GATE-2013 S K Mondal’s
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.61Q.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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GATE-2013 S K Mondal’s Question Q1 to Q25 carry one mark each (1)
Consider the matrices X (4×3) , Y(4×3) and P(2×3) . The order of [ P ( X T Y )T PT ]T will be (a) (b) (c) (d)
(2 × 2) (3 × 3) (4 × 3) (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)
(5)
lim x sin ⎛⎜ 1 ⎞⎟ is
x →0 (a) ∞
⎝ x⎠
(b) 0
If f(0) = 2 and f ( x) = value theorem are (a) 1.9, 2.2
(c) 1
(d) Non-existant
1 , the lower and upper bounds of f(1) estimated by the mean 5 − x2
(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) VT 0.21 =114 (b) VT 0.25 =87 (c) VT 0.14 =111 (d) VT 0.34 =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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GATE-2013 S K Mondal’s (a) 300
(8)
(b) 404
(c) 440
(d) 16.3
Consider the following p-v diagram. 1
P=const 2 PVn=const
P PV=cost
3
V 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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GATE-2013 S K Mondal’s (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 k f and theoretical stress concentration factor k t is (a)
k f +1 k t +1
(b)
k f -1 k t -1
(c)
(14) A cotter joint is used to transmit (a) Axial tensile load only (c) Combined axial and twisting loads 14. Ans. (d) (15)
(16)
The lead screw of a lathe with nut is a (a) Rolling pair (b) Screw pair
k t +1 k f +1
(d)
k t -1 k f -1
(b) Axial compressive load only (d) Axial tensile or compressive loads.
(c) Turning pair
(d) Sliding pair
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)
ML2 12
(b)
ML2 6
(c)
ML3 3
(d)
ML2 3
(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 (c) Support loose pieces
(b) Strengthen mounding sand (d) Remove pattern easily
S- curve is connected with (a) Combustion (b) Cutting tools
(c) Corrosion
(21)
(22)
Oxygen to acetylene ratio in case of neutral flame is (a) 0.8:1.0 (b) 1:1 (c) 1.2:1
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(d) Heat treatment
(d) 2:1
GATE-2013 S K Mondal’s (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. 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
(25)
Question Q25 to Q55 carry one mark each (26)
For the matrix
⎡3 -2 2 ⎤ p = ⎢⎢0 -2 1 ⎥⎥ ⎢⎣0 0 1⎥⎦ One of the Eigen value X is equal to -2. Which of the following is an eigen vector
⎡ −3 ⎤ ⎢ 2⎥ (b) ⎢ ⎥ ⎢⎣ −1 ⎥⎦
⎡3 ⎤ ⎢ ⎥ (a) −2 ⎢ ⎥ ⎢⎣1 ⎥⎦ (27)
⎡ 1⎤ ⎢ ⎥ (c) −2 ⎢ ⎥ ⎢⎣ 3 ⎥⎦
⎡ 2⎤ ⎢ ⎥ (d) 5 ⎢ ⎥ ⎢⎣ 0 ⎥⎦
Consider the differential equation
d 3 y 2 dy +a = cos ax dx3 dx The C.F of the above equation is
(a) c1 + ( c2 cos ax + c3 sin ax ) (b) ( c1 cos ax + c2 sin ax )
(
(c) c1 + c2 cos
a x + c3 sin ax
)
(d ) none (28)
If L defines the Laplace Transform of a function, L [sin (at) will be equal to (b) a / s 2 + a 2 (c) s / s 2 + a 2 (d) s / s 2 − a 2 (a) a / s 2 − a 2
(29)
The Fourier series expansion of a symmetric and even function, f (x) where
(
)
(
)
(
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)
(
)
GATE-2013 S K Mondal’s f (x) =1 + (2 × /π ),
−π ≤ x ≤ 0
=1 − (2 × /π ), will be ∞
(a)
0≤ x≤π ∞
∑ ( 4/π n ) (1 + cos nπ ) 2
and
2
(b)
n=1
∞
∑ ( 4/π 2n2 ) (1sin nπ )
(d)
n=1
(a) x = 1n x + C (b) y = 1n x + C
(d)
2
∑ ( 4/π n ) (1 + sin nπ ) 2 2
n=1
dy xy+y 2 = dx x2
(30) The general solution of
(c) −
2
n=1
∞
(c)
∑ ( 4/π n ) (1 − cos nπ )
x = 1n x + C y
x = x +C y
(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
B 50 kgf
A
150 kgf
(a) 100
(b) 200
(c) 64.5
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(d) 50
GATE-2013 S K Mondal’s (33)
The sling-psychrometer reads 40 ° C DBT and 28 ° C WBT. Atmospheric pressure 1.30 kg / cm 2 and partial vapour pressure is 0.03038 kg / cm 2 saturation pressure at 40 ° C is 0.075 kg / cm 2 . 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
m r
(a) (36)
1 k 2π m
(b)
1 2k 2π 3m
(c)
2k 3m
(d)
1 3m 2π 2k
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 ×10-3 kg/m.s. and specific gravity is 0.88. Calculate the shear stress on the lower plate, in Pa y U=0.3 m/s
d=0.3 mm
x
(a) 7.39 (37)
(b) 65
(c) 0.650
(d) 6.50
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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GATE-2013 S K Mondal’s For the dimensions shown, compute the maximum angular speed if there is to be no cavitation.
(a) 881 (38)
(b) 188
rad s
(c) 818
rad s
(d) 130
rad s
The x component of velocity in a steady, incompressible flow filed in the xy plane is u=A/x, where A=2 m 2 /s and x is measured in meters. Then the simplest Y-component of velocity for this flow field is? (a) v=
(39)
rad s
Ax y2
(b) v=
Ay x2
(c) v=
Ax y
(d) v=0
Determine the number of degrees of freedom of the mechanism shown in the figure below C
D
A
(40)
B
(a) 1 (b) 2 (c) 0 (d) 3 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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GATE-2013 S K Mondal’s (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 mm 2 is subjected to axial forces as shown in the figure below. ⎡ E = 105 GN
⎢⎣
L
M
50KN
N
10KN 1.2m
1.0m
The total change in length of the bar. (a) 0.4311mm (b) -0.1143m
P
20KN
80KN 0.6m
(43)
⎤ m 2 ⎥⎦
(c) 1.32m
(d) 3.25mm
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
K = 0.93+1.163 × 10-4T 2 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 σ = 5.6710-8 w/m 2 k 4
(
(a) 11.23 kW/ m
2
(b) 21.31 kW/ m
)
2
(c) 12.31 kW/ m 2
(d) 13.21 kW/ m 2
(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
h1 = 25 W /m 2 K, h 0 =10 W /m 2 K
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GATE-2013 S K Mondal’s Ta = 25 ° C (Atmospheric temperature) K (metal pipe) = 40 W/mK K (insulation) = 0.045 W/mK (a) 25cm (b) 15cm (47)
(c) 3cm
(d) 5cm
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)
(49)
Economic lot site (a) 1200
(b) 1040
Set-up time for each lot (a) 21 hrs (b) 25 hrs
(c) 4010
(d) 1090
(c) 35 hrs
(d) 10hrs
Common data question Q50 and Q51 A spherical vessel of 0.9 m3 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 Vg = 0.24 m3 /kg, h f = 721 KJ/kg, h g = 2047 KJ/kg at 4 bar Vg = 0.462 m3 /kg, h f = 605 KJ/kg, h g =2133KJ/kg at 3 bar Vg = 0.606 m3 /kg.
0.9m3 capacity
spherical vessel
valve
(50)
(51)
The mass of steam blown off? (a) 5.402 kg (b) 2.045 kg
(c) 4.025 kg
The dryness fraction of steam in the vessel after cooling? (a) 0.7 (b) 0.918 (c) 0.462
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(d) 2.122 kg
(d) 1.532
GATE-2013 S K Mondal’s 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) (53)
The maximum tensile stress in the wall. (c) 28.1 MPa (a) Zero (b) 12.8 MPa Then the maximum shearing stress in the wall. (b) 5.64 MPa (c) 6.34 MPa (a) 4.56 MPa
(d) 82.3 MPa (d) 3.67 MPa
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 ( A ) = 4 mm Traverse speed of welding: 15 cm/min Efficiency of heat input: 85 % The relation between are length ( A in mm) and arc voltage (v) is given by: V = 20 + 1.5 A (54)
Calculate the heat input into the work piece. (a) 1639 W (b) 1135 W (c) 1669 W (55) Power consumed by the machine. (a) 1230 W (b) 1820 W (c) 1930 W
(d) 1733 W (d) 1963 W
Question Q56 to Q60 carry one mark each (56)
(57)
(58)
(59)
(60)
CONTEMPORANEOUS: EVENTS:: (a) Adjacent: objects (c) Temporary: measures
(b) Modern: times (d) Gradual: degrees
LIMERICK: POEM:: (a) Motif: symphony (c) Catch: song
(b) Prologue: play (d) Sequence: sonnet
Diligent means (a) Intelligent Picturesque means (a) Photogenic
(b) Energetic
(b) Simple
(c) Modest
(c) stimulating
(d) Industrious
(d) Ugly
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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GATE-2013 S K Mondal’s 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 ⎜
(63)
⎛a⎞ ⎟ ⎝b⎠ 1 (a) 2
x −1
⎛b⎞ =⎜ ⎟ ⎝a⎠ (b)1
x −3
, then the value of x is (c ) 2
(d )
7 2
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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GATE-2013 S K Mondal’s Answer with Explanation 1. Ans. (a) 2. Ans. (a) 3. Ans. (d) 4. Ans. (b) 5. Ans. (d) 6. Ans. (a)
V1 =πDN1 =π × 0.025 × 450=35m/min V2 =πDN 2 =π × 0.025 × 710=58 m/min
Exp. T =700 × L =700 × 0.100 =280 min 1
f 0.25 0.1000 T2 =80 × =32 min 0.25
n
n
V1T1 =V2T2
V1T n =C,
or
⎛v ⎞ ln ⎜ 2 ⎟ v n= ⎝ 1 ⎠ =0.21 ⎛T ⎞ ln ⎜ 1 ⎟ ⎝ T2 ⎠
114
7. Ans. (c) Exp. P = 3400 KW,
Ns =
or
⎛ T1 ⎞ ⎛ v 2 ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ T2 ⎠ ⎝ v1 ⎠
n
H = 29.9 m,
N = 166.7 rpm
N P 166.7 × 34000 = = 440 H 5/4 (29.9)5/4
8. Ans. (a) Exp. 2 P=const n
PV =const T 1
T=const
3
S 9. Ans. (d) 10. Ans. (b) Exp. Refrigerating effect = (0.95 - 0.15) × 315 = 252 KJ/kg Compressor work = 46.7 KJ/kg
COP=
252 =5.4 46.7
11. Ans. (b)
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GATE-2013 S K Mondal’s 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-1) 2
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)
d xvx+v 2x 2 vx = ( ) dx x2
Set y = vx: i.e.
dv +v = v+v 2 dx dv x = v2 Separate variables dx dv dx And integrate: ∫ v2 = ∫ x 1 − = 1n x + C i. e. v x Re-express in terms of x, y: − = 1n x + C y x
i. e.
(subtract v from both sides)
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
σx
2
2 ⎛ t -t ⎞ ⎛ 8-4 ⎞ =⎜ p 0 ⎟ = ⎜ ⎟ =0.4356 ⎝ 6 ⎠ ⎝ 6 ⎠
The variance of time estimates given by engineer Y is 2
σy
2
2 ⎛ t -t ⎞ ⎛ 8-3 ⎞ =⎜ p 0 ⎟ = ⎜ ⎟ =0.69 ⎝ 6 ⎠ ⎝ 6 ⎠
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GATE-2013 S K Mondal’s 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 Relative humidity ( φ ) =
Pv 0.03.38 = 0.622 × Pa -Pv 1.03-0.03038
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 (since λ = 90° ) tc = 0.39 mm
rcosα = 1-rsinα t 0.2 r= = = 0.513 tc 0.39 0.513cos10 tanφ = =0.551 1-0.153sin10 orφ = 32° tanφ =
35. Ans. (b) Exp. Energy method gives
d [ K.E+P.E ] =0 dt d ⎡ 1 2 1 2 ⎤ Iθ + kx ⎥ =0 Or dt ⎢⎣ 2 2 ⎦ d ⎡1 ⎛ 3 2 ⎞ 2 1 2⎤ . ⎜ mr ⎟ θ + k. ( rθ ) ⎥ =0 ⎢ dt ⎣ 2 ⎝ 2 2 ⎠ ⎦ 2k θ+ .θ = 0 Or 3m Or
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GATE-2013 S K Mondal’s Or ωn =
fn =
Or
2k = 2πf n 3m 1 2π
2k HZ 3m
36. Ans. (c) Exp. τ =μ
du U kg 1 =μ = 0.65 × 10-3 × 0.3m/s × = 0.65 Pa dy d m.s 0.3 × 10-3m
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 = Or
38. Ans. (b) Exp. Or Or
v 2 r 2ω 2 = 2g 2g
ω =
2zg 2 × 10.0902 × 9.81 = 2 r 0.0750 rad =188 s
∂u ∂v + =0 ∂x ∂y − A ∂v + =0 x 2 ∂y ∂v A = ∂y x 2 Ay Or v= 2 +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. 2
P 1
V
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GATE-2013 S K Mondal’s Q1-2 =-45kJ, W1-2 =-82kJ Q1-2 =U 2 -U1 + W1-2 Or U 2 -U1 =37 kJ Process 2-1: W2-1 =100 kJ Q 2-1 =U1 -U 2 + W2−1 Process 1-2:
(-37)+100 = 63 kJ 41. Ans. (a) Exp. 493K Qe1 H.E W
Qe2 298K Qp1 W
H.P Qp2 273K
Amount of heat removed by
15 × 1000 × 334.5 = 58 kW 24 × 60 × 60 273 COP of R = = 10.92 298 − 273 58 = 5.3kW Work needed for R = 10.92 Heat pump Q p2 =
So power developed by the engine =5.3 kW
Qe1 =
W ηCarnot
=
5.3 =13.43kW 298 1493
42. Ans. (b) Exp. First draw FBD and then use, δ =
PL AE
43. Ans. (b) Exp. Given x= 0.7 m , T1 = 40°C, T2 = 20°C, A = 10 × 5 = 50m 2 We know that
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GATE-2013 S K Mondal’s Q = -KA
dT or Qdx = -KA dT dx
Integrating both side 0.7
20
0
40
Q ∫ dx = - ∫ 50 × (0.93+1.16310-4T 2 )dT Or Q = 1483.6 J/S 44. Ans. (c) Exp. 2
r
1
Q12 (from floor to roof) =
A1σ (T14 -T2 4 ) ⎛ 1-ε1 ⎞ 1 ⎛ 1-ε 2 ⎞ A1 +⎜ ⎜ ⎟+ ⎟ ε F ⎝ 1 ⎠ 1-2 ⎝ ε 2 ⎠ A 2
4π r 2 = 2π r 2 In this case A1 =π r , A 2 = 2 A1 =0.5, F1-2 =1 A2 2
∴ Q12 =
1 × 5.67 × 10-8 ⎡⎣ 7004 -10004 ⎤⎦
w 2 m 1-0.5 1 ⎛ 1-0.25 ⎞ + +⎜ ⎟ 0.5 0.5 1 ⎝ 0.25 ⎠ = − 12310.4 W 2 m Here -i ve sign indicates that floor gains the heat.
45. Ans. (b) Exp. standard deviation, σ = 16 = 4 For 95% probability Z = 1.65
Ts -TE =Z σ Therefore Ts = σ Z + TE = 4 × 1.65+50 57 weeks
Now
46. Ans. (c) Exp.
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GATE-2013 S K Mondal’s
metal r1 r3
q1 =
r2
insulation
2π (t g -t a ) Q = 1 An(r2 /r1 ) An(r3 /r2 ) 1 L + + + h1r1 k1 k2 h 0 r3 2π (300-25) = ⎛ 0.050 ⎞ ⎛ r ⎞ An ⎜ An ⎜ 3 ⎟ ⎟ 1 1 0.047 ⎠ 0.050 ⎠ + ⎝ + ⎝ + 25 × 0.047 40 0.045 10 × r3
The heat lost from the outer surface by convection per meter length of pipe.
q 2 = 2π r3 h 0 (50-25) = 500π r3 w/m
For steady state condition, q1 = q 2 solving we get r3 = 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
V=
T T.g 246 × 9.81 = = = 3.79 m/s 3m 3.w 3 × 5600
Maximum power Transmitted
= ( T1 – T2 ) V
⎛ T⎞ = ⎜ T1 - 1 ⎟ V K⎠ ⎝ ⎛ 1⎞ =246 ⎜ 1- ⎟ × 3.79 ⎝ 2⎠ = 0.4662 K/W 48. Ans. (b) 49. Ans. (a) Exp. U =1500 × 12 = 18000 unit/year R = Rs 30/lot I c =0.2 × 5 = Rs 1/ unit/ year
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GATE-2013 S K Mondal’s ELS =
2RU = Ic
Set up time =
2 × 30 × 18000 = 1040 I
N.S 1040 × 30 ≈ 21hrs = A 1500
50. Ans. (b) 51. Ans. (a) Exp. The mass of steam blow off: The mass of steam in the vessel ( m1 ) =
V 0.9 = =4.167kg x1 × v g 1 0.9 × 0.24
Enthalpy of steam before blowing off = h f1 +x1h fg1 =721+0.9 × 2047=2563KJ/kg. Enthalpy before blowing off = Enthalpy after blowing off
2563=605+x 2 × 2133 or x 2 =0.918 Now mass of steam in the vessel after blowing off
0.9 =2.122 kg 0.918 × 0.462 Mass of steam blown off (m) = m1 -m 2 =2.045kg. m2 =
Dryness fraction of steam in the vessel after cooling: It is constant volume cooling
x 2 v g2 (at 4 bar) = x 3 v g3 (at 3bar) 0.918 × 0.462=x 3 × 0.606 Or x 3 =0.699 =0.7 52. Ans. (b) 53. Ans. (a) Exp. (i) Stresses due to internal pressure
Pd 0.6 × 500 = = 7.5MPa 2t 2 × 20 Pd 0.6 × 500 σ2 = = = 3.75MPa 4t 4 × 20 σ1 =
and
(ii) Principal stresses due to combined effect bending moment and torque
16 ⎡ M+ M 2 +T 2 ⎤ = 5.29MPa 3 ⎣ ⎦ πd 16 σ 2′ = 3 ⎡ M- M 2 +T 2 ⎤ = -0.08MPa ⎦ πd ⎣ σ1′ =
Therefore
Maximum tensile stress (σ max ) = σ1 +σ1′ =12.79 MPa Minimum tensile stress (σ min ) = σ 2 +σ 2′ = 3.67MPa Maximum shearing stress ( τ max ) =
σ max -σ min = 4.56MPa 2
54. Ans. (c) 55. Ans. (d)
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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)
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