Memory Based Questions of GATE 2020...Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Given: W 2 = 1.1 W,f 2 = 0.9f 1 ∴ 1 2 T T = 11 11 0.9 0.9 0.8182 1.1

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Date of Exam : 1/2/2020Afternoon Session

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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

×



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



(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


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

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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


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



(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


End of Solution



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 _______.


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



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)


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



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



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.




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


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.


RHP

RVP

RQ1200 N

P

ΣMP = 0, RQ × 4 = 1200 × 5

RQ =1200

1500 N4

S× =

End of Solution



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__________.


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)



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



(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

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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)




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



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)


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



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)


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



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.


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



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.


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



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)


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


0

1

x = 1

y = x

Unit is given wrong in problem phase change.

End of Solution

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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.


( )( )/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.


End of Solution



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



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)


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

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Memory Based Questions of GATE 2020...Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Given: W 2 = 1.1 W,f 2 = 0.9f 1 ∴ 1 2 T T = 11 11 0.9 0.9 0.8182 1.1