Innddeexx · PDF fileUnit hydrographs; Khosla Theory, Well hydraulics; Jet impact, Open channel fow;Hydraulic machines, fluid properties. Easy 13 SOM/Solid Mechanics 1 M: 1

.

GATE-2016

: 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com

IInnddeexx

11.. QQuueessttiioonn PPaappeerr AAnnaallyyssiiss

22.. QQuueessttiioonn PPaappeerr && AAnnsswweerr kkeeyyss

http://www.thegateacademy.com/

.

GATE-2016 CE-SET-2

: 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com 1

ANALYSIS OF GATE 2016

Civil Engineering

GATE-2016- CE 7-Feb 9 AM-12 PM

SUBJECT NO OF

QUESTION Topics Asked in Paper

Level of Toughness

Total Marks

Enviormental Engineering

1 M: 2 2 M: 2

BOD,Air Pollution; Sound Pollution, Hardness, Ion Exchange Method, Sedimentation tank

Moderate

6

Structural 1 M: 1 Force in Truss, Kinematic indeterminacy in Moderate 5

[CATEGORY NAME]

15%

WRE(FM &hyd & Irrigation)

13%

Environmental Engineering

6%

RCC 5%

Steel Structure 5% Mechanics

2% Structural Analysis

5%

Geotechnical Engineering

14%

Transportation Engineering

7%

Surveying 7%

General Aptitude 15%

Solid Mechanics/SOM

3% CMM 3%


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GATE-2016 CE-SET-2


Analysis 2 M: 2 truss.

Water Resources Engineering

1 M: 3 2 M: 5

Unit hydrographs; Khosla Theory, Well hydraulics; Jet impact, Open channel fow;Hydraulic machines, fluid properties.

Easy 13

SOM/Solid Mechanics

1 M: 1 2 M: 1

Principal stress, Vol strain; SF & BM Easy 3

R.C.C. 1 M: 1 2 M: 2

Building materials; shear strength of RCC; modulus of rupture; Non prismatic RCC beam

Easy 5

Geomatics/Surve-ying Engineering

1 M:3 2 M:2

Photogrammetery;departure Moderate 7

Geotechnical Engineering

1 M: 4 2 M: 5

Earth Pressure; permeability& seepage;compaction; Bearing capacity

Moderate 14

Transportation Engineering

1 M: 3 2 M: 2

Geomatric design; Signal design, pavements Easy 7

* Construction Materials and Management

1 M: 1 2 M: 1

*Construction Management Tough 3

Engg. Mechanics 1 M: 0 2 M: 1

*Trusses and Frames. Tough 2

Steel Structure 1 M: 1 2 M: 2

Plastic Analysis ; Design of Tension Member

Moderate 5

Engineering Mathematics

1 M: 5 2 M: 5

Numerical Method ;Matrices;Complex Variable; Probability ; Limit & Continuty. Moderate 15

GA 1 M: 5 2 M: 5

Time & Work ;Paragraph; English fill in Blank, ;Number theory; Venn Diagram ; Mensuration& Area

Easy 15

Total 65 Moderate 100 * Indicates Questions from New Syllabus.

Faculty Feedback:

Few questions came from New Syllabus ( 2 questions from CPM, 1 question from

mechanics)

General Ability was pretty easy;

Weightage of Environment, Geotechnical and Highway was more as usual.

Construction management Structural analysis was scoring

Paper was logical. Qualifying is easy but scoring is tough. More focus was on

concepts.

Prectice Online test Series & Previous year question Series Will be beneficial .


GATE-2016

Question Paper

&

Answer Keys

.

GATE-2016 CE-SET-2


GATE 2016 Examination

Civil Engineering

Test Date: 07/02/2016

Test Time: 9:00 AM to 12:00 PM

Subject Name: CIVIL ENGINEERING

Section: General Aptitude Q NO. 1.

[Ans. C]

Q NO. 2.

[Ans. A]

Q NO. 3.

[Ans. C


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GATE-2016 CE-SET-2


Q NO. 4.

[Ans. A]

Q NO. 5.

[Ans. A]

Q NO. 6.

[Ans. D]


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GATE-2016 CE-SET-2


Q NO. 7.

[Ans. D]

Q NO. 8.

[Ans. B]

Q NO. 9.

[Ans. D]


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GATE-2016 CE-SET-2


Q NO. 10.

[Ans. C]

Section: Technical Q NO. 1.

[Ans. *] Range: 54.49 to 54.51

Q NO. 2.

[Ans. D]

Q NO. 3.

[Ans. C]


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GATE-2016 CE-SET-2


Q NO. 4.

[Ans. A]

Q NO. 5.

[Ans. D]

Q NO. 6.

[Ans. A] Kinematic inderminacy Dk = 2j − re

= 2 × 7 − 2 = 11

Q NO. 7.

[Ans. B]


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GATE-2016 CE-SET-2


Q NO. 8.

[Ans. B]

∑MA = 0 ⇒ RD × 2L − P × L = 0

⇒ RD =P

2

At joint D: ∑FY = 0

RD =P

2

45° T

RD

D

RA

L

L L

P

C B

A


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GATE-2016 CE-SET-2


⇒ T cos 45° =P

2

∴ T = P/√2

Q NO. 9.

[Ans. D]

Q NO. 10.

[Ans. C] Total float can be determined once the activity time i.e., EST, EFT, LST and LFT are known Total float, FR = LST − EST = LFT − EFT


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GATE-2016 CE-SET-2


For activity 5-7, EST = 38 EFT = 63 LFT = 63 LST = 38 FT = 0 For activity 11-12, EST = 80 EFT = 81 LFT = 81 LST = 80 FT = 0 Note: it can be seen directly that since the slack of all events are zero, there is not margin left for the occurrence of events and therefore. Maximum available line = Time required for completion of activity ∴ FT for all activities is zero.

Q NO. 11.

[Ans. D] In case of clay ultimate bearing capacity is independent of width of footing

Q NO. 12.

[Ans. C]

ic =G − 1

1 + e= (G − 1)(1 − n) ∵

1

1 + e= 1 − n

ic = (G − 1)(1 − 0.3) = (G − 1) × 0.7

G =ic

0.7+ 1 = 1.43ic + 1

38 38 63 63 70 70

80 80

81 81

38 38 63 63 70 70

8 8

1 2

0 0 1 2 3 10 11 12

4 6 8

4 6 8

A B

C

D

E F

G H

J

I

K L

1 7

30

30

25

25 7

7

7

7

3 1


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GATE-2016 CE-SET-2


Q NO. 13.

[Ans. B]

OCR =Maximum effective stress in past

Maximum effective stress in present

Maximum effective stress in present = 10 γsat = 10γw = 10 × 16 − 10 × 10 = 60 kN/m2

∴ OCR =90

60= 1.5

Q NO. 14.


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GATE-2016 CE-SET-2


[Ans. B]

ϕB =2.75

4× 100 = 68.75%

ϕP + 100 − ϕR = 31.25%

Now, ϕP =Pressure head at point P

Total head× 100

31.25 =h

4× 100

∴ h = 1.25 m

Q NO. 15.

[Ans. *] Range: 0.19 to 0.21

Sr + Sy = n 5

100+

Vw

103 × 106 × 1=

25

100

Vw = 0.2 × 109m3 = 0.2 km3

Q NO. 16.

[Ans. C]


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GATE-2016 CE-SET-2


Q NO. 17.

[Ans. D]

Q NO. 18.

[Ans. *] Range: 6.8 to 7.0 Discharge to be treated by one tank,

Q =1500

2= 750 m2/day

Surface area, A =Q

OFR=

750

20= 37.5m3

⇒ 37.5 = π ×d2

4

⇒ d = 6.91 m

Q NO. 19.

[Ans. *] Range: 380 to 390

Earth′s Surface Troposphere

Stratosphere

Mesosphere

Thermosphere Exosphere


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GATE-2016 CE-SET-2


Total water to be treated = 540 × 4 = 2160 l/day Resultant hardness required = 75 mg/l Let bypass rate be x l/day hardness of 420 mg/l Resultant hardness required = 75 mg/l

CHmix =CH1Q1 + CH2Q2

Q

75 =x × 420 + (2160 − x)0

2160

x =2160 × 75

420= 385.7 l/day

Q NO. 20.

[Ans. C] The sound pressure of the faintest sound that a normal healthy individual can hear is 20 μPa. It is taken as reference sound pressure level. A 20 μPa pressure is 0 dB on the sound pressure level scale

Q NO. 21.

[Ans. C]

Q0 = 540 I/day

1

2

3

4

Q1 = x CH1 = 420 mg/l

Bypass 75 mg/l

Softner

CH4 = 0 mg/l Q2 = 2160 − x

Q0

Q0

Q0


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GATE-2016 CE-SET-2


Q NO. 22.

[Ans. A]

Q NO. 23.

[Ans. D]

Q NO. 24.

[Ans. D]

The departure of the line ; D = l sin θ

= 10 sin 30° =10

2= 5 m

Q NO. 25.

[Ans. A]

E W

N

S

D

L l 30°


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GATE-2016 CE-SET-2


Tangent length, VT1 = R tanΔ

2

= R tan 30° = 0.577 R

Q NO. 26.

[Ans. B]

T2

V

Δ = 60°

T1

R R


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GATE-2016 CE-SET-2


Q NO. 27.

[Ans. B] Q NO. 28.

[Ans. D]

Q NO. 29.

[Ans. *] Range: 85.0 to 85.5

Q NO. 30.

[Ans. B]


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GATE-2016 CE-SET-2


Q NO. 31.

[Ans. C] Volumetric strain

∈V= (σx + σy + σz

3) (1 − 2μ)

⇒δV

V= (

σx + σy + σz

3) (1 − 2μ)

As ΔV = 0 ⇒ Either σx + σy + σz = 0 or1 − 2μ = 0

⇒ 1 − 2μ = 0 μ = 0.5

Q NO. 32.

[Ans. B]


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GATE-2016 CE-SET-2


σx + σy = σ1 + σ2

⇒ 5 + 5 = 10 + σ2 ⇒ σ2 = 0

Now, σ1 2⁄ =σx + σy

2± √(

σx − σy

2)

2

+ τxy2

∴ σ1 =5 + 5

2+ √(

5 − 5

2)

2

+ τxy2

⇒ 10 = 5 + τxy

∴ τxy = 5 MPa

Q NO. 33.

[Ans. A]

σ1

σ2 σ1

σ2 σx

σx = 5

σx = 5

σy


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GATE-2016 CE-SET-2


As there is no horizontal force, Hence HP = HS = 0

∑MQ = HP ×L

2= 0

Q NO. 34.

[Ans. C]

HS = 0

R

w kN/m

L L/2

VS VP

HP = 0 P S

Q


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GATE-2016 CE-SET-2


Given data L = 1m, A = 0.05 m2, E = 30 × 106 N/m2 Consider joint ′S′

FSQ = = FSR

2FSQ = cos 45° = W

⇒ FSQ =W

2× √2 =

W

√2

∴ FSQ =W

√2

As the truss is symmetrical

∴ FQP = FPR =W

√2(Tensile)

Now consider joint ‘Q’

FQP = FQS =W

√2

∑Fx = 0

⇒FQP

√2+

FQS

√2+ FQR = 0

⇒ FQR = W (Compressive)

∴ FQR + 100 kN (compressive)

45°

45°

FQP

FQP

FQS

FSR

W

FSQ

45° 45°

L

45°

45° 45°

45°

90°

90°

W

Q R

S

W

√2

W

√2

W

√2

W

√2


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GATE-2016 CE-SET-2


Stress in member QR

σQR =FQR

2Δ

⇒ σQR =100

2 × 1.05=

100 × 100

2 × 5= 100 kN/m2

∴ σQR = 1000 kN/m2

As the member QR consist compressive so it will go under shortening

∴ Shortening, Δ =P(Length)

2AE=

FQR. LQR

2AE

LQR = √L2 + L2 = √2L

∴ Δ =100 × 103 × √2 × 1

2 × 0.05 × 30 × 106

=100 × 103 × √2

0.1 × 30 × 106=

√2

30

∴ Δ = 0.471

Q NO. 35.

[Ans. *] Range: 64 to 66

Shear force at section X-X Vu = 100 − 5 × 10 = 50 kN Depth at section X-X

d = 400 +200

10× 5 = 500 mm = 0.5 m

Moment at section X-X Mu = 100 × 5 − 10 × 2.5 × 5 = 375 kNm

5 m

600 m 400 m

10 kN/m

Effective span = 20 m X

β

X


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GATE-2016 CE-SET-2


Design shear force at section X-X_,

Vu,design = Vu +Mu

dtan β

= 50 +375

0.5×

200

10000= 65 kN

Q NO. 36.

[Ans. *] Range: 2.9 to 3.1

BMQ = 11.25 × 150 = 1.6875 × 106N − mm

⇒σ

y=

MQ

I

Where, y =150

2= 75 mm and I =

(150)4

12

⇒ σ =1.6875 × 106 × 75

(150)4/12= 3 MPa

11.25 kN 11.25 kN

P S

Q R

11.25 kN 11.25 kN


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GATE-2016 CE-SET-2


Q NO. 37.

[Ans. C]

Tensile strength of plate in arrangement (2) will be greater than in arrangement (1) ∗ As per IS code 800: 2007 close 6.3

(0.9 Anet

fup

γml)

2

> (0.9 Anet

fup

γml)

1

(Anet)2 > (Anet)1

[(B − 2d +p2

4g) t]

2

> [(B − d)t]1

B − 2d +p2

4g> B − d

p2

4g> d

p2 > 4gd

P P P P d g

p

Figure 1 Figure 2


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GATE-2016 CE-SET-2


Q NO. 38.

[Ans. A] DS = 2 ∴ Number of plastic hinge required for complete collapse = DS + 1 = 2 + 1 = 3 Mechanism 1:

Δ =2L

3θ=

4L

3

⇒ θ = 2θ For principal of virtual work done

−2MPθ − 2MPθ − 2MPϕ − MPϕ = P (2L

3θ) = 0

⇒ 4MPθ + 3MPϕ =2PL

3θ

⇒ 8MPϕ + 3MPϕ =2PL

3ϕ

⇒ 11MP =4PuL

3

⇒ Pu =33

4

MP

L

∴ Pu = 8.25 MP

L

P

θ ϕ

ϕ θ

2MP MP

2MP 2MP

Δ

2L

3

4L

3


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GATE-2016 CE-SET-2


Mechanism 2:

2MPθ + MPθ + MPθ + MPθ = P (2L

3)

⇒ 5MP =2PL

3

⇒ Pu =15MP

2L

∴ Pu = 7.5MP

L

Q NO. 39.

[Ans. C]

P

θ θ

θ θ

2MP MP

MP MP


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GATE-2016 CE-SET-2


The critical duration is 15 days which is observed along the path P-Q-T-W-X which is the critical path of the project Note: Critical path is the one which consumes maximum amount of time

Q NO. 40.

[Ans. *] Range: 495 to 505

q =KH ⋅ Nf

Nd

K =3mm

min=

3 × 10−3m

60m/s

H = 5 m, Nf = 20, Nd = 10

q =3 × 10−3

60× 5 ×

20

10m3 sec⁄ per m length of dam

q =3 × 10−3 × 5 × 20

60 × 10× 106cm3 sec⁄ perm length of dam

q = 500 cm3/ sec per m lnegth of dam

TE = 6 TS = 7

10 20 40 50

30 70

60

80 90

TE = 0 TS = 0

TE = 2 TS = 2

P 2 4

R U 3

2

X

TE = 5 TS = 5

Q

3 5

T W

3

TE = 10 TS = 10

TE = 6 TS = 7

TE = 13 TS = 13

TE = 15 TS = 15

V

2 3

S

TE = 6 TS = 11


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GATE-2016 CE-SET-2


Q NO. 41.

[Ans. *] Range: 69 to 70.5 In c‐ ϕ soil

Earth pressure pa = kaσv − 2c √ka

ka =1 − sin ϕ

1 + sin ϕ=

1 − sin 24

1 + sin 24 = 0.4217

Note: Below water table, water pressure will not be multiplied by ka at point A σy = 1 × γb + 4γsat + 3γsat

σy = 1 × γb + (4 × γ′ + 4 × γw) + (3γ′ + 3γw)

σy = (1 × γb + 4γ′ + 3γ′) + (4γw + 3γw)

Earth Pressure at ‘A’

pa = kaσv − 2c√ka

= ka[γb + 4γ′ + 3γ′] + 4γw + 3γw − 2 × c√ka

= 0.4217[16.5 + 4(19 − 9.81) + 3(18.5 − 9.81)] + 7 × 9.81 − 2 × 25 × √0.4217 pa = 69.65 kN m2⁄

Q NO. 42.

[Ans. B]


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GATE-2016 CE-SET-2


OMC − SP > OMC − MP MDD − SP < MDD − MP

Q NO. 43.

[Ans. *] Range: 0.04 to 0.05

Discharge =200 ml

15 min=

200 cm3

15 × 60 sec=

200 × 103

900

mm3

sec

As per Darcy q = kavgiA

Water content

Standard

Proctor Test

Modified

Proctor Test

Dry

Den

sity


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GATE-2016 CE-SET-2


kavg =ΣZi

ΣZiki

[∵ Flow is normal to bedding plane]

kavg =150 + 150

1500.02

+150

k

i =Head difference

length=

600 − 300

300= 1

A = 80 cm2 = 80 × 102 mm2

q = [150 + 150

1500.02

+150

k

] × 1 × 80 × 102 =200 × 103

900

k = 0.045 mm/ sec

Q NO. 44.

[Ans. *] Range: 298 to 299 Local shear failure is occurring hence modified c and ϕ should be used

cm =2

3c =

2

3× 35 = 23.33 kN m2⁄

tan ϕm =2

3tan ϕ

2

3tan 28.63

ϕm = 20°


.

GATE-2016 CE-SET-2


For ϕm = 20°, Nc = 17.7, Nq = 7.4, Nγ = 50

⇒ Water table is at Df + B = 1.5 + 4 = 5.5 m hence no effect on bearing capacity ⇒ As per Terzaghi for strip footing qu = cNc + 8DfNq + 0.5 Bγ Nγ

qu =2

3× 35 × 17.7 + 17 + 1.5 × 7.4 + 0.5 × 4 × 17 × 5

qu = 771.7 kN m2⁄ qnu = qu − γDf = 771.7 − 17 × 1.5 qnu = 746.2 kN m2⁄ Net safe bearing capacity

qns =qnu

F=

746.2

2.5= 298.48 kN m2⁄

Q NO. 45.

[Ans. *] Range: 7.4 to 8.0

Force exerted by jet in x-direction

Water jet

30°

20

0 m

m

F

v


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GATE-2016 CE-SET-2


Fx = m[v sin θ − 0] = ρ × q × v sin θ

= (103) |π

4(0.02)2 × 10| 10 sin 30°

= 15.7079 N Taking moment about hinge

Fx ×0.2

2= F × 0.2

F = 7.85 N

Q NO. 46.


Time (1)

Ordinate of 1 hr UH

(𝐦𝟑 𝐬⁄ )

(2)

Offset ordinate

(𝐦𝟑 𝐬⁄ )

(3)

Ordinate of 2 hr DRH

(𝐦𝟑 𝐬⁄ )

(4) = (2) +(3)

Ordinate of DRH of 4 cm rainfall excess (𝐦𝟑 𝐬⁄ ) (𝟓)= (𝟒) × 𝟒 𝟐⁄

Ordinate of flood

hydrograph

(𝐦𝟑 𝐬⁄ ) (𝟔) = (𝟓) + 𝟐𝟎

10:00 am 11:00 am 12:00 pm 01:00 pm 2:00 pm 3:00 pm 5:00 pm

0 3

12 8 6 3 0

- 0 3

12 8 6 3

0 3

15 20 14 9 3 0

0 6

30 40 28 18 6 0

20 26 50 60 48 38 26 20

Ordinate of 01:00 pm = 60 m3 s⁄

Q NO. 47.

[Ans. *] Range: 0.19 to 0.21 The maximum jump (ΔZ) that can be provided is given by

y +Q2

2gA2= ΔZjump +

3

2yc

Here, A = 3 × 0.5 = 1.5 m2

yc = (Q2

gB2)

1 3⁄

= (62

9.81 × 32)

1 3⁄

= 0.7415 m


.

GATE-2016 CE-SET-2


⇒ 0.5 +62

2 × 9.81 × 1.52= ΔZjump +

3

2× 0.7415

⇒ ΔZjump = 0.20 m

Q NO. 48.

[Ans. *] Range: 6565 to 6580 Apply energy equation at the free surface of reservoir and exit of nozzle

500 = Head loss +v1

2

2g

500 = 0.05v1

2

2g+

v12

2g

√2 × 10 × 500

1.05= v1

V1 = 97.59 m sec ⁄

Water Power (W. P) =1

2mv1

2

=1

2(103) ×

π

4(0.15)2(97.59) = 8212.178 kW

Now, η0 =Shaft Power(S. P)

W. P

0.8 =Shaft Power(S. P)

8212.178

S. P = 6569.74 kW ≅ 6570 kW

Q NO. 49.

[Ans. D] Seepage velocity

Vs =v

n=

distance

time

As per Darcy, v = ki ki

n=

100 m

100 days


.

GATE-2016 CE-SET-2


i =head difference

length=

3

100

k × 3

0.15 × 100=

100

100m day⁄

k = 5 m day⁄

Q NO. 50.

[Ans. A] Ca2+ = 2.65 meq l⁄ , Mg2+ = 1.45 meq l⁄ Na+ = 2.25 meq l⁄ , K+ = 0.5 meq l⁄ HCO3

− = 3.3 meq l⁄ , SO42− = 0.6 meq l⁄

Cl− = 2.85 meq l⁄ Hardness is due to multivalent metallic cations, i.e. Ca2+ and Mg

2+

Total hardness (mg

las CaCO3) = (Total meq l⁄ ) × (eq. weight of CaCO3 in mg)

= (2.65 + 1.45) × 50 mg l⁄ as CaCO3 = 205 mg l⁄ as CaCO3

Alkalinity is due to the presence of HCO3− in this case

Alkalinity (mg l⁄ as CaCO3) = 3.3 × 50 mg l⁄ as CaCO3 = 165 mg l⁄ as CaCO3

Now, Carbonate Hardness = Min {Total Hardness, Alkalinity} = 165 mg/l

Non- carbonate hardness = Total hardness – carbonate hardness = 205 − 165 = 40 mg l⁄ as CaCO3

Q NO. 51.



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GATE-2016 CE-SET-2


L2 = L1 − |20 log (R2

R1)|

= 74 − 20 log 0.5 = 67.98 dB

Q NO. 52.

[Ans. *] Range: 275 to 277 BOD5 = L0(1 − e−0.22×5) BOD3 = L0(1 − e−0.22×3) From equation ① and ②, we get BOD5

BOD3=

(1 − e−0.22×5)

(1 − e−0.22×3)

BOD5 = 200 ×0.667

0.483

= 200 × 1.38 = 276.19 mg l⁄

Q NO. 53.

[Ans. A] Y = y1 + y2 + y3 = 0.30 + 0.25 + 0.25 = 0.80 New total lost time, L = 10 sec (given) ∴Optimum cycle time,

C0 =1.5 L + 5

1 − Y

⇒ C0 =(1.5 × 10) + 5

1 − 0.80=

15 + 5

0.20=

20

20= 100 sec

Now green times are calculated by,

G1 =y1

y2

(C0 − L) =0.30

0.80(100 − 10) = 33.75 ≅ 34 sec

G2 =y2

y(C0 − L) =

0.25

0.80(100 − 10) = 28.11 ≅ 28 sec

G3 =y3

y(C0 − L) =

0.25

0.80(100 − 10) = 28.11 ≅ 28 sec


.

GATE-2016 CE-SET-2


Q NO. 54.


Total energy lost between point ① and ② = Work done by frictional force 1

2mv1

2 −1

2mv2

2 + mg × 0.03s = f × (mg) × s

(0.278 × 100)2

2−

(0.278 × 50)2

2+ 9.81 × 0.03 × s = 0.35 × 9.81 × s

289.815 = 3.1392 s ⇒ s = 92.32 m

Q NO. 55.

[Ans. *] Range: 120 to 121

Given, H = 700 m, havg = 250

Relief distance, d = r1 − r = 112.5 − 82.40 = 30.1 mm

∴ d =hr1

H − havg [Where h is height of tower]

⇒ 30.1 =h × 112.5

700 − 250

0.03 s

s

V1 = 100 km hr⁄

V2 = 50 km hr⁄

①

②

r

H

r1

havg

h


.

GATE-2016 CE-SET-2


∴ h = 120.4 mm


Documents

Innddeexx · PDF fileUnit hydrographs; Khosla Theory, Well hydraulics; Jet impact, Open channel fow;Hydraulic machines, fluid properties. Easy 13 SOM/Solid Mechanics 1 M: 1