, θ = angle of inclination to the horizontal, Fhyd = hydrostatic force on the plate, x = distance of the point of action of Fhyd from the hinge point = 23 * 5 = 103 Fhyd = γyA, where γ = specific weight of the liquid = 9.81 * 103 y = depth of the centre of pressure from the free surface = 2.5/2 = 1.25 and A = 5 * 1. Substituting all the values in the equation, we get m = 9622.5g.

10. A large tank is filled with three liquids of densities ρ1, ρ2 and ρ3 up to heights of h1, h2 and h3 respectively. What will be the expression for the instantaneous velocity of discharge through a small opening at the base of the tank? (assume that the diameter of the opening is negligible compared to the height of the liquid column)

Answer: d [Reason:] Instantaneous velocity of discharge where h= height of the liquid column.

11. A large tank of height h is filled with a liquid of density ρ. A similar tank is half-filled with this liquid and other-halffilled with another liquid of density 2ρ as shown. What will be the ratio of the instantaneous velocities of discharge through a small opening at the base of the tanks? (assume that the diameter of the opening is negligible compared to the height of the liquid column in either of the tanks)

a) 2 : 3
b) 2 : √3
c) √2 : 3
d) √2 : √3

Answer: d [Reason:] Instantaneous velocity of discharge where h= height of the liquid column.

12. A large tank of height h is half-filled with a liquid of density ρ and other half-filled with a liquid of density 4ρ. A similar tank is half-filled with a liquid of density 2ρ and other-half filled with another liquid of density 3ρ as shown. What will be the ratio of the instantaneous velocities of discharge through a small opening at the base of the tanks? (assume that the diameter of the opening is negligible compared to the height of the liquid column in either of the tanks)

a) 1 : 1
b) 1 : 2
c) 2 : 1
d) 1 : 3

Answer: a [Reason:] Instantaneous velocity of discharge where h= height of the liquid column.

13. A large tank is filled with three liquids of densities ρ, 2ρ and 3ρ up to a height of h3 each. What will be the expression for the instantaneous velocity of discharge through a small opening at the base of the tank? (assume that the diameter of the opening is negligible compared to the height of the liquid column)