gate 2014sol(me)(evening)

8/12/2019 GATE 2014sol(ME)(Evening)

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The questions have been shared by students only. Since the questions are

purely memory based, , does not take any responsibilityregarding the correctness of questions.

In case there is any kind of discrepancy in questions please write to us at

or call us at :

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[email protected] Ph: 011-41013406, 09711853908

for rectification/deletion/updation of the same.

(Evening Session)

16 t h F e b ru a r y, 2 0 14


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Office Address : F-126, Katwaria Sarai, New Delhi - 110 016.Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406

Web : www.iesmaster.org E-mail : [email protected]

1. The value of the integral :

2 2

20

sin1 1d

cos1 1

x x xx x

is

(a) 3 (b) 0

(c) 1 (d) 2

Sol. : (b)

I =

22

20

1 sin 1

1 cos 1

x x

x xdx

=

22

20

2 1 sin 2 1

2 1 cos 2 1

x x

x x

=

22

20

1 sin 1d

1 cos 1

x xx

x x

=

22

20

1 sin 1d

1 cos 1

x xx

x x

I = I

I = 0

2. Match Group A with Group B :

Group A Group B

P. Biot number 1. Ratio of buoyancy to viscous force here

Q. Grashof number 2. Ratio of inertia force to viscous force

R. Pranolt number 3. Ratio of momentum to thermal diffusivity

S. Reynolds number 4. Ratio of internal thermal resistance to

boundary layer thermal resistance

(a) P4, Q1, R3, S2 (b) P4, Q3, R1, S2

(c) P3, Q2, R1, S4 (d) P2, Q1, R3, S4

Sol. : (a)


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3. The total number of decision variables in the objective function of an assignment problem of size n

n (n jobs and n machines) is

(a) n2 (b) 2n

(c) 2n 1 (d) n

Sol. : (d)

The total no. of decision variables in the objective functions of an assignment problem of size

n n is n.

4. Within the heat affected zone (HAZ) in a fusion welding present the work material undergoes

(a) micro structure changes but doesnt melt

(b) Neither melting and micro structure changes

(c) both melting and micro structure changes

(d) melting and retaining the original microstructure after solidification

Sol. : (a)

The area of base metal adjacent to the weld pool is called heat affected zone. Since this area is

subjected to a high temperature for considerable period of time as compared to the rest portion

of base metal, its microstructure changes and different from base metal prior to welding.

5. A f low field which has only convective acceleration is

(a) a steady uniform flow (b) an unsteady uniform flow

(c) a steady non-uniform flow (d) an unsteady non-uniform flow

Sol. : (c)

In case of steady flow local acceleration = 0

In case of uniform flow convective acceleration = 0

In case of steady uniform flow, all accelerations are zero

Hence convective acceleration exists only for steady non-uniform flow.


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8. Demand during lead time with assumed probabilities is shown below :

Demand 50 70 75 80 85

Probability 0.15 0.14 0.20 0.30 0.30

Expected demand during lead time is .......

Sol. :

Excepted demand during lead time

= 50 0.15 + 70 0.14 + 75 0.20

= 81.8

9. Moist air at 35C and 100% relative humidity is entering a psychrometric device and leaving at 25C

and 100%, relative humidity. The name of the device is

(a) Humdifier (b) Dehumidifier

(c) Sensible heater (d) Sensible cooler

Sol. : (b)

Moist air at 35C and 100% RH

25C, 100% RH

Name of device is

At 35C W1

At 25C W2

w1w2

W1 > W

2

Dehumidifier (b)

10. Two separate slab milling operation 1 and 2 are performed with identical milling cutter the depth of

cut in operation 2 is twice that in operation 1. The other cutting parameters are identical. The ratio

of maximum uncut chip thickness in operation 1 and 2 as .....

Sol. : 0.5

1 1 1

2 2 1

t d d0.5

t d 2d


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Office Address : F-126, Katwaria Sarai, New Delhi - 110 016.

Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406Web : www.iesmaster.org E-mail : [email protected]

11. If the Poissons ratio of an elastic material is 0.4, the ratio of modulus of ridigity to youngs modulus

is.....

Sol. : Modulus of rigidity

G =

E

2 1

G

E=

1

2 1 =

1 1

2 2.81.4 = 0.357

12. Laplase transform of cos t is 2 2s

s . The Laplase transform of 2te cos(4t)

is

(a) 2s 2

16s 2

(b) 2

s 2

16s 2

(c) 2s 2

16s 2

(d) 2

s 2

16s 2

Sol. : (d)

L[cos t ] = 2 2s

s

L[cos 4t] = 2s

s 16

L[2te cos4t ] = 2

s 2

(s 2) 16

13. Which one of the following is used to convert a rotational motion into a translational motion

(a) Bend gears (b) Double helical gears

(c) Worm gears (d) Rack and pinion gears

Sol. : (d)

Rotational - Translational

Rack and Pinioin

Rack

Pinion


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Sol. : (d)

(a) P (Q + R) = PQ + PR

So a is incorrect

(b) (P Q)2= (P Q) ( P Q)

= P2 PQ QP Q2

= P2PQ QP Q2

So (b) is incorrect

(c) | P + Q | = | P | + | Q |

It is not true

(d) (P + Q)2= (P + Q) (P + Q)

= P2 + PQ + QP + Q2

So (d) is correct

18. In a rolling process the maximum possible draft, defined as the difference between the initial to the

final thickness of the metal sheet, mainly depends on which pair of the following parameters ? P-

strain, Q - Strength of the work material R - Roll diameter, S - Rolls velocity, T - coefficient of friction

between roll and work

(a) Q, S (b) R, T

(c) S, T (d) P, R

Sol. : (b)

We know that maximum draft in rolling is given by

(h)max

2R

or

(h)max

= 2R

where,

: coefficient of friction between roll and work piece

R : Radius of roll.

Hence option (b) is correct.

19. A Carnot cycle refrigerator maintaining a temp of 5C the ambient air temp is 35C. The heat gained

by refrigerator at a continuous rate is 2.5 kJ/s. The power/ in watts required to pump this heat out

continuously is ...


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Sol. :

carnotCOP =2

1 2

T

T T

=273 5

35 5

= 6.7

6.7 =absorbQCOPW

6.7 =2.5

kJ sW

W

2.5 KJ/s

T25C

Qrelease

Qabsorb

W =2.5

0.3731 KJ s6.7

20. Ball bearings are rated by manufacturer for a life of 106revolution. The catalouge rating of particular

bearing is 16 kN. If the design load is 2 kN, the life of the bearing will be P 106revolution, where

P is equal to ....

Sol. :

106revolution

Catalogue rating of bearing = 16 KN

Design Load = 2 KN

Life = P 106revolution

L10

=

3C

P

life in million revolution

C = 16 kN

P = 2 kN

L10 =3

16512

2

LIfe in million hours

So P = 512


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21. The no. of independent elastic constant required to define the stress- strain relationship for an

isotropic elastic solid is ...

Sol. : Two

22. The principle of material removal in electrochemical machining is

(a) Ficks laws (b) Faradays laws

(c) kirchhoffs Laws (d) Other Laws

Sol. : (b)

Material removal in electrochemical machining is based upon Faradays law.

23. As the temp. increases, the thermal conductivity of gas

(a) increases (b) decreases

(c) remain constant (d) increase up to certain temp. and then decreases.

Sol. : (a)

In case of gases as the temperature increases the intra-molecular collisions increases which are

sole responsible for heat conduction in gases. So the conductivity of gases increases with increase

in temperature.

Hence option (a) is correct.

24. Kaplan water turbine is commonly used when the flow through its runner is

(a) axial and the head available is more than 100 m

(b) axial and the head available is less than 10 m

(c) radial and the head available on more than 100 m

(d) mixed and the head available is about 50 m.

Sol. : (b)


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25. Match the heat treatment process (Group A) and their associated effects on properties (Group B)

of medium carbon steel

Group - A Group - B

P - Tempering 1. Strengthening and grain refinement

Q - Quenching 2. Increasing toughness

P - Annealing 3. Hardening

S - Normalising 4. Softening

(a) P - 3, Q - 4, R - 2, S - 1

(b) P - 2, Q - 3, R - 4, S - 1

(c) P - 3, Q - 2, R - 4, S - 1

(d) P - 2, Q - 3 , R - 1, S - 4

Sol. : (b)

26. Steam with specific enthalpy (h) 3214 kJ/kg enters an adiabatic turbine operating at steady statewith, flow rate 10 kg/s. As it expands, at a point where h is 2920 kJ/kg, 1.5 kg/s enteracted for

heating purpose the remaining 8.5 kg/s further expands to the turbine exit, where h = 2374 kJ/kg

neglecting changes is kinetic and potential energies, the net power output (in kw) of the turbine is

2 marks

Sol. : (c)

h1

= 3214 KJ/kg

m = 10 kg/s

h2

= 2920 KJ/kg

m1

= 8.5 Kg/s

h3

= 2374 KJ/kg.

KE = 0; PE 0

work output = 1 2 31 2m m h hh h

= 10 8.53214 2920 2920 2374

= 7581 KJ/s = 7581 KW


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27. The value of the integral2 x

x y

0 0

e dydx is

(a)1

(e 1)2

(b) 1/2 (e2 1)2

(c) 1/2 (e2 e) (d) 1/2 (e - 1/e)2

Sol. : (b)

2y

0 0

e dy d x

x x =

2y

0

0

e e d x

x x

=

2

0

e e 1 d x x x

= 2

2

0

e e d x x x

=

22

0

1e e

2

x x

=4 21 1e e e e

2 2

=4 21 1e e

2 2

= 4 21

e 2e 12

= 2

21 e 12

28. A flame is subjected to a load P as shown. The frame has a constant flexural rigidity EI. The effect

of air at load is negleted. The deflection at point A due to the applied load P is

L

L

A

P


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(a)31 PL

3 EI(b)

32 PL

3 EI

(c)3PL

EI(d)

34 PL

3 EI

Sol. : (d)

A

PL, EI

L, EI

Total deflection at A = 1 2

PL

P

PL

P P

L

1= L

1

(PL)LL

EI

=

3PL

EI

P

2 =PL

3EI

3

Total deflection 1 2 =34PL

3EI


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29. A cylindrical riser of a 6 cm diameter and 6 cm height has to be designed for a sand casting mould

for producing a steel rectangular plate casting of 7 cm 10 cm 2 cm dimention having the total

solidification time of 1.36 minute. The total solidification time (in minutes) of riser is .......

Sol. : (c)

Solidification time of casting

tc =

2

VKA

1.36 =

27 10 2

K2(7 10 10 2 2 7

=

2140

K208

1.36 =

235

K52

(i)

Solidification time for riser,

tR

=

2V

KA

=

2d

K6

tR

=

26

K6

(ii)

From eq. (i) and (ii)

tR =2

1.3635

52

= 3.00

Rt 3.00 min

30. A wardrobe (mass 100 kg, height - 4m, width 2m, depth 1m) symmetric about the Y Y axis

stands on, rough level floor as shown in fig. A force P in applied at mid height on the wardrobe so


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as to toppling about point Q without slipping, what are the minimum values of the force (in newton)

and the static coefficient of friction between the floor and the wardrobe respectively ?

P

Qy

y

(a) 490.5 & 0.5 (b) 98.1 & 0.5

(c) 1000.5 & 0.15 (d) 1000.5 & 0.25

Sol. : (a)

P4

Q

1

Toppling about point Q; P = ? without slippingFBD

Taking Toruqe about 0

4P

2 = mg 1

P =100 9.81 1

2

= 490.5 N

N = P

P

N

mg

N =

P P 490.50.5

N Mg 100 9.81

31. At a work station, 5 jobs arrive every minute. The mean time spent on each job in the work-station

is 1/8 minute. The mean steady state no. of jobs in the system is ____________.


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Sol. :

Arial Rate, = 5 job minute

= 7.5 Job/min [ ]

No. of Jobs in the system =

50.6667.5 2

51 1 0.6661

7.5

32. A GO NO GO plug gauge is to be designed for measuring a hole of nominal diameter 25 mm with

a hole tolerance of 0.015 mm. Considering 10 % of work tolerance to be the gauge tolerance and

no wear condition the dimension (in mm) of the GO plug gauge as per the unilateral tolerance system

is ______________.

(a)0.0030.00324.985

(b)

0.0000.00625.015

(c)0.030.0324.985

(d)

0.0030.00024.985

Sol. : (d)

Hole size = 25 0.015 mm

Hole

Work tolerance zone

25mm

LL

Zero line

Datum line

UL

Upper limit of hole,

UL = 25.015 mm

Lower limit of hole, LL = 24.985 mm

Work tolerance = UL LL = 0.03 mm

Gauge tolerance = 10% of work tolerance

= 0.1 0.03

= 0.003 mm

For number wear condition size of GO gauge

= LL of hole + Gauge tolerance

= 24.985 + 0.003

= 24.988 mm

As per unilateral system, it can be written as + 0.003

24.985 0.000 mm

Hence option (d) is correct.


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33. Torque and angular speed data over one cycle for a shaft carrying a flywheel as shown in the fig.

The moment of inertia (in kg.m2) of the flywheel is _______________.

Torque

3000

N-m

3/2 2/20

1500

N-m

Angular speed

20 rad/s

3/2 2/20

10 rad/s

Sol. :

E =2

sI C

= max min20 0

10= =2 2

CS

= max min20 0

2= =10

E = 3000 1500 22

= 3000

3000 = 2I 10

I = 47.1238 Kg m2

34. A single degree of freedom system has a mass of 2 kg. Stiffness 8 N/m and viscous damping ratio

of 0.02. The dynamic magnification factor at an excitation frequency of 1.5 rad/s is __________.

Sol. : Single degree of freedom m= 2 kg

K = 8 N/m; = 0.02; Wn = K m 2rad s

W = 1.5 rad/s

Dynamic Magnification Factor =2 22

nn

1

ww21

ww


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=2 22

12.28

1.51.52 0.021

22

35. Two identical metal blocks L & M (specific heat : 0.4 kJ/kg-K), each having a mass of 5 kg, are

initially at 313 K. A reversible refrigerator extract heat from block L and rejects heat to block M until

the temp. of block L reaches 293 K. The final temp. (in K) of block M is _____________.

Sol. : (c)

LS = MC ln LfT

313 ... (1)

mS = MC lmmfT

313 ...(2)

universe( S) 0

But since the system is reversible

L mS S = 0

MC lnf mf

2

T T

(313)

l= 0

ln

f mf

2

T T

(313)

l=l

n1

f mf

2

T T

(313)

l= 1

Tmf

=

2

f

(313)

T

l

Tmf

= 334.36 K

36. The precedence relation and duration (in days) of activity of a project network are given in the table.

The total float (in days) of activities e & f, respectively, areActivity Predecessors Duration (days)

a 2

b 4

c a 2

d b 3

e c 2

f c 4

g d, e 5


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(a) 0 & 4 (b) 1 & 4

(c) 2 & 3 (d) 3 & 1

Sol. : (b)

a

b

d

34, 4

4

2

7, 7g

5

12, 12

4

e

f

c4, 52, 3

0, 02

2

For activity C FT= (7 4) 2 = 1

For activity f FT= (12 4) 4 = 4

37. A ladder AB of a length 5 m & weight (w) 600 N is resting against a wall. Assuming frictionless

contact at the floor (B) and wall (A), the magnitude of the force P(in newton) required to maintain

equilibrium of the ladder is _________.

w

2.5m

2.5m3 m

B

A

4 m

P

Sol. :

w

2.5m

2.5m3 m

B

A

4 m

P

N2

N1


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Since here is frictionless contact at A and B reactionless contact at A and B will be rto thesurface

W = 600N; P = ?

Thus, for frictionless wall and floor

W = N2

N1

= P

Taking torque about A.

W 2.5 cos + P 3 = N2 4

4

W 2.5 3P5

= W 4 (cos = 4/5)

2W = 3P

P =2W 2 600

3 3

= 400 N

38. Water flows through a tube of diameter 25 mm at a an average velocity of 1.0 m/s. The properties

of water are P = 1000 kg/m3, = 7.25 104N-S/m2, K = 0.62 W/mK, Pr = 4.85. Using Nu =0.023 Re0.8Pr0.4, the connective heat transfer coefficient (in W/m2K) is __________.

Sol. : d = 0.025 m

V = 1 m/s

P = 1000 kg/m3

= 7.25 104N-S/m2

K = 0.62 W/mK

Pr = 4.85

Re = 4PVd 1000 1 0.025

7.25 10

= 3.448 104

Nu = 0.023 Re0.8Pr0.4

hD

K= 0.023(34482.7586)0.8(4.85)0.4

hD

K= 184.5466


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h =184.5466 0.62

0.025

= 4567.7566 W/m2K

/ 2h 4567.7566 W m K

39. If Z in a complex variable, the value of3i

5

dz

z is _____________.

(a) 0.511 1.57 i (b) 0.511 + 1.57 i

(c) 0.511 1.57 i (d) 0.511 + 1.57 i

Sol. : (b)

3

5

dz

zi

= 3

5logz i

= log3i log 5

= 2 2 11 3

log 0 3 i tan log52 0

= log 3 + i log52

=3

log 1.575

i

= 0.511 + 1.57i

40. Two infinite long parallel plates are placed at a certain distance apart. An infinite long radiation shield

is inserted between the plates without touching any of them to reduce heat exchange between the

plates. Assume that the emmisivities of plates and radiation shield are equal, the ratio of the net

heat exchange between between the plates with and without the shield is

(a) 1/2 (b) 1/3

(c) 1/4 (d) 1/8

Sol. : (a)

For same emmisivity of plates,

qwith shield

=

without shieldq

n 1

where n= number of shield inserted.


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Here, n= 1

q

q

withshield

withoutshield

1

2

Hence option (a) is correct.

41. Consider the following statement regarding streamline (s)

I. It is a continuous line line such that the tangent at any point on it shows the velocity vector at

that point.

II. This is no flow across streamlines

III.du dy dz

u v w is the differential equation of sreamline where u, v and w are velocities in direction

x, y and z respectively.

IV. In an unsteady flow, the path of a particle is a streakline.

(a) I, II, IV (b) II, III, IV

(c) I, III, IV (d) I, II, III

Sol. : (a)

Statement (iii) is wrong because for stream linedx

u=

dy dz=

In an unsteady flow, ends points of Particle path is on a streak line

42. In a compression ignition engine, the inlet air pressure is 1 bar and the pressure at the end of

isentropic compression is 32.42 bar. The expansion ratio is 8. Assuming ratio fo specific () is 1.4the air standard efficiency (in percent) is ..........

Sol. :

r = 1 1.4 11.99932.42

=r 11 999

1 4998= =8.

.

=1

11 11

1r r

=

0 4 1 41 1 11 499811 4 12 1 4998 1

. ..

. .

= 0.5959

= 59.5938%


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43. A thin gas cylinder with an internal radius of 100 mm is subject to an internal presure of 10 MPa,

the max permissible working stress is restricted to 100 MPa. The minimum cylinder wall thickness

(in mm) for safe design must be ....

Sol. :

d = 100 mm

t = ?

p = 10MPa

Max stress =pd

2t p

d

t

for safetypd

2t

tpd

2

t10MPa 200

mm2 100 MPa

t 10 mm

tminShould be 10 mm

44. A cast iron block of 200 mm length in being shaped in a shaping machine with a depth of cut of

4 mm, feed of 0.25 mm/stroke and the tool principal cutting edge angle of 30. No of cutting strokes

per minute in 60. Using specific energy for cutting as 1.49 J/mm3the average power consumption

in watt is ...................................

Sol. : Given a cost iron block.

d

L

Direction of

cutting strokeLength, L = 200 mm

Depth of cut, d = 4 mm

Feed, f = 0.25 mm/stroke

Number of strokes = 60 / min = 1 / sec.

Material removal rate (MRR) = L d f No. of stroke/sec

= 200 4 0.25 1


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= 200 mm3/s

Average power consumption (P) = Specific energy MRR

P = 1.49 200 = 298 J/s

P 298 watt

45. A plane wall has a thermal conductivity of 1.15 w/m k// If the inner surface is at 1100C and the outer

surface is at 350C, then the design thickness in meter of the wall to maintain a steady heat flux

of 2500 w/m2should be

Sol. : Given,

k = 1.15 w/mk

T1

= 1100C

T2

= 350C

T2

T1

b

q

b : Thickness of the wall

q = 2500 w/m2

q = Tkb

2500 =

1100 350

1.15b

b = 0.345 m

b 34.5 cm

46. A bolt of major diameter 12 mm is required to clamp two steel plates. Cross sectional area of the

threaded portion of the blot is 84.3 mm2. The length of the threaded portion in grip is 30 mm, while

the length of the unthreaded portion in grip is 8 mm young modulus of material is 200 GPa the

effective stiffness (in Nm/m ) of the bolt in the clamped zone in

Sol. :

d1

= 12m

1 = 8mm A

1 =

2 2

1/4d 110.09 mm=

A2

= 84.33 mm2

2 = 30mm


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P ( x < 2) = P(x = 0) + P (x = 2)

= 5.2 5.2e 5.2 e 5.2

0.1 1.1

= 5.2e (1 5.2)

= 0.034

49. A shaft is subjected to pure torsional moment. The maximum shear stress developed in the shaft

is 100 MPa. The yield and ultimate strength of the shaft material in tension are 300 MPa and 450

MPa respectively. The factor of safety using maximum distortion energy (Von-mises) theory is

Sol. :

Max Shear stress =T

J

max shear stress =4 3

T D/2 16T100MPa= =

D D

32

As per max distortion energy theorem

T

1 =

2 =

3 = 0

2 2 2

2 3 3 11 2

1

2

2fy

F.O.S

2 221

22

2fy

F.O.S

26

2

2fy

F.O.S

F.O.S2 2

2 2

2fy fy=

6 3

F.O.S =fy / 3 300

3= =3 100

F.O.S. = 1.732


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50. A closed system contain 10 kg of saturated liquid ammonia at 10C. Heat addition required to convert

the entire liquid into saturated vapour at a constant pressure is 16.2 MJ. If the entropy of the

saturated liquid is 0.8 deg. kJ/kg K the entropy (in kJ/kg K) of saturated vapour is

Sol. :

m = 10 kg, T = 10C satinated ammonia

Q = 16.2 MJ to connect liquid into saturated vapour at constant pressure

Sl = 0.8 KJ/Kgk

SV

= ?

S =Q

mT

SV Sl = 16.2 MJ

0.00572 MJ KgKm273 10

SV

Sl= 5.72 KJ/KgK

SV2 = 6.52 KJ/KgK

51. For the truss shown in the fig. the forces F1and F2are 9 kN and 3 kN the force (in kN) in themember QS is

2

ST

1.5 3

3 3

F1 F2

R

QP

(a) 11.25 tension (b) 11.25 compression

(c) 13.5 tension (d) 13.5 compression

Sol. : (a)


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1.5 m

2 m

3 m R

9 kN

3 mP

Q

3 kN

S

3 m 1.5 m

9 kN

FQS

tan = 2 4

1.5 3

sin =4

5

cos =3

5

QSF sin = 9

FQS

= 5 459 KN4 4

FQS

= 11.25 KN (Tension)

52. Consider a velocity field v k (y i x k) , whenn K is a const. The vorticity, Z. is

(a) K (b) k

(c) k/2 (d) K/2

Sol. : (a)

V = kyi kxk u ky, w kx, v 0

Wz

=v u

x y

k0 k

53. A uniform slender rod (3 m length & 8 kg mass) is rotate in a vertical plan about a horizontal axis

1 m from its ends as shown. the magnitude of the angular acceleration a (in rad/s2) of the rod at

the position shown is


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ER1 m 2 mSol. :

Ic

=2

ML

12

I0 = Ic+ mx2

1m

1.5

2m

.5 m

C0

I0 =2

2MLMx

12

=2

28 38 0.5

12

= 8 Kg m2.

T = I

Torque T = mg x

= 8 10 0.5

= 40 Nm

.5 m

C

mg

=2

2

T 40Nm5 rad s= =

I 8Kgm

54. It is desired to avoid interference in a pair of spur gears having 20 pressure angle with increase in

pinion to gear speed ratio, the min no. of teeth on the pinion.

(a) Increase (b) decreases

(c) first increase & then decrease (d) remain unchanged.

Sol. : (a)

Spur gear having 20 pressure angle

Avoid Interference

Minimum No. of teeth of Pinia = 229

1 G sin 1G 2

As G

Denominator Decrease

Minimum No. of teeth increases


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55. Consider an ordinary differential eqn.dx

dt= 4t + 4 if x = x

0at t = 0, the increment in x calculated

usign Runge-Kutta fourth order multi step method with a step size of t = 0.2 is

(a) 0.12 (b) 0.44

(c) 0.66 (d) 0.88

Sol. : (d)

d

dt

x= 4t + 4, x(0) = x

0, t = h = 0.2

f(t,x) = 4t + 4

K1

= hf(x

0) = (0.2) (4) = 0.8

K2

= hf (0.1, x

0+ 0.4) = (0.2) (0.4 + = 0.88

K3

= hf(0.2, x0+ 0.88) = (0.2) (0.8 + 4) = 0.96

K = 1 2 3 41

K 2K 2K K 6

= 0.88

1x = 0x + K = 0x + 0.88So increment inx= 0.88

56. Which of the following option is closest in meaning. In a democracy, everybody has the freedom to

disagreewith the goverment.

(a) dissent (b) descent

(c) decent (d) decadent

Sol. : (a)

57. In a sequence of 12 consecutive odd number the sum of the first 5 no. is 425. what is the sum of

last 5 nos in the sequence.

Sol. : 495


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odd no.

2

4

6

8

x

x

x

x

x

10

12

x

x

14

16

18

20

22

x

x

x

x

x

5x + 2 + 4 + 8 + 6 = 425

x =425 20 405

815 5

Sum of last 5 nos.

5x + 14 + 16 + 18 + 20 + 22

= 5 81 + 90

= 405 + 90

= 495

58. While receiving the award the scientist said I feel vindicated, which of the following is closest in

meaning to the word vindicated.

(a) Panal (b) substantiated

(c) Appreciated (d) Chatemal.

Sol. : (b)

59. Lt f(x,y) = xnym= P, if x is doubled and y is halved the new value of f is.

(a) 2nmp (b) 2mnp

(c) 2(nm) p (d) 2(mn) p


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Sol. : (a)

f(x, y) =xnym= p

If xis doubled andyis halfed

f(x, y) =

mn n m n m n myx x y p(2 ) 2 2

2

60. After the discussion Tom said to my please revert ! He expects him to

(a) retract (b) get back to him(c) more in reverse (d) retrect.

Sol. : (b)

61. A five digit no, is formed using the digit 1, 3, 5, 7, 9 without repeating any of them. What is the

sum of all such possible five digit nos.?

(a) 6666660 (b) 6666600

(c) 666666 (d) 6666606

Sol. : (b)

62. If KCLFTSB, stands for best of luck and SHSWDG stands for good wishes. Which of the following

indicates ace of the exam ?

(a) MCHTX (b) MXHTC

(c) XMHCT (d) XMKTC

Sol. : (b)

63. A f irm producing air purif ier sold 200 units in 2012. The following pie chart represents the share of

raw materials, labour, energy, plant and machinery and transportation cost in the total manufacturing

cost of the firm in 2012. The expenditure on labour in 2012 is Rs. 4,50,000. In 2013, the raw material

expenses increase by 30% and all other expanses increases by 20%. What is the % increase in

total cost of company in 2013.


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Transportation

Labour

Material

10%

15%

20%

25%

30%

Energy

Plant andmachinery

Sol. : Let the total production cost bex in 2012.

Hence515 4.5 10

100

x

x =5

54.5 10 100 30 1015

Raw material cost in 2013 =5 201.3 30 10

100

= 7.8 105

All other expanses in 2013 = 30 105 0.8 1.2

= 28.8 105

Total cost in 2013 = (28.8 + 7.8) 105

= 36.6 105

% increase in cost in 2013 =36.6 30

10030

=

6.6

0.3 = 22%

64. Industrial consumption of power doubled from 2000-2001 to 2010-2011. Find the annual rate of

increase in percent assuming it to be uniform over the years.

(a) 5.6 (b) 7.2

(c) 10.0 (d) 12.2

Sol. : (b)


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2a0

=

11 1

0 1100

ra

r = 7.177%

65. Find the next term in the sequence : 13 M, 17Q, 19S, ___

(a) 21W (b) 21 V

(c) 23W (d) 23 V

Sol. : (c)


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