GATE 1998

GATE -1998ME: Mechanical Engineering

Read the following instructions carefully

1. Write all the answers in the answer book only.2. This question paper consists of TWO SECTIONS: A and B.3. Section A has Eight questions. Answer ALL questions in this section.4. Section B has Twenty questions. Anwer any TEN questions from this section. Ifmore number of questions

are attempted, strike off the answers not to be evaluated; else only the First Ten unscored answers will beconsidered.

5. Answers to Seeton B should start on a fresh page and should not be mixed with answers to Section A.6. Answers to questions and answers to parts of a question should appear together and should not be separated.7. In all questions of 5 marks each write clearly the important steps in your answer. These steps carry partical

credit.8. There will be no negative marking.

I SECTION-A I(100 Marks)

1. Write in your answer book at the space provided, the correct or most appropriate answer to thefollowing multiple choice questions by writing the letter - A, B, C or D -against each questionnumber. (25 x 1= 25)

n 00

1.1 For x = 6' the sum of the series L(COS xin = cos1x + cos"x + is1Wn ~3

~oo ~1Ans. (B)

cos2 2: + cos4 2: + cos6 2: + ...666

3 9 274+16+ 64 +,..00 (infiniteG.P.)

3~ 4 = 331--4

1.2 (s + 1)-1 is the Laplace transform of(A) t2

(C) e-2t

ADs. (0)

(B) t3

(0) te-t

Explanation.

By first shifting theoremI

(s + 2)2

Xl

1.3 If (/J (x) = f Ji dt, then : iso

(A) 2x2

(C) 0Ans. (A)

(B) ~(D) I

1.4 The magnitude of the gradient ofthefunctionf=xy'!! at (1,0,2) isWO ~3~8 ~ooAns. (C)

. a . a k aV.f= 1-+J-+ -ax By aziyz3 + jyz3 + k X 3xyz2

i x 0 +j x I X 23+ + k x 0 = j x 8Iv . fl =r 8.

I(A) -

6I

(B) I2I

(C) 144

Ans. (B)

Explanation. Probability that first friend is born in any month = 100% = 1.Probability that second friend is born in the same month as that of first friend

, 1(D) 24

I 1= Ix-=-.

12 121.6 A car moving with uniform acceleration covers 450 m in a 5second interval, and covers 700 m in the

next 5 second interval. The acceleration of the car is

(A) 7 m/s2

(C) 25 m/s2

Ans. (D)Explanation. For tl = 5 sees, SI = 450 m

(B) 50 m/s2

(D) 10 m/s2

and for t2 = 5 sees S2 = 700 m1

ut +-at2

2

Su +..!. x a x 25 [since initial velocity = u]2

S2Sa

u+-2

Also after first S seconds, velocity of car,v = u T at

= u + Sa.After first 5 seconds. u + Sa will be the initial velocity for next:) seconds, therefore

1 25a700 = (4+Sa)S+-ax25 = 20+25a+-2 2

(2Sa)5U+T +25a = 450+25a,

10 m/sec2•

AlternatelyConsidering the v-t graphArea ACDB = 450Area CEFD = 700Difference is the shaded regionDPQR=2S0

2S0. PD= -=SO.. S

PD 50a = tan 8 = - = - = 10

BP S1. 7 Tile buckling load for a column pinned at botll ends is 10 kN. If tile ends are fIXed, tile buckling load

cllanges to(A) 40kN(C) 5kNADs. (A)

(B) 2.5 kN(D) 20 kN

Explanation.

I case

1t2EIP e2er

e I L (for both ends pinned)Leo - (for both ends fixed)2

Perl eiI (~ r I-=-- -

PerIl ei L2 4

Pcrl 10 kN 1 IIPeril 4 Peril= 40 kN

1.8 Tile normal stesses at a point are a = 10 MPa and, a = 2 MPa,' tile shear stress at tllis point is 4MPa.x y

The maximum principal stress at this point is(A) 16 MPa (B) 14 MPa(C) II MPa (D) IO MPaAns. (C)Explanation. Maximum pri1wipal stress

(1 ~ 2 MPI

±t1<4MPI

(cr~ ;cry)+ (crx ;cry J +cr2

C02+

2)+ C0

2-2)2 +42

1.9 The ratio of lIverage shear stress to the maximum sllear stress in a beam with a square cross-section is2

(B) -3

(A) I

3(C) -

2Ans. (B)

Fd2 F.d2 3F--=--=--

81 d4 2d28.-

120'I){~&I

d V-el

shear loadside of the square section

Fd2

Fd2 2

3F"=32d2

Heat transfer coefficients for free convection in gases, forced convection in gases and vapours, andforboiling water lie, respectively in tile ranges of(A) 5 - 15,20 - 200 and 3000 - 50,000 W/m2K.(B) 20 - 50,200 - 500, and 50,000 - 105 W/m2K.(C) 50 - 100, 500 - 1000, and 105 - 106 W/m2K.(D) 20 - 100,200 - 1000, and a constant 106 W/m2K.Ans. (A)

~qmax,

1.11 If V, and a are tile nozzle exit velocity and nozzle angle in an impulse turbine, the optimum bladevelocity is given by(A) V ~ (OS '2 u

(C) VN cosex, '2

AilS. (D)

(B) VN sin 2 ex

V 2(D) _N_2

1.12 A Curtis stage, Rateau stage and a 50% reaction stage in a steam turbine are examples of(A) different types of impulse stages(B) different types of reaction stages(C) a simple impulse stage, a velocity compounded impulse stage and reaction stage(0) a velocity compounded impulse stage, a simple impulse stage and a reaction stage

Ans. (0)

1.13 The basic load rating of a ball bearing is(A) the maximum static radial load that can be applied without causing any plastic deformation of bearing

components.(B) the radial load at which 90% of the group of apparently indentical bearings run for one million revolutions

before the first evidence of failure.(C) the maximum radial load that can be applied during operation without any plastic deformation of

bearing components.(0) a combination of radial and axial loads that can be applied without any plastic deformation.

Ans. (B)

1.14 Decreasing grain size in a polycrystalline material(A) increases yield strength and corrosion resistance.(B) decreases yield strength and corrosion resistance(C) decreases yield strength but increases corrosion resistance(0) increases yield strength but decreases corrosion resistance.

Ans. (0)

1.15 Auto collimator is used to check(A) roughness(C) angle

Ans. (C)

(B) flatness(O)automobile balance.

1.16 Failure of a bead weld between a heavy steel section and a thin section is mainly due to the formation of(A) spheroidite(B) bainite(C) carbon free zone due to burning of carbon at high temperature(0) martensite.Ans. (B)

1.17 Ideal surface roughness, as measured by the maximum height of unevennes, is best achieved whe', thematerial is removed by(A) an end mill (B) a grinding wheel(C) a tool with zero nose radius (0) a ball mill.Ans. (C)

1.18 In the specification of dimensions andflts,(A) allowance is equal to bilateral tolerance(B) allowance is equal to unilateral tolerance(C) allowance is independent of tolerance(0) allowance is equal to the difference between' maximum and minimum dimension specified by the

tolerance.Ans. (C)

1.19 In machining using abrasive material, increasing abrasive grain size(A) increases the material removal rate

(B) decreases the material removal rate

(C) first decreases and then increases the material removal rate

(D) first increases and then decreases the material removal rate

1.20 With increasing temperature of intake air, IC engine efficiency(A) decreases (B) increases

(C) remains same (D) depends on other factors\

1.21 Chills are used in moulds to(A) achieve directional solidification(C) reduce freezing time

(B) reduce the possibility of blowholes(D) smoothen metal flow for reducing splatter.

Ans. (A)

Explanation. Chills are used in moulds for achieving directional solidification.

1.22 One of t/ref 0110wing statements about PRS (Periodic Reordering System) is not true. Identify.(A) PRS requires continuous monitoring of inventory levels(B) PRS is useful in control of perishable items

(C) PRS provides basis for adjustments to account for variations in demand(I" T., PRS, inventory holding costs are higher than in Fixed Recorder Quantity System

Ans. (A)

1.-. .. ·,·rp'(oryplanning, extra inventory Is unnecessarily carried to the end of the planning period when, :'....I"" of the following lot size decision policies:(A) ~.0\ -- for - lot production(B' EC0nomic Order Quantity (EOQ) lot size(C) Period Order Quantity (POQ) lot size(D) Part Period total cost balancing

Ans, (R)

1.24 In a weaving operation, the parameter to be controlled is the number of defects per 10 square yards ofmaterial. Control chart appropriatefor ,his task is(A) P-chart (B) C-chart

(C) R chart (D) X -chart

ADs. (B)

1.25 Which one of the following forecasting techniques is not suited for making forecasts for planningproduction schedules in the short range?(A) Moving average (B) Exponential moving average(C) Regression analysis (D) Delphi

Ans. (D)

Explanation. Moving, average, Exponential moving average are used for short range. Regression is usedfor short and medium range and Delphi is used for long range forecasting.

2. Write in your answer book in the space provided the correct or the most appropriate answer to thefollowing multiple choice questions by writing the letter -A, B. C or D-against each sub-questionnumber. (20 x 2 = 40) -

2 d2 Y -~dy2.1 The general solution of the differential equation x dx2 - x dx + Y = 0 is

(A) Ax + Bx2

(B) Ax + B log x(C) Ax + Bx2 log x(D) Ax + Bx log xAns. (D)

(A, B are constants)(A, B are constants)(A, B are constants)(A, B are constants)

do where D=-dz

On putting x = ez, we have(D - 1)2y 0

Whose solution is y eZ (A + Bz)or y x(A + B log x)

2.2 The bestflt line using least squares for the data (0,0), (l0, 24), (20,36) and (30,60) is(A) 2x - Y = 0 (B) 2x - Y + 4 = 0(C) 2x - Y- 4 = 0 (D) None ofthese

Ans. (D)

n = 4; ~XI = 60; ~xi = 1400, ~ft = 120,

~X/I 2760

~fl

~X/l12027606 48-anda( =-5 256 48

Y -+-x5 25

or 25y - 48x = 302.3 The maximum principal strain in a thin cylindrical tank, having a radius ul25 cm and wall thickness of

5 mm when subjected to an internal pressure of 1MPa, is (taking Young's midulus as 200 GPa andPoisson's ratio as 0.2)(A) 2.25 x 10-4(C) 2.25 x 10-6

Ans. (A)

nao + (~x,) a,

(~x) ao + (~xi) a,4ao + 60a.

60ao + 1400al

(B) 2.25(D) 22.5

pd_~2tE 4tEm

~[2-~]4tE ill

1x 106

X 50 X 10-2 (2 _ 0.2)4 X 5 x 10-3

X 200 X 109

2.25 X 10-4

2.4 A square bar of side 4 em and length 100 em is subjected to an axia/load P. The same bar is then usedas a cantilever beam and subjected to all end 10adP. The ratio of the straill energies, stored in the bar inthe second case to that stored in the first case, is(A)16 (B)400(C) 1000 (D) 2500

Ans. (D)

..!.(P)( PL)2 AE

p2 X 1002x4x4xE25p28E

"!'(P) x(PL3)

2 3EI

1 p2 x 1003

2 E-lx4x4312

15625 p2---2 E

V2 _ 15625 x ~ = 2500VI 2 25

p

~I_ 100 em _I

2.5 An Ie engine has a bore and stroke of 2 units each. The area to calculate heat loss can be taken as(A)4n (B)5n(C) 6n (D) 8nADS. (C)

Area for heat loss = ndl + (2 X n:2

) = ndl + n~2

d = diameter, and I = length of stroke.

n x (2)2Area o!' heat loss = n x 2 x 2 +---

2

2.6 An air breatlling aircraft isflying at an altitude where the air density is half the value at ground level.With reference to the ground level, the air-fuel ratio at this altitude will be

(A) ifi(C) 2

Ans. (C)

maExplanation. A/F ratio = -;;

where m. - mass of air,mr - mass of fuel

(A I F ratio)grollnd(A I F ratio)altitllde

(ma )grollnd(ma)altitude

(B) .fi(D) 4

(ma)b'TOund---- as mr is same at both places.(ma )altitude

v x PgroundV x Paltitude

_1_=2(1 \ 2)

2.7 For the data listed below for two Jou;nal bearings A and B, predict the flow conditions in the bearings

Bearing diameter Radial Surface Viscosity of Density of(m) clearance speed of lubricant lubricant

(m) shaft (m/s) (Pa-s) (kg/nr)

A 0.01 10-s 210 0.001 1000

B 0.05 10-4 10 0.01 850

(A) Iiminar in both A and B(C) laminar in A and turbulent in B

Ans. (D)

For laminar flow XFor turbulent flow X

DJ.l

(B) turbulent in both A and B(D) turbulent in A and laminar B

2 dvpy -dy

J.l

< 500> 500

0.01 m, y = 1O-sm, v = 210 m/s0.001 Pa.s, P = 1000 kglm3

850 x (10-6)2 x (~)10-5

0.01

0.05 m,0.01 Pa.s,

y = 10-4 m, v = 10 m/sp = 850 kg/m3

850 x (10-4)2 x (~)10-4

---0.01

:. Flow is laminar.Turbulent m A, and laminar in B.

2.8 The isentropic heat drop in the nozzle of an impulse steam turbine with a nozzle efficiency 0.9, bladevelocity rati 0.5, and mean blade velocity 150 mls in kJ/kg is(A) 50 (B) 40(C) 60 (D) 75

ADs. (A)u-=0.5

VI

or

where

u = 0.5 VI

U mean blade velocity, andv I Absolute velocity of steam at inlet

150-=300m/s0.5

K.E. at inlet

Nozzle efficiency

12m 2-xmxv\ =-(300) J2 20.9

(300)2 = 50kJ I kg.2 x 0.9

2.9 Air (C = 1 kJ/kg, y= 1.4) enters a compressor at a temperature of 27°C. The compressor pressure ratiopis 4. Assuming an efficiency of 80%, the compressor work required in kJ/kg is

(A)160 (B)172(C)182 (D)225ADs. (C)Explanation. For air, C

p= 1 kJ/kg

y = 1.4

T} = 27°C = 3000K4P2

Pressure ratio = PI

.-(

p )1-1 0.4P~ 1 =(4)1.4 =1.48

T2 - T) = 445.8 - 300 = 182.24.0.8 0.8

CP(T'2 - T\)1 x 182.24 kJ/kg ~ 182 kJ/kg .

..10 The dicharge velocity at the pipe exit inflgure 2.10 is

(A) J2i,H (B) ~2gh

(C) ~g(H + h) (D) 0

- ---- -.•./' --

H

1--- - - '--

Ans. (B)

A steel ball of mas 1kg and speclflc heat 0.4 kJ/kg is at a temperature of 60°C. It is dropped into 1kgwater at 20°C. Theflnal steady state temperature of water is(A) 23.5°C (B) 300C(C) 35°C (D) 40°C

Ans. (A)

Explanation. Heat lost by ball = Heat gained by wateror mb x Cb x (Tb - T) = mw x Cw x (T, - T)where subscript w stands for water and b stands for ball.

and T, steady state temperature.. 1 x 004 x (60 - T,) 1 x 4.18 x (T, - 20) [since C

w= 4.18 kJ/kg]

or 24x004T 4.18T-83.6, ,T 23049°C ~ 23.5°C,

2.12 The temperature variation under steady heat conduction across a comjJosite slab of two materials withthermal conductivities K} and K] is shown infigure 2.12. Then, which one ofthefollowing statementsholds?

(A) K, > K2

(B) K. = K2

(C) K\ = 0

(D) K\ < K2

Ans. (D)

KA~dxdt Q

where Slope = - = -dx KA

i.e., slope is more for less value of K, therefore

( :~ ) I material > (:~ lmaterial

2.13 The earth can be assumed as a uniform sphere. Suppose the earth shrinks by 1% in diameter, the newday period

(A) will not change from 24 hrs. (B) will reduce by about 2%(C) will reduce by about 1% (D) will increase by about 1%Ans. (D)

1 I 2-0)2

.!.(~MR2) 0)22 5

M [0) 2 .2D d(D) + D2 7,w.d(O)]20

2d(D) 2d(0)--+--

E D 0)

As there is no change in energy (conservation of energy)

dD dO)- --D 0)When diameter shrinks by 1% speed will increase by 1%.

2.14 Pertaining to a steam boiler, pick the correct statement among the following(A) Primary boiler heat surfaces include evaporator section, economizer and air preheater(B) Primary boiler heat transfer surfaces include evaporator section, super heater section and economizer.(C) Secondary boiler heat transfer surfaces include super heater, economizer and air preheater.(D) Primary boiler heat transfer surfaces include evaporator section, super heater section and reheat section

Ans. (C)

Explanation. Economizer, preheater, Super heater are all accessories ofthe boiler and hence forms secondaryboiler heat transfer surface.

2.15 Consider the triangle formed by the connecting rod and the crank of an IC engine as the two sides of thetriangle. If the maximum area of this traingle occurs when the crank angle is 75°, the ratio of connectingrod length to crank radius is(A) 5 (B)4(C) 3.73 (D) 3

Ans. (C)

'!'(PQ) (PR) sin A290°Area will be maximum when A

i.e. PQR is a right angled triangle.Ratio of connecting rot length to crank radius,

2.16 The difference between tensions on the tight and slack sides ofa belt drive is 3000 N. If the belt speed is15 mis, the transmitted power in k W is(A) 45 (B) 22.5(C) 90 (D) 100

Ans. (A)Explanation. Given,where

T1-T2

Tj, T2

vPower

= tensions on tight and slack side respectivelybelt speed = 15m/see

= (TJ-T)v= 3000 x 45000 watt = 45 kW.

2.17 The profile of a cam in a particular zone is given by x = .fi cos () and y =sin {},The normal to the camprofile at ()= 7l!4 is at an angle (with respect to x axis)

7t(A) -

4

7t(B) -

27t

(C) -3

Ans. (C)Explanation. sin2 e + cos2 eEquation of curve is

To find Slope at required point ( ~ , ~ ) , differentiating we s,..;(

dy2x+6y- = 0dx

dydx

-ve sign indicates that slope is - ve.

x 13y = - J3

dydx

eAngle made by normal

12x+6ytan e = J3

30°60° = 1[/3 radians.

2.18 In an orthogonal machining operation, the chip thickness and the uncut thickness are equal to 0.45 mm.If the tool rake angle is 0°, the shear plane angle is(A) 45° (B) 30°(C) 18° (D) 60°

Uncut chip thickness = 0.45 = IChip thickness after the metal is cut 0.45

shear plane angle and a = rake angle = Oc

rcosa = 1x cosO = I.1- rsina I-IsinO

45°

2.19 A strip with a cross-section 150 mm x 4.5 mm is being rolled with 20% reduction of area using 450 mmdiameter rolls. The angle subtended by the deformation zone at the roll centre is (in radian)(A) 0.01 (B) 0.02(C) 0.03 (D) 0.06

Ans. (D)Explanatin. Given, ho = initial thickness = 4.5 mm

Sectional area = 150 x 4.5 mm2

Since area is reduced by 20%, therefore

t = 0.8 x 4.5 = 3.60 ifinal area = 0.8 x 1.50 x 4.5 sq. mm. ..5 alia

Keeping width constant, final thickness

0.8 x 150x45=0.8x45 mmhi 150

R- Rcos e 4.5 - (0.8 x 4.5)..

24.5 x 0.2

or R (1- cos e)2

4.5 x 0.2or 1 - cos e

2 x225499

or cos e -500

.. e 0.063 radians.

2.20 Aflywlreel of moment of inertia 9.8 kg m2fluctuates by 30 rpmfor afluctation in energy of 1936 Joules.Tire mean speed of tireflywlreel is (in rpm)(A) 600(C) 968

Ans. (A)

(B) 900(D) 2940

mR20)2i1S

mR20) (0) I - 0)2)

9.8XO)(~~ X30)600 rpm

3. Matclr 4 correct pairs between List 1 and List llfor questions 3.1 to 3.5 below. No credit will be givenforpartial matching. Write your answers using only tire letters A to D and numbers 1 to 16. (5 x 2 = 10)

List I List II

(A) Heat to work (1) Nozzle

(B) Heat to lift weight (2) Endothermic chemical reaction .

(C) Heat to strain energy (3) Heat engine

(D) Heat to electromagnetic energy (4) Hot air balloon/evaporation

(5) Thermal radiation(6) Bimetallic strips

List I List II

(A) Sand casting (1) Symmetrical and circular shapes only(B) Plaster mould casting (2) Parts have hardened skins and soft interior(C) Shell mould casting (3) Mihimum post-casting processing(D) Investment casting (4) Parts have a tendency to warp

(5) Parts have soft skin and hard interior(6) Suitable only for non-ferrous metals

List I List II

(A) High head, low flow rate (1) Streamlined body(B) Low head, high flow rate (2) Boundary layer(C) Heat transfer (3) Orifice meter(D) Low drag (4) Centrifulgall"· .•mp

(5) Axial flew pump(6) Nusselt number

List I List II

(A) Aluminium brake shoe (1) Deep drawing(B) Plastic water bottle (2) Blow moulding(C) Stainless steel cups (3) Sand casting(D) Soft drink can (aluminium) (4) Centrifugal casting

(5) Impact extrusion(6) Upset forging

Ans. (A) -3, (B) -2, (C) -~, (D)-0

3.5 List I List II

(A) ECM (l) Plastic shear(B) EDM (2) Erosion/Brittle fracture(C) USM (3) Corrosive reaction(0) LBM (4) Melting and vapourization

(5) Ion displacement(6) Plastic shear and ion displacement

4. The radial displacement in a rotating disc is governed by the differential equation

dlu 1du U--+---- = 8xdxl X dx Xl

where u is the displacement and x is the radius.

1Ifu = 0 at x = 0, and u = 2 at x = 1, calculate the displacement at x = 2d2u 1 du U

ADs. Given, -+---- = 8xdx2

X dx x2

2 d2u dux -+x--u

dx2 dxThis reduces to {D(D - I) + D - I} U

(D2 - I) u8e3z, where x = eZ

8e3z

8 e3z

Aez+ Be-z+ -2-D -I

Aez + Be-z + e3z

BAx+-+x3

X

Ax2 + B + x4

WhenWhen

x = 0, u =0, .. O=Bx = I, u=2, .. 2=A+0+I

or A = I