GATE 2013 Test Series -1

GATE-2013

S K Mondal’s

For any query: [email protected]

Mechanical Engineering

All India Test Series

Duration: 3 Hrs Maximum Marks: 100 Read the following instructions carefully:

1. There will be a total of 65 Questions carrying 100 marks Q.1 to Q.25 (25 Questions) will carry one mark each (subtotal 25 marks).Q.26 to Q.55 (30 Question) will carry two marks each (sub-total 60 marks) Question Q.56 to Q.65 belong to General Aptitude (GA) Question 56-Q.60 will carry 1 mark each (subtotal 5 marks) and Question Q.61-Q.65 will carry-2 marks each subtotal (10 marks)

2. Question Q.48-Q.51 (2 pairs) are common data question and Question pairs (Q.52, Q.53) and (Q54, Q.55) are linked answer questions. The answer to the second Question of the linked answer to the second question of the first answer of the Pair. If the first question in the linked pair is wrongly answered or is un-attempted, then the answer to the second Question in the pair will not be evaluated

3. Fill objective Response sheet (ORS) using a soft HB Pencil. Don’t use the ORS for any rough work.

4. Answer all the questions.

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Question Q1 to Q25 carry one mark each (1) Consider the matrices (4 3) (4 3),X Y× × and (2 3)P × . The order of [ ( ) ]T T T TP X Y P will be

(a) (2 2)× (b) (3 3)× (c) (4 3)× (d) (3 4)×

(2) Following system of equation has

3x-y+2z = 1 2x-2y+3z = 2 x+2y-z = 3

(a) A unique solution (b) No solution (c) An infinite no of solution (d) Non zero solution

(3) A box contains 10 screws, 3 of which are defective. Two screws are drawn at random

with replacement. The probability that none of the two screws is defective will be (a) 100 % (b) 50 % (c) 49 % (d) None of these

(4) 1sin is0

lim xxx

⎛ ⎞⎜ ⎟→ ⎝ ⎠

(a) ∞ (b) 0 (c) 1 (d) Non-existant

(5) If f(0) = 2 and 2

1( ) ,5

f xx

=−

the lower and upper bounds of f(1) estimated by the mean

value theorem are (a) 1.9, 2.2 (b) 2.2, 2.25 (c) 2.25, 2.5 (d) None of the above

(6) In an experimental set-up for the tool life of the cutting tool, following data have been

noted No. of components Rev/min Feed in m/min machined between regrinds 700 450 0.25 80 710 0.25 What is the tool-life equation if the length of the work piece is 100 mm and the mean

diameter at the cutting edge is 25 mm? (a) 0.21VT =114 (b) 0.25VT =87 (c) 0.14VT =111 (d) 0.34VT =34

(7) A Kaplan turbine installed at Kotla (Punjab, BBMA) develops 34 MW under a head of 29.9

m. Find the specific speed of the turbine if it runs at 166.7 rpm.

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(a) 300 (b) 404 (c) 440 (d) 16.3 (8) Consider the following p-v diagram.

P

V

1 2

3

P=const

PV =constn

PV=cost

Which of the following T-S diagram is correct?

(a) (b)

(c) (d) (9) The degradation of plastics accelerated by (a) High ambient (b) Dampness (c) Corrosive atmosphere (d) Ultraviolet radiation (10) NH3 is circulated in a compression refrigeration system. NH3 enters the evaporator 0.15

dry and leaves 0.95 dry. The work done during the compression is 46.7 KJ/kg. Latent heat of NH3 is 315 KJ/kg. Calculate cop of the system

(a) 4.5 (b) 5.4 (c) 2.4 (d) 17.5

(11) A heat treatment process in which the metal is heated to a temperature near the critical point, held there for a proper time and then cooled slowly in the furnace is called

(a) Normalizing (b) Annealing (c) Tempering (d) Case hardening

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(12) The holding power of a sunk key is mainly due to the resistance offered by it in

I. Shear II. Crushing III. Tension (a) I and II only (b) II and III only (c) I and III only (d) I, II and III only

(13) The notch sensitivity q is expresses in terms of fatigue stress concentration factor

fk and theoretical stress concentration factor tk is

(a) f

t

k +1k +1

(b) f

t

k -1k -1

(c) t

f

k +1k +1

(d) t

f

k -1k -1

(14) A cotter joint is used to transmit (a) Axial tensile load only (b) Axial compressive load only (c) Combined axial and twisting loads (d) Axial tensile or compressive loads. 14. Ans. (d) (15) The lead screw of a lathe with nut is a (a) Rolling pair (b) Screw pair (c) Turning pair (d) Sliding pair (16) A thin rod of length L and mass M will have moment of inertia about an axis passing

through one of its edge and perpendicular to the rod.

(a) 2ML

12 (b)

2ML6

(c)3ML

3 (d)

2ML3

(17) Filler is used in plastics to

(a) Completely fill up the voids created due to bubble formatting during the process of forming plastics. (b) Improve plasticity and toughness. (c) Provide color, strength, impact and wear resistance and reduce cost. (d) To accelerate the condensation and polymerization.

(18) Gears are best mass produce by

(a) Milling (b) Hobbing (c) shaping (d) casting (19) Grinding is mainly used for (a) Getting better finish on already machined surface (b) Removing very small a mount of material (c) To machine hard surface (d) None of the above (20) Cores are used to (a) Make desired recess in casting (b) Strengthen mounding sand (c) Support loose pieces (d) Remove pattern easily (21) S- curve is connected with (a) Combustion (b) Cutting tools (c) Corrosion (d) Heat treatment (22) Oxygen to acetylene ratio in case of neutral flame is

(a) 0.8:1.0 (b) 1:1 (c) 1.2:1 (d) 2:1

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(23) The total number of instantaneous centers of a mechanism having 5 links is (a) 15 (b) 2 (c) 10 (d) 3

(24) An over-damped system (a) Does not vibrate at all (b) Vibrate with frequency more then natural frequency of the system. (c) Vibrate with frequency less then natural frequency of the system.

(d) Vibrate with frequency equal then natural frequency of the system. (25) Stress concentration factor in case of gears depends upon (a) Thickness of tooth at the root (b) Material of the gear (c) Pressure angle and fillet radius (d) All of the above

Question Q25 to Q55 carry one mark each (26) For the matrix

3 -2 20 -2 10 0 1

p⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

One of the Eigen value X is equal to -2. Which of the following is an eigen vector

(a) 3

21

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

(b) 3

21

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

(c) 12

3

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

(d) 250

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(27) Consider the differential equation

3

23 cosd y dya ax

dx dx+ =

The C.F of the above equation is

( )( )

( )

1 2 3

1 2

1 2 3

( ) cos sin

( ) cos sin

( ) cos sin

( )

a c c ax c ax

b c ax c ax

c c c a x c ax

d none

+ +

+

+ +

(28) If L defines the Laplace Transform of a function, L [sin (at) will be equal to (a) ( )2 2/ −a s a (b) ( )2 2/ +a s a (c) ( )2 2/ +s s a (d) ( )2 2/ −s s a

(29) The Fourier series expansion of a symmetric and even function, f (x) where

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f (x) 1 (2 / ), 0 and

1 (2 / ), 0willbe

= + × − ≤ ≤= − × ≤ ≤

xx

π ππ π

(a) ( )( )2 2

n=14/ 1 cos

∞

+∑ n nπ π (b) ( )( )2 2

n=14/ 1 cos

∞

−∑ n nπ π

(c) ( )( )2 2

n=14/ 1sin

∞

∑ n nπ π (d) ( )( )2 2

n=14/ 1 sin

∞

+∑ n nπ π

(30) The general solution of 2

2dy xy+y=dx x

(a) 1x nx C= + (b) 1y nx C= +

(c) 1x nx Cy

− = +

(d) x x Cy= +

(31) For a particular activity of a project, time estimates received from two engineers X and Y

are as follows: Optimistic time Most likely time Pessimistic time Engineer X 4 6 8 Engineer Y 3 5 8 Who is more certain about the time of completion of the job? (a) Engineer X (b) Engineer Y (c) both (d) No one

(32) Two weight of 50 kgf and 150 kgf (of two blocks) A and B respectively are connected by

a sting and frictionless and weightless pulleys as shown in figure below. The tension in the string would be

50 kgf

150 kgf

A

B

(a) 100 (b) 200 (c) 64.5 (d) 50

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(33) The sling-psychrometer reads 40°C DBT and 28 °C WBT. Atmospheric pressure

1.30 2 kg / cm and partial vapour pressure is 0.03038 2kg / cm saturation pressure at 40 °C is 0.075 2 kg / cm . Then the specific humidity (kg/kg of dry air) and relative humidity (%) is respectively. (a) 0.021, 48.4 (b) 0.189, 44.41 (c) 0.0189, 40.4 (d) 0.021, 41.2

(34) During machine of C-25 steel with 0-10-6-6-8-90-1 mm (ORS) shaped triple carbide

cutting tool, the following observation were mode: Depth of cut : 2 mm Feed : 0.2 mm/rev Chip thickness : 0.39 mm What is the value of shear angle?

(a) 11° (b) 10° (c)23° (d) 32°

(35) Find the natural frequency (Hz) of following figure. The roller rolls on the surface without. Slipping and without friction.

K

r

m

(a) 1 k

2π m (b)

1 2k2π 3m

(c) 2k3m

(d) 1 3m

2π 2k

(36) An infinite plate is moved over a second plate on a layer of liquid as shown. For small

gap width, d, we assume a linear velocity distribution in the liquid. The liquid viscosity is 0.65 -310× kg/m.s. and specific gravity is 0.88. Calculate the shear stress on the lower plate, in Pa

x

d=0.3 mm

U=0.3 m/s

y

(a) 7.39 (b) 65 (c) 0.650 (d) 6.50

(37) The U-tube shown is filled with water at T = 20°C. (Saturated pressure 2.34 kPa). It is sealed at A and open to the atmosphere at D. The tube is rotated about vertical axis AB.

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For the dimensions shown, compute the maximum angular speed if there is to be no cavitation.

(a) 881rads

(b) 188rads

(c) 818rads

(d) 130rads

(38) The x component of velocity in a steady, incompressible flow filed in the xy plane is

u=A/x, where A=2 2m /s and x is measured in meters. Then the simplest Y-component of velocity for this flow field is?

(a) v= 2Axy

(b) v= 2

Ayx

(c) v=Axy

(d) v=0

(39) Determine the number of degrees of freedom of the mechanism shown in the figure

below

D

A B

C

(a) 1 (b) 2 (c) 0 (d) 3 (40) A cylinder containing the air comprises the system cycle is completed as follows.

(i) 82 kJ of work is done by the piston on the air during compression stroke and 45 kJ of heat are rejected to the surroundings. (ii) During expansion stroke 100 kJ of work is done by the air on the piston. Calculate the quantity of heat added to the system. (a) 45 kJ (b) 37 kJ (c) 63 kJ (d) 36 kJ

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(41) An ice plant working on a reversed Carnot cycle heat pump produces 15 tonnes of ice per day. The ice is formed from water at 0°C and the formed ice is maintained at 0°C. The heat is rejected to the atmosphere at 25 °C. The heat pump used to run the ice plant is coupled to a Carnot engine which absorbs heat from a source which is maintained at 220°C and rejected the heat to atmosphere. Determine the heat input required for engine. (Enthalpy of fusion of ice=334.5kJ/kg) (a) 13.43kw (b) 33.31kw (c) 83.51kw (d) 5.3 kw

(42) A brass bar having a cross-sectional area of 1000 2mm is subjected to axial forces as

shown in the figure below. 2105GNE m⎡ ⎤=⎢ ⎥⎣ ⎦

L M N P

50KN

0.6m 1.0m

80KN 20KN

1.2m

10KN

The total change in length of the bar. (a) 0.4311mm (b) -0.1143m (c) 1.32m (d) 3.25mm

(43) An air- conditioned hall has one of its walls 0.7 m thick. The inside space is required to

be maintained at 20 °C. The thermal conductively of the wall is variable, and is given by the relation

-4 2K =0.93+1.163 10 T× Where T in °C and K in W/mk. Then the loss of heat per second through one of its walls 10m × 5m when the outside

temperature is 40 °C. (a) 4841 J/S (b) 1484 J/S (c) 8414 J/S (d)1841 J/S (44) For a hemispherical furnace, the B flat floor is at 700 K and has an emissivity of 0.5. The

hemispherical roof is at 1000K and has emissivity of 0.25. Then the net radiative heat transfer from roof to floor is ( )-8 2 4σ=5.6710 w/m k

(a) 11.23 kW/ 2m (b) 21.31 kW/ 2m (c) 12.31 kW/ 2m (d) 13.21 kW/ 2m

(45) PERT calculation yield a project length of 50 weeks, with a variance of 16. Within how many weeks would you expect the project to be completed with probability of 95 %( where Z=1.65)?

(a) 75 weeks (b) 57 weeks (c) 49 weeks (d) 47 weeks (46) A hot gas at 300 C° flows through a metal pipe of 10 cm OD and 3mm thick. It is

insulated such a way that the insulation surface temperature should not exceed 50°C. Determine the thickness of insulation required taking the following data

2 21 0h = 25 /m K, h =10 /mW W K

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Ta = 25°C (Atmospheric temperature) K (metal pipe) = 40 W/mK K (insulation) = 0.045 W/mK (a) 25cm (b) 15cm (c) 3cm (d) 5cm (47) What is the maximum power transmitted if the tension in the belt is limited to 246 N and

the belt material weight 56 N per cm? The ratio of friction tensions is 2.0. (a) 10.29 KW (b) 0.4662 KW (c) 26.46 KW (d) 66.32 KW

Common data question Q48 and Q49 The monthly requirement of a company is 1500 components. The cost of each part is Rs 5 and the cost of each set-up is Rs 30 per lot. If the carrying charge factor is 20% determine

(48) Economic lot site

(a) 1200 (b) 1040 (c) 4010 (d) 1090 (49) Set-up time for each lot (a) 21 hrs (b) 25 hrs (c) 35 hrs (d) 10hrs

Common data question Q50 and Q51 A spherical vessel of 0.9 3m capacity contains steam at 8 bar and 0.9 dryness fraction. Steam is blown off until the pressure drop to 4 bar. The valve is then closed and the steam is allowed to cool until the pressure falls to 3 bar. Assuming that the enthalpy of steam in the vessel remains constant during blowing off periods determine. at 8 bar gV = 0.24 3m /kg, fh = 721 KJ/kg, gh = 2047 KJ/kg

at 4 bar gV = 0.462 3m /kg, fh = 605 KJ/kg, gh =2133KJ/kg

at 3 bar gV = 0.606 3m /kg.

valve

spherical vessel

0.9mcapacity

3

(50) The mass of steam blown off? (a) 5.402 kg (b) 2.045 kg (c) 4.025 kg (d) 2.122 kg (51) The dryness fraction of steam in the vessel after cooling? (a) 0.7 (b) 0.918 (c) 0.462 (d) 1.532

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Linked question Q52 and Q53 A thin cylinder 500mm ID and 20mm wall thickness with closed ends is subjected simultaneously to an internal pressure of 0.60MPa, bending moment 64000 Nm and torque 16000 Nm.

(52) The maximum tensile stress in the wall. (a) Zero (b) 12.8 aMP (c) 28.1 aMP (d) 82.3 aMP (53) Then the maximum shearing stress in the wall. (a) 4.56 aMP (b) 5.64 aMP (c) 6.34 aMP (d) 3.67 aMP

Linked question Q54 and Q55 A low carbon steel plate is to be welded by the manual metal – arc welding process using a linear V-I characteristic d. c. power source. The following data are available. Open circuit voltage of the power source: 62 volts short circuit current (for the setting used):130 A Are length ( ) = 4 mm Traverse speed of welding: 15 cm/min Efficiency of heat input: 85 % The relation between are length ( in mm) and arc voltage (v) is given by: V = 20 + 1.5

(54) Calculate the heat input into the work piece. (a) 1639 W (b) 1135 W (c) 1669 W (d) 1733 W (55) Power consumed by the machine. (a) 1230 W (b) 1820 W (c) 1930 W (d) 1963 W

Question Q56 to Q60 carry one mark each (56) CONTEMPORANEOUS: EVENTS::

(a) Adjacent: objects (b) Modern: times (c) Temporary: measures (d) Gradual: degrees

(57) LIMERICK: POEM::

(a) Motif: symphony (b) Prologue: play (c) Catch: song (d) Sequence: sonnet

(58) Diligent means (a) Intelligent (b) Energetic (c) Modest (d) Industrious (59) Picturesque means (a) Photogenic (b) Simple (c) stimulating (d) Ugly (60) Two pipes A and B can fill a tank in 24 minute and 32 Minutes. Respectively if both the

pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 minutes

(a) 10 (b) 8 (c) 12 (d) 18

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Question Q61 to Q65 carry two marks each (61) If a sum on compound interest become three times in four years, then with the same

interest rate, the sum will become 27 times in (a) 8 years (b) 12 years (c) 24 years (d) 36 years

(62) If 1 3

,x xa b

b a

− −⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

then the value of x is

1 7( ) ( )1 ( ) 2 ( )2 2

a b c d

(63) Complete the series 2, 3, 3, 5, 9, 12, 17...... (a) 32 (b) 30 (c) 24 (d) 26 (64) The greatest number which will divide 410,751 and 1030 leaving a reminder 7 in each

case is (a) 29 (b) 31 (c) 17 (d) 37

(65) When the price of a pressure cooker was increased by 15 %, the sale of pressure cooker

decreased by 15 %. What was the net effect on the sales? (a) 15 % decrease (b) no effect (c) 2.15 % increase (d) 2.25 % decrease

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Answer with Explanation 1. Ans. (a) 2. Ans. (a) 3. Ans. (d) 4. Ans. (b) 5. Ans. (d) 6. Ans. (a)

Exp.

1 1

2 2

1

2

V =πDN =π 0.025 450=35m/minV =πDN =π 0.025 710=58m/min

L 0.100T =700 =700 =280minf 0.25

0.1000T =80 =32 min0.25

× ×× ×

× ×

×

nn n 1 2

1 1 2 22 1

2

1

1

2

T vV T =V T or =T v

vlnv

or n= =0.21TlnT

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠

n1V T =C, 114

7. Ans. (c) Exp. P = 3400 KW, H = 29.9 m, N = 166.7 rpm

s 5/4 5/4

N P 166.7 34000N = = = 440H (29.9)

×

8. Ans. (a) Exp.

T

S

1

2

3

P=constPV =constn

T=const

9. Ans. (d) 10. Ans. (b)

Exp. Refrigerating effect = (0.95 - 0.15)× 315 = 252 KJ/kg Compressor work = 46.7 KJ/kg

COP= 25246.7

=5.4

11. Ans. (b)

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12. Ans. (a) 13. Ans. (b) 15. Ans. (b) 16. Ans. (d) 17. Ans. (c) 18. Ans. (b) 19. Ans. (a) 20. Ans. (a) 21. Ans. (d) 22. Ans. (b) 23. Ans. (c)

Exp. ( )n n-12

24. Ans. (a) 25. Ans. (d) 26. Ans. (d) 27. Ans. (a) 28. Ans. (b) 29. Ans. (b) 30. Ans. (c)

Exp. RHS = quotient of homogeneous functions of same degree ( = 2)

Set y = vx: i.e. ( )2 2

2d xvx+v xvx =dx x

i. e. 2dvx +v = v+vdx

Separate variables 2dvx = vdx

(subtract v from both sides)

And integrate: 2dv dx=v x∫ ∫

i. e. 1 1nx Cv

− = +

Re-express in terms of x, y: 1x nx Cy

− = +

31. Ans. (a) Exp. The degree of uncertainly (or otherwise) is indicated by the variance of the time

estimates. The variance of time estimates given by engineer X is

2 2p 02

x

t -t 8-4= = =0.43566 6

σ ⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

The variance of time estimates given by engineer Y is 2 2

p 02 t -t 8-3= = =0.696 6

σ ⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

y

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Thus, the variance of time estimates given by Y is more. Since greater the variance, greater will be uncertainty, engineer X’s time estimation have more certainty. 32. Ans. (c)

Exp. 150g - 2T = 150.a T – 50g =50× 2a Acceleration of 50kg mass will be double than acceleration of 150kg mass.

33. Ans. (c)

Exp. Specific humidity (ω ) = 0.622 v

a v

P 0.03.38=0.622P -P 1.03-0.03038

×

Relative humidity (φ ) =0.03038 40.4%0.07520

=

34. Ans. (d) Exp. From tool designation α=10 ,λ =90° ° d = 2 mm, t=f=0.2 mm (sinceλ =90° ) tc = 0.39 mm

rcosαtan =1-rsinα

φ =

0.2 0.5130.39

trtc

= = =

0.513cos10tan = =0.5511-0.153sin10

or =32

φ

φ °

35. Ans. (b) Exp. Energy method gives

[ ]d K.E+P.E =0dt

Or 2 2d 1 1I + kx =0dt 2 2

θ⎡ ⎤⎢ ⎥⎣ ⎦

Or ( )22 2d 1 3 1. mr θ + k. rθ =0dt 2 2 2⎡ ⎤⎛ ⎞

⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Or 2kθ+ .θ=03m

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Or 2k= = 2π3m

ωn nf

Or 1 2k= H

2 3mπn Zf

36. Ans. (c)

Exp. -3-3

du U kg 1τ=μ =μ =0.65 10 0.3m/s =0.65Pady d m.s 0.3 10 m

× × ××

37. Ans. (b) Exp. Atmosphere pressure = 101.325 kPa Vapour pressure = 2.34 kPa Negative pressure needed for cavitation = 98.985 kPa≡ 10.0902m of H2O

Then z = 2 2 2v r=

2g 2gω

Or ω = 2

2zg 2 10.0902 9.81=r 0.0750

× ×

=188rads

38. Ans. (b)

Exp. 0u vx y∂ ∂

+ =∂ ∂

Or 2 0A vx y− ∂

+ =∂

Or 2

v A=y x∂∂

Or 2

Ayv= +c (simplest component if c=0)x

39. Ans. (c) Exp. The mechanism, as shown in Fig. has five links and six equivalent binary joints (because

there are two binary joint at B and D, and two ternary joint at A and C, i.e. l = 5 and j = 6 ∴ n = 3(5-1) -2× 6 = 0 40. Ans. (c)

Exp.

P

V

2

1

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Process 1-2: 1-2 1-2Q =-45kJ,W =-82kJ 1-2 2 1 1-2Q =U -U + W Or 2 1U -U =37 kJ Process 2-1: 2-1W =100kJ 2-1 1 2 2 1Q =U -U + W − (-37)+100 = 63 kJ

41. Ans. (a) Exp.

493K

298K

273K

H.E

H.P W

W

Qe1

Qe2

Qp1

Qp2

Amount of heat removed by

Heat pump 2p

15 1000 334.5Q = =58kW24 60 60× ×

× ×

COP of R =273 10.92

298 273=

−

Work needed for R =58 =5.3kW

10.92

So power developed by the engine =5.3 kW

e1Carnot

W 5.3Q = = =13.43kW298η 1-493

42. Ans. (b)

Exp. First draw FBD and then use, PLδ=AE

43. Ans. (b) Exp. Given x= 0.7 m , 2

1 2T = 40 C, T = 20 C, A = 10 5 = 50m° ° × We know that

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Q = -KA dTdx

or Qdx = -KA dT

Integrating both side 0.7 20

-4 2

0 40

Q dx = - 50 (0.93+1.16310 T )dT×∫ ∫

Or Q = 1483.6 J/S 44. Ans. (c) Exp.

r 1

2

12Q (from floor to roof) = 4 4

1 1 2

1 2 1

1 1-2 2 2

A (T -T )1-ε 1-ε A1+ +ε F ε A

σ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

In this case 2

2 21 2

4 rA = r , A = = 2 r2ππ π

11-2

2

A =0.5, F =1A

-8 4 4

212

1 5.67 10 700 -1000=

1-0.5 1 1-0.25+ + 0.50.5 1 0.25

⎡ ⎤× × ⎣ ⎦∴⎛ ⎞⎜ ⎟⎝ ⎠

wQ m

212310.4=− Wm

Here -i vesign indicates that floor gains the heat. 45. Ans. (b)

Exp. standard deviation, σ= 16 = 4 For 95% probability Z = 1.65

Now sT -T =σ

E Z

Therefore sT = σ Z + ET = 4 1.65+50 57 weeks× 46. Ans. (c) Exp.

GATE-2013

S K Mondal’s


r3 r2

r1

metal

insulation

g a1

2 1 3 2

1 1 1 2 0 3

2 (t -t )Qq = = 1 n(r /r ) n(r /r ) 1L + + +h r k k h r

π

=3

3

2 (300-25)0.050n n

1 10.047 0.050+ + +25 0.047 40 0.045 10 r

rπ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

× ×

The heat lost from the outer surface by convection per meter length of pipe. 2 3 0 3q = 2 r h (50-25) =500 r w/mπ π

For steady state condition, 1 2q = q solving we get 3r = 0.08m = 8cm ∴ Thickness of insulation = 8 - 5 = 3cm

47. Ans. (b) Exp. Maximum tension in the belt = 246 N Ratio of tensions, k = 2

For maximum power, velocity V is given by

T T.g 246 9.81V = = = =3.79 m/s

3m 3.w 3 5600×

×

Maximum power Transmitted

( )1 2

11

= T – T V

T= T - VK

1=246 1- 3.792

=0.4662 K/W

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞ ×⎜ ⎟⎝ ⎠

48. Ans. (b) 49. Ans. (a) Exp. U =1500× 12 = 18000 unit/year

R = Rs 30/lot cI =0.2× 5 = Rs 1/ unit/ year

GATE-2013

S K Mondal’s


ELS = c

2RU 2 30 18000 1040I I

× ×= =

Set up time =N.S 1040 30= 21hrsA 1500

×≈

50. Ans. (b) 51. Ans. (a) Exp. The mass of steam blow off:

The mass of steam in the vessel ( 1m ) =11 g

V 0.9= =4.167kgx v 0.9 0.24× ×

Enthalpy of steam before blowing off =

1 1f 1 fgh +x h =721+0.9 2047=2563KJ/kg.× Enthalpy before blowing off = Enthalpy after blowing off 2 22563=605+x 2133 or x =0.918× Now mass of steam in the vessel after blowing off

20.9m = =2.122 kg

0.918 0.462×

Mass of steam blown off (m) = 1 2m -m =2.045kg. Dryness fraction of steam in the vessel after cooling: It is constant volume cooling

2 32 g 3 gx v (at 4 bar)= x v (at 3bar)

30.918 0.462=x 0.606× × Or 3x =0.699=0.7 52. Ans. (b) 53. Ans. (a) Exp. (i) Stresses due to internal pressure

1 aPd 0.6 500σ = = =7.5MP2t 2 20

××

and 2 aPd 0.6 500σ = = =3.75MP4t 4 20

××

(ii) Principal stresses due to combined effect bending moment and torque

2 21 a3

16σ = M+ M +T =5.29MPdπ

⎡ ⎤′⎣ ⎦

2 22 a3

16σ = M- M +T =-0.08MPdπ

⎡ ⎤′⎣ ⎦

Therefore Maximum tensile stress max 1 1 a(σ )=σ +σ =12.79 MP′

Minimum tensile stress min 2 2 a(σ )=σ +σ =3.67MP′

Maximum shearing stress ( ) max minmax a

σ -στ = = 4.56MP2

54. Ans. (c) 55. Ans. (d)

GATE-2013

S K Mondal’s


56. Ans. (a) 57. Ans. (c) 58. Ans. (d) 59. Ans. (a) 60. Ans. (b) 61. Ans. (b) 62. Ans. (c) 63. Ans. (d) 64. Ans. (b) 65. Ans. (d)

GATE 2013 Test Series -1

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