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Multiple choice question for engineering

Set 1

1. If there is no exchange of heat between system and surrounding where system comprises of a compressible fluid but the heat is generated due to friction, the process is an adiabatic.
a) True
b) False

View Answer

Answer: b [Reason:] For process to be adiabatic, there is no heat exchange and no heat generation within fluid.

2. For a compressible fluid, if there is no change in specific volume at constant temperature, what type of process it is?
a) Isothermal process
b) Adiabatic Process
c) Polytropic process
d) None of the mentioned

View Answer

Answer:a [Reason:] As, specific volume remains constant, density remains constant. Therefore for given temperature there is no change in volume. hence, the process is isothermal.

3. If the fluid is incompressible, do thermodynamic properties play an important role in its behaviour at varying temperature and pressure?
a) Yes
b) No
c) Depends on the fluid
d) None of the mentioned

View Answer

Answer: b [Reason:] If fluid is incompressible there is not much change in observed properties with variation in temperature and pressure. Hence, no perceivable change.

4. If for same temperature and pressure change, the value of bulk modulus is compared for isothermal process and adiabatic process, which one would be higher?
a) Isothermal process
b) Adiabatic process
c) Value is constant for both the processes
d) None of the mentioned

View Answer

Answer: b [Reason:] For isothermal process K=p For adiabatic process K=kp where K=Bulk modulus k=Polytropic constant p=Pressure.

5. The value of gas constant is same for all the gases
a) True
b) False

View Answer

Answer: b [Reason:] The value of gas constant depends on molecular weight. As the molecular weight is different, gas constant will be different.

6. Calculate the pressure exerted by 9 kg of air at a temperature of 20℃ if the volume is 0.8m3. Assuming ideal gas laws are applicable.
a) 946 kN/m2
b) 1892 kN/m2
c) 1419 kN/m2
d) None of the mentioned

View Answer

Answer: a [Reason:] Ideal gas Law: PV=nRT n=M/m P=(9*8314*293)/28.97=946 kN/m2.

7. A gas weighs 16 N/m3 at 30℃ and at an absolute pressure of 0.35 N/mm2. Determine the gas constant.
a) 708.23
b) 354.11
c) 531.17
d) 1062.34

View Answer

Answer:a [Reason:] R=P/(ρ*T)=3500000*9.81/16*303=708.23.

8. A cylinder of 0.8 m3 in volume contains superheated steam at 70℃ and .4 N/m2 absolute pressure. The superheated steam is compressed to .3 . Find pressure and temperature.
a) 0.74 N/m2, 422.3℃
b) 1.48 N/m2, 422.3℃
c) 0.74 N/m2, 844.6℃
d) 1.48 N/m2, 844.6℃

View Answer

Answer: a [Reason:] For polytropic process,

P2=(v1/v2)n *P1 =(0.8/0.3)1.3 * 0.4 ……..(for superheated stream n=1.3) =.74 N/m2 T1=P1v1/nR=422.3℃.

9. Determine the compressibility of an incompressible fluid, if the pressure of the fluid is changed from 70 N/m2 to 130 N/m2. The volume of the liquid changes by 0.15 percent.
a) 0.0025 m2/N
b) 0.0050 m2/N
c) 0.0070 m2/N
d) 0.0012 m2/N

View Answer

Answer :a [Reason:] Compressibility=1/Bulk Modulus =1/K K=(dp*V/dv) =60/0.15 =400 Compressibility=.0025.

10. What is the variation of cp, cv and k in case of gases when the temperature increases?
a) cp and cv decreases with temperature, and k increases
b) cp and cv increase with temperature, and k decreases
c) cp and cv increase with temperature, and k increases
d) cp and cv decreases with temperature, and k decreases

View Answer

Answer:b [Reason:] cp is molar heat capacity at constant pressure. As temperature is increased, enthalpy increases, heat capacity increases. Same is for cv, cp is molar heat capacity at constant volume. However cp-cv=R and cp/cv = R Hence, as cp, cv increases R decreases.

Set 2

1. A solid cylinder of diameter 5.0 m has a height of 6.0 m. Find the meta-centric height of the cylinder if the specific gravity of the material of cylinder 0.45 and it is floating in water with its axis vertical. State whether the equilibrium is stable or unstable.
a) -0.29 m
b) -0.61 m
c) -1.16 m
d) 0.14 m

View Answer

Answer: b [Reason:] BG=Centre of pontoon – Centre of immersed portion=0.3-0.45*0.3=1.65 Metacentric height=I/∀ -BG I=π*r⁴=π*2.5⁴ ∀=π*r*r*h=π*2.5*2.5*6 Metacentric height=-0.61.

2. A solid cylinder of 15 cm diameter and 40 cm long, consists of two parts made of different materials. The first part at the base is 1.5 cm long and of specific gravity=6.5. The other part of the cylinder is made of the material having specific gravity 0.75. State, if the it can float vertically in water.
a) It will float
b) It will not float
c) Data insufficient
d) None of the mentioned

View Answer

Answer: a [Reason:] AG=(weight of base*distance of C.G from base point A) + (weight of upper part*distance of C.G from point A)/ )weight of base + weight of upper part) = 14.52 By principle of buoyancy, Weight of cylinder = Weight of water displaced h=38.625 AB=19.31 BG=14.25-19.31= -4.79 GM= Metacentric height=I/∀ -BG = 6.16 As metacentric height is positive, it will float.

3. A wooden cylinder of sp.gr. = 0.6 and circular in cross-section is required to float in oil(sp.gr. = 0.90). Find the L/D ratio for the cylinder to float with its longitudinal axis vertical in oil, where L is the height of cylinder and D is its diameter.
a) L/D<9/16
b) L/D<3/4
c) L/D<2/3
d) None of the mentioned

View Answer

Answer: b [Reason:] By principle of buoyancy, Weight of cylinder = Weight of water displaced h=2L/3 AG=L/2 AB=L/3 BG=AG-AB=L/6 GM= Metacentric height=I/∀ – BG=3D2/32L-L/6

For stable equilibrium, GM should be positive GM>0 i.e L/D<3/4.

4. A cylinder(uniform density distribution) of radius 3.0 m has a height of 9.0 m. The specific gravity of the material of cylinder 0.85 and it is floating in water with its axis vertical. State whether the equilibrium is stable or unstable.
a) Stable
b) Unstable
c) Insufficient Data
d) None of the mentioned

View Answer

Answer: a [Reason:] BG=Centre of pontoon – Centre of immersed portion=0.3-0.45*0.3=1.65 Metacentric height=I/∀ -BG I=π*r⁴=π*3⁴ ∀=π*r*r*h=π*3*3*9 Metacentric height=0.325.

5. If the magnitude of dimension of a rectangular wooden block is length>breadth>height, then for it to float on the water, it should be immersed in what manner?
a) It should be immersed vertically such that length is partially immersed
b) It should be immersed horizontally such that breadth is partially immersed
c) It should be immersed such that height is partially immersed
d) None of the mentioned

View Answer

Answer: b [Reason:] When it is immersed in such a manner where height is partially immersed, its stability is most as moment of inertia is most about that axis.

6. When body is completely or partially immersed in a fluid, how much its weight be distributed for it to be in stable equilibrium.
a) Around the lower part
b) Around the upper part
c) Is independent of weight distribution
d) None of the mentioned

View Answer

Answer: a [Reason:] When the weight distribution is around the lower part, the centre of gravity is at lower portion and hence below the centre of buoyancy which is condition for stable equilibrium.

7. In unstable equilibrium what is the relation between forces?
a) Buoyancy force= Weight of body
b) Buoyancy force > Weight of body
c) Buoyancy force < Weight of body
d) None of the mentioned

View Answer

Answer: a [Reason:] Fb=W and the the centre of buoyancy is below the centre of gravity.

8. The floating body is said to be in unstable equilibrium if the metacentre is below the centre of gravity.
a) True
b) False

View Answer

Answer: b [Reason:] The floating body is said to be in unstable equilibrium if the metacentre is above the centre of gravity.

9. The floating body is said to be in neutraL equilibrium if the metacentre is above the centre of gravity.
a) True
b) False

View Answer

Answer: b [Reason:] The floating body is said to be in unstable equilibrium if the metacentre coincides with the centre of gravity.

10. In stable equilibrium for completely submerged bodies what is the relation between forces?
a) Buoyancy force= Weight of body,the centre of buoyancy is below the centre of gravity.
b) Buoyancy force=Weight of body, the centre of buoyancy is above the centre of gravity.
c) Buoyancy force < Weight of body
d) None of the mentioned

View Answer

Answer: b [Reason:] Fb=W and the the centre of buoyancy is above the centre of gravity.

Set 3

1. What is the reduction in crest length due to each end contraction?
a) 0.1H
b) 0.2H
c) 0.15H
d) 0.25H

View Answer

Answer: a [Reason:] Francis through his experiment had derived this empirical relation.

2. In Francis formula, the effective length is –
a) L-0.2H
b) L-0.4H
c) L-0.3H
d) L-0.1H

View Answer

Answer: a [Reason:] In Francis formula, the effective length is L-0.2H.

3. In Francis empirical expression for discharge, the relation between head of water and discharge is
a) Q is directly proportional to H
b) Q is directly proportional to H1.5
c) Q is directly proportional to H2.5
d) Q is directly proportional to H0.5

View Answer

Answer: b [Reason:] In Francis empirical expression for discharge, the relation between head of water and discharge is Q is directly proportional to H1.5.

4. In Bazin’s formula, the discharge is inversely proportional to the length of weir.
a) True
b) False

View Answer

Answer: b [Reason:] In Bazin’s formula, the discharge is directly proportional to the length of weir.

5. The head of water over a rectangular weir is 38 cm. The length of the crest of the weir end contraction suppressed is 1.3 m. Find the discharge using the Francis formula.
a) 0.56 m3/s
b) 0.75 m3/s
c) 0.85 m3/s
d) 0.69 m3/s

View Answer

Answer: a [Reason:] Q = 1.84*L*H1.5 = 0.56 m3/s.

6. The head of water over a rectangular weir is 28 cm. The length of the crest of the weir end contraction suppressed is 1.27 m. Find the discharge using the Francis formula.
a) 0.346 m3/s
b) 0.556 m3/s
c) 0.788 m3/s
d) 0.225 m3/s

View Answer

Answer: a [Reason:] Q = m * L * (2*g)0.5 * H1.5 m = ⅔ * Cd = 0.405 + 0.003/H = 0.405 + 0.003/0.28 Q = 0.346 m3/s.

7. The head of water over a rectangular weir is 26 cm. The length of the crest of the weir end contraction suppressed is 1.25 m. Find the discharge using the Francis formula.
a) 0.304 m3/s
b) 0.502 m3/s
c) 0.350 m3/s
d) 0.625 m3/s

View Answer

Answer: a [Reason:] Q = 1.84*L*H1.5 = 0.304 m3/s.

8. The head of water over a rectangular weir is 28 cm. The length of the crest of the weir end contraction suppressed is 1.27 m. Find the discharge using the Francis formula.
a) 0.346 m3/s
b) 0.556 m3/s
c) 0.788 m3/s
d) 0.225 m3/s

View Answer

Answer: a [Reason:] Q = m * L * (2*g)0.5 * H1.5 m = ⅔ * Cd = 0.405 + 0.003/H = 0.405 + 0.003/0.28 Q = 0.346 m3/s.

9. Find the discharge over a cipolletti weir of length 1.5 m when the head over the weir is 0.85 m. Take Cd = 0.61.
a) 2.12 m3/s
b) 1.25 m3/s
c) 2.5 m3/s
d) 1.5 m3/s

View Answer

Answer: a [Reason:] Q = ⅔ * Cd * L * (2g)0.5 * H1.5 = 2.12 m3/s.

10. Find the discharge over a cipolletti weir of length 1.3 m when the head over the weir is 0.65 m. Take Cd = 0.60.
a) 2.12 m3/s
b) 1.21 m3/s
c) 2.5 m3/s
d) 1.5 m3/s

View Answer

Answer: b [Reason:] Q = ⅔ * Cd * L * (2g)0.5 * H1.5 = 1.21 m3/s.

Set 4

1. Find the position of centre of buoyancy for a wooden block of width 3.5 m and depth 1 m, when it floats horizontally in water. The density of wooden block id 850 kg/m3 and its length 7.0 m.
a) 0.95
b) 0.85
c) 1.05
d) 1.65

View Answer

Answer: b [Reason:] Weight of the block=ρ*g*Volume=850*9.81*7*3.5*1=204.29 kN Volume of water displaced= Weight of water displaced/weight density of water = 20.825 m3. h=20.825/3.5*7=0.85 m.

2. A stone weighs 450 N in air and 200 N in water. Compute the volume of stone.
a) .025 m3
b) .05 m3
c) .075 m3
d) None of the mentioned

View Answer

Answer: a [Reason:] Weight of water displaced=Weight of stone in air – Weight of stone in water =250 Volume of water displaced=Volume of stone=250/9.81*1000=0.025 m3.

3. A stone weighs 650 N in air and 275 N in water. Compute its specific gravity.
a) 1.73
b) 2.45
c) 3.46
d) 0.865

View Answer

Answer: a [Reason:] Weight of water displaced=Weight of stone in air – Weight of stone in water =375 Volume of water displaced=Volume of stone=375/9.81*1000=0.038 m3 Density of stone= mass/volume=650/9.81*0.038=1733 kg/m3 specific gravity= Density of stone/Density of water=1.73.

4. A body of dimensions 2.7 m * 3.8 m * 2.5 m, weighs 2500 N in water.Find its weight in air.
a) 254.12 kN
b) 508.25 kN
c) 101.65 kN
d) 127.06 kN

View Answer

Answer: a [Reason:] Weight of stone in air = Weight of water displaced+Weight of stone in water = 9.81*1000*2.7*3.8*2.5+2500=254.12 kN.

5. Find the density of metallic body which floats at the interface of mercury of sp.gr 13.6 and water such that 40 % of its volume is sub-merged in mercury and 60% in water.
a) 6040 kg/m3
b) 12080 kg/m3
c) 24160 kg/m3
d) 3020 kg/m3

View Answer

Answer: a [Reason:] Total Bouyant force=Force of bouyancy due to water+Force of bouyancy due to mercury For equilibrium, Total bouyant force= Weiht of body 1000*9.81*0.6*V + 13.6*1000*9.81*0.4*V=ρ*g*V ρ=6040 kg/m3.

6. What is the principal cause of action of buoyant force on a body submerged partially or fully in fluid?
a) Displacement of fluid due to submerged body
b) Development of force due to dynamic action
c) Internal shear forces mitigating external forces
d) None of the mentioned

View Answer

Answer: a [Reason:] The principal cause of action of buoyant force on a body submerged partially or fully in fluid is the force equal in magnitude to the weight of the volume of displaced fluid.

7. How can relatively denser object be made to float on the less dense fluid?
a) By altering the shape.
b) By altering the forces acting on the object
c) By altering the shear forces acting on the object
d) None of the mentioned

View Answer

Answer: a [Reason:] By changing the shape of an object it can be made to float on a fluid even if it is denser than that fluid. This principle is used in ship building.

8. What happens to the buoyant force acting on the airship as it rises in the air?
a) Buoyant force increases
b) Buoyant force decreases
c) Buoyant force remains constant
d) Buoyant force first increases then shows decrease

View Answer

Answer: b [Reason:] Buoyant force acting on the airship decreases as it rises in the air as air at higher altitude becomes rarer and its density decreases.

9. As a balloon rises in the air its volume increases, at the end it acquires a stable height and cannot rise any further.
a) True
b) False

View Answer

Answer: a [Reason:] As balloon rises in air, pressure acting on it reduces and therefore its volume increases. Also, a rising balloon ceases rising when it and the displaced air are equal in weight.

10. Submarines use principle of ‘neutral buoyancy’ to go into the water.
a) True
b) False

View Answer

Answer: a [Reason:] To dive, the submarine tanks are opened to allow air to exhaust, while the water flows in. When the weight has been balanced so the overall density of the submarine is equal to the water around it, it has neutral buoyancy and hence will go down.

Set 5

1. Find the discharge of water flowing over a rectangular notch of 1.5 m length when the constant head over the notch is 275 mm. Take Cd = .60
a) 400 lit/s
b) 465 lit/s
c) 385 lit/s
d) 575 lit/s

View Answer

Answer: c [Reason:] Q = 2/3 * L * √2g * H1.5 = .67 * 1.5 * √19.62 * .2751.5 = .385 m3/min.

2. The head of water over a rectangular notch is 900 mm. The discharge is 300 litres/s. Find the length of the notch, when CD =.62
a) .192 m
b) .250 m
c) .205 m
d) .175 m

View Answer

Answer: a [Reason:] L = 1.5 * Q / (Cd * √2g * H 1.5) = 1.5 * .3 / (.62 * √19.62 * .91.5) = .192 m.

3. Find the discharge of water flowing over a rectangular notch of 1.3 m length when the constant head over the notch is 255 mm. Take Cd = .62
a) 400 lit/s
b) 465 lit/s
c) 385 lit/s
d) 575 lit/s

View Answer

Answer: a [Reason:] Q = 2/3 * L * √2g * H1.5 = .67 * 1.3 * √19.62 * .2551.5 = .385 m3/min.

4. The head of water over a rectangular notch is 700 mm. The discharge is 200 litres/s. Find the length of the notch, when CD =.63
a) .125 m
b) .265 m
c) .250 m
d) .200 m

View Answer

Answer: a [Reason:] L = 1.5 * Q / (Cd * √2g * H1.5) = 1.5 * .2 / (.62 * √19.62 * .71.5) = .125 m.

5. Find the discharge over triangular notch of angle 50° when the head over the V notch
a) .93 m3/min
b) 1.45 m3/min
c) .88 m3/min
d) .90 m3/min

View Answer

Answer:a [Reason:] Q = 8/15 * √2g * H1.5 * tan(x) Here, x is the angle. = 8/15 * √19.62 * .221.5 * tan(50) = .93 m3/min.

6. The expression for discharge for a right angled notch is more complex than rectangular notch.
a) True
b) False

View Answer

Answer: b [Reason:] : The expression for discharge for a right angled notch is easier than rectangular notch.

7. The results of which are more accurate; rectangular notch or triangular weir.
a) Rectangular notch
b) Triangular weir
c) Both are equally accurate
d) Rectangular weir

View Answer

Answer: b [Reason:] : The results of triangular notch are more accurate for low discharge.

8. What is main reading required in calculation for rectangular notch or weir.
a) H
b) x, x is angle
c) L
d) None of the mentioned

View Answer

Answer: a [Reason:] : H i.e height is main reading required in calculation for rectangular notch or weir.

9. We need to obligatorily have ventilation in triangular notch.
a) True
b) False

View Answer

Answer: b [Reason:] : We need not obligatorily have ventilation in triangular notch.

10. Rectangular notch may or may not have ventilation.
a) True
b) False

View Answer

Answer: b [Reason:] : Rectangular notch must have ventilation.