India's Best Institute for GATE, IES, PSUs and IRMS ......GATE 2021 Regular Weekend Early Start... • Extra Edge... 1 Year/2Years Classroom Courses BATCH COMMENCEMENT DATES India’s

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 : 1/2/2020Forenoon 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. | Forenoon Session

GENERAL APTITUDE

Q.1Q.1Q.1Q.1Q.1 8 + 88 + 888 + ........... Sum of the series is

(a) ( )80 810 1

81 9n n− + (b) ( )80 8

10 181 9

n n− −

(c) ( )81 810 1

80 9n n− + (d) ( )81 8

10 180 9

n n− −

Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)

End of Solution

Q.2Q.2Q.2Q.2Q.2 Build : Building : : Grow : __________(a) Grown (b) Growth(c) Grow (d)

Ans.Ans.Ans.Ans.Ans. ()()()()()Options are incomplete

End of Solution

Q.3Q.3Q.3Q.3Q.3 If P is coded as αα, Q as αβ then R and S in terms of α, β is coaded as?(a) βα, ββ (b) αβ, ββ(c) αα, βα (d) αα, ββ

Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)

End of Solution

Q.4Q.4Q.4Q.4Q.4 Jofra archer is a bowler. He is ______ then accurate(a) more fast (b) less faster(c) faster (d) fast


Jofra Archer is a bowler. He is more fast than accurate.

when different qualities of the same noun are compared, we use more + positive degreeadjective,not comparative degree adjective. Hence use of 'faster' is incorrect here.

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



Q.5Q.5Q.5Q.5Q.5 A bar graph is shown in which number of students in 4 schools are given and also numberof students passing in respective school is also given; then calculate average passingpercentage of students.

500600

700

400

240

350

455

250

Attended Passed

Ans.Ans.Ans.Ans.Ans. (58.86)(58.86)(58.86)(58.86)(58.86)

Average percentage of passing students = Totalno.of passedstudents

Totalno.of students

=240 350 455 250

100500 600 700 400

+ + + ×+ + +

=1295

100 58.862200

× =

End of Solution

MECHANICAL ENGINEERING

Q.1Q.1Q.1Q.1Q.1 For a bead of 1 mm diameter, initial temperature of 100°C and surrounding temperatureof 20°C. After 4.35 seconds, temperature is 28°C. Calculate heat transfer coefficient,ρ = 8500 kg/m3.

Ans.Ans.Ans.Ans.Ans. (299.95)(299.95)(299.95)(299.95)(299.95)r = 0.5 mm; cp = 400 J/kgk; ρ = 8500 kg/m3; t = 4.35 sec

∵VA

=

3

2

43

34

R R

R

π=

π

∵T TT T

∞

∞

−−i

= p

hArC Ve− × τ

⇒28 20

100 20−−

=

·30000.5 8500 400 4.35

h

e⎛ ⎞−⎜ ⎟⎝ ⎠× × ×

⇒ h = 299.95 W/m2K

End of Solution



Q.2Q.2Q.2Q.2Q.2 LMTD of counter flow heat exchanger:

102°C

42°C 65°C

25°C

Oil enters in a concentric heat exchanger at 102°C and exit at 65°C and water enters at25°C and exits at 42°C, What is the logarithmic mean temperature difference of heatexchanger.

Ans.Ans.Ans.Ans.Ans. (49.33)(49.33)(49.33)(49.33)(49.33)Thi = 102°C, The = 65°CTc,i = 25°C, Tce = 42°C

LMTD = 1 2

1

2ln

T TTT

Δ − Δ⎛ ⎞Δ⎜ ⎟Δ⎝ ⎠

; ΔT1 = 102° – 42° = 60°C;ΔT2 = 65° – 25°C = 40°C;

=60 40 20

60 0.4054ln

40

− =⎛ ⎞⎜ ⎟⎝ ⎠

= 49.33°C

End of Solution

Q.3Q.3Q.3Q.3Q.3 Match the following:List-IList-IList-IList-IList-I List-IIList-IIList-IIList-IIList-II

A. Prandlt number 1.Bouyancy forceViscous force

B. Grashoff number 2. Momentum diffusivity

Thermal diffusivity

C. Reynold number 3.Inertia force

Viscous force

D. Nusselt number 4.Convective Heat TransferConductive Heat Transfer

AAAAA BBBBB CCCCC DDDDD(a) 2 1 3 4(b) 1 2 3 4(c) 4 3 2 1(d) 3 1 2 4


End of Solution



Q.4Q.4Q.4Q.4Q.4 Froude number is the ratio of(a) Viscous force and inertia force (b) Inertia force and buoyancy force(c) Viscous force and buoyancy force (d) Inertia force and gravity force

Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)

End of Solution

Q.5Q.5Q.5Q.5Q.5 System is a massless rod, hinged at one end and on the other end a disk of mass

‘m’ and radius 4L⎛ ⎞

⎜ ⎟⎝ ⎠ is attached as shown in figure. Assuming no-slip condition. Find

the natural frequency of the system.

L2

L2 L

42km

Radius =

k

(a)3

2km

(b) 3km

(c) (d)


End of Solution

Q.6Q.6Q.6Q.6Q.6 For given flat face follower motion given byy(0) = 4(2πθ – θ2) 0 ≤ θ ≤ 2πIf radius of curvature due to contact stresses should not be less than 40 mm then determinethe minimum radius of base circle?

Ans.Ans.Ans.Ans.Ans. (8.52)(8.52)(8.52)(8.52)(8.52)Flat face followerDisplacement equation:

y = ( )24 2πθ − θ

dydθ

= V = ( )4 2 2π − θ

= ( )8 π − θ (For y to be max 0dyd

=θ

⇒ θ = π)

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



a = 8dvd

= −θ

(Rcurvature)Min = RBase + (y + a)min

40 = RBase + [4(2πθ – θ2) – 8]Min

40 = RBase + [4π2 – 8]RBase = 40 + [4π2 – 8]

= 40 – 31.478= 8.5215 mm

End of Solution

Q.7Q.7Q.7Q.7Q.7 If one mass is deattached. Find unbalanced force.

m = 10 gm

m = 10 gm

m = 10 gmm = 10 gm

ω = 10 rad/sr = 100 mm


m = 10 gm

m = 10 gm

m = 10 gmm = 10 gm

ω

ω = 10 rad/s, r = 100 mm = 0.1 m

If one mass is deattached then

m = 10 gm

m = 10 gm

m = 10 gmm = 10 gm

F

F F



Now, unbalance tone, F = mrω2

= 2100.1 (10)

1000× ×

=10 0.1 100

1000× ×

= 0.1 N Answer

End of Solution

Q.8Q.8Q.8Q.8Q.8 What will be the value of l for crank-rocker mechanism?

S

P Q

R600

300

400

l

(a) 350 (b) 80(c) 200 (d) 300


SR

QP

600

400

300l

For Crank-Rocker mechanism, shortest link must be crank and adjacent to fixed as wellas grashoffs law must be satisfied.

If l = 80 mm

then shortest will be = 80 mm

as well as (80 + 600) < (400 + 300)

680 < 700

There for law is satisfied.

⇒ l = 80 mm

End of Solution



Q.9Q.9Q.9Q.9Q.9 A simple arrangement of block of mass ‘m’ attached with spring ‘k’ and damper ‘c’ isunder the influence of force F = F0 cos ωt. If ω ≠ 0 then the value of ω for which amplitudetransmitted is F0 is

(a)km

ω = (b)2km

ω =

(c)2km

ω = (d) 2km

ω =


m

S C

F F tUN = cos 0 ω

Given,FT = F0

Then transmissibility

∈ = 0

0 01TF F

F F= =

2

22 2

21

21

n

n n

⎛ ⎞πω+ ⎜ ⎟ω⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞ω ξω− +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ω ω⎝ ⎠ ⎝ ⎠⎝ ⎠

= 1

22

1n

⎛ ⎞ξω+ ⎜ ⎟ω⎝ ⎠ =

22 22

1n n

⎛ ⎞⎛ ⎞ ⎛ ⎞ω ξω− +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ω ω⎝ ⎠ ⎝ ⎠⎝ ⎠2

1n

⎛ ⎞ω− ⎜ ⎟ω⎝ ⎠ = ±1

Taking (–ve) sign

⇒2

1n

⎛ ⎞ω− ⎜ ⎟ω⎝ ⎠ = 2

⇒2

n

⎛ ⎞ω⎜ ⎟ω⎝ ⎠ = 2

⇒n

ωω

=2

2 2 nk

m= ω =

End of Solution



Q.10Q.10Q.10Q.10Q.10 If ω1 = 100 rad/s, ω2 = 110 rad/s is and ΔKE = 1.05 kJ. The moment of inertia of flywheelin kg-m2.

Ans.Ans.Ans.Ans.Ans. (1)(1)(1)(1)(1)

ΔKE = ( )2 21

12 Lω − ωI

∴ 1.05 × 103 × 2 = ( )2 2110 100−I

I = 1 kg.m2

Q.11Q.11Q.11Q.11Q.11 z = x + iy, f(z) = x2 – y2 + iψ (x, y), find the value of imaginary part of f(z) at z =1 + i(a) (b)(c) (d)

Ans.Ans.Ans.Ans.Ans. (2)(2)(2)(2)(2)F (z) = x2 + y2 + φ ψ (x, y)

φ = (x2 – y2)

dψ =d dd dyd dy

ψ ψ+x

x

=d dd dy

dy d− φ φ

+xx

= –(–2y)dx + 2xdyψ = 2xy

At Z = 1 + i = x + iyx = 1, y = 1

So, ψ = 2 × 1 × 1 = 2

End of Solution

Q.12Q.12Q.12Q.12Q.12( )

( )

1

11

1lim1

C

Cx

ee

− −

− −→

−−

x

xx

(a)2

1CC+

(b)1 C

C+

(c)1

CC+

(d)12

CC

+

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)

By L Hospital rule,



=

( ) ( )( )( ) ( ) ( )( )

1

1 11

1 1lim

1 0 1

C

C Cx

e C

e e C

− −

− − − −→

− − −

− − × − −

x

x xx

=( )

( ) ( )

1

1 11lim

C

C Cx

Cee Ce

− −

− − − −→

−− −

x

x xx

=1 1 1 1 1

C C CC C C

− −= =

− − × × − − +

End of Solution

Q.13Q.13Q.13Q.13Q.13 x and y are exponential distribution function of random variable. If, z = x + y and meanof x and y are 0.5 each, variance of z = 0.The co-relation between x and y is ________.

Ans.Ans.Ans.Ans.Ans. (–1)(–1)(–1)(–1)(–1)

X ~ E (λ1); mean =1

1 0.5=λ

⇒ λ1 = 2

Variance, x = 21

1 10.25

4= =

λ

Y ~ E (λ2); Mean =2

10.5=

λ

⇒ λ2 = 2Variance, y = 0.25

Given Var (Z) = Var(x + y) + Var (y) + 2 COV (x, y)0 = 0.25 + 0.25 + 2 COV (x, y)

COV (x, y) =0.5

0.252

− = −

Correlation, ρ =( )

( ) ( ), 0.25

10.25 0.25y

COV y −= = −

σ σx

x

End of Solution

Q.14Q.14Q.14Q.14Q.14 Solve the following integration by simpson’s 1/3 rule,1.4

1

0.6d h−

= =∫I x x x,

1.4 14

0.6b a

h nn− += = =,



( )

0 1 2 3 4

1 0.4 0.2 0.8 1.4

1 0.16 0.04 0.64 1.92

y y y y y

f

− −− −

xx


By simpson’s 1/3 rules,

I = ( ) ( ) ( )4 1 3 24 23 oh

y y y y y⎡ ⎤+ + + +⎣ ⎦

= ( ) ( ) ( )0.61 1.92 4 0.16 0.64 2 0.04

3⎡ ⎤− + + − + +⎣ ⎦

I = 0.2 [0.92 + 1.92 + 0.08] = 0.584

End of Solution

Q.15Q.15Q.15Q.15Q.15 Find the area under the curve, y = [x], x ε [1, 4](a) 4 (b) 1(c) 3 (d) 6


1 2 3 4

1

2

3

12

3

Area = 1 × 1 + 1 × 2 + 1 × 3

= 1 + 2 + 3 = 6

End of Solution

Q.16Q.16Q.16Q.16Q.16 Laplace inverse of ( ) ( )2 2

1F s

s=

+ ω

(a) ( )2

11 sin t− ω ω

ω(b) ( )2

11 sin t− ω

ω

(c)1

cos tωω

(d)1

sin tωω


End of Solution



Q.17Q.17Q.17Q.17Q.17 For 0 ≤ x ≤ 1 and n > 1. graph of y = x1/m and y = xm will be

(a)

y = x1/m

y = xm

(b)y = x

1/m

y = xm

(c)y = x

1/my = xm (d)

y = x1/m

y = xm


End of Solution

Q.18Q.18Q.18Q.18Q.18 A vector, ( ) ( ) ( )3/2 3/2 3/22 2 2 2 2 2 2 2 2

y zF j k

y z y z y z= + +

+ + + + + +

x ix x x

. F ds⋅∫ ∫�

is

obtained over hollow sphere of raddi 2 m and 1 m. What is the ratio of inner radi toouter raddi.(a) 2 π (b) 4 π(c) 3 π (d) 0


F�

=3rr

�

F∇ ⋅�

= 3 0rr

∇ ⋅ =�

1

ˆS

F NdS⋅∫∫�

= 4

By div theorem,

1

ˆS

F NdS⋅∫∫�

= ( ) 0R

F d dy dz∇ ⋅ =∫∫∫ x�

End of Solution



Q.19Q.19Q.19Q.19Q.19 Which of the following is not analytic function?(a) e z (b) sinz(c) logz (d) z2


End of Solution

Q.20Q.20Q.20Q.20Q.20 It two real valued square matrices of same order are multiplied, then product will be?(a) associative (b) commutative(c) can’t be evaluated (d) will be a positive matrix


End of Solution

Q.21Q.21Q.21Q.21Q.21 What is crystalline structure of γ-iron?(a) BCC (b) HCP(c) HCT (d) FCC


α - iron - BCC → α - Ferrite

γ - iron - FCC → Austenite

δ - iron - BCC → δ - Ferrite

End of Solution

Q.22Q.22Q.22Q.22Q.22 Match the following:EffectsEffectsEffectsEffectsEffects Heat treatmentHeat treatmentHeat treatmentHeat treatmentHeat treatment

A. Hardening 1. TemperingB. Softening 2. QuenchingC. Toughing 3. AnnealingD. Strengthening 4. Normalising

AAAAA BBBBB CCCCC DDDDD(a) 2 1 3 4(b) 2 3 1 4(c) 4 1 3 2(d) 2 4 1 3


1 - C decrease hardness and increase ductility and toughness.

2 - A increases in wear resistance strength and hardness.

3 - B remove or relieve strains or stresses induced by cold working.

4 - D improved mechanical properties (tensile strength, ductility).

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



Q.23Q.23Q.23Q.23Q.23 In a rolling operation, thickness of strip is reduced to 15 mm from initial thickness of20 mm. If width is 120 mm and roller radius is 450 mm. Calculate the Roll seperating

force if strength of strip is 210 MPa and σ0 = ( )Strength

1

nT

n∈

×+

. Where, n = Strain

hardening coefficient = 0.25.


Given: ho = 20 mm, hf = 15 mm, ∈T = 15

ln ln 0.287682 0.287720

f

o

hh

⎛ ⎞= = − ≈ −⎜ ⎟⎝ ⎠

oσ =0.25210 0.2877

123.04 MPa1 1 0.25

nTk

n× ∈ ×

= =+ +

F = ( )123.04 450 20 15 120No R h bσ Δ × = × × − ×

= 700355.96 N≈ 700.356 kN

End of Solution

Q.24Q.24Q.24Q.24Q.24 For basic size of 50 mm, hole tolerance of 0.002 mm, allowance of 0.003 mm and shafttolerance of 0.001 mm. Determine the maximum hole size according to shaft basissystem.


Using diagram method

0.002 mm

0.003 mm

0.001 mmZero line

UL

of h

ole

BS =

50.

000

mm

UL of hole = BS + 0.003 + 0.002 mm

= 50.005 mm

End of Solution



Q.25Q.25Q.25Q.25Q.25 What is the resultant shear load in B?

BA

CD

400 mm

50 mm

50 mm

50 mm50 mm

10 kN


Primary shear force, FP = 10

2.5 kN4

=

Secondary shear force Fs

( ) ( )2 250 2 50 2

50 2sF ⎡ ⎤+⎢ ⎥⎣ ⎦ = 10 × 400

50 2 4sF × × = 10 × 400

FP

FS

45°B

Resultant force on BFB = (2.5 + 14.142 sin 45) + (14.142 cos 45)2

= 16.0 kN

End of Solution

Q.26Q.26Q.26Q.26Q.26 A disc-plate type clutch having outer radius 250 mm and inner radius 50 mm. The torquetransmitted by new clutch is equal to torque transmitted by the old clutch, then find

the ratio of force of new clutch to old clutch? [μ same in both cases] new

old?

FF

=

Ans.Ans.Ans.Ans.Ans. (0.871)(0.871)(0.871)(0.871)(0.871)T = μW × Rm

Tnew = μ × W × Rnew

Tnew =( )3

0new 2 2

0

23

R RW

R R−

μ × ×−

3i

i



Told = μ × W × Rold

Tnew = Told

( )3 30

new 2 20

25

R RW

R R−

×−

i

i=

( )0old 2

R RW

+× i

new

old

WW = 0.871

End of Solution

Q.27Q.27Q.27Q.27Q.27 Two helical gears of helix angle 30° and pressure angle 20° is at middle of which issupported by bearing at both ends shaft is subjected to(a) Bending stress in two plane, torsional stress(b) Bending stress in two plane, axial stress and torsional stress(c) Torsional stress and axial stress(d) Axial and bending stresses


End of Solution

Q.28Q.28Q.28Q.28Q.28 Which of the above component will fail according to maximum shear stress theory?

Given permissible shear stress = 20 MPa.

1.P = 80 kN

(4 × 4) cm

2.P = 80 kN

(4 × 4) cm

3.d = 4 cm M

4.

d = 4 cm T = 64 Nmπ

(a) 1 and 2 (b) 1 only(c) 2 and 3 (d) 1 and 4

Ans.Ans.Ans.Ans.Ans. ()()()()()Question incomplete

End of Solution

Q.29Q.29Q.29Q.29Q.29 A beam as shown below having hinge at B. What is the reaction at C.

3 m3 m

50 kN

A

BC

10 kN/m

4 m(a) 20 kN (b) 40 kN(c) 30 kN (d) 50 kN



Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)In BC, ΣMB = 0

RC × 4 = 10 × 4 × 2RC = 20 kN

End of Solution

Q.30Q.30Q.30Q.30Q.30 Which member is subjected to zero force?

B D E

AG C

P

HF45°

(a) BG, EH, CD, BE, DE (b) BG, EH, CD, AB, EF(c) BG and EH only (d) BG, EH and CD


End of Solution

Q.31Q.31Q.31Q.31Q.31 A ball of 250 BHN is used to shear a sheet metal. The clearance of die and punch is1 mm. If hardness is increased to 400 BHN then what is the clearance?(a) 1.25 mm (b) 1.5 mm(c) 1 mm (d) 0.75 mm


Tensile strength ∝ BHN

[Formula available in my PPT slide NO. 7, hardness test material science]

∴ Shear strength ∝ BHN

and C = 0.0032t τ

C ∝ Shear strength BHN∝

2

1

CC =

( )( )

2

1

BHN

BHN

or C2 =( )( )

21

1

4001 1.265 mm

250

BHNC

BHN× = × =

End of Solution



Q.32Q.32Q.32Q.32Q.32 In rolling process, initial thickness 40 mm and final thickness is 20 mm. If roller diameteris 200 mm. Find coefficient of friction and projected length.


h0 = 40 mm, hf = 20 mm, Δh = 40 – 20 = 20 mm, D = 200 mm, R = 100 mm

Projected length, L = 100 20 2000 44.7213 mmR hΔ = × = =

Δh = μ2R∴ 20 = μ2⋅100

∴ 0.2 = μ2

μ = 0.4472

End of Solution

Q.33Q.33Q.33Q.33Q.33 In a rankine cycle, workdone by turbine is 903 kJ/kg and workdone of pump is 3 kJ/kg.Specific steam consumption in kg/kW-hr is __________?


(WD)turbine = 903 kJ/kg

(WD)pump = 3 kJ/kg

Specific steam consumption = ? (kg/kW-hr)

SSC =3600

T CW W−

=3600

4 kg/kW-hr903 3

=−

End of Solution

Q.34Q.34Q.34Q.34Q.34 Overall pressure ratio of 2 stage adiabatic compressor is 6. At the inlet of first compressor,temperature is 300 K and pressure is 100 kPa. Determine the intermediate pressure.(a) 244.94 kPa (b) 300 kPa(c) 250 kPa (d) 344.5 kPa


3

1

PP = 6 ⇒ P3 = 600 kPa

Intermediate pressure = 3 1 600 100 244.94 kPaP P = × =

End of Solution

Q.35Q.35Q.35Q.35Q.35 Air is passing through a nozzle of inlet and exit area of 0.1 m2 and 0.02 m2 respectively.Inlet pressure of air is 0.36 MPa (gauge). If after pasing through nozzle, air is strikinga vertical wall as shown. Determine the gauge pressure at B. (Assumeing density tobe constant).



A1

A2

B

Ans.Ans.Ans.Ans.Ans. (0.375)(0.375)(0.375)(0.375)(0.375)On applying continuity equation,

m� = 1 1 1 2 2 2A v A vρ ⋅ ⋅ = ρ ⋅ ⋅

⇒ 0.1V1 = 0.02V2

⇒ V2 = 1 110

52

V V=

Now on applying Bernoulli between 1 and 2 section

21 1

12P V

zg g

+ +ρ = 2P

gρ

0 22

22V z

g+ +

⇒23

10.36 101.21 9.81 2

Vg

× +× =

2125

2Vg

⇒ V1 = 4.98 m/s⇒ V2 = 24.89 m/sOn applying Bernoulli between 2 and 3 sections

2Pgρ

0 22

22V z

g+ + =

23 3

2P Vg g

+ρ

0

3z+

P3 =2 22 1.21 (24.89)

375 Pa2 2

g Vg

ρ ⋅ ×= =

= 0.375 kPa (gauge)

End of Solution

Q.36Q.36Q.36Q.36Q.36 In a axial turbine, diameter of tip is 4 m and diameter of hub is 2 m. If discharge is100 m3/s and speed of turbine is 300 rpm find the blade outlet angle of turbine.Dhub = 2 m, Dtip = 4 m, N = 300 rpm, Dmean = 3 m, Q = 100 m3/s

β

VfVr

u(a) 12.67°C (b) 12.33°C(c) 11.67°C (d) 11.33°C




Q = 2 2( )4 tip hub fD D vπ

− ⋅

⇒ 100 = 2 2(4 2 ) 10.61m/s4 fvπ

− × =

V =3 300

47.2160 600DN pπ × ×

= =

tanβ = fVu

⇒ β =1 10.61

tan 12.6747.21

− ⎛ ⎞ =⎜ ⎟⎝ ⎠

End of Solution

Q.37Q.37Q.37Q.37Q.37 In T-s diagram, constant volume curve and constant pressure curve intersect at point,find ratio of the slope of constant pressure to constant volume?

(a) p

v

C

C(b)

p

p v

CC C+

(c) v

p

CC

(d) None of these


T

S

V C =

P C =

If V = C

Tds = du pdv+

Tds = dudu = WdT [∵ V = C]

V

dTds

⎛ ⎞⎜ ⎟⎝ ⎠ =

v

TC

If P = C



Tds = dh – Vdpdh = CPdT [∵ P = C]

Tds = CPdT

P

dTds

⎛ ⎞⎜ ⎟⎝ ⎠ =

P

TC

Ratio,( )( )

/

/P

V

dT ds

dT ds=

( )( )

/

/P VP

V PV

T C CT C C

=

End of Solution

Q.38Q.38Q.38Q.38Q.38 Value of Joule Thomson coefficient for ideal gas is _________(a) positive (b) zero(c) negative (d) indeterminate

Ans.Ans.Ans.Ans.Ans. (0)(0)(0)(0)(0)Value of joule thompson coefficient for ideal gas,

μ =h

TP

∂⎛ ⎞⎜ ⎟⎝ ⎠∂

= 0

End of Solution

Q.39Q.39Q.39Q.39Q.39 1 kg of ideal gas at temperature 127°C is expanded at constant pressure when till itsvolume is doubled. For ideal gas, R = 0.287 kJ/kgK. Find work transfer is process in kJ?(a) 110 (b) 120.5(c) 114.8 (d) 115.3


m = 1 kg, T1 = 127°C = 400 k; P = C, V2 = 2V1; R = 0.287 kJ/kgK

1 2

P

V

W = ( )1 2 1P V V−

= ( )1 1 12P V V− = P1V1 = mRT1

= 114.8 kJ

End of Solution



Q.40Q.40Q.40Q.40Q.40 What is the brake power of Otto cycle 10 kW heat is added, given rc = 8, nm = 80%?(a) 5.64 kW (b) 4.52 kW(c) 6.32 kW (d) 3.64 kW


1

WQ = 0.5647

W = 10 × 5647 = 5.647 kW

η =1 0.4

1 11 1 0.5647

8crγ −− = − =

BP = ηm × W = 0.8 × 5.617

= 4.5175 kW

End of Solution

Q.41Q.41Q.41Q.41Q.41 In a tank barrier which act as a cantilever separates water as shown.

4 m

1 m

Barrier

Aρ = 1000 kg/m3, g = 10 m/s2

What is the net bending moment in barrier. Take unit width.(a) 90 kN-m (b) 80 kN-m(c) 100 kN-m (d) 50 kN-m


1/3 m

4/3 m

5 kN

80 kN

A

F1 = 1gAρ ⋅ x

= 1000 × 10 × (1 × 4) × 2 = 80 kN

F2 = 2gAρ ⋅ x

= 1000 × 10 × (1 × 1) × 0.5 = 5 kN

CP1 =1 4

4 m3 3

× =



CP2 =1 1

1 m3 3

× =

So, MA =4 4

80 53 3

× − ×

= 100 kN-m

End of Solution

Q.42Q.42Q.42Q.42Q.42 Which of the following is correct for cost slope in crashing?

(a)Crash cost Normal cost

Cost slope =Normal time Crash time

−−

(b)Crash cost

Cost slope =Normal time Crash time−

(c)Normal cost

Cost slope =Normal time Crash time−

(d)Crash cost Normal cost

Cost slope =Crash time

−


End of Solution

Q.43Q.43Q.43Q.43Q.43 Production rate is 4 pieces/hr and operation time is 60 min. If operation time is decresedby 30%, then keeping prduction rate constant, work in process will be(a) increase by 30% (b)(c) (d)


End of Solution

Q.44Q.44Q.44Q.44Q.44 Velocity distribution in pipe is given as V = 2

2

4D

C r⎡ ⎤

− −⎢ ⎥⎣ ⎦

. The rate of kinetic energy

at section AB is proportional to Dn. What is the value of n.

A

B




v =2

2 2 2

4D

C r C r R⎡ ⎤ ⎡ ⎤− − = − −⎢ ⎥ ⎣ ⎦⎣ ⎦

v =2

2 2 22

1r

C R r CRR

⎡ ⎤⎡ ⎤− = −⎢ ⎥⎣ ⎦ ⎣ ⎦

0K E

⎛ ⎞⎜ ⎟⎝ ⎠ = 3

0

Rv dAρ∫

=

323 6

20

1 ·2 ·R r

C R r drR

⎡ ⎤ρ − π⎢ ⎥

⎣ ⎦∫

=

323 6

20

2 1 · ·R r

C R r drR

⎡ ⎤ρ π −⎢ ⎥

⎣ ⎦∫

=6 2 2

3 66 2 2

0

32 1 1 · ·

R r r rC R r dr

R R R

⎡ ⎤⎡ ⎤ρ π − − −⎢ ⎥⎢ ⎥

⎢ ⎥⎣ ⎦⎣ ⎦∫

=6 2 4

3 66 2 4

0

3 32 1 · ·

R r r rC R r drR R R

⎡ ⎤ρ π − − −⎢ ⎥

⎣ ⎦∫

=2 2 2 2

3 6 32

2 8 4 2R R R R

C R⎡ ⎤

ρ π − − +⎢ ⎥⎣ ⎦

=2 2

3 6 2 32

8 4R R

C R R⎡ ⎤

ρ π − −⎢ ⎥⎣ ⎦

=2

3 628

RC R

⎡ ⎤ρ π ⎢ ⎥

⎣ ⎦

=3

828

C Rρ π

=3

88

28 (2)

C Dρ π×

×n = 8 Answer

End of Solution



Q.45Q.45Q.45Q.45Q.45 Mohr circle is represented by

10 kPa

10 kPa

10 kPa

10 kPa

(a) A point 10 cm away from origin on σ axis(b) A point 10 cm away from origin on τ axis(c) A circle with origin as center and 10 cm radius(d) A circle with 10 cm radius and at 10 cm from origin in σ axis


End of Solution

Q.46Q.46Q.46Q.46Q.46 Initially rod is unstretched, and velocity is given 1.5 m/s. What is the value of velocity(m/s) when rod is stretched by 0.4 m?

ν

m = 2 kg, Stiffness = 5 N/m(a) 1.26 (b) 1.06(c) 0.26 (d) 1.36

Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)Energy conservation,

21

12

mV = 2 20

1 12 2

mV k+ x

⇒ 2 × 1.52 = 2 202 5 0.4V× + ×

V0 = 1.360 m/s

End of Solution

Q.47Q.47Q.47Q.47Q.47 A 25 mm × 25 mm slot is milled having length of 300 mm using face and end mill cutterwidth 25 mm, diameter 100 mm and 20 teeth cutting speed is given as 35 m/min, feedper tooth is 0.1, approach and over travel are 5 mm, what is the machining time of slot?



Ans.Ans.Ans.Ans.Ans. (83.485)(83.485)(83.485)(83.485)(83.485)V = πDN

or 35 = π × 0.100 × N ⇒ N = 111.408 rpm

t =L A O

fZN+ +

= 300 5 5

min0.1 20 111.408

+ +× ×

= 1.391 min= 83.485 sec

End of Solution

Q.48Q.48Q.48Q.48Q.48 A square frame ABCD with only 1 diagonal member AC which is 4 times more elasticallystiff than 4 side members. Which member will buckle?

45°

P

A

B C

D

P

45°

l, EI

l, EI

l, EI

l, EI

Ans.Ans.Ans.Ans.Ans. ()()()()()Data is incomplete

End of Solution

Documents

India's Best Institute for GATE, IES, PSUs and IRMS ......GATE 2021 Regular Weekend Early Start... • Extra Edge... 1 Year/2Years Classroom Courses BATCH COMMENCEMENT DATES India’s