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Interview MCQ Set 1

1. Among the following, production through which material is not possible in a crucible?
a) Glass
b) Metal
c) Pigment
d) Oils

View Answer

Answer: d [Reason:] Crucibles are known for operating under high temperature conditions. These crucibles can be employed for the production of glass, metal and pigment, but not oils.

2. Which of the following metals or alloys cannot be used for making crucibles?
a) Nickel
b) Zirconium
c) Mercury
d) Platinum

View Answer

Answer: c [Reason:] For the making of crucibles, nickel and zirconium have been employed in the recent times. Platinum had been used for this purpose in the earlier times, owing to its inertness, but mercury cannot be used for making it.

3. The crucible process is used for production of which material?
a) Pig iron
b) Cast iron
c) Tool steel
d) Gray iron

View Answer

Answer: c [Reason:] The crucible process is used for the production of tool steel. Hammered iron which had rich carbon content in it was first used for producing tool steel.

4. Which of the following is not a property of a crucible?
a) Retention of thermal conductivity
b) Resistance to chemical attack
c) Retention of electrical conductivity
d) Resistance to erosion

View Answer

Answer: c [Reason:] It is very necessary for the crucible to have retention of thermal conductivity, to possess resistance to chemical attacks and to erosion. But, there is no need for retention of electrical conductivity.

5. What is the maximum pressure which the isostatic presses can take for making crucibles?
a) 890 atm
b) 1270 atm
c) 1000 atm
d) 1100 atm

View Answer

Answer: c [Reason:] Isostatic presses are one of the most advanced methods of making a crucible. For forming, these presses can work up to a pressure of 1000 atmospheres.

6. Which of the following does not hold valid for an isostatic press?
a) Consistent mixture
b) High pressure operation
c) High density of product
d) Wet mix used

View Answer

Answer: d [Reason:] An isostatic press is considered to be one of the most advanced methods of making a crucible. The entire operation is at high pressure conditions. The mixture used is consistent, because of which the product formed has high density. But the mixture that is taken, is completely dry.

7. Which of the following cannot be counted among the advantages of a crucible?
a) Uniform heating
b) Low flexibility
c) Low installation cost
d) No contamination of charge

View Answer

Answer: b [Reason:] Melting various materials through crucibles is an old process. This process has the advantage of uniform heating and no contamination of charge by the product obtained from combustion. The operation is inexpensive, and the flexibility too is high.

8. Production of crucibles is inexpensive.
a) True
b) False

View Answer

Answer: a [Reason:] Production of crucibles is a very inexpensive process. It does not include much capital and the installation costs are also not very high.

9. Which of the following is not a raw material for making a crucible?
a) Clay
b) Wax
c) Graphite
d) Silicon carbide

View Answer

Answer: b [Reason:] For making a crucible, various types of clays are used, carbon in the form of graphite is used, since, graphite has higher melting point and also silicon carbide is used, but not wax.

10. Crucibles do not possess refractoriness.
a) True
b) False

View Answer

Answer: b [Reason:] Crucibles work at conditions of very high temperature, hence it is extremely important for the crucible to possess the ability to withstand high temperature, that is, refractoriness.

Interview MCQ Set 2

1. In cycloidal gears contact area is
a) Comparatively smaller
b) Comparatively larger
c) Can’t be determined
d) None of the listed

View Answer

Answer: b [Reason:] Convex flank on one tooth meets with concave on the other thus increasing the contact area.

2. Involute gears have greater contact area as compared to cycloidal gears.
a) True
b) False

View Answer

Answer: b [Reason:] There is mating of two convex surfaces and hence lesser contact area.

3. Cycloidal teeth consist of
a) Hypocycloid curve
b) Epicycloid gear
c) Both hypocycloid curve and epicycloid curve
d) None of the mentioned

View Answer

Answer: c [Reason:] It consist of both and thus are hard to manufacture.

4. Pressure angle remains constant in case of involute profile.
a) True
b) False

View Answer

Answer: a [Reason:] The common normal always passes through the pitch point and thus maintain the constant inclination.

5. Pressure angle is _____ in case of cycloidal teeth.
a) Constant
b) Variable
c) zero
d) None of the listed

View Answer

Answer: b [Reason:] Cycloidal teeth consist of two profiles.

6. Velocity ratio is the ratio angular velocity of driving gear to that of driven gear.
a) True
b) False

View Answer

Answer: a [Reason:] Velocity ratio is simply the angular velocities ratio.

7. Velocity ratio and transmission ratio are the same thing.
a) True
b) False

View Answer

Answer: b [Reason:] Transmission ratio is measured between first and last gear.

8. Contact ratio is always
a) =1
b) >1
c) <1
d) Can’t be determined

View Answer

Answer: b [Reason:] Some overlapping is essential for continuous transfer of power.

9. Product of diametric pitch and circular pitch is?
a) π
b) 1/π
c) None of the listed
d) 2

View Answer

Answer: a [Reason:] CP=πd/z and circular pitch=z/d.

10. Diameteral pitch is 5, then calculate module of the gear.
a) 0.2
b) 0.4
c) 5
d) 10

View Answer

Answer: a [Reason:] Module is the inverse of diameteral pitch.

Interview MCQ Set 3

1. Superheated R-134a at 0.5 MPa, 20°C is cooled in a piston-cylinder at constant temperature to a final two-phase state with quality of 50%. The refrigerant mass is 5 kg, and during the process 500 kJ of heat is removed. Find the necessary work.
a) -67.9 kJ
b) -77.9 kJ
c) -87.9 kJ
d) -97.9 kJ

View Answer

Answer: c [Reason:] Energy Eq.: m(u2 -u1) = 1Q2 – 1W2 = -500 – 1W2 State 1: T1,P1, v1 = 0.04226 m3/kg ; u1 = 390.52 kJ/kg => V1 = mv1 = 0.211 m3 State 2: T2 , x2 ⇒ u2 = 227.03 + 0.5 × 162.16 = 308.11 kJ/kg, v2 = 0.000817 + 0.5 × 0.03524 = 0.018437 m3/kg => V2 = m(v2) = 0.0922 m3 work = -500 – m(u2 – u1) = -500 – 5 × (308.11 – 390.52) = -87.9 kJ.

2. Air at 600 K flows with 3 kg/s into a heat exchanger and out at 100°C. How much (kg/s) water coming in at 100 kPa, 20°C can the air heat to the boiling point?
a) 0.37 kg/s
b) 0.17 kg/s
c) 0.27 kg/s
d) 0.57 kg/s

View Answer

Answer: c [Reason:] C.V. :Heat Exchanger, No external heat transfer and no work. Writing the Steady State Energy Equation (SSEE) and putting values, we get the water flow rate at the exit is 0.27 kg/s.

3. Nitrogen gas flows into a convergent nozzle at 200 kPa, 400 K and very low velocity. It flows out of the nozzle at 100 kPa, 330 K. If the nozzle is insulated find the exit velocity.
a) 681.94 m/s
b) 581.94 m/s
c) 481.94 m/s
d) none of the mentioned

View Answer

Answer: d [Reason:] C.V.: Nozzle; steady state; one inlet and exit flow; insulated so it is adiabatic. SSEE: h1 + 0 = h2 + [(V2)2] / 2 [(V2)2] = 2(h1 – h2) = 2Cp(T1 – T2) = 2 × 1.042 (400 – 330) = 145.88 kJ/kg = 145 880 J/kg V2 = 381.94 m/s.

4. A steam turbine has an inlet of 2 kg/s water at 1000 kPa, 350°C and velocity of 15 m/s. The exit is at 100 kPa, x = 1 and very low velocity. Find the specific work.
a) 382.3 kJ/kg
b) 482.3 kJ/kg
c) 582.3 kJ/kg
d) 682.3 kJ/kg

View Answer

Answer: b [Reason:] SSEE is W/m = (h1 – h2) + [(V1)2 – (V2)2]/2 + g(z1 – z2) here z1=z2 and V2=0 hence w = (h1 – h2) + [(V1)2]/2 h1 = 3157.65 kJ/kg, h2 = 2675.46 kJ/kg wT = 3157.65 – 2675.46 + ½ (152/1000) = 482.3 kJ/kg.

5. A steam turbine has an inlet of 2 kg/s water at 1000 kPa, 350°C and velocity of 15 m/s. The exit is at 100 kPa, x = 1 and very low velocity. Find the power produced.
a) 664.6 kW
b) 764.6 kW
c) 864.6 kW
d) 964.6 kW

View Answer

Answer: d [Reason:] SSEE is W/m = (h1 – h2) + [(V1)2 – (V2)2]/2 + g(z1 – z2) here z1=z2 and V2=0 hence w = (h1 – h2) + [(V1)2]/2 h1 = 3157.65 kJ/kg, h2 = 2675.46 kJ/kg wT = 3157.65 – 2675.46 + ½ (152/1000) = 482.3 kJ/kg thus power produced = (2 kg/s)(482.3 kJ/kg) = 964.6 kW.

6. 10kg of water in a piston-cylinder exists as saturated liquid/vapour at 100 kPa, with a quality of 50%. It is now heated till the volume triples. The mass of piston is such that a cylinder pressure of 200 kPa will float it. Find the heat transfer in the process.
a) 23961 kJ
b) 24961 kJ
c) 25961 kJ
d) 26961 kJ

View Answer

Answer: c [Reason:] m(u2 − u1) = 1Q2 − 1W2 Process: v = constant until P = Plift , then P is constant. State 1: Two-phase; u1 = 417.33 + 0.5 × 2088.72 = 1461.7 kJ/kg and v1 = 0.001043 + 0.5 × 1.69296 = 0.8475 m3/kg State 2: v2, P2 ≤ Plift => v2 = 3 × 0.8475 = 2.5425 m3/kg ; Interpolate: T2 = 829°C, u2 = 3718.76 kJ/kg => V2 = mv2 = 25.425 m3 1W2 = P(lift)(V2 −V1) = 200 × 10 (2.5425 − 0.8475) = 3390 kJ 1Q2 = m(u2 − u1) + 1W2 = 10×(3718.76 − 1461.7) + 3390 = 25961 kJ.

7. A 1L capsule of water at 150°C, 700 kPa is placed in a larger insulated (otherwise evacuated) vessel. The capsule breaks resulting which its contents fill the entire volume. If the final pressure is not to exceed 125 kPa, find the vessel volume?
a) 115 L
b) 125 L
c) 135 L
d) 145 L

View Answer

Answer: a [Reason:] m2 = m1 = m = V/v1 = 0.916 kg Process: expansion with 1Q2 = 0, 1W2 = 0 Energy: m(u2 – u1) = 1Q2 – 1W2 = 0 ⇒ u2 = u1 State 1: v1 = vf = 0.001091 m3/kg; u1 = uf = 631.66 kJ/kg State 2: P2 , u2 ⇒ x2 =(631.66 – 444.16)/2069.3 = 0.09061 v2 = 0.001048 + 0.09061 × 1.37385 = 0.1255 m3/kg V2 = m(v2) = 0.916 × 0.1255 = 0.115 m3 = 115 L.

8. A vertical cylinder fitted with a piston contains 5 kg of R-22 at 10°C. Heat is transferred causing the piston to rise until the volume has doubled. Additional heat is transferred until the temperature inside reaches 50°C, at which point the pressure inside the cylinder is 1.3 MPa. Find the work done.
a) 34.1 kJ
b) 44.1 kJ
c) 54.1 kJ
d) 64.1 kJ

View Answer

Answer: a [Reason:] Process: 1 -> 2 -> 3 As piston floats, pressure is constant (1 -> 2) and the volume is constant for the second part (2 -> 3). So we have: v3 = v2 = 2 × v1 State 3: (P,T) v3 = 0.02015 m3/kg, u3 = 248.4 kJ/kg v1 = 0.010075 = 0.0008 + x1 × 0.03391 => x1 = 0.2735 u1 = 55.92 + 0.2735 × 173.87 = 103.5 kJ/kg State 2: v2 = 0.02015 m3/kg, P2 = P1 = 681 kPa this is still 2-phase Work = P1(V2 – V1) = 681 × 5 (0.02 – 0.01) = 34.1 kJ.

9. A 250L rigid tank contains methane at 1500 kPa, 500 K. It is now cooled down to 300K. Find the heat transfer.
a) –402.4 kJ
b) –502.4 kJ
c) –602.4 kJ
d) –702.4 kJ

View Answer

Answer: b [Reason:] Assume ideal gas, P2 = P1 × (Τ2 / Τ1) = 1500 × 300 / 500 = 900 kPa m = P1V/RT1 =(1500 × 0.25)/(0.5183 × 500) = 1.447 kg u2 – u1 = Cv (T2 – T1) = 1.736 (300 – 500) = –347.2 kJ/kg 1Q2 = m(u2 – u1) = 1.447(-347.2) = –502.4 kJ.

10. A rigid container has 2kg of carbon dioxide gas at 1200 K, 100 kPa that is heated to 1400 K. Find the heat transfer using heat capacity.
a) 231.2 kJ
b) 241.2 kJ
c) 251.2 kJ
d) 261.2 kJ

View Answer

Answer: d [Reason:] Energy Eq.: U2 – U1 = m (u2- u1) = 1Q2 − 1W2 Process: ∆V = 0 ⇒ 1W2 = 0 For constant heat capacity we have: u2- u1 = Cv (T2- T1) 1Q2 = mCv (T2- T1) = 2 × 0.653 × (1400 –1200) = 261.2 kJ.

11. A piston cylinder contains 3kg of air at 20°C and 300 kPa. It is now heated up in a constant pressure process to 600 K. Find the heat transfer.
a) 941 kJ
b) 951 kJ
c) 961 kJ
d) 971 kJ

View Answer

Answer: a [Reason:] Ideal gas PV = mRT P2V2 = mRT2; V2 = mR T2 / P2 = 3×0.287×600 / 300 = 1.722 m3 Process: P = constant, work = ⌠ PdV = P (V2 – V1) = 300 (1.722 – 0.8413) = 264.2 kJ Energy equation: U2 – U1 = 1Q2 – 1W2 = m(u2 – u1) Q2 = U2 – U1 + 1W2 = 3(435.097 – 209.45) + 264.2 = 941 kJ.

Interview MCQ Set 4

1. What do we get on equating the first and second TdS equations?
a) Cp-Cv = T*(∂T/∂p)*(∂V/∂T)
b) Cp-Cv = T*(∂p/∂T)*(∂V/∂T)
c) Cp+Cv = T*(∂p/∂T)*(∂V/∂T)
d) none of the mentioned

View Answer

Answer: b [Reason:] This is the relation we get on equating first and second TdS equations.

2. Consider the equation Cp-Cv = -T*(∂p/∂V)(∂V/∂T)2 , which of the following is correct?
a) (∂V/∂T)2 is always positive
b) (∂p/∂V) for any substance is negative
c) (Cp-Cv) is always positive
d) all of the mentioned

View Answer

Answer: d [Reason:] From this we can conclude that, Cp is always greater than Cv.

3. When do we have the condition Cp=Cv?
a) as T approaches 0K, Cp tends to approach Cv
b) when (∂V/∂T)=0, Cp=Cv
c) both of the mentioned are correct
d) none of the mentioned are correct

View Answer

Answer: c [Reason:] These facts come from the equation Cp-Cv = -T*(∂p/∂V)(∂V/∂T)2.

4. For an ideal gas,
a) Cp-Cv = R
b) Cp-Cv = mR
c) Cp=Cv
d) all of the mentioned

View Answer

Answer: b [Reason:] This comes from the ideal gas equation, pV=mRT.

5. The volume expansivity and isothermal compressibility is defined as
a) volume expansivity = (1/V)*(∂V/∂T) at p and isothermal compressibility = (-1/V)*(∂V/∂T) at T
b) volume expansivity = (1/V)*(∂V/∂T) at p and isothermal compressibility = (-1/V)*(∂V/∂T) at T
c) volume expansivity = (1/V)*(∂V/∂T) at p and isothermal compressibility = (-1/V)*(∂V/∂T) at T
d) none of the mentioned

View Answer

Answer: a [Reason:] These two terms are used for better representation of the original equation.

6. The equation Cp-Cv = -T*(∂p/∂V)(∂V/∂T)2 can also be expressed as
a) Cp-Cv = T*V*(isothermal compressibility)2 / (volume expansivity)
b) Cp-Cv = T*V*(isothermal compressibility) / (volume expansivity)
c) Cp-Cv = T*V*(volume expansivity)2 / (isothermal compressibility)
d) Cp-Cv = T*V*(volume expansivity) / (isothermal compressibility)

View Answer

Answer: c [Reason:] This comes from the equation Cp-Cv = -T*(∂p/∂V)(∂V/∂T)2 when we use volume expansivity and isothermal compressibility in it.

7. At constant entropy, the two TdS equations give us the relation
a) Cp+Cv = 0
b) Cp=Cv
c) Cp-Cv = mR
d) Cp/Cv = ɣ

View Answer

Answer: d [Reason:] This relation is obtained on dividing the two TdS equations.

8. The slope of an isentrope is ____ the slope of an isotherm on p-v diagram.
a) less than
b) greater than
c) equal to
d) less than or equal to

View Answer

Answer: b [Reason:] This comes from the fact that ɣ>1.

9. Work done in reversible and isothermal compression is ____ the work done in reversible and adiabatic compression.
a) equal to
b) greater than
c) less than
d) less than or equal to

View Answer

Answer: c [Reason:] We get this from the p-v diagram for compression work in different reversible processes.

10. Isothermal compression requires minimum work.
a) true
b) false

View Answer

Answer: a [Reason:] This is because work in isothermal is less than the work in adiabatic process and that of polytropic process lies in between these values.

11. Which of the following relation gives ɣ .
a) 1/(isothermal compressibility *adiabatic compressibility)
b) isothermal compressibility * adiabatic compressibility
c) adiabatic compressibility / isothermal compressibility
d) isothermal compressibility / adiabatic compressibility

View Answer

Answer: d [Reason:] The adiabatic compressibility is defined as (-1/V)*(∂V/∂p).

Interview MCQ Set 5

1. Among spur gear and helical gear, which has smooth engagement and thus lesser noise?
a) Helical Gears
b) Spur Gears
c) Both have equal noises
d) Can’t be determined

View Answer

Answer: a [Reason:] There is a gradual pick up of load in helical gears and hence smooth operation.

2. There is same type of tooth meshing in helical and spur gear.
a) True
b) False

View Answer

Answer: b [Reason:] In spur gears contact occurs along entire face width which leads to impact condition while in helical contact begins from a single point and then there is gradual increase in load.

3. Among the normal module and transverse module, which one has greater value?
a) Normal Module
b) Transverse Module
c) Both have equal module
d) Insufficient information

View Answer

Answer: b [Reason:] Normal Module=Transverse modulexCos(helix angle).

4. Below is a representation of a helical gear. Blue lines in the figure signify?
tough-machine-design-questions-q4
a) Pinion right handed helical teeth
b) Pinion left handed helical teeth
c) Both right and left handed
d) Symmetric helical teeth

View Answer

Answer :a [Reason:] The lines are slopping downwards to the right side.

5. Below is a representation of a helical gear. Blue lines in the figure signify?
tough-machine-design-questions-q5
a) Pinion right handed helical teeth
b) Pinion left handed helical teeth
c) Both right and left handed
d) Symmetric helical teeth

View Answer

Answer: b [Reason:] The lines are slopping upwards to the right side.

6. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the transverse module.
a) 4.3mm
b) 4.1mm
c) 3.9mm
d) 3.7mm

View Answer

Answer: a [Reason:] m=4/Cos(22⁰).

7. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the transverse pressure angle in degrees.
a) 17.9
b) 20.4
c) 19.6
d) 18.4

View Answer

Answer: b [Reason:] tanᾰ=tan(19⁰)/Cos(22⁰).

8. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the axial pitch.
a) None of the listed
b) 34.2mm
c) 33.4mm
d) 29.6mm

View Answer

Answer: c [Reason:] p=πx(transverse module)/tan(22).

9. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the pitch circle diameter of pinion.
a) 64.7mm
b) 52.6mm
c) 56.6mm
d) 68.8mm

View Answer

Answer: a [Reason:] d=zxm/Cos(22).

10. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the pitch circle diameter of the gear.
a) 172.6mm
b) 142.6mm
c) 180.3mm
d) 202.4mm

View Answer

Answer: a [Reason:] d=zxm/Cos(22).

11. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the centre distance.
a) 125.4mm
b) 132.6mm
c) 118.65mm
d) 145.4mm

View Answer

Answer: c [Reason:] C=Sum of diameter of pinion and gear/2.

12. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate addendum circle diameter of the pinion.
a) 59.2mm
b) 72.7mm
c) 65.4mm
d) None of the listed

View Answer

Answer: b [Reason:] D(a)=m[z/Cos(22) + 2].

13. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the dedendum circle diameter of the pinion.
a) 64.5mm
b) 54.7mm
c) 59.2mm
d) None of the listed

View Answer

Answer: b [Reason:] D(f)=m[z/Cos(22) – 2.5].

14. The direction of tangential component for a driving gear is same to the direction of rotation.
a) True
b) False

View Answer

Answer: b [Reason:] The direction is opposite and not same.

15. If tangential component of force on tooth is 200N and helix angle is 25⁰, calculate the axial component of the force.
a) 200N
b) 302.5N
c) 93.26N
d) 215.6N

View Answer

Answer: c [Reason:] P(a)=200xtan(25).