a jugad gate 2007 2001 1995 iit kanpur set questions.pdf

8/9/2019 A JUGAD GATE 2007 2001 1995 IIT Kanpur set Questions.pdf

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GATE -1995

ME: Mechanical Engineering

1. Choose anwer and indicate by writing only the corresponding' capital letter (A, B, C, D as the case

may be) (15 xl =15)

1.1 Tlte mU/lufacturing area of a plat is dividell into four quadrants. Four macltines Illl'Veto he located, one

i/l eaclt tjuandTtlllt. Tlte total numher ofpossihle layouts is

(A) 4

(8) 8

(C) 16

(D) 24

Ans. (D)

Explanation. In quadrant I, we can locate anyone of the four machines. (i.e.) we can allocate quadrant I in

4 ways. Thereafter quadrant 1 \ in 3 ways, thereafter quadrant 1 1 \ in 2 ways. No further choice for quadrant IV.

... Total number of possible layouts = 4 x 3 x 2 = 24.

1.2 Tlte fUltctimt f(>..)= I x + lion tlte illterval{-2, 0 / is(A) continuous and differentiable .(B) continuous on the intergal but not differentiable at all pomts

(C) neither continous nor differentiable

(D) differentiable but not continous

Ans. (B)


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Explanation. ttx) = I x {. I I .The graph y =I x + 1 I is the well known "V" graph which is continuous everywhere in particular in [-2,0] but not diffrentiable at x = -1.

1.3 if V if a dijJerelltiable vector fUIlctioll ami f is (I slIjjiciellt differelltiable scalar fUIlctioll, thell curl

if 1 1 i,· equal to

(A)(grad f) x (V) + (fcurl V)

(C)fcurl (V)

Ans. (A)

(8) 6

(D) (grade f) x (V)

o - 0 - 0-i x-(fV) + jx -(fV) + k x -(fV)

o x o y o z

[ o V D f - ]

i x f D x + D x V + two similar terms

[. o V . - o V

1 [ o f D f

D r ] -f I X - + .I x oV + k x - + ix -- + j - + k _ x Va x D z J D x o y D z

f curl V + (grad f) x V .

1.4 If the demaml for (IIIitem is doubled ami the orderillg cost halved, the ecollomic order qU(lIItily

(A) remains unchanged (B) increases by a factor of .fi(C) is doubled (D) is halved

Ans. (A)

J 2 ~ DOrdering cost

Demand

Unit holding cost.1.5 III PERT, the (Iistribuitoll of activity times is asnlllled to be(A) Normal (B) Gamma

(C) Beta (D) Exponential Ans. (C)

1.6 Stati,,'tic(11qlltlli(I' colltrol was developed by

(A) Frederick Taylor

(C) George Dantzig

Ans. (8)

(B) Water shewhart

(D) W.E. Deming


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E\phmation. Statistical quality control was developed by Dr. Walter Shewhart all American sd(~n1i~tduring II World War.

1.7 Amollg the COlIl'el1tiollalmachillillg pml.'t!.ue:., nli/ximum .Vh!cilic ellergy i3'COII.\U/W',;im

(A) turning (Bldrilling

(C) planing (D) grinding

Ans. (0)

1.8 Cutting power cOII.5tlmptioll ill turnillg call be sigllificalltly relluced by

(A) increasing rake angle of the tool (8) increasing the cutting angks 01 the tool

(C) widening the nose radius of the tool (D) increasing the clearance angle

Ans. (A)

Explanation. Increasing rake angle reduces the force on the tool and thus pG"V'~' CerLd((,,; I":'. i,:,',ll:ced

).9 Plaill millillg of miltl steel plate prot/uces(A) irregular shaped discontinuous chips

(C) contiuous chips withuot built up edge

(B) regular sharwd discontinuous chip';

(D) jOl fHed chips

J..~ A fluid is .mid to be Newtolliall whell the shear stress i~

\(A) directly proportional to the velocity gradient

1(8) inversely proportional to the velocity gradient//

(C) independent of the velocity gradient

(0) none of the above

Ans. (A)

Explanation. According to Newtonian's law of viscosity

du . du, hi' d' d' h hr = , = ~t -, where - IS t e ve oClty gra Icnt an ,IS t e s ear stress,dy dy

J.I) COllsider a refrigerator (/nd a heat pump workillg 011the reversell Carnot cycle bet~ cell tlli! same

temperature limits. Which of the fOllowing is correct.

(A) COP of refrigerator = COP of heat pump

(8) COP of refrigerator =COP ofheat pump -I- 1

(C) COP of refrigerator = COP of heat pump-I

(0) COP of refrigerator = inverse COP of heat pump

1. 12. COIlStlllltpressure lines in the superheated region l~lthe iHollier diagram will have

(A) a positive slope (B) a negative slope

(C) zero slope (D) both positive and negative slope

AilS. (A)

).)3 A test specimen is .\'tressed sliglltly beyond the yield p(jim allli '"e11 unloaded. Its yield strength will

(A) decrease (8) increase

(C) remains same (D) becom~s equal to ultimate tensile strength


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True

a l r • • •

y\-...._--,,,,,'(I.

I II,

I ,,,

t i : I. 'tI '

• •'I ','.:

1.14 A key cO/meeting aflange coupling to a sltaft i.~likely to fail in

(A) shear (8) tension

(C) torsion (D) bendingAns. (A)

Explanation. Shear is the dominant stress on the key.

~~_iteeti01l ~~ ",,~t.tiOJl

SHAFT

1.15 A frere bar (If lengtlt I uniformly Iteated from O °C to a temperature tOe. a is tlte coefficient of linear expansion and E is tlte modulus of elasticity. Tlte stres.~ill tlte bar i.~

(A) a tE (8) a tE/2

(C) zero (D) None of the above

Ans. (C)

Explanation. Ends are not constrained. It is a free expansion problem. Hence there is no stress in the

member.

2. Each blank ( ) is to be suitably filled in using one of the given options. In the answer book write the

question number and the answer only. Also, no explanations for the answer are to be given.

(IOxI=IO)

2.1 Tlte layout witlt a Itigher material handling effort is a layout. (product/process)

Ans. process


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2.2 Checkillg tile {liameter of a IIole using GO-NO-GO gauges i.••w, example of illspectio .., by .

(variables/at trib utes)

Ails. attributes

2.3 Macllille tool structures are made for IIigll proce!J'scapability. (tough/strong/rigid)

Ans. rigid

2.4 For plw1I1illg tile procuremellt or production of dependent demand itenu, tile technique most suitable

is (MRP/EOQ)

Ails. MRP

2.5 IIIful(I' de~'eloped laminar flow ill a circular pipe, tile IIead lo.\'s{Iue to friction is direct(r proportiollal

to (mean veloci(l'/square of the mean velocity)

Ans. square of mean velocity

2.6 In mliabatic flow witll friction, tlte stagllatiolltemperature along (l streamline .

(increases/decreases/remain.\' constant)

Ans. remains constant

2.7 /11tlte case of a refrigeration system undergoing an irreversible cycle, ¢iis «0/ =0/ >0)Ans. < 0

2.8 Tile slle{lYforce ill a beam subjected to pure positive hemling is (positive/zero/negative)

AilS. zero

2.9 /n a single {Iegree {~/free{lom vibration system, tlle tilulamped natural frequency is to/tlulIl tile

damped natural frequency. (greater than/equal/les.\)

AilS. greater than

2.) 0 An aeroplane make.'i ClIIalf circle towards left. The engine rum clockwise wilen viewed from tile rellr.

Gyroscopic effect Oil tile aeroplalle causes tile 1I0.'ieto (lift/dip)

Ans. lift

3. State TRUE or FALSE witllout copying tlte questions. Give no reason(s) for your answer.

(10 x ) = 10)

w3.1 71,e Laplace transform of cos wi is ---

02 +a/'Ans. False

Explanation. Laplace transform of cos wt is -~.

8 -w

3.2 A differential equation ofform dy = f(x,y) is IIomogelleous iftllefunctiollf(x,y) depends ollly Oil tile{L~

y x ratio - or - .

x Y

AilS. True

Explanation. Since equation is defined as Homogeneous, f (x, y) = F ( ~) or G (~ )


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3.3 Roth the Iii'!' (lIntn: ilIid tll..: dead celltre of cenlre lathes are hardened and lubricated for reelucing fril"f;oil and wear.

An~.hli,c

Expl:w

•'-.4 IHamuI,J wh,'cls shoulJ nu: Or :plaoatinn, 0::l\ !l,ird marni:Jls shouid l,e !'{Gund , · n diamond wheels. (e.g,) carbide and ceramic cuttinglook sun mlk'rials like steel, ledd the whed quickly,

3.5 Slip gauges are calibrated by outside micrometers.

AilS, Ld:,~'

Exphw:2.bm Accura'.y of calibration source should be atleast one order hight:r than that of the instrument.

I-Ien,.;~.'litsidc micrometrc:, callnot be used for the calibration of slip gauges,

3.6 Comider (f Rankine (vde with superheat. If tile nUL\-imum pressure in tile cycle is' increa.,ed witllout

dllwgin;,' tlte mt/ximum tcmpNilture and the minimum pressure, the drynes,'1fracthm Of.HCiil1l iifter litei.\'el1fropj,' ",'(I'l!.;HiOIl Will increase.

Ails. False

Explanation. p', > PJ' Dryness fraction after isentropic expansion decreases.

I ! ~ --- - -- c _ ~~t I -1 1

4'__

3.7 The specific speed of an impulse Itydraulic turhine will be greater titan tite specific jpeed of a reactiontype hydraulic turhine.

Ans. False

Explanation. Specific speed of impulse hydraulic turbine 10 -- 35 rpm

Sp\:cific speed of a reaction hydraulic turbine 300 - 1000 rpm

3.8 Self locking in power screw is better achieved by decreasing tlte helix angle and increasing the coeJficiemof fricl;oil.

Ans. True

E, planation. For self locking, angle of friction should be greater than helix angle of screw.

3.9 Interference ill a pair of gears is avoided, if tlte addendum circle.s of botlt the gears imcT!J'ectcommon

fallgem 10 the base circles witltin the points oftqange/l(:y,


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3. to I" afour-.'ilroke engine, tlte secondary imbalance 1In"(} frcqut!11cy equal In four times engine v,ce/!.

Ans. False

Explanation. Frequency of secondary imbalance will be two times the engine speed.

Li~t I List II i(Prob/em arem) (Techniques) I '

t-----t----------(A) JIT 1. CRAFl

(B) Computer assisted layout 2. PERT I(C) Scheduling 3. Johnson's rule

(D) Simulation 4. K,mbans

5. EQQ rule

6. Monte Cario

4.2 --------L-is-t-I--------.r-----L-i-st-I-t-------l

l(Manufacturing Processes) (Conditiom)(A) Finish turning 1. Backlash eliminator

(B) Forming 2. Zero rake

(C) '1 hread clltting 3. Nose radiu:,ing

(0) Down milling 4. Lmv ~pced

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

4.3 i List II (Measuring instruments)--

(A) Talysurf

(8) Telescopic gauge

(C) Transfer callipcrs

(D) Autoc0l1imator

Ans. (A) -4, (8) --3, (C)-I, (D) -2

List II

(Appltcatol!1s)------1

L T-slots

2. Flatness

3. Internal diametcr

4. Roughness

--,-'lList 1\

(Cvcles)

1. Constant volume heat addition and constant volume heat rejection

2. Constant pressure hcat addition and constant volume heat rejection

1. Constant pressure heat ,;ddition and constant pressure heat rejection

4. Heat additIOn at con:;t;lt1t volume foilowed by heat addition at constant

temperalUre

.5 H~'lt ~,jectionat constant volume followed by heat rejection at constant

temper ature I

List I(Heat Eliginc:~)

(A) Gas Turbme

(B) Petrol Engme

(C) Stirling Engine

(D) Diesel Engine


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List I List II

(Gear types) (Applications)

(A) Worm gears I. Parallel shafts

(B) Cross helical gears 2. Non-parallel, intersecting shafts

(C) Bevel gears 3. Non-parallel, non-intersecting shafts

(D) Spur gears 4. Large speed ratios

5. Choose the correct answer and indicate by writing only the con \:sponding capital letter (A, B, C, D

as the case may be) (15 x 2 =30). x3 -cos x

5.1 1 1 m - - - - - equalx-.co x2 +(sin x)2

(A) 00

( C) 2

Ans. (A)

Explanation. cos x and sin x are finite whatever x may be

. x3._ cos xhm-----

x~oo x2 + (sin x)2

x3

J i m - , , ' = 00.X~O()X'

(B) 0

(D) does not exist

5.2 Among thefollowing, the pair of vectors orthogonal to each other is

(A) [3.4,7], [3,4,7] (8) [1,0,0], [I, 1,0]

(C) [I. 0, 2], [0, 5, 0] (D) [I, I, I], [-I, -I, -I]

Ans. (C)

Explanation. (ai' b j• c) and (a2, b2, c2) are orthogonal only if

a ja2, b1 b1, C 1 C 2 =o .Only C fits

[I, 0, 2] x [0, 5, 0] = 0 + 0 + 0 = 0

5.3 The function f(x) =xJ - 6x2 + 9x + 25 has

(A) a maxima at x =I and a minima at x = 3

(8) a maxima at x =3 and a minima at x = I(C) no maxima, but a minima at x = 3

(D) a maxima at x = I, but no minima

Ans. (A)

Explanation. f(x) = 3(x2 - 4x + 3), f '(x) =6(x) = 0

at x = I and x = 3

f '(I)=-ve; f '(3)=+ve

5.4 The .mlution to the differential equationf"(x) + 4j'(x) + 4f(x) =0 is

(A) fl(x) = e-2x

(8) f/x) = e2x, f 2

(x) = e2x

(C) fl(x) = e2

" f 2(x) '; = e-2x (D) fl(x) = e-2x, f

2(x) = e-X

Ans. (C)


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(D2 + 4D + 4)y = 0, where, y = f(x)

102 + 410 + 4 = 0

or (10 + 2)2 = 0, repeated roots.

y = e-2x (A + Bx)

Particular solutions are Ae-2xand Bx e-2xfor all arbitrary constants A and B.

Here A = 1, B = I.

5.5 11 , tlrefollowing FORTRAN statements, what is tl,e value of X at exit?

X 1

DO 10 I= 1,3,2

X = 2*X**2+1

10 CONTINUE

(A)3 (B) 5

(C) 19 (D) 101

ADS. (C)

Explanation. Two iterations will be carried out by the DO loop.

X = (2 x 12) + 1 = 3X = (2 x )2) + 1 = 19.

5.6 A stop watch time study on an operator witlr aperformance rating of 120 yielded a time of 2 minute.~. If

allowances of 10% of the total available time are to be given, tire standard time of tire operation is

(A) 2 minutes (B) 2.4 minutes

(C) 2.64 minutes (D) 2.67 minutes

Ans. (C)

Explanation. Standard time of the operation = Basic time + Allowances

Observed time x Rating factor

100

2 x 120 .= --- =2.4 mmutes

10 0

=10% of total available time

= 2.4 x J . Q . . =2.4 minutes100

=2.4 + 0.24 =2.64 minutes

(A) gd[ha - (H - h) A]

(B) gdHA

(C) GdHa

(D) gd (H - h) A

Explanation. Pressure at the piston

Force acting on piston

Hdg

HdgA.

rictionless ApIston of area =


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,~:'J '·,.ii,'. I : : , : , , ' /I,,(FlIClltsill the x alld y directiolls are givell by

2 3 4~I= t,.x)') - x22 , v = xy - - Y

4i be va!!:" "i •ror:1 possihle flow field involving an incompressible fluid is

4(8) --

3

4IT) .-" '3

Ans. (Dl

Explanation. FlJr steady state incompressible flow. the continuity cquatins

a u (Iv

- + - - - 0, must be satisfied i 'lx ; :'y

07 2 xy - 3y. JDy

" \) 1 ~:yy _. 3y' =0

r _ .= 3

5.() n"e J.i/omaie oj an idea/1:11s is tf,rott/ed from an initial pre,'isure of 0.5 MPa to O./ MPa. The i"itit,1tcmpcrlUi;rc {l 3{Jf}K. 'Ihi' i'lltropy change of the ulliverse is

( ,\ ) 1138 kJ/K (8)401.3 kJ/K

(C) 0 ft4~6 kJK (D) -0,0446 kJ/K

T, 1\S S .- Cp,lV In-=-- RII/n-"" - 'I - T

I PI

...R In~£u \)

I

0.1 kJ-8.314/n- =13.38-.

0.5 K

For an idt~al ga~ change in enthalpy is a function of tern perature alone and change in enthalpy ofa throttling

roccss IS zero.

10 A heaf re:,eHoir at 90{) K is brought into contact with the ambient at 300 Kfor a short time. During this;'1,.,iO(I 9(IfJtJ k.l of heat i,'i10,'itby the heat reservoir. The total Ion in t,vai/ability due to this proce,'i.'ii.'i

( A ) U W O O k .J (8) 9000 kJ

Ie) 6000 lei (D) None of the above

An~. (01

F"r'arwliofl Ihe availability of a thermal reservoir t r equivalent to the work output of a Carnot heatI.~ngine operating between the reservoir and the environment. Here as there is no change in

the temperatures of source (reservoir) or the sink (atmosphere), the initiat and final

availabilities are same, Hence there is no loss in availability.

1 t III order to bit", / kilogram of CIlJ comletely, the minimum number of kilogram!>'of oxygen "eeded i,'i

(take atomic weight~ of II, Ca"d 0 as I, /2 and /6 respectively).


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(A) 3

(C) 5

Ans. (B)

Explanation. CH4 + 2C

2 ~ CO

2 + 2Hp

16 kg 64 kg

16 kg of CH4

requires 64 kg of oxygen.

:. 1 kg of CH4

requires 4kg of oxygen

5.12 A spring scale indicates a tension T in the right hand cable of the pulley system shown in Fugure 5. J 1.

Neglecting the mass of the pulleys and ignoringfriction between the cable and pulley the mass mis

(A) 2T

g

(B)4

( 0) 6

(C) 4T

g

Ans. (C) Fill. 512

Explanation.

Referring Figure,

mg = = 4T

4Tm

g

5.13 Fig. 5.J 3 shows a quick return mechanism. The crank OA rotan clockwise uniformly.

OA =2cm,

OO=4cm.

(A) 0.5

(B) 2.0

(C) .fi

(D) 1

ADS. (B)

o~III

I

II


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·Forward strokeReturn stroke

240

120=2.

5. t 4 Tile deflection of a spring witlllO active turns under lliolid of 1000 N is 10 min. Tile spring i.~m(ule illto

two pieces eacll of 10 active coils and placed in pllrallel under tile same load. Tile deflection of this

system is

(A) 20 mm (8) 10mm

(C) 5 mm (0) 2.5 mm

Ans. (0)

Explanation. Stiffness before cutting

Stiffness after cutting into two

1000 = 100 N! mm.10

100 x 2 =200 N/mm.

Load 1000/2

Stiffness 200

t 5. Tile arm OA of an epicyclic gear train sllown in Fig. 5.15 revolves counter clockwise about 0 witll (Ill

angular velocity of 4 rad/s. Botll gears are of same size. Tire angular velocity of gear C, if tile sun gear B is fIXed, is

(A) ~ rad/s

«(;;)10 rad(s

:;':t'Ans..(8)' ,

Fix arm A.

Give GAerotation to 8

Multipl)'QY X c . '..;'

Addy

=>.!, '.Angular velocity of gear

(8) 8 rad/s

(0) 12 rad/s

o

o +xy x+y

__x.

y- x

O .

~&q/~~c&ccw)--4 rad/sec (cw)

y - x = 4 - (--4) = 8 rad/s

a jugad gate 2007 2001 1995 iit kanpur set questions.pdf

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