Memory Based Questions of
GATE 2020Mechanical Engineering
www.madeeasy.in
Corporate Office: 44-A/1, Kalu Sarai, New Delhi - 110016 | Ph: 011-45124612, 9958995830
Delhi | Hyderabad | Noida | Bhopal | Jaipur | Lucknow | Indore | Pune | Bhubaneswar | Kolkata | Patna
Detailed Solutions
Scroll down to view
Date of Exam : 1/2/2020Afternoon Session
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 2
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.1Q.1Q.1Q.1Q.1 In MRP, if inventory cost is very high and setup cost is zero, which of the following lotsizing approach should be used?(a) fixed period quantity for period (b) EOQ(c) base stock level (d) lot for lot
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)
End of Solution
Q.2Q.2Q.2Q.2Q.2 CustomerA B C D E→ → → → →
Initial price = `120After every stage price increased by 25%. Calculate the cost when the product in thehand of customer.(a) (b) `366.210(c) `292.96 (d)
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)Cost = 1.255 × 120 = Rs. 366.210
End of Solution
Q.3Q.3Q.3Q.3Q.3 Consider the following welding processes: arrange the following in increasing order ofHAZ1. Arc welding2. MIG3. Submerged arc welding4. Laser beam welding(a) 1 - 2 - 3 - 4 (b) 1 - 4 - 2 - 3(c) 3 - 2 - 4 - 1 (d) 4 - 3 - 2 - 1
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)
End of Solution
Q.4Q.4Q.4Q.4Q.4 There are two identical shaping machine S1 and S2. In S2 the width of workpiece isincreased by 10 percent and feed is decreased by 10 percent with respect to S1. Ifall other condition remains same then ratio of total time per pass in S1 and S2 will be:
Ans.Ans.Ans.Ans.Ans. (0.8182)(0.8182)(0.8182)(0.8182)(0.8182)
In shaper total time of total time of machining
T =( )1
1000
W mf
+
∴ T ∝Wf
1
2
TT
= 1 2
2 1
W fW f
×
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 3
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Given: W2 = 1.1 W, f2 = 0.9f1
∴ 1
2
TT
= 1 1
1 1
0.9 0.90.8182
1.1 1.1W f
W f× = =
End of Solution
Q.5Q.5Q.5Q.5Q.5 Keeping all the parameter identical the compression ratio (CR) of an air std. diesel cycleis increased from 15 to 21. Take ratio of specific heats = 1.3 and cut-off ratio of thecycle ρ = 2. The difference between the new and old efficiencies is ________ %.
Ans.Ans.Ans.Ans.Ans. (4.79)(4.79)(4.79)(4.79)(4.79)Keeping all the parameter identical
(CR)1 = 21(CR)2 = 15
ρ = 2γ = 1.3
η =( ) 1
1 1 11
1
r
rr γ −
⎡ ⎤ρ −− × ⎢ ⎥ρ −⎣ ⎦
(ηnew|CR = 21) – (ηold|CR = 15) = ( ) ( )1 1
1 2
1 1 1 11 11
r
r r rγ − γ −
⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞ρ − ⎢ ⎥− − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ρ −⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
= ( ) ( )1.3
1.3 1 1.3 1
2 1 1 1 12 1 1.3 15 21− −
⎡ ⎤⎛ ⎞⎛ ⎞− ⎢ ⎥−⎜ ⎟⎜ ⎟ ⎜ ⎟− ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
= [ ]1.32 1
0.04260 4.79%1.3
⎛ ⎞− × =⎜ ⎟⎝ ⎠
End of Solution
Q.6Q.6Q.6Q.6Q.6 The given graph is between 3 variable x, y and z. Which of following is incorrect?
xy
z
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 4
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
(a) when y is constant, x is inversely proportional to z.(b) when x is constant, z is inversely proportional to y.(c) when z is constant, x is inversely proportional to y.(d) None of these
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)
End of Solution
Q.7Q.7Q.7Q.7Q.7 An inventory model is as shownDemand = 10,000 unit/yearCarrying cost = Rs. 4 /unit/yearOrder cost = Rs. 300/orderBack order cost = Rs. 25/unit/year
Assume as soon as order Q is received, the back order is supplied to customer. Whatis maximum amount of inventory reached?
Q
Qs
Ans.Ans.Ans.Ans.Ans. (1137)(1137)(1137)(1137)(1137)
Maximum inventory level, D = 10000 units/year
Holding cost, Ch = `4/unit/year
Back order, Cb = `25/unit/year
Ordering cost, Co = `300/unit/year
EOQ =( )2 10000 300 25 42
*4 25
o
bh
h b
DCQ
CC
C C
× × × += =
×⎛ ⎞× ⎜ ⎟+⎝ ⎠
=2 10000 300
254
29
× ×
×
Q* = 1319.0905 units
(Q – S)*Ch = S × Cb
S* = ( )* 1319.090 4
29h
h b
Q CC C
×⎡ ⎤= ⎢ ⎥+ ⎣ ⎦ = 181.943 units
Maximum inventory level= Q* – S* = 1319.0905 – 181.943 = 1137.14697 units
End of Solution
Location : Delhi Centre
300-350 Hrs of comprehensive classes.Dynamic test series in synchronization with classes.Well designed comprehensive Mains workbooks.Special sessions to improve writing skills, time management & presentation skills.
Conventional Questions Practice Programme
Admission open
Classroom Course
Batches from th18 Feb, 2020
Admission open
300-350 Hrs of comprehensive recorded sessions.Convenience of learning at your own pace. Physical study materials will be provided at your address.
Conventional Questions Practice ProgrammeOnline
ClassesBatches from
th25 Feb, 2020
15 Tests | Mode : Oine/OnlineTest series will be conducted in synchronisation with subjects taught in the classes.Exactly on the UPSC pattern and standard.Contains Repeat Topics and New Topics to maintain continuity in study.Facility to cross check the evaluated answer sheet & access to the top scorer copy.Time bound evaluation of answer sheets with feedback.
Mains Test
SeriesBatches from
th15 Mar, 2020
India’s Best Institute for IES, GATE & PSUsESE 2020Main Exam
www.madeeasy.in
Streams: CE ME EE E&T
Admission open
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 5
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.8Q.8Q.8Q.8Q.8 Circular roller of radius ‘R’ is pure rolling without slip at P and Q. Calculate angular velocityof disc.
v
Q
P
2v
R
(a)vR
(b)2vR
(c)2vR
(d)32
vR
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)
v
Q
P
2v
wV0
For pure rolling condition P and Q velocity should be zero.Assume roller moves with linear velocity with V0 and rotate with ω angular velocitySo, for P:
V0 + V = RωV0 – Rω = –v ...(i)
For Q, V0 + Rω = 2 V ...(ii)Solve Equation (i) and (ii),
ω =32
vR
End of Solution
Q.9Q.9Q.9Q.9Q.9 At both the surfaces, coefficient of static friction is μ. Draw FBD of the body
F
W
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 6
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
(a)
F
W
N2N2
μN2
(b)
(c) (d)
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)
End of Solution
Q.10Q.10Q.10Q.10Q.10 m1 = 10 kgm2 = 30 kgAfter collision, both masses get stuck. Find the maximum compression of spring.
Vo = 1 m/s
m1 m2
1 m 0.25 mFrictionless
Ans.Ans.Ans.Ans.Ans. ()()()()()∵ There is no friction, so no energy loss.
21 0
12
m V = 2max
12
k ⋅ x
where compression will be maximum, velocity of combined blocks will be zero.
⇒ 10 × 12 = 2maxk ⋅ x
xmax =10k
Data is not complete.
End of Solution
Q.11Q.11Q.11Q.11Q.11 The recent measures to improve the output would ______ the level of production to oursatisfaction.(a) speed (b) decrease(c) equalize (d) increase
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)
End of Solution
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 7
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.12Q.12Q.12Q.12Q.12 A fair coin tossed 20 times. The probability that head will appear exactly 4 times in the1st ten tosses, and tail will appear exactly 4 times in the next ten tosses is _______.
Ans.Ans.Ans.Ans.Ans. (0.0420)(0.0420)(0.0420)(0.0420)(0.0420)
Fair coin tossed 20 times
First 10 times probability that head will appear exactly 4 times
P(x = 4) = 4
4 61 1102 2C
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
next 10 times probability that tail will appear exactly 4 times
P(y =4) = 4
4 61 1102 2C
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Total probability, P = P(x) × P(y)
= 4
24 61 110
2 2C
⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
=
22 10
10 10
10 9 8 7 1 20.0420
4 3 2 1 2 2
⎡ ⎤× × ×⎡ ⎤× = =⎢ ⎥⎢ ⎥× × ×⎣ ⎦ ⎣ ⎦
End of Solution
Q.13Q.13Q.13Q.13Q.13 A mould cavity of 1200 cm3 volume has to be filled through a sprue of 10 cm lengthfeeding a horizontal runner. Cross-section area at the base of sprue is 2 cm2. Consideracceleration due to gravity as 9.81 m/s2. Neglecting fractional losses due to molten metalflow, the time taken to fill the mould cavity is ______ sec.
Ans.Ans.Ans.Ans.Ans. (4.28)(4.28)(4.28)(4.28)(4.28)1
2
3
Spruehs
hc
ht
Assuming top gate: A3 = Ag = As
Neglecting height of pouring basin (hc)
ht = hs + hc = hs
tf =1200
4.28 s2 2 981 10g g
VA v
= =× ×
End of Solution
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 8
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.14Q.14Q.14Q.14Q.14 Phase diagram does not represent(a) heat transfer rate(b) temperature at which phase change takes place(c) phase transformation rate(d)
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)
End of Solution
Q.15Q.15Q.15Q.15Q.15 Initial tool geometry (ASA) 0 - 9 - re - rs - ce - 0 - RAfter some time 0 - 9 - re - rs - ce - 30 - RSpecific energy consumption (e) = V0 (t)
0.4 where t = uncut chip thickness
Initial cutting force is FC1 and final cutting force is FC2 find the value of 2 1
2100C C
C
F FF
⎛ ⎞−×⎜ ⎟⎝ ⎠ .
Ans.Ans.Ans.Ans.Ans. (5.022)(5.022)(5.022)(5.022)(5.022)We know, specific energy consumption,
(e) = 1000CF
fd
f = f sinλλ = 90 = Cs ⇒ λ1 = 90 – 30 = 60°
λ2 = 90 – 0 = 90°
Now, e = V0(t)0.4 =
0.40( sin )
1000CF
V ffd
λ =
∴ FC = V0(f sinλ) × 1000 × fdFC ∝ (sinλ)0.4
∴2 1
2100%C C
C
F FF−
× =
0.42
1
sin1 100%
sin
⎡ ⎤⎛ ⎞λ⎢ ⎥− ×⎜ ⎟λ⎝ ⎠⎢ ⎥⎣ ⎦
=
0.4sin90
1 100% 5.022%sin60
⎡ ⎤⎛ ⎞°⎢ ⎥− × =⎜ ⎟⎝ °⎠⎢ ⎥⎣ ⎦
End of Solution
Q.16Q.16Q.16Q.16Q.16 Module, m = 1.25, teeth on pinion = 20, teeth on gear = _____, pressure angle = 20°.Maximum allowable stress = σ0 = 400 MPa, Lewis form factor = y = 0.322, tooth thickness= __________.Find maximum power that can be transmitted?
Ans.Ans.Ans.Ans.Ans. ()()()()()Teeth of gear is not received. So will be calculated after response sheet.
End of Solution
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 9
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.17Q.17Q.17Q.17Q.17 The spectral distribution of radiation from a block body at T1 = 3000 K has a maximumwavelength λmax. The body cools down to a temp T2. If the wavelength correspondingto the maximum wavelength at T2 is 1.2 times of the original wavelength at Tmax, thenthe temperature T2 is __________ k. (round off to the nearest integer)
Ans.Ans.Ans.Ans.Ans. (2500)(2500)(2500)(2500)(2500)State-1: Spectral distribution
T1 = 3000 K(λ1) = λmax
State-1: T2 = ?(λ2) = 1.2λmax
As we know, from Wien’s displacement lawλT = Constant
λ1T1 = λ2T2
λmax × 300K = 1.2 λmaxT2
T2 = 2500 K
End of Solution
Q.18Q.18Q.18Q.18Q.18 Equation of motion of spring-mass damper is given as2
2
39 10sin5
d dt
dtdt+ + =x x x
The damping factor for the system is _________.
Ans.Ans.Ans.Ans.Ans. (0.5)(0.5)(0.5)(0.5)(0.5)By comparing equation of motion,
2ξωn = 3ωn
2 = 92ξωn = 3
2ξ = 1ξ = 0.05
End of Solution
Q.19Q.19Q.19Q.19Q.19 Total qualitative inversion of Grashof linkage if all the joints are revolute?(a) 4 (b) 3(c) 2 (d) 1
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)Total qualitative inversion when Grashof’s law satisfied.(i) Crank-crank mechanism when shortest link at the fixed position.(ii) Crank-rocker mechanism when shortest link at adjacent to the fixed position.(iii) Rocker mechanism when shortest link opposite to the fixed link that is coupler
position.
End of Solution
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 10
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.20Q.20Q.20Q.20Q.20 If number of teeth in sun and planet is equal and speed of ring (ωR) and speed of sunis (ωS) and direction of gear P and R is same then arm speed is given by
P
S R
A
C
C = CentreS = SunP = PlanetR = RaceA = Arm
(a)3
4 4SR ωω
+ (b)1 34 4R Sω + ω
(c) (d)
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)Let teeth of sun gear Ts = T
Ts = Tp = TRR = rs + 2rp
TR = T + 2T = 3TAs we know,
y + s = ωS
3y −
x= ωR
On subtraction,43x
= ωS – ωR
x =3 34 4S Rω − ω
y + x = ωS
Speed of arm, y =3 34 4S S Rω − ω + ω
=34 4
SR
ω⎛ ⎞ω +⎜ ⎟⎝ ⎠
End of Solution
Q.21Q.21Q.21Q.21Q.21 In a machine member, stress is varying asσ = 350 sin (8πt) in MPaFOS = 3.5 find maximum endurance limit.
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 11
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Ans.Ans.Ans.Ans.Ans. (1225)(1225)(1225)(1225)(1225)
m v
yt eS Sσ σ
+ =1
FOS
∵ σm = 0, σv = 350
⇒350
eS=
13.5
Se = 350 × 3.5 = 1225 MPa
End of Solution
Q.22Q.22Q.22Q.22Q.22 What is the increasing order of HAZ1. arc welding2. MIG3. Submerged are welding4. Laser beam welding(a) 1 – 2 – 3 – 4 (b) 1 – 4 – 2 – 3(c) 3 – 2 – 4 – 1 (d) 4 – 3 – 2 – 1
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)
End of Solution
Q.23Q.23Q.23Q.23Q.23 A beam of negligible mass hinged at support P and has a roller at Q.
1200 N
5 m 5 m
Q
4 m
P R
Point load at R = 1200 N, then find magnitude of reaction force at Q.
Ans.Ans.Ans.Ans.Ans. (1500)(1500)(1500)(1500)(1500)
RHP
RVP
RQ1200 N
P
ΣMP = 0, RQ × 4 = 1200 × 5
RQ =1200
1500 N4
S× =
End of Solution
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 12
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.24Q.24Q.24Q.24Q.24 One kg of air in closed system is undergoing a reversible process from P1 = 1 bar [abs],T1 = 27°C to P2 = 3 bar [abs] and T2 = 127°C. If R = 287 J/kgK, V = 1.4, then calculatethe change in entropy in J/kgK.
Ans.Ans.Ans.Ans.Ans. (–26.2)(–26.2)(–26.2)(–26.2)(–26.2)
At state -1:
P1 = 1 bar, T1 = 300 K, P2 = 3 bar, T2 = 400 K, R = 287 J/kgK, r = 1.4
Change in entropy, ΔS = 2 2
1 1
ln lnpT P
c RT P
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=400 3
1005ln 287ln300 1
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= –26.18 ≈ –26.2 J/kgK
End of Solution
Q.25Q.25Q.25Q.25Q.25 A Carnot cycle working between 27°C and –3°C then ratio of COPref to COPheat pump is__________.
Ans.Ans.Ans.Ans.Ans. (0.9)(0.9)(0.9)(0.9)(0.9)
COP of refrigerator, COPref = 270
930
=
COP of heat pump, COPheatpump = 300
1030
=
COPref /COPheat pump =9
0.910
=
End of Solution
Q.26Q.26Q.26Q.26Q.26 A cantilever beam, a load is acting at the free end and spring is also connected at freeend. What is the transverse deflection under load in terms of E, I and W ?
W
K N/m
(a)3
33
WL
E KL+I(b)
(c) (d)
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 13
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)
W
K N/m
A B
FS
FS
Let deflection at B will be δSo, FS = Kδ
SFK
= δ
For cantilever beam and end load (W – FS)
δ =3 3
3 3SWL F L
E E−
I I
=3 3
3 3WL K L
E Eδ
−I I
3
13KLE
⎡ ⎤δ +⎢ ⎥⎣ ⎦I
=3
3WL
EI
δ =
3
33
33
WLE
E KLE
⎛ ⎞⎜ ⎟⎜ ⎟
+⎜ ⎟⎜ ⎟⎝ ⎠
III
=3
33
WL
E KL
⎛ ⎞⎜ ⎟+⎝ ⎠I
End of Solution
Q.27Q.27Q.27Q.27Q.27 Diameter = 2r, both ends are fixed and Gas gauge pressure is p and t is thicknessof pipe. Find maximum increase in temperature for which there is no leakage of gas.
t
α, , E, poisson’s ratiopCoefficient of linear expansion
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 14
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
(a)14
rPE t
⎛ ⎞ν −⎜ ⎟⎝ ⎠ α (b)12
rPE t
⎛ ⎞ν +⎜ ⎟⎝ ⎠ α
(c)1 1
2 2pd
E t⎛ ⎞ν −⎜ ⎟⎝ ⎠α (d)
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)∈long + ∈thermal = 0
⇒ ( )1long hoop T
Eσ − νσ + αΔ = 0
⇒1
04pd pd
TE t t⎛ ⎞− ν + αΔ =⎜ ⎟⎝ ⎠∂
⇒ αΔT =1
2 4pd pd
E t tν⎛ ⎞−⎜ ⎟⎝ ⎠
ΔT =1 1
2 2pd
E t⎛ ⎞ν −⎜ ⎟⎝ ⎠α
[d = 2r]
End of Solution
Q.28Q.28Q.28Q.28Q.28 A helical spring of stiffness k, if wire diameter spring diameter and number of coils areall doubled, then knew is equal to(a) 16 k (b) 8 k
(c) k (d)2k
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)We know that,
k =4
38
Gd
D N
knew =4 4
3 3(2)
8 (2) 2G d
D N×
× ×
=4
38
Gdk
D N=
End of Solution
New Batches
ESE 2021GATE 2021
Regular Weekend
Early Start... • Extra Edge...
1 Year/2YearsClassroom Courses
www.madeeasy.in
BATCH COMMENCEMENT DATES
India’s Best Institute for IES, GATE & PSUs
DELHI
Patna : 24-02-2020
Lucknow : 20-02-2020
Bhopal : 16-01-2020
Indore : 20-02-2020
Pune : 10-02-2020
Hyderabad : 16-03-2020
Bhubaneswar : 23-01-2020
Kolkata : 25-01-2020
Jaipur : 16-02-2020
Delhi and Noida Rest of India
Evening : thME : 16 Jan, 2020thCE : 30 Jan, 2020thEE, EC : 20 Jan, 2020
Morning : thCE, ME : 12 Feb, 2020thEE : 18 Feb, 2020thEC : 6 Apr, 2020
stCE : 1 Feb, 2020thME : 9 Feb, 2020
thEE : 18 Feb, 2020ndEC : 22 Feb, 2020
WEEKEND BATCHES
REGULARBATCHES
DELHI
NOIDAth18 Jan, 2020
CE, ME, EE, EC
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 15
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.29Q.29Q.29Q.29Q.29 A vessel containing water is placed in ambient atmospheric conditions at 25°C, 1 bar
Water
Water vapour
Valve
25°C, 1 atm 80°C, 1 atm
Oven
Then it is transferred to an oven at 80°C, 1 bar, till it attain equilibrium. What will happenimmediately after opening the valve inside oven?(a) The water vapour will condense(b) The water vapour will escape out(c) The hot air will enter into vessel(d) Nothing will happen, everything would be in equilibrium
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
End of Solution
Q.30Q.30Q.30Q.30Q.30 If D1 > D2 then
x1
D1
α α
x2
D2
α α
(a)1 2
1 2sin
D D−α =
−x x (b)1 2
1 2cos
2D D−
α =−
x x
(c)( )
( ) ( )1 2
1 2 1 2sin
2D D
D D−
α =− − −x x (d)
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 16
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)
x1
B
α
O
AD12
OA = x1 – 1
2D
AB = 1
2D
sinα =ABOA
as ∠ABO = 90° ...(i)
sinα = 1
1 1
/2/2
DD−x
Similarly,
x2
B1
α
O1
A1
D22 90°
O1A1 = x2 – 2
2D
A1B1 = 2
2D
sinα = 1 1 2
1 1 2 2
/2/2
A B DO A D
=−x
...(ii)
For eq. (i) and (ii)
sinα =1
1 1
/2/2
DD−x = 2
2 2
/2/2
DD−x
= ( )
( ) ( )1 2
1 2 1 2 /2
D D
D D
−− − −x x
=( )
( ) ( )1 2
1 2 1 22
D D
D D
−− − −x x
End of Solution
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 17
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.31Q.31Q.31Q.31Q.31 A bolt has to be made at the end of rod of d = 12 mm by localised forging (upsetting).Length of unsupported portion of rod is 40 mm. To avoid buckling of rod, a closed forgingoperation to be performed at max die diameter of ______mm.
Ans.Ans.Ans.Ans.Ans. ()()()()()
Data is incomplete. Solution will be given after the response sheet.
End of Solution
Q.32Q.32Q.32Q.32Q.32 In a CNC machine G00 command is given and in travel point P (200, 300) to Q(300, 600)Two separate stepper motor are connected to stage with a lead screw of 0.5 mm pitch,motor rotates with 800 rpm. What is the time of travel in min?
100200
100 200
1000 Q(300, 600)
P(200, 300)
Ans.Ans.Ans.Ans.Ans. (0.75)(0.75)(0.75)(0.75)(0.75)
100
200
600
600
500400
300
200100
Q(300, 600)
Δy = 300 mm
300
400
500
y
x
Δx = 100 mmP (200, 300)
For Δ x = 100 mm movement
t x =100
min0.5 800PN
Δ =×
x = 0.25 min
ty =300
0.5 800y
PNΔ =
× = 0.75 min
For this complete movement 0.75 min time needed. x-axis motor and y-axis motor bothwill work 0.25 min then x-axis motor will stop and y-axis motor will run 0.5 min more.Therefore total time for this operation will be 0.75 min.
End of Solution
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 18
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.33Q.33Q.33Q.33Q.33 Calculate the tooling cost for per piece of bar of length 250 mm and diameter 25 mm.Feed is 0.2 mm/rev. For maximum production rate if tool change time is 1 min, Taylorequation VT 0.2 = 24 and tool required cost is Rs. 20(a) (b)(c) (d)
Ans.Ans.Ans.Ans.Ans. ()()()()()
T0 =1 1 0.2
1 4 min0.2C
nT
n− −⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Optimum speed, (V0) = 0.20
2418.1886 m/min
4n
C
T= =
Machining Time (Tm) =LfN
Cost of operation = ( )0
mm m A C m
TC T C T C
T× + + × ×
Some more data needed, will be confirmed after response coming.
End of Solution
Q.34Q.34Q.34Q.34Q.34 Given below is the representation of surface roughness symbol. Match the following
(vi)(v)
(iv)
(iii)(i) – (iii)
(i) (ii) (iii) (iv) (v) (vi)(a) P Q S R T U(b) P Q R S U T(c) P Q R S T U(d) R S P Q T UWhere, P = Maximum waviness height; Q = maximum width R = Roughness width cutoff;S = Maximum roughness width; T = Maximum roughness height; U = Minimum roughnessheight.
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
End of Solution
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 19
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.35Q.35Q.35Q.35Q.35 Initially pressure inside cylinder is 105 kPa, the piston moves for 8 cm against atmosphericand another 8 cm against a spring of stiffness k = 2 × 105 N/mm. Find total work done
8 cm
8 cm
P1 = 105 kPa
Patm = 100 kPa, k = 2 × 105 N/mm, Piston area = 300 cm2.
Ans.Ans.Ans.Ans.Ans. (544)(544)(544)(544)(544)
Area of piston = 300 cm2
k = 12.5 kN/m
W1-2 = P1 × A × x= 105 × 300 × 104 × 8 × 10–2
= 0.252 kJ
P2 = 105 kPa
P2 = 8 cm
x = 8 cm
P1 = 105 kPa
3
2
1
kx
3
2
V2 V3V
P
P3A = P2A + kx
P3 =2
2 4
12.5 8 10105
2 300 10
kP
−
−× ×+ = +×
x = 138.33 kPa
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 20
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
W2-3 = ( ) ( ) ( )2 3 3 21 1
105 138.332 2
P P V V A+ × − = + × x
= ( ) 4 21243.33 300 10 8 10
2− −× × × ×
W2-3 = 0.2919 kJ
∴ Wtot = W1-2 + W2-3 = 0.5439 kJ = 543.91 ≈ 544 J
End of Solution
Q.36Q.36Q.36Q.36Q.36 Total student are 5000 and girl are 1500
Art20%
Art30%
Management = 15%
Students Girl
Find ratio of boys registered in art to girl registered in management.
Ans.Ans.Ans.Ans.Ans. (2.44)(2.44)(2.44)(2.44)(2.44)
Ratio =Total student in art Girls in arts
Girls in management−
=5000 0.2 1500 0.3 1000 450
2.441500 0.15 225× − × −
= =×
End of Solution
Q.37Q.37Q.37Q.37Q.37 Directional derivative of f(x) = xyz in direction of i – 2y + 2k at (–1, 1, 3) is:
Ans.Ans.Ans.Ans.Ans. (7/3)(7/3)(7/3)(7/3)(7/3)
Directional derivative = ( ) ˆpf n∇ ×
Here, ∇f = yzi + xzj + xyk(∇f)(–1, 1, 3) = 3i – 3j – k
n̂ =ˆ 2 2 2 2
31 4 4
n j k j kn
− + − += =+ +
i i�
∴ Directional derivative = [ ]1 73 6 2
3 3+ − =
End of Solution
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 21
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.38Q.38Q.38Q.38Q.38 Given that function is analytical f(x) = ux i + v(x y)
Where, ux = x3y2 – xy3. Find v(x, y) = ?(a) (b)(c) (d)
Ans.Ans.Ans.Ans.Ans. ((((()))))u(x , y) is not properly given. So, we cannot guess the answer.
End of Solution
Q.39Q.39Q.39Q.39Q.39 Let I be a 100 multidimensional matrix, and E is set of its distinct real eigen value, thennumber of elements in E ______?(a) (b)(c) (d)
Ans.Ans.Ans.Ans.Ans. ()()()()()
Data is incomplete. Solution will be given after the response sheet.
End of Solution
Q.40Q.40Q.40Q.40Q.40 Let
212
0 0
,y
y dyd= =
= ∫ ∫x
x
I x x then I may also be expressed as,
(a)1 1
2
1y y
y d dy= =∫ ∫x
x x (b)1 1
2
1y y
y d dy= =∫ ∫x
x x
(c)1
2
0 0
y
yy d dy
= =∫ ∫x
x x (d)1
2
0 0
y
yy d dy
= =∫ ∫x
x x
Ans.Ans.Ans.Ans.Ans. ()()()()()
0
1
x = 1
y = x
Unit is given wrong in problem phase change.
End of Solution
UPPSCCombined State
Engineering Services Exam, 2019
Assistant Engineer | Total Posts : 692
Admission open
650 Hrs of comprehensive course.General Studies and Hindi covered.Exclusive study materials will be provided as per requirement of UPPSC.
thCommencing from 10 Feb, 2020 | Classes at Delhi and Lucknow
CourseClassroom
Streams: CE, ME, EE
Admission open
Quality questions with detailed solutions. Comprehensive performance analysis. Tests on standard and pattern as per UPPSC examination.
rd20 Tests | Commencing from 23 Feb, 2020
(Online/Offline)Test Series
Streams: CE, ME, EE
Admission open
Technical theory books with practice questions. Previous years' solved papers. Practice books (MCQ) for technical subjects. General Studies theory book with practice questions. Hindi book with practice questions.
Enrollment open
CoursePostal
Streams: CE, ME, EE
Admission open
Useful for those candidates who are not able to join classroom programme. 650 Hrs of quality classes at your doorstep. Flexibility to learn at your own pace. Physical study materials will be dispatched at your address.
thCommencing from 10 Feb, 2020
ClassesLive/Online
Streams: CE, ME, EE
India’s Best Institute for IES, GATE & PSUs
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 22
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.41Q.41Q.41Q.41Q.41 ( )( )/2
0
8 4cos 8t dtπ
+∫ What is the absolute error in % if integral is evaluated using
trapezoidal rule only by considering Ist and last term.
Ans.Ans.Ans.Ans.Ans. (5.187)(5.187)(5.187)(5.187)(5.187)
( )( )/2
0
8 4cos 8y t dtπ
= +∫By trapezoidal method
1
0 / 2
12 8
oy y
π
y1 = 12 oh
y y+⎡ ⎤⎣ ⎦
y1 = [ ]12 8 15.7072 2π + =×
By integration:
y2 = ( ) [ ]/2
/20
0
8 4cos 8 4 sin2
dπ
ππ⎛ ⎞+ = +⎜ ⎟⎝ ⎠∫ x x x
= 8 4 16.5662π + =
Absolute error =16.566 15.707
0.05187 5.187%16.566
− = =
End of Solution
Q.42Q.42Q.42Q.42Q.42 The sum of normally distributed random variable x and y if(a) normally distributed, only if x and y have same mean.(b) normally distributed, only if x and y have same standard deviation.(c) normally distributed, only if x and y are independent.(d) always normally distributed.
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)
End of Solution
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 23
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.43Q.43Q.43Q.43Q.43 Solution of 2
2 1,d y
ydt
− = which satisfies 00
0tt
dyy
dt==
= = in Laplaces domain is
(a) ( )1
1s s + (b) ( )1
1s −
(c) ( )1
1s s − (d) ( )( )11 1s s s+ −
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)Apply Laplace transform on both sides,
L[Y ′′] – L[Y] = L[1]
s2Y(s) – sY(0) – sY ′(0) – Y(s) = 1s
Y(s) =1
( 1)( 1)s s s− +
End of Solution
Q.44Q.44Q.44Q.44Q.44 A matrix P is decomposed into its symmetric part ‘S’ and skew symmetric part ‘V ’ if
4 4 24 3 7/2 , symmetric matrix2 7/2 2
S−⎛ ⎞
⎜ ⎟= ⎜ ⎟⎝ ⎠
, 0 2 32 0 7/ 2 ,skew symmetric matrix3 7/ 2 0
V−⎛ ⎞
⎜ ⎟= ⎜ ⎟− −⎝ ⎠
.
then matrix P is
(a)
2 9/2 11 81/4 112 45/2 73/4
− −⎛ ⎞⎜ ⎟−⎜ ⎟−⎝ ⎠
(b)
4 2 56 3 71 0 0
−⎛ ⎞⎜ ⎟⎜ ⎟−⎝ ⎠
(c)
4 2 56 3 71 0 2
−⎛ ⎞⎜ ⎟⎜ ⎟−⎝ ⎠
(d)
4 6 12 3 05 7 2
−⎛ ⎞⎜ ⎟− −⎜ ⎟− − −⎝ ⎠
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)For matrix P its symmetric part is
S =2
TP P+
For matrix P its skew symmetric part is
V =2
TP P−
P = S + V =
4 2 56 3 71 0 2
−⎛ ⎞⎜ ⎟⎜ ⎟−⎝ ⎠
End of Solution
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | www.madeeasy.in Page 24
Memory Based Questions of GATE 2020Mechanical Engg. | Afternoon Session
Q.45Q.45Q.45Q.45Q.45 Water (density 1000 kg/m3) flows through on inclined pipe. The velocity, pressure andelevation at section are VA = 3.2 m/s, PA = 186 kPA and ZA = 24.5 m respectively, andthose at section B are VB = 3.2 m/s, PB = 260 kPa and ZB = 9.1 m respectively. Ifacceleration due to gravity is 10 m/s2 then the head lost due to friction is_______ m.(round off to one decimal place)
Ans.Ans.Ans.Ans.Ans. (8)(8)(8)(8)(8)
VA = 3.2 m/sPZ
A
A
= 186 kPa = 24.5 m
g = 10 m/s2
ρ = 1000 kg/m3
VB = 3.2 m/sPZ
B
B
= 260 kPa= 9.1 m
A
B
DatumEnergy head at A :
EA =2 3 2
3186 10 3.2
24.5 43.612 m2 2 1010 10
A AA
P VZ
g g×+ + = + + =
ρ ××
Energy head at B:
EB =2 3 2
3
260 10 3.29.1 35.61m
2 2 1010 10B B
BP V
Zg g
×+ + = + + =ρ ××
EA > EB, so flow from A to B
Friction head, hf = EA – EB = 8 m