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CONTENTS VOL 1

ENGINEERING MECHANICS

AM 1 Equilibrium of Forces AM 3

AM 2 Structure AM 40

AM 3 Friction AM 81

AM 4 Virtual Work AM 117

AM 5 Kinematics of Particle AM 128

AM 6 Kinetics of Particles AM 157

AM 7 Plane Kinematics of Rigid body AM 190

AM 8 Plane Kinetics of Rigid body AM 206

STRENGTH OF MATERIALS

SM 1 Stress and Strain SM 3

SM 2 Axial Loading SM 41

SM 3 Torsion SM 86

SM 4 Shear Force and Bending Moment SM 118

SM 5 Transformation of Stress and Strain SM 179

SM 6 Design of Beams and Shafts SM 226

SM 7 Deflection of Beams and Shafts SM 270

SM 8 Column SM 315

SM 9 Energy Methods SM 354

THEORY OF MACHINES

TM 1 Analysis of Plane Mechanism TM 3

TM 2 Velocity and Acceleration TM 20

TM 3 Dynamic Analysis of Slider - Crank and Cam TM 38

TM 4 Gear - Trains TM 59

TM 5 Fly Wheel TM 91

TM 6 Vibration TM 109

MACHINES DESIGN

MD 1 Static and Dynamic Loading MD 3

MD 2 Joints MD 22

MD 3 Shaft and Shaft Components MD 54

MD 4 Spur Gears MD 71

MD 5 Bearings MD 88

MD 6 Clutch and Brakes MD 105

CONTENTS VOL 2

FLUID MECHANICS

FM 1 Basic Concepts and Properties of Fluids FM 3

FM 2 Pressure and Fluid Statics FM 33

FM 3 Fluid Kinematics & Bernouli Equation FM 80

FM 4 Flow Analysis Using Control Volumes FM 124

FM 5 Flow Analysis Using Differential Method FM 172

FM 6 Internal Flow FM 211

FM 7 External Flow FM 253

FM 8 Open Channel Flow FM 289

FM 9 Turbo Machinery FM 328

HEAT TRANSFER

HT 1 Basic Concepts & Modes of Heat-Transfer HT 3

HT 2 Fundamentals of Conduction HT 34

HT 3 Steady Heat Conduction HT 63

HT 4 Transient Heat Conduction HT 94

HT 5 Fundamentals of Convection HT 114

HT 6 Free and Force Convection HT 129

HT 7 Radiation Heat Transfer HT 155

HT 8 Heat Exchangers HT 181

THERMODYNAMICS

TD 1 Basic Concepts and Energy Analysis TD 3

TD 2 Properties of Pure Substances TD 28

TD 3 Energy Analysis of Closed System TD 52

TD 4 Mass and Energy Analysis of Control Volume TD 76

TD 5 Second Law of Thermodynamics TD 106

TD 6 Entropy TD 136

TD 7 Gas Power Cycles TD 166

TD 8 Vapor and Combined Power Cycles TD 199

TD 9 Refrigeration and Air Conditioning TD 226

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CONTENTS VOL 3

MANUFACTURING PROCESS

INDUSTRIAL ENGINEERING

OPERATION RESEARCH

CONTENTS VOL 4

ENGINEERING MATHEMATICS

EM 1 Linear Algebra EM 1

EM 2 Differential Calculus EM 24

EM 3 Integral Calculus EM 46

EM 4 Directional Derivatives EM 67

EM 5 Differential Equation EM 79

EM 6 Complex Variable EM 103

EM 7 Probability and Statistics EM 123

EM 8 Numerical Methods EM 142

VERBAL ANALYSIS

VA 1 Synonyms VA 1

VA 2 Antonyms VA 16

VA 3 Agreement VA 26

VA 4 Sentence Structure VA 37

VA 5 Spellings VA 58

VA 6 Sentence Completion VA 87

VA 7 Word Analogy VA 111

VA 8 Reading Comprehension VA 135

VA 9 Verbal Classification VA 148

VA 10 Critical Reasoning VA 153

VA 11 Verbal Deduction VA 168

QUANTITATIVE ANALYSIS

QA 1 Number System QA 1

QA 2 Surds, Indices and Logarithm QA 14

QA 3 Sequences and Series QA 28

QA 4 Average, Mixture and Alligation QA 44

QA 5 Ratio, Proportion and Variation QA 59

QA 6 Percentage QA 75

QA 7 Interest QA 89

QA 8 Time, Speed & Distance QA 99

QA 9 Time, Work & Wages QA 112

QA 10 Data Interpretation QA 126

QA 11 Number Series QA 145

SOLVED PAPER

SP 1 Engineering Mathematics SP 3

SP 2 Engineering Mechanics SP 65

SP 3 Strength of Materials SP 90

SP 4 Theory of Machines SP 138

SP 5 Machine Design SP 189

SP 6 Fluid Mechanics SP 218

SP 7 Heat Transfer SP 265

SP 8 Thermodynamics SP 303

SP 9 Refrigeration and Air-Conditioning SP 358

SP 10 Manufacturing Engineering SP 375

SP 11 Industrial Engineering SP 448

SP 12 General Aptitude SP 496

SM 1STRESS AND STRAIN

Common Data For Q. 1 and 2A long wire of tungsten ( 190 /kN mT

3g = ) hangs vertically from a high-altitude balloon, is shown in figure.

SM 1.1 If the ultimate strength (or breaking strength) is 1500 MPa, the greatest length that it can have without breaking, is(A) 3950 m (B) 7900 m

(C) 1975 m (D) 790 m

SM 1.2 If the same wire hangs from a ship at sea ( 10 / )kN msea water3g = , the greatest

length is(A) 8300 m (B) 2075 m

(C) 7500 m (D) 3750 m

Common Data For Q. 3 and 4The 650 N load is applied along the centroidal axis of the member as shown in figure. Take 60cq = .

SM 1.3 The resultant internal normal and shear forces in the member at section a a- , which passes through point A, is(A) N 0= , V 0= (B) 50 NN = , 650 NV =(C) N 0= , 650 NV = (D) 650 NN = , V 0=


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SM 1.4 The resultant internal normal and shear forces in the member at section b b- , which passes through point A, is(A) 325 NN = , 563 NV = (B) 650 NN = , 563 NV =(C) 563 NN = , 325 NV = (D) 325 NN = , 1126 NV =

SM 1.5 In the figure shown, link BC of 6 mm thickness is made of a steel with a 450 MPa ultimate strength in tension. If the structure is being designed to support a 20 kN load P with a factor of safety of 3, its width w should be

(A) 13.9 mm (B) 55.6 mm

(C) 27.8 mm (D) 41.7 mm

SM 1.6 In figure shown, the two-member frame is subjected to the distributed loading. Member CB has a square cross section of 35 mm on each side and take 8 /kN mw =. The average normal stress and average shear stress acting at section b-b , are

(A) 4.41MPas = , 5.88 MPat = (B) 11.76 MPas = , 4.41MPat =(C) 8.82 MPas = , 5.88 MPat = (D) 5.88 MPas = , 4.41MPat =

Common Data For Q. 8 and 9A solid bar of circular cross section has a hole of diameter /d 4 drilled laterally through the center of the bar as shown in figure below. The allowable average tensile stress on the net cross section of the bar is allows .

SM 1.7 The formula for the allowable load Pallow that the bar can carrying in tension, is(A) . d0 27 allow

2# s (B) . d0 54 allow

2# s

(C) . d0 675 allow2# s (D) . d0 54 allow# s


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SM 1.8 If the bar is made of brass with diameter 40 mmd = and 80 MPaallows = , the value of Pallow is(A) 86.5 kN (B) 70 kN

(C) 172 kN (D) 35 kN

SM 1.9 An axial load P is supported by a short 250 0.67W # column of cross-sectional area 8580 mmA 2= and is distributed to a concrete foundation by a square plate as shown in figure. If the average normal stress in the column must not exceed 150 MPa and the bearing stress on the concrete foundation must not exceed 12.5 MPa, the side a of the plate which will provide the most economical and safe design is

(A) 103 mm (B) 321 mm

(C) 8.6 mm (D) 160 mm

SM 1.10 The column shown in figure, is subjected to an axial force of 8 kN at its top. What is the average normal stress acting at section a -a ?

(A) 1.82 MPa (B) 3.64 MPa

(C) 0.91MPa (D) 2.73 MPa

SM 1.11 A round bar of 10 mm diameter is made of aluminum alloy, as shown in figure. When the bar is stretched by axial forces P , its diameter decreases by 0.016 mm. The magnitude of the load P is (Take 72 GPaE = , 0.33n = , 480 MPaYs = )

(A) 27.4 kN (B) 54.8 kN

(C) 13.7 kN (D) . kN37 7

SM 1.12 A steel bar of length 2.5 m with a square cross section 100 mm on each side is subjected to an axial tensile force of 1300 kN as shown in figure. The increase in





volume of the bar is (Take 250 MPaYs = , 200 GPaE = , .0 3n = )

(A) 8112 mm3 (B) 4868 mm3

(C) 3245 mm3 (D) 6490 mm3

Common Data For Q.14 and 15Three steel plates, each 16 mm thick, are joined by two 20 mm diameter rivets as shown in the figure.

SM 1.13 If the load 50 kNP = , the largest bearing stress acting on the rivets is(A) 39 MPa

(B) 156 MPa

(C) 78 MPa

(D) 117 MPa

SM 1.14 If the ultimate shear stress for the rivets is 180 ,MPa what force Pu is required to cause the rivets to fail in shear ? (Disregard friction between the plates.)(A) 170 kN (B) 57 kN

(C) 226 kN (D) 113 kN

SM 1.15 The small block of 5 mm thickness is shown in figure. If the stress distribution at the support developed by the load varies as shown, the force F applied to the block and the distance d to where it is applied, respectively, are

(A) 220 mm (B) 110 mm

(C) 165 mm (D) 55 mm

SM 1.16 The bar has a cross-sectional area of (400 10 )m6 2#

- . If it is subjected to a uniform axial distributed loading along its length and to two concentrated loads as shown in figure, the average normal stress in the bar as a function of x for





0 0.5 mx< # , is

(A) . . MPax47 5 20 0-^ h (B) 67.5 MPax

(C) . . MPax47 5 20 0+^ h (D) 27.5 MPax

Common Data For Q. 18 and 19In the figure shown, a hollow box beam ABC of length L is supported at end A by a 20 mm diameter pin that passes through the beam and its supporting pedestals. The roller support at B is located at distance /L 3 from end A.

SM 1.17 If load P is equal to 10 kN, the average shear stress in the pin is(A) 15.9 MPa (B) 31.8 MPa

(C) 63.6 MPa (D) 7.95 MPa

SM 1.18 If the wall thickness of the beam is equal to 12 mm, the average bearing stress between the pin and the box beam will be(A) 41.7 MPa (B) 125.1MPa

(C) 83.4 MPa (D) 20.85 MPa

SM 1.19 Rods AB and BC shown in figure, have diameters of 4 mm and 6 mm, respectively. The vertical load of 8 kN is applied to the ring at B . If the average normal stress in each rod is equivalent then this stress will be





(A) 237 MPa (B) 316 MPa

(C) 474 MPa (D) 158 MPa

Common Data For Q. 20 amd 21A steel plate ( /kN m77 3g = ) of dimensions 2.5 1.2 0.1 m# # is hoisted by a cable sling that has a clevis at each end as shown in figure. The pins through the clevises are 18 mm in diameter and are located 2.0 m apart. Each half of the cable is at an angle of 32c to the vertical.

SM 1.20 For above conditions, the average shear stress avert in the pins will be(A) 8.9 MPa (B) 6.7 MPa

(C) 13.4 MPa (D) 26.8 MPa

SM 1.21 The average bearing stress bs between the steel plate and the pins is(A) 22.7 MPa (B) . MPa15 2

(C) 7.57 MPa (D) 30.3 MPa

SM 1.22 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown in figure. If the average normal stress must not exceed 150 MPa in either rod, the smallest allowable values of the diameters d1 and d2 are





(A) 45.2 mmd1 = , 20.1 mmd2 = (B) 22.6 mmd1 = , 40.2 mmd2 =(C) 20.1 mmd1 = , 45.2 mmd2 = (D) 40.2 mmd1 = , 22.6 mmd2 =

SM 1.23 Members AB and AC of the truss as shown, consist of bars of square cross section made of the same alloy. It is known that a 20 mm square bar of the same alloy was tested to failure and that an ultimate load of 120 kN was recorded. If a factor of safety of 3.2 is to be achieved for both bars, the required dimension of the cross section of the bar AB is

(A) 27 mma = (B) 12 mma =(C) 13.5 mma = (D) 6 mma =

SM 1.24 The two steel members are joined together using a 60c scarf weld as shown in figure. The average normal and average shear stress resisted in the plane of the weld are

(A) 8 MPaavgs = , 4.62 MPaavgt = (B) 4.62 MPaavgs = , 8 MPaavgt =(C) 4.62 MPaavgs = , 16 MPaavgt = (D) 16 MPaavgs = , 4.62 MPaavgt =

SM 1.25 A steel pipe of 300 mm outer diameter is fabricated from 6 mm thick plate by welding along a helix which forms an angle of 25c with a plane perpendicular to the axis of the pipe. If a 250 kN axial force P is applied to the pipe, the normal and shearing stresses in directions respectively normal and tangential to the weld are





(A) 18.5 MPa- , 17.28 MPa (B) 37.1MPa- , 34.56 MPa

(C) 18.5 MPa- , 34.56 MPa (D) 37.1MPa- , 17.28 MPa

SM 1.26 A 6 kN load is supported by two wooden members of 75 125mm mm# uniform rectangular cross section which are joined by the simple glued scarf splice as shown in figure. The normal and shearing stresses in the glued splice respectively, are

(A) 565 kPa, 206 kPa (B) 282 kPa, 206 kPa

(C) 565 kPa, 103 kPa (D) 282 kPa, 103 kPa

SM 1.27 In the figure shown, the wooden members A and B are to be joined by plywood splice plates which will be fully glued on the surface in contact. If the clearance between the ends of the members is to be 8 mm and the average shearing stress in the glue is not to exceed 800 kPa, the smallest allowable length L will be

(A) 308 mm (B) 150 mm

(C) 300 mm (D) 292 mm





SM 1.28 In the figure shown, the frame is subjected to the distributed loading of 2 /kN m. What is the required diameter of the pins at A and B if the allowable shear stress for the material is 100 MPaallowt = ? Both pins are subjected to double shear.

(A) 2.6 mmd = (B) 7.8 mmd =(C) 5.2 mmd = (D) 10.4 mmd =

SM 1.29 A specially designed wrench is used to twist a circular shaft by means of a square key that fits into slots (or keyways) in the shaft and wrench as shown in the figure. The shaft has diameter ,d the key has a square cross section of dimensions b b# and the length of the key is c . The key fits half into the wrench and half into the shaft (i.e., the keyways have a depth equal to /b 2). When a load P is applied at distance L from the center of the shaft, the formula for the average shear stress avert in the key is( Hints : Disregard the effects of friction, assume that the bearing pressure between the key and the wrench is uniformly distributed)

(A) bc d b

PL2 +^ h

(B) bc d b

PL22

+^ h

(C) bc d b

PL23

+^ h (D)

bc d bPL

24

+^ h

SM 1.30 The two wooden members shown in figure supports a 20 kN load, are joined by plywood splices fully glued on the surface in contact. The ultimate shearing stress in the glue is 2.8 MPa and the clearance between the members is 8 mm. If





a factor of safety of 3.5 is to be achieved, the required length L of each splice is

(A) 216 mm (B) 208 mm

(C) 200 mm (D) 104 mm

SM 1.31 A torque T0 is transmitted between two flanged shafts by means of four 20 mm bolts as shown in figure. The diameter of the bolt circle is 150 mmd = . If the allowable shear stress in the bolts is 90 MPa, the maximum permissible torque will be

(A) 16.96 kN m- (B) 8.48 kN m-

(C) 12.72 kN m- (D) 4.24 kN m-

SM 1.32 The cross section of an aluminium tube serving as a compression brace in the fuselage of a small airplane is shown in the figure. The outer diameter of the tube is 25 mmd = and the wall thickness is 2.5 mmt = . If the factors of safety with respect to the yield stress and the ultimate stress are 4 and 5 respectively, the allowable compressive force Pallow is (Take 270 MPaYs = , 310 MPaus = )

(A) 9.5 kN (B) 12.0 kN

(C) 11.0 kN (D) 13.7 kN

SM 1.33 In the figure shown, a long steel wire ( 77.0 /kN m3g = ) hanging from a balloon carries a weight W at its lower end. The 4 mm diameter wire is 25 m long. The tensile yield stress for the wire is 350 MPaYs = and a margin of safety against yielding of 1.5 is desired. The maximum weight Wmax that can safety be carried is (Include the weight of the wire in the calculations.)





(A) 1783 N (B) 1711 N

(C) 1759 N (D) 1735 N

SM 1.34 What is the smallest dimensions of the circular shaft and circular end cap if the load it is required to support is 150 kNP = ? The allowable tensile stress, bearing stress and shear stress is ( ) 175 MPat allows = , ( ) 275 MPab allows = and

115 MPaallowt = .

(A) 15.8 mmd1 = , 26.4 mmd3 = , 44.6 mmt =(B) 26.4 mmd1 = , 44.6 mmd3 = , 15.8 mmt =(C) 44.6 mmd1 = , 26.4 mmd3 = , 15.8 mmt =(D) 44.6 mmd1 = , 15.8 mmd3 = , 26.4 mmt =

SM 1.35 The assembly shown in figure, consists of three disks A, B and C are used to support the load of 140 kN. The allowable bearing stress for the material is ( ) 350 MPab allows = and allowable shear stress is 125 MPaallowt = . The smallest diameter d1 of the top disk, the diameter d2 within the support space and the diameter d3 of the hole in the bottom disk are





(A) 27.6 mmd1 = , 22.6 mmd2 = , 35.7 mmd3 =(B) 22.6 mmd1 = , 35.7 mmd2 = , 27.6 mmd3 =(C) 22.6 mmd1 = , 27.6 mmd2 = , 35.7 mmd3 =(D) 35.7 mmd1 = , 22.6 mmd2 = , 27.6 mmd3 =

SM 1.36 In the structure shown, an 8 mm diameter pin is used at A and 12 mm diameter pins are used at B and D . The ultimate shearing stress is 100 MPa at all connections and the ultimate normal stress is 250 MPa in each of the two links joining B and D . If an overall factor of safety of 3.0 is desired, the allowable load P is

(A) 7.7 kN (B) 14.04 kN

(C) 3.97 kN (D) 3.72 kN

SM 1.37 The bar shown in figure, is held in equilibrium by the pin supports at A and B. The support at A has a single leaf and therefore it involves single shear in the pin and the support at B has a double leaf and therefore it involves double shear. The allowable shear stress for both the pins is 125 MPaallowt = . If 1 mx = and

12 /kN mw = , the smallest required diameter of pins A and B are (Neglect any axial force in the bar.)

(A) 957 mmdA = , 20.6 mmdB = (B) 10.3 mmdA = , 9.57 mmdB =(C) 19.14 mmdA = , 10.3 mmdB = (D) 9.57 mmdA = , 10.3 mmdB =

SM 1.38 Two plates, each 3 mm thick, are used to splice a plastic strip as shown below. If the ultimate shearing stress of the bonding between the surface is 900 kPa and

1500 NP = , the factor of safety with respect to shear will be

(A) 2 (B) 3.6

(C) 5.4 (D) 1.8





SM 1.39 The cable shown in figure has a specific weight g (weight/volume) and cross-sectional area A. If the sag s is small, so that its length is approximately L and its weight can be distributed uniformly along the horizontal axis, the average normal stress in the cable at its lowest point C is

(A) sL2

2

s g= (B) sL

8s g=

(C) sL8

2

s g= (D) sL4

2

s g=

SM 1.40 An elastomeric bearing pad consisting of two steel plates bonded to a chloroprene elastomer, is subjected to a shear force V during a static loading test as shown in figure. The pad has dimensions 150 mma = , 250 mmb = and the elastomer has thickness 50 mmt = . When the force V equals 12 kN, the top plate is found to have displaced laterally 8.0 mm with respect to the bottom plate.

The shear modulus of elasticity G of the chloroprene is(A) 0.5 MPa (B) 1MPa

(C) 4 MPa (D) 2 MPa

SM 1.41 A metal bar AB of weight W is suspended by a system of steel wires arranged as shown in the figure. The diameter of the wire is 2 mm and the yield stress of the steel is 450 MPa. The maximum permissible weight Wmax for a factor of safety of 1.9 with respect to yielding, is





(A) N685 (B) 10 N28

(C) 1370 N (D) 2740 N

Common Data For Q. 42 and 43In figure shown below, link BD consists of a single bar 30 mm wide and 12 mm thick. Each pin has a 10 mm diameter.

SM 1.42 If 0cq = , the maximum value of the average normal stress in link BD is (A) zero (B) 72 MPa

(C) 24 MPa (D) 48 MPa

SM 1.43 If 90cq = , the maximum value of the average normal stress in link BD is (A) 83 MPa (B) 125 MPa

(C) 42 MPa (D) 44.5 MPa

SM 1.44 The rigid beam AC shown in figure, is supported by a pin at A and wires BD and CE . If the load P on the beam causes the end C to be displaced 10 mm downward, the normal strain developed in wires CE and BD are

(A) .0 00025CEe = , 0.0107BDe = (B) .0 0025CEe = , .0 00107BDe =(C) .0 025CEe = , 0.0107BDe = (D) .0 00107CEe = , .0 0025BDe =

SM 1.45 The rigid beam shown in figure, is supported by a pin at A and wires BD and CE . If the load P on the beam is displaced 10 mm downward, the normal strain developed in wires CE and BD are





(A) .1 43 10CE3

#e = - , .3 58 10BD3

#e = -

(B) .1 43 10CE3

#e = - , .1 79 10BD3

#e = -

(C) .1 79 10CE3

#e = - , .1 43 10BD3

#e = -

(D) .2 86 10CE3

#e = - , .1 79 10BD3

#e = -

SM 1.46 A steel pipe is to carry an axial compressive load 1200 kNP = as shown in figure. A factor of safety of 1.8 against yielding is to be used. If the thickness t of the pipe is to be one-eighth of its outer diameter, the minimum required outer diameter dmin is (Take 270 MPaYs = )

.

(A) 153 mm (B) 76.5 mm

(C) 114.75 mm (D) 38.25 mm

Common Data For Q. 47 and 48A circular aluminum tube of length 400 mmL = is loaded in compression by forces P as shown in figure. The out-side and inside diameters are 60 mm and 50 mm, respectively. A strain gage is placed on the outside of the bar to measure normal strains in the longitudinal direction.

.

SM 1.47 If the measured strain is 550 10 6#e = - , the shortening d of the bar is

(A) 0.220 mm (B) 2.20 mm

(C) 0.022 mm (D) 1.10 mm

SM 1.48 If the compressive stress in the bar is intended to be 40 MPa, the load P should be(A) 17.35 kN (B) 34.6 kN

(C) 69.4 kN (D) 52.0 kN





Common Data For Q. 49 and 50A suspender on a suspension bridge consists of a cable that passes over the main cable is shown in figure and supports the bridge deck, which is far below. The suspender is held in position by a metal tie that is prevented from sliding downward by clamps around the suspender cable. Let P represent the load in each part of the suspender cable and q represent the angle of the suspender cable just above the tie. Also, let allows represent the allowable tensile stress in the metal tie.

SM 1.49 The minimum required cross-section area of the tie is

(A) tanA Pmin

allowsq= (B) /tanA P2min allowq s=

(C) /cotA P2min allowq s= (D) cotA Pmin

allowsq=

SM 1.50 If 130 kNP = , 75cq = and 80 MPaallows = , the minimum area will be(A) 6064 mm2 (B) 12130 mm2

(C) 435 mm2 (D) 870 mm2

SM 1.51 An elastomeric bearing pad consisting of two steel plates bonded to a chloroprene elastomer, is subjected to a shear force V during a static loading test as shown in figure. The pad has dimensions 150 mma = , 250 mmb = and the elastomer has thickness 50 mmt = . When the force V equals 12 kN, the top plate is found to have displaced laterally 8.0 mm with respect to the bottom plate.

The shear modulus of elasticity G of the chloroprene is(A) 0.5 MPa (B) 1MPa

(C) 4 MPa (D) 2 MPa

SM 1.52 Part of a control linkage for an airplane consists of a rigid member CBD and a





flexible cable AB . Originally the cable is unstretched. If a force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 /mm mm, the displacement of point D is

(A) 21.9 mm (B) 4.38 mm

(C) 43.8 mm (D) 8.76 mm

Common Data For Q. 53 and 54The material distorts into the dashed position is shown in figure.

SM 1.53 The average normal strains xe , ye and the shear strain xyg at A are(A) 0x ye e= = , 0.0798 radxyg =(B) 0xe = , .0 00319ye = , 0.0798 radxyg =(C) .0 00319xe = , 0ye = , 0.0798 radxyg =(D) .0 00319x ye e= = , 0.0798 radxyg =

SM 1.54 The average normal strain along line BE is(A) 0.179 /mm mm- (B) 0.0179 /mm mm

(C) 0.0179 /mm mm- (D) 0.179 /mm mm

SM 1.55 In the figure shown, the bar is originally 300 mm long when it is flat. It is subjected to a shear strain defined by . x0 02xyg = , where x is in millimeters. It is distorted into the shape shown, where no elongation of the bar occurs in the x direction. The displacement yD at the end of its bottom edge will be





(A) 2.03 mm (B) 4.06 mm

(C) 1.015 mm (D) 3.045 mm

Common Data For Q. 56 and 57The steel wires AB and AC support the 200 kg mass. The allowable axial stress for the wires is 130 MPaallows = . Take the unstretched length of AB to be 750 mm and 200 GPaEst = .

SM 1.56 The required diameter of each wire is(A) 3.23 mmdAC = , 7.08 mmdAB = (B) 3.54 mmdAC = , 3.23 mmdAB =(C) 3.23 mmdAC = , 3.54 mmdAB = (D) 6.46 mmdAC = , 3.54 mmdAB =

SM 1.57 What is the new length of wire AB after the load is applied ? (A) 749.51 mm (B) 750.49 mm

(C) 751 mm (D) 749.00 mm

SM 1.58 The plug shown in figure has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve are 50 mm long. What is the axial pressure p , that must be applied to the top of the plug to cause it to contact the sides of the sleeve ? (Take 5 MPaE = , .0 45n = )

(A) 926.25 kPa (B) 370 kPa

(C) 741 kPa (D) 555.75 kPa

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GATEMECHANICAL ENGINEERING

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Multiple Choice QuestionsGATE Mechanical Engineering Vol 2, 1e

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PREFACE

This book doesn’t make promise but provide complete satisfaction to the readers. The market scenario is confusing and readers don’t find the optimum quality books. This book provides complete set of problems appeared in competition exam as well as fresh set of problems.

The book is categorized into units which are sub-divided into chapters. The aim of the book is to avoid the unnecessary elaboration and highlights only those concepts are techniques which are absolutely necessary. Again time is crucial factor both from the point of view of preparation duration and time taken for solving each problem in the book are those which take the least distance to the solution.

But however to make a comment that the book is absolute for GATE preparation will be an inappropriate one. The theory for the preparation of the examination should be followed from the standard books.

Authors

SYLLABUS

ENGINEERING MATHEMATICSLinear Algebra: Matrix algebra, Systems of linear equations, Eigen values and eigen vectors.

Calculus: Functions of single variable, Limit, continuity and differentiability, Mean value theorems, Evaluation of definite and improper integrals, Partial derivatives, Total derivative, Maxima and minima, Gradient, Divergence and Curl, Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green’s theorems.

Differential equations: First order equations (linear and nonlinear), Higher order linear differential equations with constant coefficients, Cauchy’s and Euler’s equations, Initial and boundary value problems, Laplace transforms, Solutions of one dimensional heat and wave equations and Laplace equation.

Complex variables: Analytic functions, Cauchy’s integral theorem, Taylor and Laurent series.

Probability and Statistics: Definitions of probability and sampling theorems, Conditional probability, Mean, median, mode and standard deviation, Random variables, Poisson,Normal and Binomial distributions.

Numerical Methods: Numerical solutions of linear and non-linear algebraic equations Integration by trapezoidal and Simpson’s rule, single and multi-step methods for differential equations.

APPLIED MECHANICS AND DESIGNEngineering Mechanics: Free body diagrams and equilibrium; trusses and frames; virtual work; kinematics and dynamics of particles and of rigid bodies in plane motion, including impulse and momentum (linear and angular) and energy formulations; impact.

Strength of Materials: Stress and strain, stress-strain relationship and elastic constants, Mohr’s circle for plane stress and plane strain, thin cylinders; shear force and bending moment diagrams; bending and shear stresses; deflection of beams; torsion of circular shafts; Euler’s theory of columns; strain energy methods; thermal stresses.

Theory of Machines: Displacement, velocity and acceleration analysis of plane mechanisms; dynamic analysis of slider-crank mechanism; gear trains; flywheels.

Vibrations: Free and forced vibration of single degree of freedom systems; effect of damping; vibration isolation; resonance, critical speeds of shafts.

Design: Design for static and dynamic loading; failure theories; fatigue strength and the S-N diagram; principles of the design of machine elements such as bolted, riveted and welded joints, shafts, spur gears, rolling and sliding contact bearings, brakes and clutches.

FLUID MECHANICS AND THERMAL SCIENCESFluid Mechanics: Fluid properties; fluid statics, manometry, buoyancy; control-volume analysis of mass, momentum and energy; fluid acceleration; differential equations of continuity and momentum; Bernoulli’s equation; viscous flow of incompressible fluids; boundary layer; elementary turbulent flow; flow through pipes, head losses in pipes, bends etc.

Heat-Transfer: Modes of heat transfer; one dimensional heat conduction, resistance concept, electrical analogy, unsteady heat conduction, fins; dimensionless parameters in free and forced

convective heat transfer, various correlations for heat transfer in flow over flat plates and through pipes; thermal boundary layer; effect of turbulence; radiative heat transfer, black and grey surfaces, shape factors, network analysis; heat exchanger performance, LMTD and NTU methods.

Thermodynamics: Zeroth, First and Second laws of thermodynamics; thermodynamic system and processes; Carnot cycle. irreversibility and availability; behaviour of ideal and real gases, properties of pure substances, calculation of work and heat in ideal processes; analysis of thermodynamic cycles related to energy conversion.

Applications: Power Engineering: Steam Tables, Rankine, Brayton cycles with regeneration and reheat. I.C. Engines: air-standard Otto, Diesel cycles. Refrigeration and air-conditioning: Vapour refrigeration cycle, heat pumps, gas refrigeration, Reverse Brayton cycle; moist air: psychrometric chart, basic psychrometric processes. Turbomachinery: Pelton-wheel, Francis and Kaplan turbines — impulse and reaction principles, velocity diagrams.

MANUFACTURING AND INDUSTRIAL ENGINEERINGEngineering Materials: Structure and properties of engineering materials, heat treatment, stress-strain diagrams for engineering materials.

Metal Casting: Design of patterns, moulds and cores; solidification and cooling; riser and gating design, design considerations.

Forming: Plastic deformation and yield criteria; fundamentals of hot and cold working processes; load estimation for bulk (forging, rolling, extrusion, drawing) and sheet (shearing, deep drawing, bending) metal forming processes; principles of powder metallurgy.

Joining: Physics of welding, brazing and soldering; adhesive bonding; design considerations in welding.

Machining and Machine Tool Operations: Mechanics of machining, single and multi-point cutting tools, tool geometry and materials, tool life and wear; economics of machining; principles of non-traditional machining processes; principles of work holding, principles of design of jigs and fixtures

Metrology and Inspection: Limits, fits and tolerances; linear and angular measurements; comparators; gauge design; interferometry; form and finish measurement; alignment and testing methods; tolerance analysis in manufacturing and assembly.

Computer Integrated Manufacturing: Basic concepts of CAD/CAM and their integration tools.

Production Planning and Control: Forecasting models, aggregate production planning, scheduling, materials requirement planning.

Inventory Control: Deterministic and probabilistic models; safety stock inventory control systems.

Operations Research: Linear programming, simplex and duplex method, transportation, assignment, network flow models, simple queuing models, PERT and CPM.

GENERAL APTITUDEVerbal Ability: English grammar, sentence completion, verbal analogies, word groups, instructions, critical reasoning and verbal deduction.

Numerical Ability: Numerical computation, numerical estimation, numerical reasoning and data interpretation.

CONTENTS

FLUID MECHANICS


FM 2 Pressure and Fluid Statics FM 33

FM 3 Fluid Kinematics & Bernouli Equation FM 80

FM 4 Flow Analysis Using Control Volumes FM 124

FM 5 Flow Analysis Using Differential Method FM 172

FM 6 Internal Flow FM 211

FM 7 External Flow FM 253

FM 8 Open Channel Flow FM 289

FM 9 Turbo Machinery FM 328

HEAT TRANSFER

HT 1 Basic Concepts & Modes of Heat-Transfer HT 3

HT 2 Fundamentals of Conduction HT 34

HT 3 Steady Heat Conduction HT 63

HT 4 Transient Heat Conduction HT 94

HT 5 Fundamentals of Convection HT 114

HT 6 Free and Force Convection HT 129

HT 7 Radiation Heat Transfer HT 155

HT 8 Heat Exchangers HT 181

THERMODYNAMICS

TD 1 Basic Concepts and Energy Analysis TD 3

TD 2 Properties of Pure Substances TD 28

TD 3 Energy Analysis of Closed System TD 52

TD 4 Mass and Energy Analysis of Control Volume TD 76

TD 5 Second Law of Thermodynamics TD 106

TD 6 Entropy TD 136

TD 7 Gas Power Cycles TD 166

TD 8 Vapor and Combined Power Cycles TD 199

TD 9 Refrigeration and Air Conditioning TD 226

***********

FM 1BASIC CONCEPTS AND PROPERTIES OF FLUIDS

Common Data For Q. 1 and 2In an automobile tire the pressure is 245 kPa and the air temperature is 298 K. The volume of tire is 0.050 m3 and gas constant of air is 0.287 kPa- /m kgK3 .

FM 1.1 The pressure in the tire at air temperature of 32 K2 when volume of tire is constant, will be (A) 336 kPa (B) 26 kPa

(C) 310 kPa (D) 1854.02 kPa

FM 1.2 What amount of air should be come out to obtain pressure to its original value at same temperature ?(A) 0.1812 kg (B) 0.1672 kg

(C) 0.014 kg0 (D) 0.3484 kg

FM 1.3 Consider Carbon dioxide at 1 atm2 and 400 Cc . What will be the density of Carbon dioxide and cp at this state and the new pressure when the gas is cooled isentropically to 150 Cc ? (For Carbon dioxide .k ��= and �� m s �R 2 2= )

(A) 0.797 /kg m3ρ = , 4�.�kg �

�cp−

= , ��kPap2 =

(B) 1.3 10 /kg m4 3ρ #= - , ��kg �

�cp−

= , ��5.5 kPap2 =

(C) 7.97 /kg m3ρ = , ��kg �

�cp−

= , ��5.5 kPap2 =

(D) 7.97 /kg m3ρ = , ��kg �

�cp−

= , ��5.5 �Pap2 =

FM 1.4 A Cane of beverage contains 455 ml of liquid. The mass of cane with liquid is 0.369 kg while an empty cane weighs 0.1 3 N9 . What will be the specific weight, density and specific gravity of liquid respectively ?(A) 0.977 /kN m3, 99.6 /kg m3, .0 0996

(B) 9.77 /kN m3, 996 /kg m3, .0 996

(C) 9.77 /N m3, 996 /kg m3, .9 96

(D) 97.7 /kN m3, 996 /kg m3, .0 996

FM 1.5 The specific gravity of a gas contained in a tank at the temperature of 25 Cc is 2 10 3#

− . If the atmospheric pressure is 10.1 kPa , the gage pressure is(A) 70 kPa (B) 7 kPa

(C) 0.7 kPa (D) 70 kPa

FM 1.6 Consider steam at state near the saturation line : ( , )p T1 1 (1.31 , 250 )MPa Cc= , 4�� .��)m s �andR ksteam

2 2−= = . If the steam expands isentropically to a new

pressure of 414 kPa, what will be the density 1ρ and the density 2ρ ?(A) 5.44 / , 5.04 /kg m kg m1

32

3ρ ρ= = (B) 2.28 / , 5.44 /kg m kg m13

23ρ ρ= =

(C) 5.44 / , 2.28 /kg m kg m13

23ρ ρ= = (D) 5.04 / , 5.44 /kg m kg m1

32

3ρ ρ= =

FM 1.7 A 30 m3 cylinder contains Hydrogen at 25 Cc and 200 kPa What amount of


GATE MCQ Mechanical Engineering (4-volumes) by NODIA and COMPANY



Hydrogen must be bled off to maintain the Hydrogen in cylinder at 20 Cc and 600 kPa ? ( 0.2968 . / . )kPa m kg KR 3=(A) 271.35 kg (B) 206.99 kg

(C) 478.34 kg (D) 64.36 kg

FM 1.8 Wet air with 100% relative humidity, is at 30 Cc and 1 atm. If �� m s �Rair� �

−=, Rwater 461 /m s K2 2

−= and vapor pressure of saturated water at 30 Cc is 4242 Pa, what will be the density of this wet air using Dalton’s law of partial Pressures ?(A) 1.12 /kg m3 (B) 1.09 /kg m3

(C) 0.03 /kg m3 (D) 1.147 /kg m3

FM 1.9 In a formula one race, at the start of the race the absolute pressure of a car tire is 362.5 kPa and at the end of the race the absolute pressure of car tire is measured to be 387.5 kPa . If the volume of the tire remains constant at 0.022 m3 then percentage increase in the absolute temperature of the air in the tire is(A) 6.9% (B) 69%

(C) 0.69% (D) Not increased

FM 1.10 A compressed air tank contains 24 kg of air at a temperature of 80 Cc . If the reading of gage mounted on the tank is 300 kPa, what will be the volume of tank in m3 ?(A) 404 (B) 4.04

(C) 0.404 (D) 40.4

FM 1.11 A small submersible moves in 30 Cc water ( 4.242 kPapv = ) at 2-m depth, where ambient pressure is 133 kPa. Its critical cavitation number is .Ca 0 2�. . At what velocity will cavitation bubbles form ?(A) 22.72 m/s (B) 32.66 m/s

(C) Zero (D) 32.13 m/s

FM 1.12 What will be the speed of sound of steam at 150 Cc and 400 kpa? (k = 1.33, R = 461 /m s K2 2

− )(A) 50.9 m/s (B) 509 m/s

(C) 30.3 m/s (D) 303 m/s

FM 1.13 A liquid has a weight density of 9 /N m268 3 and dynamic viscosity of . /N s m131 5 2−

. What will be the kinematic viscosity of the liquid in /secm2 ?(A) 0.0139 (B) 1.39

(C) 0.139 (D) 13.9

FM 1.14 A 72 m long and 30 m diameter blimp is approximated by a prolate spheroid

whose volume is given by v LR32 2p= . The weight of 20 Cc gas within the blimp

for (a) helium at 1.1 atm and (b) air at 1.0 atm, is ( 2077 / �m sRHe

2 2−= , Rair 287 /m s K2 2

−= )(A) 60.97kNWHe = , 401.1kNWair = (B) 401.1kNWHe = , 6.97kNWair =(C) 6.2kNWHe = , 40.9kNWair = (D) 40.9kNWHe = , 6.2kNWair =

FM 1.15 The oil having viscosity of 4.56 10 /N s m2 2# −

− , is contained between two parallel plates. The bottom plate is fixed and upper plate moves when a force F is applied. If the distance between the stationary and moving plates is 2.54 mm and the area of the upper plate is . m0 129 2, what value of F is required to translate


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the plate with velocity of 1 / secm ?(A) . N2 32 (B) 23.2 N

(C) 232 N (D) 0.232 N

FM 1.16 A thin moving plate is separated from two fixed plates by two fluids of different viscosity as shown in figure below. If the contact area is A, the force required for the flow to be steady laminar viscous flow, is

(A) F h h VA�

�

2

2m m= +; E (B) F h h VA

2

2

�

�m m= −: D

(C) F h h VA2

2

�

�m m= −; E (D) F h h VA

�

�

2

2m m= +: D

FM 1.17 A large movable plate is located between two large fixed plates. Two fluids having the different viscosities are contained between the plates. If the moving plate has a velocity of 6 /secm , what will be the magnitude of the shearing stresses on plate 1 and plate 2 respectively, that act on the fixed plates ?

(A) 10 /N m2, 15 /N m2 (B) 20 /N m2, 15 /N m2

(C) 15 /N m2, 15 /N m2 (D) 15 /N m2, 20 /N m2

FM 1.18 A thin flat plate of area A is moved horizontally between two plates, one stationary and one moving with a constant velocity Vm as shown in figure below. If velocity of flat plate is Vp and dynamic viscosity of oil is μ, the force must be applied on the plate to manage this motion is

(A) A hV

hV Vp p m

1 2μ + -

; E (B) ( )�A V V hp m 2μ -

(C) hAVp

1

μ (D)

( )A h

Vh

V Vp p m

1 2μ - -

; E

FM 1.19 A Newtonian fluid having the specific gravity of 0.91 and Kinematic viscosity of 4 10 / secm4 2#

− , flows over a fixed surface. The velocity profile near the surface





is given by the relation:

Uu sin y

2dp= a k

What will be the magnitude of the shearing stress developed on the plate in term of U and δ ?

(A) 0.571 /N mU 2

δ (B) 5.71 /N mU2δ

(C) 5.71 /N mU 2

δ (D) 0.571 /N mU2δ

FM 1.20 A 50 30 20cm cm cm# # block of 15 kg mass is to be moving at a constant velocity of 0.8 /m s on an inclined plane. If a 0.8 mm thick oil film with a dynamic viscosity of 0.006 Pa s− is there between the block and inclined plane, what amount of force is required in x -direction ? ( 10 / )m sg 2=

(A) 55 N (B) 55.55 N

(C) 6.42 N (D) 414.75 N

FM 1.21 A closed rectangular container is half filled with water at 45 Cc . If the air in remaining half section of container is completely escaped. The absolute pressure in the escaped space at same temperature (saturation pressure of water at 45 Cc is9.593 kPa) is(A) P P> saturation (B) P P< saturation

(C) P Psaturation= (D) Not determined

FM 1.22 Consider two parallel plates as shown in figure below. If the fluid is glycerin (ρ1264 /kg m3= , μ 1.5 /N s m2

−= ) and the distance between plates is 9 mm. What will be the shear stress required to move the upper plate at � �m sV = and the Reynolds number respectively ?

(A) 100 , 4Pa 60 (B) 10 , 4Pa 600

(C) 10000 , 4.Pa 6 (D) 1000 , 46Pa

FM 1.23 The velocity profile in a pipe flow is given by ( � )u u r R� n n�= − , where r is the

radial distance from the centre. If the viscosity of the fluid is μ then the drag force applied by the fluid on the pipe wall in the direction of flow across length L





of the pipe is ( )radius of circular pipeR = .

(A) n u L�π μ (B) �n u R�μ(C) n u L2 0πμ (D) n u2 0πμ

FM 1.24 Consider air at 20 Cc with 1.8 10 sPa5μ # -= - . Its viscosity at 400°C by (a) The Power-law (n=0.7) (b) the sutherland law (S = 110 K) respectively, are(A) �.��1 10 � , 1.� 10 �� s �� sp s

� �μ μ# #- -= =- -

(B) �.��1 10 � , �.�� 10 �� s �� sp s� �μ μ# #- -= =- -

(C) �.�� 10 � , �.��1 10 �� s �� sp s� �μ μ# #- -= =- -

(D) 1.� 10 � , �.��1 10 �� s �� sp s� �μ μ# #- -= =- -

FM 1.25 Consider a block of mass m slides down on an inclined plane of a thin oil film as shown in figure below. The film contact area is A and its thickness is h . The terminal velocity V of the block is

(A) sinV Amgh

mq= (B) cosV A

mghm

q=

(C) sinV hmgA

mq= (D) cosV h

mgAm

q=

FM 1.26 A thin layer of glycerin flows down on an inclined plate of unit width with the velocity distribution:

Uu h

yhy�

�

�

= −

If the plate is inclined at an angle α with the horizontal, the expression for the surface velocity U will be





(A) sinU h��

gma= (B) sinU h

�

�

m ag=

(C) sinU h�

�

mg a= (D) sinU h� �

m ag=

FM 1.27 A shaft of 8.0 cm diameter and 30 cm length is pulled steadily at �� m sV = through a sleeve of 8.02 cm diameter. The clearance is filled with oil of 0.003 /m s2ν = and . . .S G ��= , the force required to pull the shaft is ( 998 / )kg mw

3ρ =(A) 793 N (B) 795 N

(C) 79.3 N (D) 7.95 N

FM 1.28 Match List I (Properties of fluids) with List II (Definition/ Result) and select the correct answer using the codes given below :

List-I List-II

a. Ideal fluid 1. Viscosity does not vary with rate of deformation

b. Newtonian fluid 2. Fluid of zero viscosity

c. /μ ρ 3. Dynamic viscosity

d. Mercury in glass 4. Capillary depression

5. Kinematic viscosity

6. Capillary rise

Codes a b c d(A) 1 2 4 6(B) 1 2 3 4(C) 2 1 3 6(D) 2 1 5 4

FM 1.29 Match List I (Fluid properties) with List II (Related terms) and select the correct answer using the codes given below :

List-I List-II

a. Capillarity 1. Cavitation

b. Vapour pressure 2. Density of water

c. Viscosity 3. Shear forces

d. Specific gravity 4. Surfaces Tension

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

FM 1.30 The hydrogen bubbles have diameter �.�1 mmD - . Assume an ‘‘air-water” interface at 30 Cc and surface tension 0.0712 /N mσ = . What will be the excess pressure within the bubble ?(A) 1.42 kPa (B) 2.85 kPa

(C) 28.5 kPa (D) 14.2 kPa

FM 1.31 The surface tension in a rain drop of 3 mm diameter is 7.3 10 /N m2#

− . The





excess pressure inside the rain drop is(A) 973.3 Pa (B) 97.33 Pa

(C) 9.73 Pa (D) 97.33 kPa

FM 1.32 A shower head emits a cylindrical water jet of diameter 0.73 mm into air. The pressure inside the jet is approximately 300 Pa greater than the air pressure. What will be the surface tension of water ?(A) 0.0365 N/m (B) 0.73 N/m

(C) 0.365 N/m (D) 0.073 N/m

FM 1.33 A thin wire ring of 6 cm diameter is lifted from a 20 Cc water surface. How much lift force is required if 0.0728 /N mσ = ?(A) 0.274 N (B) 0.0274 N

(C) 0.137 N (D) 0.0137 N

FM 1.34 A 4 mm diameter glass tube is immersed in water and mercury. The temperature of the liquid is 20 Cc and the values of the surface tension of water and mercury at 20 Cc in contact with air are 0.0734 /N m and 0.51 /N m, respectively. The angle of contact for water is zero and that for mercury is 128c. What will be the capillary effect for water and mercury in millimeters, respectively ?(A) 4.60, 3.82 (B) 2.35, 7.48

(C) 3.82, 4.60 (D) 7.48, 2.35

FM 1.35 The system shown in figure below is used to estimate the pressure inside the tank by measuring the height of liquid in the 1 mm diameter tube. The fluid is at 60 Cc . What will be the capillary rise if the fluid is (a) water ( 0.0662 /N mσ =, 983 /kg m2ρ = , 0c,θ ) and (b) Mercury ( 0.47 /N mσ = , 13500 /kg m3ρ = ,

130c,θ ) ?

(A) 0.0275mhw = , 0.0456 mhm =− (B) 0.0275 mhw =− , 0.00�1mhm =(C) 0.0275 mhw = , 0.00�1 mhm =− (D) 0.0137 mhw = , 0.00456 mhm =−

FM 1.36 A glass tube of 4.6 mm diameter is inserted into milk and milk rises upto 3.5 mm in the tube. If the density of milk is 960 /kg m3 and contact angle is 15c, the surface tension of milk is(A) 0.2315 /N m (B) 0.025 /N m

(C) 0.0236 /N m (D) 0.02315 /N m

FM 1.37 A liquid film suspended on a rectangle wire frame of one movable side of 12 cm. What amount of surface tension is required if the movable side of frame is to be moved with 0.018 N ?(A) 0.075 /N m (B) 0.00432 /N m

(C) 0.055 /N m (D) 0.75 /N m





FM 1.38 In figure shown, a vertical concentric annulus with outer radius ro and inner radius ri is lowered into the fluid of surface tension σ and contact angle 45< cθ. If the gap is very narrow, what will be the expression for the capillary rise h in the annulus gap ?

(A) ( )

coshg r r�

o i�r

s q=−

(B) ( )cosh

g r r�

o irs q= −

(C) ( )cosh

g r ro irs q= − (D)

( )cosh

g r r�

o i� �r

s q=−

FM 1.39 A solid cylindrical needle of diameter 1.6 mm and density 7824 /kg m3 may float on a liquid surface. Neglect buoyancy and assume a contact angle of 0c. What will be the surface tension σ ?(A) 0.0772 N/m (B) 0.154 N/m

(C) 0.772 N/m (D) 0.0154 N/m

Common Data For Linked Answer Q. 40 and 41A Frustum-shaped body is rotating at a uniform angular velocity 200 /rad sω = in a container. The gap of 1.2 mm on all sides between body and container is filled with oil of viscosity 0.1 Pa s− at 2 C0c .

FM 1.40 The power required at the top surface to maintain this motion is

(A) hD

24

2 3πμω (B) hD

32

2 4πμω

(C) hD

4

2 4πμω (D) hD

16

2 2πμω

FM 1.41 The reduction in power required at the top surface when oil viscosity is 0.0078 Pa s− at 8 C0c , will be(A) 5.29 W (B) 67.824 W

(C) 62.533 W (D) No reduction





FM 1.42 A fluid of surface tension 0.0728 /N mσ = and contact angle 0cθ = is filled between 0.75 mm apart two parallel plates as shown in figure. If the density of fluid is 998 /kg m3ρ = , the capillary height h will be

(A) 2 mm (B) 10 mm

(C) 20 mm (D) 1 mm

FM 1.43 A 56 kg block slides down on a smooth inclined plate. A gap of 0.1 mm between the block and plate contains oil having viscosity 0.4 /N s m2

− . If the velocity distribution in the gap is linear and the area of the block in contact with the oil is 0.4 m2, the terminal velocity of the block is(A) 0.03125 /m s (B) 0.3125 /m s

(C) 3.125 /m s (D) 0.03125 /mm s

FM 1.44 Two 50 cm long concentric cylinders are mounted on a shaft. The inner cylinder is completely submerged in fluid and is rotating at 200 rpm and the outer cylinder is fixed. The fluid film thickness between two cylinders is 0.1 cm2 and outer diameter of the inner cylinder is 20 cm. If the torque transmitted by the shaft to rotate inner cylinder is 0.8 N, the viscosity of the fluid is

(A) 0.0173 /N s m2− (B) 0.0231 /N s m2

−

(C) 0.173 /N s m2− (D) 0.0346 /N s m2

−

FM 1.45 A layer of water having the viscosity of 1.2 10 /N s m3 2# −

− flows down on inclined fixed surface with the velocity distribution as given by:





Uu h

yhy�

�

�

= −

If the velocity of water � �secmU = and ��mh = , what will be the magnitude of the shearing stress that the water exerts on the fixed surface in /N m2 ?

(A) 7.20 (B) 0.720

(C) .7 2 10 3#

− (D) .0 072

FM 1.46 A 2.5 mm diameter aluminum sphere ( 2700 /kg m3ρ = ) falls into an oil of density 875 /kg m3. If the time to fall 75 cm is 48 s then the oil viscosity is(A) 0.0589 /kg m s−

(B) 0.589 /kg m s−

(C) 0.397 /kg m s−

(D) 0.0397 /kg m s−

FM 1.47 Consider a concentric shaft fixed axially and rotates inside the sleeve. If the shaft of radius ri rotates at /rad sω inside the sleeve of radius r0 and length L and the applied Torque is T, what will be the relation for the viscosity μ of the fluid between shaft and sleeve ?

(A) ( )

r LT r r2

i

i�

0μπω

= - (B)

( )��

T � �2

�

03

0μπω

= -

(C) ( )

��T � �2 �

�3

0μπω

= - (D)

( )��

T � �2 �

�3

0μπω

= +

FM 1.48 The velocity profile for laminar one-dimensional flow through a circular pipe is given as ( ) ( � )u r u r R�max

2 2= − , where R is the radius of the pipe and r is the radial distance from the centre of the pipe. If an oil at 40 Cc flows through a 15 m long pipe with 0.0�mR = and maximum velocity of � �m sumax = , what will be the friction drag force applied by the fluid on inner surface of the pipe when

0.0010 /kg m sμ -= ?

(A) 0.0942 N (B) 0.942 N

(C) 0.856 N (D) 0.916 N





FM 1.49 A 1 m diameter cylindrical tank has a length of 5 m long and weight 125 N. If it is filled with a liquid having a specific weight of 10.9 /kN m3, the vertical force required to give the tank an upward acceleration of 2.75 / secm 2 is(A) 550 kN (B) N55

(C) 5. N5 (D) 55 kN

FM 1.50 A cylindrical rod of diameter D , length L and density sρ falls due to gravity inside a tube of diameter Do . The clearance, ( )D D D<<o − is filled with a film of viscous fluid ( , )ρ μ .The expression for terminal fall velocity would be

(A) ( )

V gD D D

�sor m= −

(B) ( )

VgD D D

�s o

mr= +

(C) ( )

Vg D D�s o

mr= −

(D) ( )

VgD D D

�s o

mr= −

FM 1.51 The belt as shown in figure below moves at steady velocity of 2.5 /m s and skims the top of a tank of oil SAE 30 W ( 0.29 / )kg m sμ -= at 20 Cc with �mL = ,

�0 cmb = and �cmh = . What power P in watts is required to remain belt in motion ?

(A) 11 Watts (B) 44 Watts

(C) 109 Watts (D) 1.1 Watts

FM 1.52 Two balls of Steel and Aluminum can float on water due to surface tension effect. The density of steel and aluminium balls are to be 7800 /kg m3 and 2700 /kg m3, respectively. Which metal ball would have maximum diameter to float on water at 2 C0c and what will be the diameter of that ball when surface tension of water at 20 Cc is 0.073 /N m ?(A) steel, 4. mm1 (B) Aluminium, 2.4 mm

(C) Aluminium, 4.1 mm (D) Steel, 2. mm4

FM 1.53 For a cone-plate viscometer of radius �cmR = , the angle 3cθ = and the gap is filled with liquid as shown in figure. If the viscous torque .��T �= and rotation rate is 94.2 /rad s, the liquid viscosity will be

(A) 0.0116 /kg m s−

(B) 0.116 /kg m s−

(C) 0.193 /kg m s−

(D) 0.0193 /kg m s−

FM 1.54 A solid cone of base r0 and initial angular velocity 0ω is rotating inside a conical seat as shown in figure below. If there is no applied torque and air drag is





neglected, the cone’s angular velocity ω is

(A) exp sinr t

mh5

30

03ω ω

μθ= -; E (B) exp sinmh

r t35

003

ω ω θμ= -; E

(C) exp sinmhr t

35

004

ω ω θμ= -; E (D) exp sinmh

r t35

002

ω ω θπμ= -; E

FM 1.55 The rotating-cylinder viscometer as shown in figure below shears the fluid in a narrow clearance ( )R r RΔ = - with a linear velocity distribution in the gap. If the driving torque measured is T and the bottom friction is included then the expression for μ is

(A) ( �)( )

R L RT r R

��μπω

=+-

(B) ( / )

( )� � ��

2 43μπω

=+

-

(C) ( / )

( )� � ��

2 42μπω

=+

- (D)

( / )( )

� � ��

2 43μπω

=-

-

FM 1.56 For a 300 mm long sliding lubricated bearing, the viscosity of oil is 0.008 /kg m s− during steady operation at 80 Cc . The average oil film thickness between the shaft and journal is 1.2 mm. If shaft of 80 mm diameter is rotated at 750 rpm, the amount of torque needed to overcome bearing friction would be(A) 0.0063 N m− (B) 0.063 N m−

(C) 0.63 N m− (D) 6.3 N m−

FM 1.57 0.063 N m−= A disk of radius �cmR = , rotates at 1200 . .r p m inside an oil container of viscosity 0.29 /kg m sμ -= as shown in figure below. The oil film thickness is �mmh = . If the velocity profile is linear and neglecting shear on the outer disk edges, the viscous torque on the disk is





(A) 0.716 N m− (B) 6.83 N m−

(C) 0.0716 N m− (D) 14.3 N m−

FM 1.58 A soap bubble of diameter D� coalesces with another bubble of diameter D� to form a single bubble D� with the same amount of air. For an isothermal process, D� as a function of ,D D� �, patm and surface tension σ is(A) � ( � ) ( � )p D D p D D p D Da a a�

��

��

��

��

��s s s+ = + + +

(B) � ( � ) ( � )p D D p D D p D Da a a��

��

��

��

��

��s s s+ = + − +

(C) � ( � ) ( � )p D D p D D p D Da a a��

��

��

��

��

��s s s− = − + −

(D) � ( � ) ( � )p D D p D D p D Da a a��

��

��

��

��

��s s s+ = + + −

FM 1.59 A skater of mass m moving at constant speed Vo , suddenly stands stiff with skates pointed directly forward and allows herself to coast to a stop. If blade length is L, water film thickness h , water viscosity μ and blade width is b then how far will she travel (on two blades) before she stops ?

(A) x LbV mho

m= (B) x V mhLb�

o

m=

(C) x LbV mh�

o

m= (D)x V mhLb

o

m=

FM 1.60 Two thin flat plates are tilted at an angle φ and placed in a tank of surface tension σ and contact angle θ as shown in figure below. At the free surface of the liquid in the tank, the distance between two plates are L and width is b into the paper. What will be the expression for σ in terms of other variables ?

(A) ( )

( )cos

tangbh L h2

σ θ φρ φ= -

- (B)

( )cos

tangh L h2σ φ

ρ φ= -

(C) ( )

( )cos

tangh L h2

σ θ φρ φ= -

- (D)

( )( )cos

tangh L h2

σ θ φρ φ= +

+

***********





SOLUTIONS

FM 1.1 Option (A) is correct.

We have p� 310 ,kPa= �.�� mv��= �.��aR = - /m kgK3

T� 298 K= and ��T�=Treating air as an Ideal gas, the final pressure in the tire from the ideal gas law,

Tp v

�

� � Tp v

�

� �=

p� TT p

�

��#= 298

323 310#= 336 kPa= ( )tancons tv v� �=

FM 1.2 Option (C) is correct.Amount of air needs to be bled off to restore pressure ( 310 )kPap2 = is

m m m� �Δ = -

m RTp v

��

�= .. 0.1812 kg0 287 298

310 0 050#

#= =

and m RTp v

��

�= .. 0.1672 kg0 287 323

310 0 050#

#= =

Hence mΔ 0.1812 0.1672 0.014 kg= − =

FM 1.3 Option (C) is correct.

We have p� 10 1013250atm Pa= = T� 400 400 273 673C Kc= = + =From ideal gas law

ρ ( ) ( )

�.��m�g

RTp

��

�

��

#= = =

cp .. ( )

��g �

�kkR

� �� #

−= − = − =

For gas cooled isentropically to �� C �T� c= = , the formula is

pp

�

� TT �k k

�

��

=−

b l

p� ��ap TT

��

..k k

��

��

��

# #= =− −

b bl l 135.5 kPa=

FM 1.4 Option (B) is correct.

Specific weight γ Volume of fluidWeight of fluid= g v

mgv

Wγ ρ= = =

Volume of fluidTotal weight weight of Cane= −

. . . .mg355 10

0 153355 10

0 369 9 81 0 1536 6

# #

#= − = −− −

. 9.77 /kN m355 10

3 476

3

#= =−

Density ρ . ��.�� g m �g mg �� -

g= = =





Specific gravity �.S G ��

waterrr= = =

FM 1.5 Option (D) is correct.We have . � ��.S G �

#= − , �� (�� ) �� T �c= = + = , �� ap �.atm =Density of gas ρ . . Density of �aterS G #= 2 10 1000 2 /kg m3 3

# #= =−

From gas equation p RTr= 2 287 298 171 kPa# #= = (absolute pressure)

Also pabsolute p patmospheric gage= + pgage 171 101= − 70 kPa=


For ideal gas 1ρ RTp

�

�= ( )

. . 5.44 /kg m461 273 250

1 31 10461 5431 31 106 6

3

##

##= + = =

For isentropic expansion

TT

�

� pp k

k

�

��

=−

b l

T� ��.

T pp

�� .

.

kk

��

��

�

� ��

#

## #= =

− −

b cl m 393 K=

Now 2ρ RTp

�

�= 2.28 /kg m461 393414 103

3

##= =

FM 1.7 Option (D) is correct.We have �� ,mv �= �� ,��ap�= �� C �T� c= = + = �� ,��ap�= �� C �T� c= = + =The initial mass of Hydrogen in cylinder

m RTp v

��

�= . 271.35 kg0 2968 298800 30

##= =

Final mass of Hydrogen in cylinder

m RTp v

��

�= . 206.99 kg0 2968 293600 30

##= =

Thus the amount of Hydrogen that must be bled off is

mΔ . .271 35 206 99= − 64.36 kg=

FM 1.8 Option (D) is correct.Dalton’s law of Partial Pressure is

ptotal p p vm R T v

m R Tair watera

aw

w= + = +

or mtotal m m R Tp v

R Tp v

a wa

a

w

w= + = + For an ideal gas

Since ptotal � ��atm Pap pair water= + = = pair 101325 101325 4242 97083 Papwater= − = − =

Now ρ vm m

R Tp

R Tpa w

a

a

w

w= + = +

( )287 30 273

97083461 303

4242# #

= + + 287 30397083

461 3034242

# #= +

1.147 /kg m3=





FM 1.9 Option (A) is correct.We have ��.� ��ap�= ��.� ��ap�= �. ��mv v v ��

�= = =From ideal gas law,

Tp v

�

� � Tp v

�

� �=

TT

�

� pp

�

�= v v� �=

T pp T�

�

��#= .

. .T T362 5387 5 1 0691 1#= =

Increase in temperature T T� �= − 1.069 (1.069 1)T T T1 1 1= − = − . T0 069 1=or 6.9% of T1

FM 1.10 Option (B) is correctWe have ��gm = , �� (�� ) �� T c= = + = , pgage 300 kPa=

From gas equation ρ RTpabsolute=

( )

RTp p

�� .atm gage

�

#

#= + = + 3.96 /kg m3=

Thus, Volume of tank v mr= . 4.04 m3 96

16 3= =

FM 1.11 Option (D) is correct

By definition Cacritical 0.25( )

Vp p2 a v

2r= = −

.0 25 ( )

V9982 133000 4242

2#

= −

V .( )

998 0 252 133000 4242

#= −

32.13 /m s=

FM 1.12 Option (B) is correctThe ideal gas formula Predicts:

Speed of sound a . ( )kRT �� # #, = +

. 509 /m s1 33 461 423# #= =

FM 1.13 Option (C) is correct.We have 9 /N m268 3γ = , . /Ns m131 5 2μ =Weight density γ gr=

ρ . . /kg m9 819268 944 75 3= =

Kinematic viscosity ν ..

944 75131 5

rm= =

0. . /secN m kg139= 0. / .secm139 2=

FM 1.14 Option (A) is correct.The volume of blimp is

v ( )R L32

32 15 722 2# # #p p= = 33929 m3=

From the ideal gas law, the respective densities of helium and air





(a) Heρ � ( )

�� R Tp

��

He

He �

#

#= = =

(b) airρ �� R Tp

��

air

air �

# #= = =

Then the respective gas weights are

WHe �� gvHer # #= = = Wair �� gvairr # #= = =


We have .5 10 /Ns m4 6 2 2μ #= - , . �y �� #= − , � .sec�V �= , . �A �� =

When force F is applied on the plate, shear force comes in the action.

F A#t= yV A# #m= y

Vτ μ=

4.56 10.

0.1292 54 10

123

## # #= −

− . N2 32=

FM 1.16 Option (D) is correct.Assuming a linear velocity distribution on each side of the plate.

F A A1 2t t= + hV A h

V A11

22

m m= +b bl l

hV

hV A h h VA

�

�

�

�

�

�

�

�m m m m= + = +; :E D

FM 1.17 Option (C) is correct.From Newton’s law of viscosity

τ dydu

yUm m= =

1τ . .0 02 0 0086

#= 15 /N m2=

and 2τ . .0 01 0 0046

#= 15 /N m2=

FM 1.18 Option (A) is correct.The magnitudes of shear forces acting on the upper and lower surfaces of the plate are

F ,shear upper A A dydu

,w uppert m= =

A hV

hAV�p p

� �m m= − =

F ,shear lower A A dydu

,w lowert m= =

( )

A hV Vp m

�m= −

Both shear forces are acting in the opposite direction of motion of the plate,





therefore from force balancing

F F F� �shear upper shear lower= +

� �

hAV

hA V Vp p m

� �

m m= + −

A hV

hV Vp p m

� �m= + −

; E

FM 1.19 Option (A) is correct.We have �.��. .S G = , 4 10 / .secm4 2ν #= -

From Newton’s law of viscosity (at the surface of plate)

( )y 0τ = dydu

y �

m==

c m ...(i)

dydu

y �=c m cosU y

2 2 y 0

pd

pd=

=a k: D U

2pd=

From equation (i) τ U2#nr pd= μ νρ=

( . . 1000)S G U2n p

d# # # #=

( . ) . U4 10 0 91 1000 1 574# # # # # d= −

0.571 /N mU 2

d=

FM 1.20 Option (B) is correct.We have �� ,�gm = V 0.8 / ,m s= 0.006 Pa sμ -= y 0.8 8 10mm m4

#= = −

The force balance from figure gives

��FxΣ = � �� cos sinF F F�shear Nc c− − = ...(i)

��FyΣ = �� cos sinF F WN shearc c− − = ...(ii)

Weight W �� Nm g# #= = =

and Fshear As st=

(�.��) (�.� �.�) .A yV

� ��

s �#

m # # #= = −

Fshear 0.9 N=

Equation (ii) gives FN 20

( 20 )cossinF Wshear

cc= +

. 159.95cossin N

200 9 20 150#

cc= + =

By substituting the value of Fshear and FN in equation (i),we get

F cos sinF F�� shear Nc c= + . .cos sin0 9 20 159 95 20# #c c= + 55.55 N=





FM 1.21 Option (C) is correct.We haveThe saturation pressure of water at 45 9.593C kPac =When air is fully escaped, the space is filled with vapor and the container have a two-phase mixture of saturated water vapor.Then vapor pressure �.��aP P (�� )Cv saturation= =c

FM 1.22 Option (D) is correct.Shear stress is given by

τ .. ��aL

V��

�� #m= = =

and the Reynolds Number is

Re ..VL

�� # # bm

r= =


Velocity profile u ( � )u r R� n n�= −

We know that the wall shear stress in pipe flow

wτ drdu

r Rm=−

=

u drd

Rr� n

n

r R�m=− −

=: D u

Rnr

n

n

r R�

�

m=− − −

=; E R

u n�m=

Then the drag force applied by the fluid on the pipe wall becomes

F Aw wt= ( )Rn u R L n u L� ��

�# #m p pm= =

FM 1.24 Option (B) is correct.(a) From the Power-law for air

pμ TT n

00

m= b l 1.8 10 3.221 10 /kg m s293673 .

50 7

5# # # −= =− −

b l

(b) From the sutherland law

sμ ( / ) ( )

T ST T T S.

00

1 50m= ++

; E

1.8 10( )

( / ) ( )673 110

673 293 293 110.5

1 5#

#= ++−

= G

3.225 10 /kg m s5# −= −

FM 1.25 Option (A) is correct.Assume a linear viscous velocity distribution in the film below the block. Then a force balance in x - direction gives:

FxΣ sinW Aq t= −

�sinW hV A maXq m= − = =: D

or sinW θ hV Am=

V sin sinA

hWA

mghm

qm

q= = W mg=

FM 1.26 Option (C) is correct.Due to the flow of glycerin, shear force acting in the opposite direction to this





flow. The FBD is shown below.

In equilibrium condition

�FxΣ = sinW α l �# #t= b �= sinmg α l#t=

sinvgρ α l#t= m Vr=

sing vgγ α l#t= gρ γ=

�sinl hγ α# # # l#t= τ sinhg a= ...(i)From the Newton’s law of viscosity, shear stress at the plate ( )� 0=

τ dydu

y �

m==

c m hU

hUy� �

y�

�

m= −=

; E hU�m= ...(ii)

From equation (i) and (ii), we get

hU�μ sinhg a=

U sin�2

2

mg a=

FM 1.27 Option (A) is correct.Assuming a linear velocity distribution in the clearance, the force is balanced by resisting shear stress in the oil.

F ( )A RV D Lwall it m pD #= = b l

F R RV D L

i

i

�

m p= − ...(i)

For the given oil

μ ( )�� rn r n# #= =

0.8 998 0.003 2.63 /kg m s7# # −= =Then by substituting in equation (i), we get

F ( . . )

. . . .0 0401 0 0400

2 63 0 4 0 08 0 3# # # #p= − 79 . 793 N2 79 b=

FM 1.28 Option (D) is correct

List-I List-II

a. Ideal fluid 2. Fluid of zero viscosity

b. Newtonian fluid 1. Viscosity does not vary with rate of deformation

c. /μ ρ 5. Kinematic viscosity

d. Mercury in glass 4. Capillary depression.

So , correct pairs are a-2, b-1, c-5, d-4.





FM 1.29 Option (D) is correct.

List-I List-II

a. Capillarity 4. Surface tension

b. Vapour pressure 1. Cavitation

c. Viscosity 3. Shear forces

d. Specific gravity 2. Density of water

So, correct pairs are a-4, b-2, c-3, d-2.

FM 1.30 Option (C) is correctFor a droplet or bubble with one spherical surface

pΔ R2s=

.. .

0 005 102 0 0712

5 102 0 0712

3 6#

#

#

#= =− −

28480 28.5Pa kPa-=

FM 1.31 Option (B) is correct.We have 3 ��d = , 7.3 10 /N m2σ #= -

We know that surface tension on liquid droplet is given by the relation,

p d4s= .

3 104 7 3 10

3

2

#

# #= −

−

97.33 Pa=

FM 1.32 Option (D) is correctFor a liquid cylinder, the internal excess pressure is

pΔ Rs=

σ p RD #= ( . )

200 20 00073

#=

.200 0 000365#= 0.073 /N m=

FM 1.33 Option (B) is correctThere are two surface, inside and outside the ring. So the total force measured is

F ( )D D2 2sp ps= = 2 0.0728 (0.06)p# # #= 0.0274 N=

FM 1.34 Option (D) is correct.We have 4 4 ��mm md 3

#= = −

The capillary effect is given by the equation,

h cosg d

4# #rs q=

where σ = Surface tension in /N m

θ = Angle of contactCapillary effect for water

σ 0.0734 /N m= , 0cθ = , ρ 1000 /kg m3=

h .. cos

1000 9 81 4 104 0 0734 0

3# # #

# c= −

7.48 10 7.48m mm3#= =−

Capillary effect for mercury





σ 0.51 /N m= , 128cθ = ρ �� S G #= 13.6 1000 13600 /kg m3

#= =

h .

. cos13600 9 81 4 10

4 0 51 1283

# # #

# # c= −

2.35 10 m3#=− − 2.35 mm=−

Here the negative sign indicates the capillary depression.

In magnitude h 2.35 mm=

FM 1.35 Option (C) is correct.(a) For water, capillary rise

hw . ( . )

.cos cosgD

��

# ## # c

rs q= = 0.0275 m=

(b) For Mercury

hm . ..cos cos

gD�

��

# ## # c

rs q= = 0.00912 m=−

Here negative sign shows the capillary depression.


We have 960 / ,kg m3ρ = �. �.� ��mm mD � �#= = −

R D�= . 1.9 10 m2

3 8 10 33#

#= =−

−

�.�mmh = 0.0025 ,m= 15contact angle cφ =The surface tension of milk

milkσ . . .cos cosgRh

2 2 15960 9 81 1 9 10 2 5 103 3

## # # # #

cfr= =

− −

0.02315 /N m=


We have ��cmb = , 0.12 ,m= �.��F =From the surface tension force relation,

sσ ( . )

.b

F2 2 0 12

0 018#

= = 0.075 /N m=





FM 1.38 Option (B) is correct.From the figure above, the force balance on the annular fluid is

Force in vertical direction Weight of fluid film= ( )cos r r2 2o i#σ θ π π+ ( )g r r ho i

� �r p#= −

h ( )cos

g r r�

o irs q= −


The needle “dents” the surface downward and the surface tension forces are upward as shown in figure. Then a vertical force balance gives:

Vertical forces Weight of needle=

cos L2 #σ θ g D L��#r p=

cos2σ θ g D�

�

r p=

2σ g D�

�

r p= 0 0 1cos"c cθ = =

σ g D� �

�

#r p=

. . ( . )8

7824 9 81 3 14 0 0016 2# # #=

0.0772 /N m=

FM 1.40 Option (B) is correct.The wall shear stress anywhere on the surface of the frustum at a distance r from the axis of rotation is

wτ drdu

hV

hrm m mw= = =

The shear force on the area dA,

dF dA hr dAwt m w= =

Torque dT rdF hr dA

�

mw= =

T h r dAA

�mw= #The shaft power required at top surface is

P ,shaft top T h r dAA

�#w w mw= = #

h r dAA

��mw= # ...(i)

For the top surface dA rdr2p=

Hence P ,shaft top ( )h r r dr��

r

D��

�

�mw p==

#





h r dr hr� ��

� �

r

D

r

D��

�

� � �

�

�pmw pmw#= =

= =: D# h

D32

2 4pmw=

FM 1.41 Option (C) is correct.By putting the value in expression of shaft power at top (20 )Cc ,

P ,shaft top ( . ). ( . ) ( ) ( . )

hD

32 32 0 00123 14 0 1 200 0 122 4 2 4

#

# # #pmw= =

67.824 W=The power is proportional to viscosity. Thus the power required at 80 Cc is

P , , Cshaft top ��c P , ,C

CCshaft top

20

8020m

m#=

c

cc

.. 67.824 5.29 W0 1

0 0078#= =

Therefore, the reduction in the required power input at 80 Cc is

P P, , , ,C Cshaft top shaft top�� −c c . .67 824 5 29= − 62.533 W=

FM 1.42 Option (C) is correctWith b the width of the plates into the paper, the capillary forces on each wall together balance the weight of fluid held above the free surface.

Weight of fluid Surface tension force= ( . )g h b��# # #ρ ( )cosb2# s q=

or h ( . )cos

g ��#rs q=

. ( . ). cos

998 9 81 0 000752 0 0728 0# ## # c= 0.020 20m mm, =

FM 1.43 Option (A) is correct.We have ��gm = , �.�mmy = , �.�mA �= , 30cα = , 0.4 /Ns m2μ =The FBD of the block shown below.

In equilibrium condition

FxΣ 0= sinW ��c At=

sinmg ��c yV A#m= film thicknessy =

V sinA

mgy ��#

cm= W mg=

. ..

0 4 0 410 10 10 0 54

## # #=

− 0.03125 / secm=

FM 1.44 Option (A) is correct.We have �� ,cm mL = = �� ,rpmN = �.�� .��cm mh = = �� ,cmD = �� .� �.�� ,cm mR = = = �.��T =Torque transmitted by the shaft





T h R RL hR L� ��

�

# #mw p pmw= = ...(i)

and ω 20.94 /rad sN60

260

2 200# #p p= = =

From equation (i),we get

μ . . ( . )

. ( . )R L

T h2 2 3 14 20 94 0 075 1

0 8 0 00123 3

# # # #

#

pw#= =

0.0173 /N s m2−=

FM 1.45 Option (D) is correct.We have 1.2 10 /Ns m3 2μ #= - , � �secmU = , �.�mh =From Newton’s law of viscosity

τ dydum= ...(i)

At the fixed surface (at y �= )

dydu h

UhUy� �

y�

�

= −=

; E hU�=

From equation (i) τ hU�

#m= . .1 2 10 0 12 33

# ##= − 0.072 /N m2=

FM 1.46 Option (C) is correct.According to stokes law

μ DLW t3net #p= ...(i)

The net weight of the sphere in the fluid is

Wnet ( ) ( )g v g D6sphere fluid fluid sphere fluid

3

# # #r r r r p= − = −

( ) . ( . )2700 875 9 81 6 0 0025 3# # #

p= − 1.46 10 N4#= −

Then from equation (i),we get

μ ( . ) ( . )

1.46 103 0 0025 0 75

484

# # #

#

p#=

−^ h 0.397 /kg m s−=

FM 1.47 Option (C) is correctAssuming a linear velocity distribution inside the annular clearance, the shear stress is

τ rV

r rr

i

i

�m m wDD= = − ...(i)

This stress causes a force

dF ( )dA r d Lit t q= = ...(ii)The torque of this force about the shaft axis is

dT r dFi= ...(iii)Put equation (i), (ii) and (iii) together

T ( )r dF r r Ld r r rr r Ldi i i i

i

ii

�

�

��

�

# #t q m w q= = = −p p

# ##

r rr L d r r

r L di

i

i

i

�

�

�

�

�

�

�

�mw q mw q= − = −p p

# # r rr L�

i

i

�

�pmw= −

μ ( )

r LT r r2 i

i3

0

pw= −





FM 1.48 Option (B) is correct.The velocity profile is given by

( )u r uRr�max �

�

= −c m

The shear stress at pipe surface is expressed as

sτ drdu u dr

dRr�max

r R r R�

�

#m m=− =− −= =

; E Ru� maxm=

Then the friction drag force

FD ( )A Ru RL� �max

s st m p= = �A RLs p=

FD Lu4 maxpm= ...(i)By substituting the given values in equation (i), we get

FD . ( . ) ( )4 3 14 0 0010 15 5# # # #= 0.942 N=


In the figure WT 125tanWeight of k N= = WL Weight of liquid mg vg vr g= = = =where γ 10.9 /specific weight of liquid kN m3= =

WL . (1) 510 9 10 43 2

#p

# # #= 42.8 kN=

From the Newton’s law of motion in vertical direction

FyΣ may= F W WV T L− − may=

��F ��V − − . 2.759 81125 42800= +b l

F ��V − 12033= FV 54958 55N kN-=

FM 1.50 Option (A) is correct.At terminal velocity, the rod weight should equal the viscous drag.

W Viscous Drag=

g vs #ρ ( )/D D

V DL2o

#m p= −; E

g D L�s�

# #ρ π ( )/D D

V DL2o

m p= −

V ( )gD D D8

s o

mr= −





FM 1.51 Option (C) is correctThe power is the viscous resisting force times the belt velocity.

P Viscous resisting force Velocity#= A Voil belt beltt # #=

hV b L V# # #m= b ^l h V b h

L�m=

By substituting the values, we get

P . ( . ) ( . ) .0 29 2 5 0 9 0 0642

# # #= 108.75 109W W,=


We have �� ,�g msteel�ρ = �� ,�g maluminum

�ρ = �.�� mwaterσ =From surface tension force relation,

F Ds sp s= and ��W mg vg g D�r r p= = =

When the ball floats Fs W= D sπ σ ��g D�r p=

D g�s

rs=

For Steel Dsteel ( ) .( . )

�.� �� mg�

��

steel

s �

#

#rs

#= = = −

2.4 mm=

For Aluminum Daluminum .( . )

�.� �� mg�

��

aluminium

s �

# #

#r

s#= = = −

4.1 mm=Hence Daluminum D> steel

So aluminum ball would be larger in size.

FM 1.53 Option (C) is correctFor any radius r R# , the liquid gap is tanh r q= . Then

dT dA rt= w

.tan cosrr r dr r�m q

w p q= a bk l cosL drq=

T sin sin�� 23

2�2

0

3

qpwm

qpwm= =#

μ sin�

�23

3pwq=





Substituting the numerical values, we get

μ ( . ) ( . )

. sin2 94 2 0 06

3 0 157 33

# #

# # cp

= 0.193 /kg m s−=

FM 1.54 Option (D) is correct.At any radial position r r< � on the cone surface and instantaneous rate ω dT r dAt= w

sinr hr r dr�#m w p q#= 9 :C D sinh r dr� �

#qmw p=

Torque T sinh r dr�r

�

�

�

wmw p= # sinh

r2

04

qpmw= ....(i)

Since for cone T I dtd

�w=− mr dt

d103

02#

w=− For cone I mr103

0 02=

Then from equation (i),

mr dtd

103

02 w− sinh

r2

04

qpm w=

Separating the variables and integrating both the sides,

d�

ωω

ω

ω#

sinmr hr dt

3 210 t

02

04

0#

#

qpm= − #

or ω exp sinmhr t

35

002

w qpm= −; E

FM 1.55 Option (B) is correct.For the fluid in the annular region

TA RdF R dA R RR RLd

�

�

# # #t m w qD= = =p

b l###

RR L� �pmw

D=

Now Tbottom r dA r Rr rdr�

R

�t m w pD= = a k# # R r dr� R

�

�

pwmD= #

RR

42 4pwm

D=

Ttotal RR L

RR�

�� pwm pwm

D D= +

μ ( / )R L R

T R2 43pw

D=+

μ ( / )

( )R L RT r R

2 43pw=

+−

FM 1.56 Option (B) is correct.We have �� . ,mm mL �= = 0.008 / . ,kg m sμ = �.� �.��mm mtfilm = = �� .�� ,mm mD = = ��rpmN =

Torque is given by T Areat Rfilm

�mw# #=

T tR RL t

R L� �film film

� �

#mw p pmw= = �A RLs p=

T ( � )

tN R L� � ��

film

�#pm p= N

602ω π=

T tNR L

604

film

2 3

#

p m=





By substituting numerical values

T .. ( . ) .

60 0 00124 0 008 750 0 04 0 32 3

#

# # # # #p=

FM 1.57 Option (A) is correct.At any r R# , the viscous shear on both sides of the disk

totalτ 2 2 hrt mw

# #= =

and viscous force

dF dAtotal #t= w (� )hr rdr�mw p#=

Then viscous torque

dT .dF r= .hr dr r� �pmw= h

r dr� �pmw=

Integrating both the sides

T .h r dr�r

R�

�

pmw==

# hR

hR�

��

#pmw pmw= =

Substituting the numerical values

T ( . )

. ( . )N0 001 60

0 29 2 0 05 4

#

# # # #p p= N60

2ω π=

( . ). ( . )0 001 60

2 0 29 1200 0 052 4

#

# # # #p= 0.716 N m−=

FM 1.58 Option (A) is correctThe masses remain the same for an isothermal process of an ideal gas

m m� �+ m�= v v1 1 2 2ρ ρ+ v3 3r=

or RTp D RT

p D� ��

��

��π π

# #+9 9 9 9C C C C RTp D�

��p

#= 9 9C C

,RTp v D�

�ρ π= =

� �RT

p rD RT

p rD

��

��

a a��

��s p s p

# #+ + +

; 9 ; 9E C E C �

RTp r

D�

�a �

��s p

#= +; 9E C

P P R�

a�s− =

The temperature cancels out, and we may clean up and rearrange as follows

�p D Da ��

��s+ ( 8 ) ( 8 )p D D p D Da a2

322

13

12s s= + + +

FM 1.59 Option (C) is correct

The skate bottom and the melted ice are like two parallel plates.





τ hVm=

F A hVLbt m= =

Using F ma= to find the stopping distance

FxΣ F hVLb ma m dt

dV�x

m=− =− = = The ‘2’ is for two blades

Separate and integrate once to find the velocity

VdV

V

V

o

# mhLb dt�t

�

m=− #

or log VV

eo

: D mhLb t2m=−

or V V eo mhLb

t2

=m−

Integrate once again to find distance

x Vdt V e dto mhLb

t

�

2

�= =

3 3 m−

# #

LbV mh2

o

m=

FM 1.60 Option (C) is correct

Consider the right side of the liquid column, the surface tension acts tangent to the local surface that is along the dashed line at right. This force has magnitude F bs= as shown. Its vertical component is ( )cosF θ φ- as shown. There are two plates, therefore the total vertical force on the liquid column is

Fvertical ( )cosb2s q f= −Then the vertical force holds up the entire weight of liquid column between plates, which is

W ( )tangbh L hr f= −Set W equal to F, we get

( )cos�2σ θ φ- ( )tangbh L hr f= −

or σ ( )

( )cos

tan�� 2 q f

r f= −−

***********

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