2

1. In a reaction turbine, the enthalpy drop in a stage is 60 units.The enthalpy drop in the moving blader is 32 units. Then, thedegree of reaction will be ________.

2. If the speed of engine varies between 390 rpm and 410 rpm ina cycle of operation, then the coefficient of fluctuation ofspeed will be ________.

3. A rolling disk of radius ‘R’ and mass ‘M’ is connected to oneend of a linear spring of stiffness ‘K’ as shown in figure. Thenatural frequency of oscillation is given by:

M

RK

(a)2Kw3M

(b)KwM

(c)Kw

2M(d)

2KwM

4. Bending moment distribution in a built beam is shown infigure.

AB

C

DE

The shear force distribution in the beam is shown by:

(a) A CE (b) A C

E

(c)A E (d)

A

C

E

5. In the mechanism shown below, link 3 her,

1

2

3

45

(a) curvilinear translation and all points in it trace outidentical cycloids

(b) curvilinear translation and all points in it trace outidentical involutes

(c) linear translation and all points in it trace out identicalhelics

(d) linear translation and all points in it trace out identicalidentical ellipse.

6. In an air-standard diesel cycle, r = compression ratio, is thefuel cut-off ratio and is the adiabatic index (Cp/Cv).

(a) r

P 111 .P 1

(b)

1

r 1

P 111 .P 1

(c) r 1

P 111 .P 1

(d)

1P 111 .P 1r

7. A centrifugal pump operating at 1000 rpm develops a head of30 m. If the speed is increased to 2000 rpm and the pumpoperates with the same efficiency, the head developed bypump will be _______.

8. A 320 cm high vertical pipe at 150°C wall temperature is in aroom with still air at 10°C. This pipe supplies heat at the rateof 8 kw into the room air by natural convection. Assuminglaminar flow, the height of pipe needed to supply 1 kw onlywill be ________.

9. Which one of the following is the steady flow energyequation for a boiler?

(a)2 21 2

1 21 1

v vh h

2g 2g

(b) Q = h2 – h1

(c)2 21 2

1 21 1

v vh Q h

2g 2g

(d) s 2 1w h h Q10. A spur gear transmits 10 kw at a pitch line velocity of 10m/s

driving gear hav a diameter of 1 m. The tangential forcebetween the driver and follower and transmitted torque willbe ________.

11. A motor has a wheel base of 280 cm and the pivot distance offront stub axles is 140 cm. When the outer wheel is turnedthrough 30°, angle of turn of the inner front wheel for correctsheering will be ________.

12. An aeroplane travel at 400 km/h at sea level where thetemperature is 15°C. The velocity of aeroplane at the samemach number at an altitude where a temperature of – 25°C isprevailing will be ________.

13. In an SI engine, combustion stage 1 takes 1 ms and combustionstage II takes 1.5 ms when the engine runs at 1000 rpm. Ifstage I time duration is independent of engine speed, thenthe additional spark advance necessary (if engine speed is

3

doubled) will be ________.14. A circular cylinder of 400 mm diameter is rotated about its

axis in a stream of water having a uniform velocity of 4m/s.When both of the stagnation points coincide, the lift forceexperienced by the cylinder is ________.

15. In a system, metabolic rate = M, work done by man = w, rateof convective, radiative, evaporative heat losses = Q, rate ofheat storage = s. Then heat exchange between man and hisenvironment is:(a) M + W = Q + S (b) M – W = Q – S(c) M + W = Q – S (d) M – W = Q + S

16. During steady flow compression process of a gas with massflow rate of 2 kg/s, increase in specific enthalpy is 15kJ/kgand decreases in K.E is 2 kJ/kg. The rate of heat rejection toenvironment is 3 kw. The power needed to drive a compressorwill be ________.

17. A journal bearing supports a shaft which is rotating at aspeed of 2000 rpm. Clearance to radius ratio, bearing pressureand viscosity of lubricant are 1/130, 2.7 mPa and 40 × 10–4 Pasrespectively. Then the sommerfeld number will be ________.

18. The pressure difference of two gases in two vessels is beingmeasured by a vertical U-tube manometer. The reading istaken as 10 cm. Then the pressure difference between thegases will be ________.

19. Consider the following conditions for heat transfer:where, t = thermal boundary layer thickness

= boundary layer velocityPt = Prandtl number

1. t(x) = (x) if Pr = 12. t(x) >> (x) if Pr << 13. t(x) << (x) if Pr >> 1Which of these conditions apply for convective heat transfer?(a) 1, 2 (b) 2, 3(c) 1, 3 (d) 1, 2, 3

20. Masses A1, A2 and 8 kg are attached to a shaft in parallelplanes as show in figure. If the shaft is rotating at 100 rpm,the mass A2 will be ________.

8 kg

50 cm100 cm

A1A2

50 cm

50 cm

50 cm

21. An automatic machine accomplishes five operations and theyperform operations in 110, 130, 140, 145 and 170 seconds ateach of its work stations. Then the cycle time of work stationwill be ________.

22. The assembly of Hole-shaft specification is described asfollows:

(i) Hole of size = 0.0800.00027.000 mm

(ii) Shafts of size = 0.0400.02027.000 mm

Then, the maximum possible clearance in the assembly willbe ________.

23. In a rolling operation, thickness of sheet is reduced from 40mm to 25 mm by using 750 mm diameter rolls which are rotatingat 140 rpm. Then the roll strip contact length will be ________.

24. Two identical springs labelled (i) and (ii) are arranged asshown in figure given below.

F(i) (ii)

If spring constant is ‘k’ for each of the spring, then the strainenergy stored in spring (i) is:(a) F2/2K (b) F2/4K(c) F2/8K (d) F2/16K

25. In a counterflow heat exchanger, the hot fluid is cooled from110°C to 80°C by a cold fluid which gets heated from 30°C to60°C. LMTD for heat exchanger is ________.

26. When the Brayton cycle working in the pressure limits ofP1 and P2 is reversed and operated as a refrigerator, the idealvalue of CoP for such a cycle will be:

(a) 12 1P / P 1 (b)

12 1

1

P / P 1

(c)1/

2 1

1

P / P 1(d)

12 1

1

P / P 1

27. Air conditioning has to be done for a hail whose RSH = 50 kwand RLH = 50kw. There are no other sources of heat additionor leakager. The value of RSHF will be ________.

28. A penstock of 10 m diameter carries water under a pressurehead of 100 m. If the wall thickness is 9 mm, the tensile stressin pipe wall will be ________. (Take g = 10 m/s2)

29. The relation 2 2

2 20

x y for an irrotational flow is termed

as :(a) Navier-Stokes equation(b) Laplace equation(c) Reynold’s equation(d) Euler’s equation

30. A linkage is shown in figure in which links ABC and DEF areternary links, whereas AF, BE and CD are binary links.

A

B

C

D

E

F

The degree of freedom of linkage when link ABC is fixed is:

4

(a) 0 (b) 1(c) 2 (d) 3

31. Match List-I with List-IIList-I List-II

A. Saddle key 1. strong in shearB. Woodruff key 2. withstands tension in

one directionC. Tangent key 3. transmission of power

through fr ictionalresistance

D. Kennedy key 4. semi-circular in shapeCodes:

A B C D(a) 3 4 1 2(b) 4 3 2 1(c) 4 3 1 2(d) 3 4 2 1

32. When a weight of 100 N falls on a spring of stiffness 1 kN/mfrom a height of 2 m, the deflection caused in the first fall is________.

33. If the spindle speed is 80 rpm and cross-feed is 0.3 mm/rev,then the time taken to face a workpiece of 72 mm diameter willbe ________.

34. The spot welding is done on two thick steel sheets of 3 mm.The current is 4500 A for 0.15S on each spot. The othernecessary data is given are follows:Diameter of electrode = 6 mmResistance = 375 µDensity of material = 8700 kg/m3

Melting specific energy =2.5 Mj/kgThen the amount of heat generated and heat dissipated willbe _______.

35. The specification for a hole is 0.0600.00054 mm . The mating

shaft has a clearance fit with minimum clearance of 0.04 mm.If the tolerance on the shaft is 0.08 mm, then the maximumclearance in mm between the shaft and the hole will be________.

36. In an operation of NC work table, the following data is given:Pitch of a lead screw = 9 mmGear ratio = 8 : 1Number of pulse generated = 58 pulses/rev.Distance moved by the table = 230 mmFeed rate = 430 mm/min.Then the number of pulses received by the control systemfor purpose of verification for the movement of table throughenact distance of 230 mm will be ________.

37. In an assembly line, the rate of arrival of products is 0.8 parts/minute. Processing time with mean is 1.8 min. with exponentialdistribution. Then the probability, that a random part arrivalfinds that there are 9 parts already exist in the system, will be________.

38. Eight bolts are to be used for fixing the cover plate of a cylindersubjected to a maximum load of 980.175 kN. If the designstress for the bolt material is 315 N/mm2, then the diameter of

each bolt will be _________.39. A circular solid shaft is subjected to bending moment of 400

kN-m and a twisting moment of 300 kN-m. On the basis ofmaximum principle stress theory, the direct stress is andaccording to the maximum shear stress theory, the shear stress

is . Then the ratio of direct stressshear stress will be ________.

40. The inter arrival times at a tool crib are exponential with anaverage time of 10 min and length of service time is assumedto be 6 min. Then the probability that a person arriving at thebooth will have to wait equal to –

41. The humidity ratio of atmospheric air at 28°C DBT and760 mm of Hg is 0.016 kN/m2. Then the partial pressure ofwater vapour will be ________.

42. The angle of rotation of a rigid body is given by = 0 +At +Bt2. In which, A, B, are constants and t is time in seconds. If

= 4 rad/sec at t = 0 and w = 2 rad/sec after 1 second fromstart, then the values of constants will be ________.

43. In a reversible isothermal expansion process, the fluidexpands from 10 bar and 2 m3 to 2 bar and 10 m3, during theprocess and the heat supplied is 100 kw. Then the work doneduring the process will be ________.

44. For flow over a flat plate, the hydrodynamic boundary layerthickness is 0.5mm. The dynamic viscosity is 25 × 10–6 Pa-s,specific heat is 2 kJ/kg-k and thermal conductivity is 0.05 w/mk. Then, the thermal boundary layer thickness will be________.

45. A close coiled helical spring is made up of 5 mm diameter wirecoiled to 50 mm mean diameter. Maximum shear stress in springunder the action of an axial force is 20 N/mm2. The maximumshear stress in a spring made of 3 mm diameter wire coiled to30 mm mean diameter under the action of force is ________.

46. The arm of a radial drilling machine is being raised at a speedof 3.9 m/min. by single start square threads of 6 mm pitch and30 mm diameter. The speed of screw will be ________.

47. A ball bearing is characterized by basic static capacity = 11000N and dynamic capacity = 18000 N. This bearing is subjectedto equivalent static load = 5500N. Then the bearing loadingratio and life in million revolution will be ________.

48. If the annular wheel of an epicyclic gear train has 100 teethand the planet wheel has 20 teeth, the number of teeth on thesun-wheel is ________.

49. In a cotter joint, the width of the cotter at the centre is 50 mmand its thickness is 12mm. The load acting on the cottor is 60KN. Then the shearing stress induced in the cotter joint willbe ________.

50. A time standard for a data entry operator is to be specified. Ajob in rated at 140%, it takes 40 seconds to enter each recordand the allowances are 20%. The normal time will be________.

51. In a 1 ton capacity water cooler, water enters at 30°C at therate of 200L/h. If specific heat of water = 4.18 KJ/kg-k, thenthe outlet temperature of water is ________.

5

52. In a strained material, normal stresses on two mutuallyperpendicular planes are x and y (both are alikes) and shearstress is xy, and one of the principal stress will be zero onlyif

(a) x yxy 2

(b) xy x y

(c) xy x y (d) 2 2xy x y

53. The index of expansion of dry saturated steam flowingthrough a nozzle is 1.135, then the critical pressure ratio ofthe following steam will be ________.

54. Maximum angular velocity of connecting rod with a crank toconnecting rod ratio 1 : 5 for a crank speed of 3000 rpm will bearound ________.

55. Plane stress at a point in a body is described by principalstresses 3 and . Then, the ratio ( n/ man) on the plane ofmaximum shear stress is ________. (where n = normal stress,

max = maximum shear stress).56. In orthogonal cutting, cutting force and thrust force are 1000

N and 500 N respectively. If the rake angle of tool is zero, thecoefficient of friction in chip-tool interface will be ________.

57. In a small engineering project, for an activity, the optimistictime is 2 min, the most likely time is 5 min and the pessimistictime is 8 min, then the expected time of activity will be________.

58. The pressure drop for a relatively low Reynolds number flowin a 600 mm, 30 m long pipeline is 70 KPa. Then the wall shearstress will be ________.

59. A cube with a side length of 1 cm is heated uniformly 1°Cabove the room temperature and all the sides are free toexpand. If thermal expansion coefficient is °C, then increasein volume of the cube will be ________.

60. The length of a cylindrical shaped casting is equal to diameter.The solidification time for cylindrical casting is 25 min. If theaspect ratio and material is kept same, then solidification timefor a 8 timer heavier cylinder will be ________.

61. In a blanking operation, the maximum punch load in 6 × 106 N.If the percentage penetration is 40% and thickness of plate is9mm, then the work done during the shearing operating willbe ________.

62. Actual demand for the first period is 92 units and forecastwas 96 units. If exponential smoothing factor is 0.5, then theforecast for the next period will be ________.

63. The distance covered by a body of mass 2 kg with respect totime is given as:s = 6t3 – 4t2 + 9where, S = distance, t = timeThen, the acceleration and the driving force of the body after9 seconds will be ________.

64. A body of mass 8 kg is supported on a spring of stiffness 200

N/m and has a dashpot connected to it which produces aresistance of 0.003 N at a velocity of 2 cm/s. After 6 cycles,the reduction ratio of amplitude of vibration will be ________.

65. The values of enthalpy of starting of compression, at end ofcompression and at the end of condensation are 85 KJ/kg,185 KJ/kg and 210 kJ/kg respectively. Then the value of COPof vapour compression refrigeration system will be ________.

66. In a study to estimate the idle time of a machine, out of 100random observations, the machine is found idle on 40observations. If confidence level is 95% and accuracy is ± 5%,then the total random observations required are _________.

67. What is the vertical distance of centre of pressure below thecentroid of plane area?

(a) GIA.h

(b) GI .sinA.h

(c)2

GI .sinA.h

(d)2

G2

I .sin

A.h68. In a multiple disc clutch, the axial intensity of pressure is not

to exceed 0.2 MPa. The inner radius of the discs is 100 mmand is half the outer radius. Then the axial force per pair ofcontact surfaces in N will be ________.

69. A truss structure is shown in figure given below whichsupports a load 18 KN at point ‘S’. Then the load is themember PQ will be ________.

P Q R S

T

3 m

6 m U

3 m 3 m3 m

18 KN

4 m

70. The velocity of a particle is given as :v = 9 – 0.06 x m/sIf x = 0, t = 0, then the distance travelled (when it comes torest) and acceleration (at t = 0) will be ________.

71. A heat engine is supplied with 2700 KJ/min of heat at 750°C.Heat rejection with 1000 KJ/min takes place at 99°C. Then thetype of heat engine will be:(a) Ideal (b) Irreversible(c) Impossible (d) Practical

72. When the pressure on a given mass of liquid is increasedfrom 3MPa to 3.5 MPa, the density of liquid increases from500 kg/m3. Then the average value of bulk modulus of theliquid over the given pressure range will be ________.

73. There are two bodies of equal masses are rotating along acircular path of radius R1 and R2 with same speed. The ratioof their centripetal forces would be equal to:

6

(a) R2/R1 (b)2

1

RR

(c)2

22

1

R

R(d)

21

22

R

R74. Which one of the following is the value of helix angle for

maximum efficiency of a square threaded screw.(a) 45° + (b) 45° – (c) 45° – /2 (d) 45° + /2

75. In a flat belt drive, the belt can be subjected to a maximumtension T and centrifugal tension TC. The condition fortransmission of maximum power is:

(a) cT T (b) cT 3T(c) T = 2TC (d) T = 3TC

76. A cantilever of length L, moment of inertia I, young’s modulusE carries a concentrated load w at the middle of its length.The slope of cantilever at free end is:

(a)wL2EI

(b)wL4EI

(c)wL8EI

(d)wL16EI

77. Which one of the following cast irons consists of carbon inrosette form?(a) White cast iron (b) Grey cast iron(c) Malleable cast iron (d) Nodular cast iron

78. A hydraulic pressure have a ram of 20 cm diameter and aplunger of 5 cm diameter. The force required at the plunger tolift a weight of 16 × 104 N will be ________.

79. In a reaction turbine, stage enthalpy drop in the stator bladeris 4.62 kJ/kg and in the rotor blade is 2.38 kJ/kg. Then, thedegree of reaction will be ________.

80. A single parallel fillet weld of total length L and weld size hsubject to a tensile load P, will have the design stress:

(a) Tensile and equal to P

0.707Lh

(b) Tensile and equal to P

Lh

(c) Shear and equal to P

0.707Lh

(d) Shear and equal to P

Lh81. Which one of the following Mohr’s circles represents the

state of pure shear?

(a) O (b) O

(c) O (d)

82. The static deflection of a shaft under a flywheel is 4 mm. If g= 9.8 m/s2, then the initial speed in rad/s will be ________.

83. The modulus of elasticity for a material is 200 GN/m2 andPoisson’s ratio is 0.25. Then the modulus of rigidity will be________.

84. In a tool life test, doubling the cutting speed reduces the tool

life to 1 th8

of the original. The Taylor’s tool life index is:

(a)12

(b)13

(c)14

(d)18

85. If critical path of a project is 20 months with a standarddeviation 4 months, the probability that the project will becompleted in 24 months is ________.

86. Which one of the following is clearance fif?

(a) 0.015 0.010500.005 0.000H50 h

(b) 0.010 0.025500.000 0.015H50 h

(c) 0.015 0.025500.000 0.005H50 h

(d) 0.010 0.030500.000 0.005H50 h

87. The working temperature in evaporator and condenser coilsof a refrigerator are –23°C and 27°C respectively. The COP ofa refrigerator is 0.8 of the maximum COP. If the power input is1 kw, then, the refrigeration effect produced will be ________.

88. A mass ‘m’ vibrates on a frictionless platform as shown infigure. The combined spring constant will be:

m

K1

K1

K2

K2

(a) K1 + K2 (b) 2(K1 + K2)

(c)1 2

1 2

K KK K (d)

1 2

1 2

2 K KK K

89. In a new temperature scale i.e. °P, the boiling and freezingpoints of water at one atmosphere are 100°P and 300°Prespectively. Correlate this scale with the centigrade scale,the reading of 0°P on the centigrade scale will be ________.

7

90. Consider the phase diagram of a certain substance as shownin figure, Match List I with List II

DE

F

G

Pressure

Temperature

List I List IIA. Vaporization 1. FEB. Fusion 2. EGC. Sublimation 3. EDCodes

A B C(a) 1 3 2(b) 1 2 3(c) 3 2 1(d) 3 1 2

91. Heat is conducted through a 10 cm thick wall at a rate of 30 w/m2 when the temperature difference across the wall is 10°C.The thermal conductivity of the wall be ________.

92. For an air-conditioning system, the outdoor and indoor drybulb temperature are 45°C and 25°C respectively. The spaceto be air conditioned is 20 m × 30 m × 5 m and infiltration isestimated to be one air change. If the density and specificheat of air are 1.2 kg/m3 and 1.02 kj/k°C, then the sensibleheat load due to infiltration is _________.

93. A wooden rectangular block of length l is made to float inwater with its axis vertical. The centre of gravity of the floatingbody is 0.15 above the centre of buoyancy. Then the specificgravity of the wooden block will be _______.

94. The cavitation number of any fluid machinery is defined

2P P

v / 2 where, P = absolute pressure, p = density v =

free stream velocity, then the symbol p denotes:(a) Static pressure of fluid(b) dynamic pressure of fluid(c) vapour pressure of fluid(d) Shear stress of fluid

95. The order of values of thermal efficiency of otto, diesel anddual cycler, when they have equal compression ratios andheat rejection, is given by

(a) otto diesel dual

(b) diesel dual otto

(c) dual diesel otto

(d) otto dual diesel

96. Water (CP = 4kJ/kgk) is fed to a boiler at 30°C, the enthalpy ofvaporization at atmospheric pressure in the boiler is 24000kJ/kg, the steam coming from the boilder is 0.9 dry. The netheat supplied in the boiler will be _________.

97. A straight bar is fixed at A and B as shown in figure. A load P= 120 N is acting at point c. Then the reactions at the end willbe ________.

A B

L 2L

C P = 120 N

98. Steel shuts are to welded by spot welding.The following data is providedSteel sheet thickness = 3 mmCurrent = 6000 AmpTime for flow of current = 0.185Diameter of electrodes =8 mmResistance =250Density of metal = 8000 kg/m3

Specific energy for melting = 1.8 MJ/kgBased on the above data, the heat generated will be________.

99. The specification of a shaft is provided as 0.0600.00045 mm .

Minimum clearance and tolerance are 0.02 mm and 0.05 mmrespectively. The maximum clearance between hole and shaftwill be ________.

100. An industry consumes 4000 units of a product on annualbasis. Delivery lead time is 9 days. The reorder point to achieveoptimum inventory (in number of units) will be ________.

101. Match List-I with List-IIList-I List-II

A. Laser welding 1. Uniting large area sheetsB. Friction welding 2. Repairing large partsC. Ultrasonic welding 3. Welding a rod to flat

surfaceD. Explosive welding 4. Fabrication of nuclear

reactor components5. Welding very thin mate-

rialsCodes

A B C D(a) 5 4 3 2(b) 1 4 2 5(c) 1 3 4 2(d) 5 3 4 1

8

HINTS & SOLUTIONS1. 0.533

Given: Enthalpy drop in a stage = 60 units.Enthalpy drop in moving blades = 32 units

Degree of reaction = Enthalpy drop in moving blades

Enthalpy drop in a stage

=3260 = 0.533

2. 0.05Given: Variation in speed = 410 – 390 = 20 rpm

Mean speed = 410 390 800

2 2 = 400 rpm.

Coefficient of fluctuation of speed

= Variation in speed 20

Mean speed 400 = 0.05

3. (a) Applying Newton’s method,

I KyR

I KyR 0

2I K R 0

on solving, we get 2K 03M

M

R

Ky

M

here, 2Kw3M

4. (a)L/2

P

A

L

EC

L/2

SF A ECSF

SF Shear Force

5. (a) In the mechanism shown in figure, there in curvilineartranslation and all points in it will trace out identicalcycloids.

6. (c) Thermal efficiency for air standard diesel cycle

1th r

P 111P 1

7. 120 mGiven: Case I

Speed (N1) = 1000 rpmHead (H1) = 30 m

Case IISpeed (N2) = 2000 rpmHead (H2) = ?

As we know that for centrifugal pump,2N H N H

2 21 12 2 2

N H 30 1000 1N H H 2000 4

H2 = 30 × 4 = 120 m8. 40 cm

Given: Height of pipe (H1) = 320 cmTemperature difference ( T) = 140°CRate of Heat transfer (Q1) = 8 kwRate of Heat transfer haded (Q2) = 1 kwQ = HdA T (According to Newton’s Law of colling)

Here, on applying 1 12 2

Q HQ H

[For laminar flow]

22

8 320 320H 40cm1 H 8

9. (b) Stead flow energy equation for a boiler is given as:

2 1Q h h10. 500

Given: Power (P) = 10 kw = 10 × 103 wVelocity (v) = 10 m/s

Diameter (d) = 1mPower (P) = Torque × angular speed

= T × wwhere, Ft = tangential force

P = Ft × r × w

Ft =3P P 10 10 1000N

rw v 10Torque transmitted = Ft × r

= 11000 500N m2

9

11. 39.11Given: Wheel base (A) = 280 cm

Pivot distance (B) = 140 cmOuter wheels’s turn ( ) = 30°angle of turn of inner wheel, ( ),

B 140cot cot cot 30 cotA 280

1cot cot 30 1.732 0.5 1.232

1cot 1.23 = 39.11

12. 103.22 m/sGiven: Velocity of aeroplane (va) = 400 km/h

= 111.11 m/stemperature (t) = 273 + 15 = 288°K

Mach number = 111.11 111.11

Rt 1.4 288 287

= 111.11 0.327340.17

when temperature = – 25°C = – 25 + 273 = 248 K

Mach number = 0.327 = v

1.4 248 287= 0.327 × 315.66= 103.22 m/s

13. 6°Given: Time taken by stage I (T1) = 1 msTime taken by stage II (T2) = 1.5 msSpeed of engine = 1000 rpmSpark advance time = T2 – T1 = 1.5 – 1

= 0.5 msNow, If the speed of engine is doubled = 2000 rpmthen, additional spark advance necessary

= 2 180 0.5 2000

60 1000= 6°

14. 10048 N/mGiven: diameter (d) = 400 mm = 400 × 10–3 mvelocity (v) = 4 m/s

Lift force (L) = 22

1 pv Ac2

where, p = air densityv = velocity

22 31 1L pv d 1000 4 400 102 2

= 10048 N/m15. (b) For thermal comfort, It is a situation which is maintained

when the heat generated by the human metabolism isallowed to dissipate, thus it maintains thermalequilibrium including its surroundings.

16. 29 kwGiven: Mass flow rate (M) = 2kg/sIncrease in specific Enthalpy ( H) = 15 kJ/kgDecrease in K.E. = 2 kJ/kg

Rate of heat rejection (Qrej) = 3 kwNow, writing the steady flow energy equation,

w= 2rej

1M H M V Q2

= 2 × 15 – 2 × 2 + 3= 30 – 4 + 3 = 29 kw

17. 711.11

Given: Speed (N) = 2000 rpm = 2000 rps

60clearance to radius ratio (c/r) = 1/120viscosity of lubricant ( ) = 40 × 10–4Pa.s

Bearing pressure (P) = 2.7 mPa

Now, sommerfelt number (s) = 2r N

c P

= 2 4 200120 40 10 711.11

6 2.718. 980 N/m2

Given: Reading of manometer (H) = 10 cm = 0.1mDensity of water ( water) = 1000 kg/m3

gravity (g) = 9.8 m/s2

Pressure difference between gases,

difference gHP = 1000 × 9.8 × 0.1 = 980 N/m2

19. (d) As we know that

rviscous diffusion ratePrandtl number Pthermal diffusion rate

For Pr = 1, = tFor Pr >> 1, > tFor Pr << 1, < t

Here, 3 rt

P

where, = boundary layer thickness (velocity)t = thermal boundary layer thickness

20. 2.67 kgGiven: Mass (M) = 8 kgUsing the following equation,M × R1 × w2 × 50 = M2 × R2 × w2 × (50 + 100)8 × 50 × 50 × w2 = M2 × 50 × w2 × (150)

28 50 50M 2.67kg50 150

21. 170 sec.Maximum time taken to perform the operations out of fiveoperations = 170 seconds.The cycle time is the highest time taken by an individualwork station so the cycle time = 170 seconds.

22. 0.1 mmUpper limit for hole = 27.080Lower limit for shaft = 27.000 – 0.020 = 26.98Now, the maximum clearance = 27.080 – 26.98

= 0.1 mm

10

23. 75 mmGiven: Initial thickness (ti) = 40 mmFinal thickness (tf) = 25 mmRoll diameter (d) = 750 mm

Radius (r) = 750

2 = 375 mm

Roll strip contact length = i fr t t

= 375 40 25= 75 mm

24. (c) Force in spring (i) = Fi = F/2Force in spring (ii) = Fii = F/2

Deflection of spring ( 1) = F

2KDeflection of spring ( 2) = F/2K

Strain energy = 1 F F2 2K 2

= 2F

8K25. 50oC

Given: Temperature difference ( T1) = 110 – 60 = 50°C Temperature difference ( T2) = 80 – 30 = 50°C

Now, Logarithmic Mean Temperature Difference:

2

1 2

1T

T TLMTD

Tin

As, 1 2T T , But, 12

Ta

T

22 2 T a 1a T TLMTD

in a in a

Now, As, LMTD 1, then LMTD = T2 = 50°C

26. (c) CoP = Total effect of refrigeration

Total work supplied

= 1 /

21

1

P 1P

27. 0.5Given: Latent heat generation (RLH) = 50 kw.

Sensible heat generation (RSH) = 50 kw.

Room sensible heat factor (RSHF) = SHTH

R 50R 50 50

= 50

100 = 0.5

28. 555.6 × 106

Given: Diameter of pipe (d) = 10 mPressure Head (H) = 100 mWall thickness (t) = 9 mm = 9 × 10–3 m

Tensile stress ( ) = Pressure P diameter d

2 thickness t

where, P = ghP = 1000 × 10 × 100P = 106 m

6 10 4 6

310 10 10 10 10

2 9 2 92 9 10 = 555.6 × 106 MPa

29. (b) Relation 2 2

2 20

x y is known as Laplace equation

for an irrotational flow.30. (a) Given: Number of links (n) = 5

Number of lower pairs (lp) = 6Equation of Degree of freedom (DoF) = 3(n – 1) – 2lp – hpwhere, hp = Number of higher pairsDoF= 3(5 – 1) – 2 × 6

= 3 × 4 – 2 × 6= 12 – 12 = 0

31. (d) Saddle key = Transmission of power through frictionalresistance.Wood-ruff key = Semi-circular in shapeTangent key = withstands tension in one direction.Kennedy key = Strong in shear.

32. 0.621Given, weight (w) = 100NStiffness (k) =1kN/mHeight (h) = 2m

Potential Energy (P.E.) =mg w× 100

+ + +mgh w×h 100×2 200Nm 100

Kinetic Energy (K.E) = 221 1K 10002 2

From law of conservation of energy,21000 200 100

2500 200 10025 2 0

= 0.62133. 1.5 min

Given: Spindle speed = 80 rpmCross-feed = 0.3 mm/revDiameter of work piece = 72 mm

Depth of cut = 722

= 36 mm

Depth of cut 36Time takenSpeed feed 80 0.3

= 1.5 min

34. 1843.9J

11

Given: I = 4500 AmpR = 375 t = 0.15 s

Heat developed/generated (Qg) = I2Rt= (4500)2 × 375 × 10–6 × 0.15= 1139.06 J

Heat dissipated (Qd) =2DP 2t 4

4

=

236

6 108700 2 0.15 2.5 10

4= 1843.9 J

35. 0.18 mm

Given: Hole specification : 0.0600.00054 mm

Minimum clearance = 0.04 mmTolerance on shaft = 0.08 mmHole size = 54.000 – 54.060Maximum shaft size = 54.000 – 0.04 = 53.96 mmMinimum size of shaft = 53.96 – 0.08 = 53.88 mm

Maximum clearance = 54.060 – 53.88= 0.18 mm

36. 1482 PulsesGiven: Distance moved = 230 mm

Pitch = 9 mmNumber of pulses = 58 pulses/rev.Number of pulses received

= Distance moved ×Number of Pulses

Pitch

= 230 58 1482.22 1482 Pulses9

37. 0.0871Rate of arrival ( ) = 0.8 Parts/min.

1.82

= 0.90/min.

So, the probability (P) = 0.8 0.890.9

probability of arriving random parts while 9 parts already

exists = nP 1 P

= 0.8910 (1 – 0.89)= 0.0871

38. 22.3 mmGiven: Maximum Load = 980.175 KNNumber of bolts = 8

Load on one bolt = 3980.175 10

8 = 122.5 × 103N

Design stress = 315 N/mm2

Let ‘D’ be the diameter for design stress.

Design stress = 315 =LoadArea =

3

2122.5 10

D4

3

2122.5 10 4315

D

32 122.5 10 4D 495.4

315

D = 22.3 mm.39. 1.8

Given: Bending moment (M) = 400 kN-mTwisting moment (T) = 300 kN-m

Direct stress ( ) = 2 2

316 M M Td

Shear stress ( ) = 2 2

316 M Td

Ratio of ( to ) = 2 2

2 2M M T

M T

= 2 2

2 2

400 400 300

400 300 = 1.8

40. 0.6

Probability (P) =

where, customer per unit time/ arrival at average rate. customer served at average rate.

Given: = 6 persons/hour = 10 persons/hour

Probability (P) = 6

10 = 0.6

41. 2.51 kN/m2.Given: Humidity ratio (y) = 0.016 kN/m2

Humidity ratio (y) = wa w

P0.62198

P P

where, Pw = Partial pressure of water vapour.Pa = Atmospheric pressure of moist air.

w

w

Py 0.62198 0.016

100 P

12

ww

P 0.016100 P 0.62198

Pw = 2.51 kN/m2.42. – rad/s

Given: Angle of rotation of a given rigid body = 0 + At + Bt2

on differentiating, = A + 2BtAt, t = 0, = , A = 4 rad/sAt, t = 1, = ,then, 2 = 4 + 2B

2B = 2 –4 = – 2B = – rad/s

42. 100 kwGiven: Heat supplied (Q) = 100 kw.From first law of thermodynamics,

U = Q – wQ = u + wwhere, w = work done by system.

u = Internal energyHere, u = m × cv × tfor isothermal process, temperature remains constant.i.e., t = 0

u = 0Q = w = 100 kw

44. 0.5 mmGiven: Hydrodynamic boundary layer thickness( x) = 0.5 mmDynamic viscosity ( ) = 25 × 10–6 Pa-sSpecific heat = 2 kJ/kg-kthermal conductivity (k) = 0.05 w/mk.

pr

Cviscour diffusion ratePrandtl number Pthermaldiffusion rate k

6r

2 25 10 1000P 10.05

Now, using the following equation,3 rthermalx x P

where, x = Hydrodynamic boundary layer thickness.

thermalx = thermal boundary layer thickness.

3thermal0.5 x 1

thermalx 0.5mm

45. 55.56 N/mm2

Shear stress of closed coil helical spring.

38wD

dwhere, w = axial load.

D = mean coil diameter.d = diameter of spring wire.

Given: D1 = 50 mm, d1 = 5 mm, 1 = 20 N/mm2

D2 = 30, d1 = 3 mm,

31 1 1 2

32 2 21

d8w D8w Dd

31 232 21

dD20Dd [w1 = w2]

3

32 2

320 50 20 0.36305

22

20 55.56 N / mm0.36

46. 650 rpmGiven: Speed = 3.9 m/min

Pitch of screw = 6 mmDiameter = 30 mm

Speed of screw = Speed

Pitch of screw = 33.9 10 650rpm

647. 3.5 × 106 revolutions

Given: Dynamic capacity (Cr) = 18000 NEquivalent static load (Pr) = 5500 NBasic static capacity = 11000 N

Loading ratio (L) = rr

C 18000 3.3P 5500

Life = 3

36 6rr

C10 3.3 10

P = 3.5 × 106 revolutions

48. Given : Teeth on annular wheel (TA) = 100Teeth on planet wheel (TP) = 20TA = 2TP + TS [where TS = Teeth on sun wheel]100 = 2 × 20 + TS TS = 100 – 40 = 60

49. Shearing stress induced in cotter joint, (P) = 2b × t × shearwhere, P = Load acting, shear = shear stresst = thicknessb = widthGiven: P = 60 KN = 60 × 103 N , t = 12 mm, b = 50 mm60 × 103 = 2 × 50 × 12 × shear

3shear

60 102 50 12

= 50 N/mm2

50. Normal time = Average time Rating of performance

100

= 40 140

100= 56 seconds

51. Given: temperature difference ( t) = (30 – t2)

13

mass flow rate (m) = 200 L/h = 200 L / S

3600specific heat of water (c) = 4.18 KJ/kg-kOne ton of refrigeration (ITR) = 3.517 kwNow, using following equation

2200q mC t 4.18 30 t

36003.517 = 0.232 (30 – t2)15.15 = 30 – t2 t2 = 30 – 15.15 = 14.85 15 C

52. (c) Using the following equation for principal stresses,

2x y x y 2

1,2 xy2 2

Given: 1 or 2 = 0

2x y x y 2

xy02 2

2 2x y x y2

xy2 2

2 2 2 2x y x y x y2

xy x y4

4

xy x y

53. Given: expansion index (K) = 1.135

Critical pressure ratio =

KK 12

K 1

=

1.1351.135 12

1.135 1

= 8.410.937= 0.58

54. Given: Crank to connected ratio (r) = 1:5 or 1/5Crank speed (h) = 3000 rpm

Angular velocity (w) = 2 h 2 3.14 300060 60

= 314

Connecting rod angular velocity (wc) =w cos

r

c maxw , when cos = 1

then, c maxw = w 314r 115

= 2.73

55. Given : Principal stresses: 1 = 3 , 2 =

Normal stress ( n) = 1 2 3 4 22 2 2

Maximum shear stress ( max) = max min2

= 3 2

2 2

N 2Ratio 2

56. Coefficient of friction at chip-tool interface ( )

tan = th

cut

Thrust force FCutting force F

Given : Fth = 500 N, Fcut = 1000 N, = 0500 1tan 0

1000 2

1 0.52

57. Given Optimistic time (to) = 2 min.Most likely time (tm) = 5 min.Pessimistic time (tp) = 8 min.

Expected time of activity (E) = o m pt 4t t

6

= 2 4 5 8

6 = 306 = 5 min.

58. Wall shear stress ( wall) = f hh rRu

y Lwhere, hf = Head lossr = specific weightRh = hydraulic radius

wall = 70 × 100059. As we know that,

change in dimension ( l) = l l Twhere, T = change in temperature

l =coefficient of thermal expansionl = side length

Given: l = 1 cm, T = 1°C, l = Change in side length ( l) = 1 × × 1 = Change/increase in volume = 3 cm3

60. Given: Length of cylindrical casting = diameter of casting= D

Volume of cylinder (Vc) = r2h = 2 3D DD

2 4Surface area of cylinder (AC) = 2 r(r + h)

= D D2 D2 2 =

2 2 23D 6 D 3 D24 4 2

Ratio of VC/AC =

33

2 2

D2 D D4

63 D 4 3 D2

(Diameter)8 times hevier = 3 8D 2D

As we know that, solidification time (t) = 2VB

A

14

when B = constant2 2

2 2 1 11 22 2

2 11 1 2 2

V / A V / At tt tV / A V / A

2 2

2 2 22D16 4Dt 25 25

DD16t2 = 100 mins.

61. Given: Maximum punch load (Pmax) = 8 × 106 N

Penetration by percent ( x) = 40% = 40

100Plate thickness (t) = 9 mm = 9 × 10–3 mWork done during shearing operation (w)= Pmax × Dx × t

= 6 3408 10 9 10100

= 2 4288 10 2.9 10 J62. Given: Actual demand for first period (Dt–1)= 92 units

Forecast (St–1) = 96 units.Exponential smoothing factor ( ) = 0.5Forecast for next period (St) = Dt–1 + (1 – )St–1

= 0.5 × 92 + (1 – 0.5) 96= 46 + 48 = 94 units

63. Given: Distance travelled, S = 6t3 – 4t2 + 9Mass (m) = 2 kgOn differentiating equation two times, we get

Acceleration (a) = 2

2d Sdt

2ds 18t 8tdt ,

2

2d s 36t 8dt

a of t = 9, (a)9 = 36(9) – 8 = 316 m/s2

Now, Force acting on the body (F) = ma= 2 × 316= 632 N

64. Given: mass (m) = 8 kg, spring stiffness (k) = 200 N/m.

Damping coefficient (c) = 0.0030.02 = 0.15 N-s/m

frequency of vibration (w) = K 200M 8

= 5 rad/s

Critical damping (cc) = 2w × m = 2 × 5 × 8 = 80N-s/m

Damping ratio ( ) = c

c 0.15c 80 = 0.0019

Now, logarithmic decreament ( )

= 2 2

2 2 3.14 0.0019

1 1 0.0019

= 0.011932 0.011932

1 0.00000361 0.999

= 0.011932 0.0119

0.999

= 1 1

n 1 7

x x1 1In In 0.0119n x 6 k

In 17

xx = 0.0119 × 6 = 0.0714

17

xx = e0.0714 = 1.074

65. Given: Enthalpy at compression start (Hc) = 85 KJ/kgEnthalpy at end of compression (He) = 185 KJ/kgEnthalpy at end of condensation (Hcon.) = 210 KJ/kg

vapour compressionsystem

COP = e c

con. e

H H 185 85H H 210 185

= 10025 = 4

66. (b) Given: confidence level (c) = 95%

Accuracy (A) = 5% = 5

100 = 0.05

Occurrence of events (Po) = 0cA

where, Po = 400 observations out of 100

= 40

100 = 0.4

o oo

P 1 P 0.4 1 0.4N N

2 0.4 0.60.40.05 N

0.4 40 0.24 / N1600 0.29 3840.16

N NN = 2400

67. (c) The vertical distance of centre of pressure below the

centroid of plane area will be 2

GI sinA.h

68. Given : Pressure (P) = 0.2 MPaInner radius (R1) = 100mmOuter radius (R2) = 200mmAxial force (F) = Pressure × Area

= 2 22 10.2 R R

= 2 20.2 3.14 200 100

= 0.2 3.14 100 300 = 18840 N69. Using method of sections, cut three members PQ, QT and

TU by reaction (1) - (1) and considering the equilibrium ofright portion of truss.

15

P Q R S

T

3 m

6 m U

3 m 3 m3 m

18 KN

4 m

1

1

Taking moment about T,FPQ × QT = 18 × 6

PQ18 6 18 6FQT 4 = 27 KN

70. Given: Velocity of particle (v) = 9 – 0.06x.(particle at rest) At v = 0, 9 – 0.06x = 0 x = 150 m

Now, Acceleration (A) = dv dx0.06dt dt

= – 0.06 × velocity

At t = 0, dxvdt

= 9 m/s

So, Acceleration (A) = – 9 × 0.06 = – 0.54 m/s2

71. (c) Given: heat supplied (qs) = 2700 KJ/minTemperature (Ts) = 550 = 823°KHeat rejection (qs) = 1000 KJ/minTemperature (Tr) = 99°C = 372°k

s r

s

q q 2700 1000Efficiencyq 2700

= 0.63 or 63%

Efficiency for ideal cycle ( ideal) = 823 372

823= 54.8%

Av, Efficiency ( ) is more than Ideal efficiency,So heat engine is impossible.

72. Given: Initial pressure (Pi) = 3MPaFinal pressure (Pf) = 3.5 MPaInitial density i = 500 kg/m3

Final density ( f) = 504 kg/m3

Bulk modulur (K) = PP

where, = density, p = change in pressure= Pf – Pi = 3.5 – 3= 0.5 MPa

= 504 – 500 = 4 Kg/m3

0.5K 5004

= 62.5 MPa

73. (d) Ratio of centripetal force = 2

1

2

RR

74. (c) Helix angle for maximum efficiency of a square threadedscrew = (45 – /2).

75. (d) Writing the equation for ratio of tensions in belt,

12

Te

T

where, T1 = Tension in slack sideT2 = Tension in tight side

Power transmission (P) = (T1 – T2) × v

= 11

TT ve

= 11T 1 v

e

Or, 2 µ1P T AV 1 e

For maximum power,

21

dP T 3 AV 1 e 0dv

21 CT 3 AV 3T

where, TC = centrifugal tension

76. (c) Slope of cantilever at free end = 2wL

8EIwhere w = Load, L = length, E = Young’s modulur ofElasticity, I = Moment of inertia.

77. (d) Nodular cast iron consists of carbon in rosette form. Itis a grey cast iron with lamellar graphite. It contains0.04 to 0.06% magnesium, cerium. Properties dependon size, number and arrangement of graphite bollos incast iron.

78. Given: Diameter of ram (d1) = 20 cm = 0.2 mDiameter of plunger (d2) = 5 cm = 0.05 m.Weight of liquid (w) = 16 ×1 04 NForce required (f) = ?

Pressure exerted on ram (P1) 1

Force / Load wArea A

= 4

1

16 10A

Pressure on Plunger (P2) = 2

fA

Now, equating the above two pressure,P1 = P2

21 2 1

Aw f f wA A A

GATE 101 Milestone Problems WithSolution On Mechanical Engineering

Publisher : Faculty Notes Author : Panel Of Experts

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