gate 2014 ce answer key

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GATE-2014

Detailed Solutions

Civil Engineeringof

Morning Session

Expert Opinion

Dear Students,

The Questions of GATE 2014 are based on fundamental

and basic concepts of the syllabus. There is no ambiguity

and misprint noticed till now, however, it is an observation based

on students feedback.

The level and standard of GATE 2014 questions are relatively easier

than the exam of GATE 2013. There are 3 important observations

made by me about GATE 2014 exam.

1. The GATE 2014 exam is conducted in two seating i.e. morning session and afternoon

session. The question papers of both seatings are different. The difficulty level of questions

are nearly same and due care has been taken to balance both the papers, however small

differences are certainly there. The morning session paper seems to be little easier by 2

to 5%, however, it varies on the perception of person to person also.

The average marks of both the papers should be equated and necessary scaling criteria

should be adopted for this purpose.

2. The GATE 2014 cut-off will be nearly same as that of last year, perhaps it may be little

lesser than that of GATE 2013. GATE-2013 cutoff was 33 marks. Though the paper of GATE

2013 was tougher and number of students were less, 6 marks questions were wrongly

framed and hence, these 6 marks were awarded to all the candidates, which was certainly

a kind of bonus.

Therefore expected cut-off for GATE 2014 may be between 30 to 34 marks (General category).

It may be noted that the following formulae is used to evaluate GATE cut-off marks.

GATE Cutoff =

Total Marks obtained by all the candidates

Total number of candidates

3. The topper’s marks in GATE 2013 was nearly 83 in which 6 marks of bonus to all are

included. In my opinion topper’s marks in GATE 2014 will be between 80 to 85.

GATE cutoff <| 25 Marks

DisclaimerDear Students, MADE EASY has taken due care in collecting the data and questions. Since ques-tions are submitted by students and are memory based, therefore the chances of error can not be ruled out. Therefore MADE EASY takes no responsibility for the errors which might have incurred.

If any error or discrepancy is recorded then students are requested to inform us at: [email protected]

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GATE-2014 Exam SolutionsCivil Engineering (Morning Session)

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Section - I (Civil Engineering)One Mark Questions

Q.1 For a saturated cohesive soil, a tri-axial test yields the angle of intervalfriction (φ) as zero. The conducted test is(a) Consolidated Drained (CD) test(b) Consolidated undrain (CU) test(c) Unconfined compression (UC) test(d) Unconsolidated undrain (UU) test

Ans. (d)

Q.2 The possible location of shear centre of the channel section shown below is

P Q R S

(a) P (b) Q(c) R (d) S

Ans. (a)

Q.3→∞

+⎛ ⎞⎜ ⎟⎝ ⎠x

x sinxlimx is equal to

(a) –∞ (b) 0(c) 1 (d) ∞

Ans. (c)

Put x = 1h as x→∞ ⇒ h→0

→∞

+⎛ ⎞⎜ ⎟⎝ ⎠x

x sinxlimx = →

⎛ ⎞+⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

h 0

1 1sinh hlim 1

h



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= →

⎛ ⎞⎜ ⎟+ =⎜ ⎟⎜ ⎟⎝ ⎠

h 0

1sinhlim 1 11

h

Q.4 A conventional flow duration curve is a plot between(a) Flow and % time flow is exceeded(b) Duration of flooding and ground level elevation(c) Duration of water supply in a city and proportion of area recurring

supply exceeding this duration.(d) Flow rate and duration of time taken to empty of a reservoir at that

flow rate.Ans. (a)

Q.5 A steel section is subjected to a combination of shear and bending action.The applied shear force is V and shear capacity of the section is Vs. Forsuch a section, high shear force (as per IS 800-2007) is defined as(a) V > 0.6 Vs (b) V > 0.7 Vs(c) V > 0.8 Vs (d) V > 0

Ans. (a)Clause 9.2.1 IS 800:2007.

Q.6 In reservoir with an uncontrolled spillway the peak of the plotted outflowhydrograph(a) Lies outside the plotted inflow hydrograph.(b) Lies on the recession limb of the plotted inflow hydrograph.(c) Lies on the peak of the inflow hydrograph.(d) is higher than peak of the plotted inflow hydrograph.

Ans. (b)

Inflow hydrograph

Recession limb

Outflow hydrograph



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Q.7 If y = 5x2 + 3 than the tangent at x = 0 and y = 3(a) passes through x = 0, y = 0 (b) has a slope +1(c) is parallel to x-axis (d) has a slope of –1

Ans. (c)y = 5x2 + 3

dydx = 10 x

⇒( )0,3

dydx = 10 × 0 = 0

Tangent

y

x

(0, 3)

⇒ tangent is parallel to x-axis.

Q.8 The dimension for kinematic viscosity is

(a) LMT (b) 2

LT

(c)2L

T (d) MLT

Ans. (c)The SI unit of kinamatic viscosity is m2/sec.∴ dimension of kinamatic viscosity [ν] = L2/T.

Q.9 The following statements are related to temperature stress developed inconcrete pavement slab with four edge (without any restrain)P : The temperature stress will be zero during both day and night time if

the pavement slab is considered weight less.Q. : The temperature stress will be compressive at the bottom of the slab

during night time if the self weight of the pavement slab is considered.R : The temperature stress will be compressive at the bottom of the slab

during day time if the self weight of the pavement slab is considered.



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The true statement(s) is(are)(a) P only (b) Q only(c) P and Q only (d) P and R only

Ans. (c)The temperature stress will be tensile at the bottom of the slab during theday time if the self weight of the pavement slab is considered.

Q.10 An incompressible homogeneous fluid flowing steadily in a variable dia pipehaving the large and small dia as 15 cm and 5 cm respectively. If velocityat the section of 15 cm dia portion of the pipe is 2.5 m /sec, the velocityof fluid (in m/s) at section falling in 5 cm portion of the pipe is _________.

Sol.

π ×2(15) 2.54

= π ×2(5) V4

⇒ V = 2.5 × 9 = 22.5 m/s

Q.11 The monthly rainfall chart based on 50 years of rainfall in Agra is shownin the following figure which of the following are true ? (K percentile is thevalue such that K % of data fall below that value)

100200300400500

600650700

800 650

400

200100

50

600

Rainfall (mm)

Feb Mar Apr May June July Aug Sep Oct Nov Dec

(i) On average it rains more in July than in Dec.(ii) Every year, the amount of rainfall in August is more than that in

January.



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(iii) July rainfall can be estimated with better confidence than Feb. rainfall.(iv) In Aug, there is at least 500 mm of rainfall.(a) (i) and (ii) (b) (i) and (iii)(c) (ii) and (iii) (d) (iii) and (iv)

Ans. (b)

Q.12 The potable water is prepared from turbid surface water by adopting thefoil treatment square.(a) Turbid surface water → Coagulation → Flocculation → Sedimentation →

Filtration → Disinfection → Storage and supply(b) Turbid surface water → Disinfection → Flocculation → Sedimentation →

Filteration → Coagulation → Storage and supply(a) Turbid surface water → Filteration → Sedimentation→ Disinfection →

Flocculation → Coagulation(a) Turbid surface water → Sedimentation → Flocculation → Coagulation →

Disinfection → Filteration

Ans. (a)

Q.13 The minimum value of 15 minutes peak hour factor on a section of a roadis(a) 0.1 (b) 0.2(c) 0.25 (d) 0.33

Ans. (c)15 min. peak hr factor is used for traffic intersection design

PHF =15

(V/4)V

V = ⎛ ⎞⎜ ⎟⎝ ⎠

veh.Peak hourly volume inhr.

V15 = Maximum 15 minimum volume within the peak hr.(veh.)Maximum value is 1.0 and minimum value is 0.25Normal range is 0.7 – 0.98 = 0.25



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Q.14 Some of the non-toxic metal normally found in natural water are(a) Arsenic, Lead, Mercury (b) Calcium, Sodium, Silver(c) Cadmium, curomium, copper (d) Iron, Mangnese, Magnesium

Ans. (d)

Q.15 The degree of disturbances of a sample collected by sampler is expressedby a term called the area ratio. If outer diameter and inner dia of sampleare D0 and Di respecgively, the area ratio is

(a)−2 2

0 i2i

D DD (b)

−2 2i 0

2i

D DD

(c)−2 2

0 i20

D DD (d)

−2 2i 0

20

D DD

Ans. (a)

Q.16 The degree of static indeterminacy of a rigid jointed frame PQR supportedas shown is

y

S145°R

EI

Q90°

EIP x

(a) 0 (b) 1(c) 2 (d) 3

Ans. (a)Ds = DSe + Dsi

= (re – 3) + 3C – rr

= (4 – 3) + 3 × 0 – 1= 0



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Q.17 The action of negative friction on the pile is to(a) Increase the ultimate load on the pile(b) Reduce the allowable load on the pile(c) Maintain the working load on the pile(d) Reduce the settlement

Ans. (b)

Q.18 The ultimate collapse load (Wu) in terms of plastic moment Mp by kinematicapproach for a propped cantilever of length L with W acting at its mid spanas shown in fig would be

W

L/2 L/2

(a) p2ML (b) p4M

L

(c) p6ML (d) p8M

L

Ans. (c)

W

θ θ

θ θ

Mp

Mp Mp

Wu

L/2 L/2

From principal of virtual work

− θ − θ − θ + θp p p uLM M M W2 = 0

⇒ Wu = p6ML



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Q.19 Match the following:Group I Group II

P. Alidade 1. Chain SurveyQ. Arrow 2. LevellingR. Bubble tube 3. Plant table surveyingS. Stedia hair 4. Theodolite(a) P – 3, Q – 2, R – 1, S – 4(b) P – 2, Q – 4, R – 3, S – 1(c) P – 1, Q – 2, R – 4, S – 3(d) P – 3, Q – 1, R – 2, S – 4

Ans. (d)P – 3, Q – 1, R – 2, S – 4

Q.20 The sum of eigen value matrix [M] is

When⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

215 650 795[M] 655 150 835

485 355 550

(a) 915 (b) 1355(c) 1640 (d) 2180

Ans. (a)Sum of eigen values = trace of matrix

= 215 + 150 + 550 = 915

Q.21 The probability density function of evaporation E on any day during a yearin watershed is given by

f (E) = ⎧ ≤ ≤⎪⎨⎪⎩

1 0 E mm/day50 Otherwise

The probability that E lies in between 2 and 4 mm/day in a day in watershedis (in decimal)

Sol.

f (E) =⎧ ≤ ≤⎪⎨⎪⎩

1 0 E mm/day50 Otherwise



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P (2 < E < 4) = ( ) [ ]= =∫ ∫4 4

42

2 2

1 1f E dE dE E5 5

= ( )− =1 24 25 5 = 0.4

Q.22 A box of weight 100 kN shown in the figure to be lifted without swinging.If all the forces are coplanar, the magnitude and direction (θ) of force Fw.r.t. x axis is ________.

90 kN

30°

y

45°θ

40 kNF

x

100 kN

(a) F = 56.389 kN and θ = 28.28°(b) F = –56.389 kN and θ = –28.28°(c) F = 9.055 kN and θ = 1.1414°(d) F = –9.055 kN and θ = –1.1414°

Ans. (a)For no swinging ∑Fhorizontal = 0

90 kN

30°45°

θ

40 kNF

x

100 kN

⇒ 90 cos 30° = 40 cos 45° + F cos θ49.658 = F cos θ

F cos θ from option (a) = 56.389 cos 28.28° = 49.658 kN

Q.23 The amount of CO2 generated in kg while completely oxidizing one kg ofCH4 is ________.



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Sol.

+4 216g

CH 2O → +2 2O44g

CO 2H

⇒ 16 g of CH4 when completely oxidized leads to 44 g of CO2

⇒ 1 kg of CH4 when completely oxidized leads to × =44 116 22.75kg CO

Q.24 While designing for a steel column of Fe250 grade the base plate restingon a concrete pedestal of M20 grade, the bearing strength of concrete(N/mm2) in LSM of design as per IS 456: 2000 is ..........

Sol.Permissible bearing stress = 0.45 fck

= 0.45 × 20 = 9 N/mm2

Q.25 Given J = ⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎣ ⎦

13 2 12 4 2 and K = 21 2 6 1

then product KT JK is _______.

Sol.

J =⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

3 2 12 4 21 2 6

, K = ⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

121

KT JK = [ ]⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥− ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎣ ⎦

13 2 11 2 1 2 4 2 2

1 2 6 1

= [ ]⎡ ⎤⎢ ⎥− = + + =⎢ ⎥⎢ ⎥−⎣ ⎦

16 8 1 2 6 16 1

123

Two Marks Questions

Q.26 Three rigid bucket are of identical height and base area. Further assumethat each of these buckets have negligible mass and are full of water. Theweight of water in these bucket are denoted by W1, W2, W3 respectively.Which of the following option are correct.



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h h h

(a) W2 = W1 = W3 and F2 > F1 > F2

(b) W2 > W1 > W3 and F2 > F1 > F3

(c) W2 = W1 = W3 and F1 = F2 = F3

(d) W2 > W1 > W3 and F2 = F1 = F3

Ans. (d)Bucket → identical height

→ identical base area

h h hh

A A A21 3

⇒ W2 > W1 > W3

Force on the base in each case will be equal to = γ w h AHence, F1 = F2 = F3

Q.27 If the following equation establishes equilibrium in slightly bent position.

+2

2d y py

EIdx = 0

the mid-span deflection of a member shown in figure isy

xP

M

EIy

LN

P

[a is the amplitude constant for y]

(a)π⎛ ⎞= −⎜ ⎟⎝ ⎠

1 2 xy 1 a cosP L (b)

π⎛ ⎞= −⎜ ⎟⎝ ⎠1 2 xy 1 a sinP L

(c)π

=a sin n xy

L (d)π

=′a cos n xy

L



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Ans. (c)

2

2d ydx

= − ×P yEI

= –m2y∴ Solution of above differential equation is

y = a sin mx + b cos mxat x = 0, y = 0⇒ b = 0at x = L, y = 0⇒ 0 = sin mL⇒ mL = nπ

⇒ m = πnL

∴ y = nπxasinL

Q.28 A rectangular beam of 230mm width and effective depth = 450 mm, isreinforced with 4 bars of 12 mm diameter. The grade of concrete is M 20,grade of steel is Fe 500. Given that for M 20 grade of concrete, the ultimateshear strength τuc = 0.36 N/mm2 for steel percentage of = 0.25, andτuc = 0.48 N/mm2 for steel percentage = 0.5. For a factored shear force of45 kN, the diameter (mm) of Fe 500 steel 2 legged stirrups to be used atspacing of 325 mm should be(a) 8 (b) 10(c) 12 (d) 16

Ans. (a)

230

450

4 × 12φ

τuc = 0.36 N/mm2 [For M20] for × =stA 100 0.25bd

τuc = 0.48 N/mm2 [For M20] for × =stA 100 0.5bd

Factored SF = 45 kN = Vu



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We have to calculate the dia of Fe 500 2-L egged stirrup to be used at aspacing of 325 mm c/c

τv =×

= =×

2uV 45 1000 0.4348 N/mmbd 230 450

% tensile steel =( )π×

× =×

24 124 100 0.437%230 450

τc = ( )+ × − = 20.120.36 0.437 0.25 0.45N/mm0.25

Since τv – τc < 0⇒ Min shear reinforcement is required⇒ Min shear reinforcement is given by

sv

v

AbS =

y

0.40.87 f

Asv =( ) ( )× v

y

0.4 S b0.87 f

Since we limit fy to 415 N/mm2 hence,

Asv = ( )π × ×× φ = =

×2 20.4 325 2302 82.814 mm

4 0.87 415φ = 7.26 mm

adopt φ = 8 mm

Q.29 16 MLD of water is flowing through a 2.5 km long pipe of diameter 45 cm.The chlorine at the rate of 32 kg/d is applied at the entry of this pipe sothat disinfected water is obtained at the exit. These is a proposal to increasethe flow through the pipe to 22 MLD from 16 mLD. Assume the dilutioncoefficient n = 1. The minimum amount of chlorine (in kg per day) to beapplied to achieve the same degree of disinfection for the enhanced flowis(a) 60.5 (b) 4.4(c) 38 (d) 23.27

Ans. (a)In the disinfection process we have the relationship,

tCn = Kwhere t = time required to kill all organism



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c = concentration of disinfectantn = dilution coefficientk = constant

⇒ n1 1t C = n

2 2t C

in our case n = 1⇒ t1C1 = t2C2

t1 =1

Lv

L = length of pipe ; V1 = velocity of flow

t1 =1

LQ / A

t1 =1

LAQ

C1 = 1

1

W ,Q where W1 = weight of disinfectant per day ; Q1 = discharge per

day

⇒ × 1

1 1

LA WQ Q = × 2

2 2

LA WQ Q

⇒ W2 = ⎛ ⎞× = ×⎜ ⎟⎝ ⎠

222

121

Q 22W 32 kg /day16Q

= 60.5 kg/day

Q.30 A rectangular channel flow have bed slope of 0.0001 width = 3 m coefficientn = 0.015, Q = 1 m3/sec given that normal depth of flow ranges between 0.76 mand 0.8 m. The minimum width of throat (in m) that is possible at a givensection while ensuring that the prevailing normal depth does not exceedalong the reach upstream of the concentration is approximately, equal to(assume negligible loss)(a) 0.64 (b) 0.84(c) 1.04 (d) 1.24

Ans. (b)n = 0.015Q = 1 m3/s

Normal depth of flow between 0.76 m to 0.8 m.



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If prevailing normal depth of flow is not exceeded, there must not bechocking of the section or there must be just chocking.Thus the width of the section should be such that for the prevailing specificenergy there should be critical flow at the contracted section

i.e.⎛ ⎞⎜ ⎟⎝ ⎠

1/323 q2 g = EC = Einitial

⇒

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

1/32

min

QB3

2 g= Einitial

Let is now calculate Einitial

Q = 2/3 1/20

1 AR S2

⇒ 1 = ( ) ( )⎛ ⎞⎜ ⎟⎝ ⎠+

2/31/21 3y3y 0.0001

0.015 3 2y

⇒ y = 0.78 m

⇒ Einitial = +2

2qy

2gy

=( )

1⎛ ⎞⎜ ⎟⎝ ⎠

+ =× ×

2

230.78 0.7893 m

2 9.81 0.78

⇒

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

1/32

min

QB3

2 g = 0.7893

⇒( )( )

2/3

2/31/3min

132 g B = 0.7893

Bmin = 0.836 m

Q.31 A levelling is carried out to established the reduced level (RL) of point Rwith respect to the bench mark (BM) at P. The staff reading taken are givenbelow



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− − −− − − −

− −

Staff station BS IS FS RLP 1.655Q 0.95 1.5R – 0.75

If RL of P is + 100 m, then RL (m) of R is(a) 103.355 (b) 103.155(c) 101.455 (d) 100.355

Ans. (c)HI = RL + BS

and RL = HI – FS

− − −− − − −

− − −

Staff station BS IS FS RL HI RLP 1.655 101.655 100Q 0.95 1.5 102.205 103.155R – 0.75 101.455

∴ RL of R = 101.455 m

Q.32 A given cohensionless soil has emax = 0.85, emin = 0.5. In the field, the soilis compacted to a mass density of 1800 kg/m3 at water content of 8%. Takethe mass density of water as 1000 kg/m3 and GS = 2.7.(a) 56.43 (b) 60.25(c) 62.87 (d) 65.41

Ans. (d)emax = 0.85emin = 0.5ρfield = 1800 kg/m3 at water content = 8%

ρw = 1000 kg/m3

GS = 2.7Relative density, ID = ?

ρ = ( ) ( )ρ + ×=

+ +wG 1 w 2.7 1000 1.081 e 1 e

⇒ 1 + e =× ×2.7 1000 1.08

1800⇒ e = 0.62

⇒ ID =−

×−

max

max min

e e 100e e



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=−

×−

0.85 0.62 1000.85 0.5

= 65.714%

Q.33 Then tension and shear force (both in kN) at both joints as shown beloware respectively

435

P = 250 kNu

(a) 30.3 and 20 (b) 30.33 and 25(c) 33.33 and 20 (d) 33.33 and 25

Ans. (d)

θ P cos =u θ 4 Pu5

P cos =u θ 3 Pu5

Pu

tanθ = 34

cosθ = 45

sinθ = 35

Tension in each bolt = ×

u4P5 6

××

4 2505 6 = 33.33 kN

Shear in each bolt = ×=× ×

u3P 3 2505 6 5 6

= 25 kN



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Q.34 For a sample of water with the ionic composition shown below, the Carbonateand Non-carbonate hardness concentration (in mg/l as CaCO3) respectivelyare.

0meq/l

meq/l

Ca2+

HCO3–

Mg2+

SO42–

Na+

4 5 7

3.5 7

(a) 200 and 500 (b) 175 and 75(c) 75 and 175 (d) 50 and 200

Ans. (c)Carbonate hardness = 3.5 × 10–3 g-eq [if NCH is present sodium alkalinitywill be absent i.e. NaHCO3 absent]

= −× ×l

33

50g3.5 10 as CaCO

= 175 mg/l as CaCO3

Non carbonate hardness = total hardness-carbonate hardnessTotal hardness = 5 × 50 mg/l as CaCO3 [total hardness is due to Ca2 + andMg2+]

= 250 mg/l as CaCO3⇒ NCH = 250 – 175 = 75 mg/l as CaCO3

Q.35 Group-I Group-IIP. Curve J 1. No apparent heaving of soil around the footing.Q. Curve K 2. Rankine passive zone develops imperfectlyR. Curve L 3. Well defined slip surface extends to ground

surface.

Load

Settlement

J K L

(a) P-1, Q-3, R-2 (b) P-3, Q-2, R-1(c) P-3, Q-1, R-2 (d) P-1, Q-2, R-3



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Ans. (d)L → General shear failureK → Local shear failureJ → Punching shear failure

Q.36 A horizontal jet of water with its cross section area 0.0028 m2 hits a fixedvertical plate with a velocity of 5 m/s. After impact the jet split symmetricallyin a plane parallel to the plane of the plate. The force of impact (in N) ofthe jet on the plate is(a) 90 (b) 80(c) 70 (d) 60

Ans. (c)

5 m/s

area of jet = 0.0028 m2

Force on plate = (ρwaV)V= ρwaV2 = 1000 × 0.0028 × (5)2

= 70 N

Q.37 In a simply supported beam of length L four influence line diagram for shearat a section located at a distance of L/4 from the left support marked(P, Q, R, S) are shown below the correct, ILD is

P.

0.75

L/4 3L/4

0.25

Q.

0.6

L/4 3L/4

0.6

R.

0.5

L/4 3L/4

0.5

S.L/4 3L/4



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(a) P (b) Q(c) R (d) S

Ans. (a)

BA

X

1 unit

1 unit

X

+

–

0.75

0.25

L/4 3L/4

ILD for SF at X-Xi.e., option (a)

Q.38 A long slope is formed in a soil with shear strength parameter C′ = 0, φ′ = 34°.Firm strate lies below the slope and it is assumed that water table mayoccasionally rise to the surface, with seepage taking place parallel to theslope. Use γsat = 18 kN/m3 and γw = 10 kN/m3. maximum slope angle (indegree) to ensure the factor of safety 1.5. Assuming a potential failuresurface parallel to the slope would be(a) 45.3 (b) 44.7(c) 12.3 (d) 11.3

Ans. (d)

C = 0′φ = 34°

γsat = 18 kN/m3

γw = 10 kN/m3

β



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Page21

FOS =γ φγ β

sub

sat

tan.tan

⇒ tan β =⎛ ⎞γ φ⎜ ⎟γ⎝ ⎠

sub

sat

tan.FOS

⇒ tan β =γ φγ

sub

sat

tan.1.5

tan β = ( )−× °

×18 10

tan 3418 1.5

⇒ β = 11.30°

Q.39 For the truss shown below, the member PQ is short by 3 mm. The magnitudeof the vertical displacement of joint R in mm is ________.

R

P Q

3m

4m4m

Sol.PQ is short by 3 mmWe have to find out vertical displacement of joint R in mm

Δ R = ∑u(λ)Let in apply unit load at R as shown below

R

P Qθ

1/2 1/2

uPQ

uPR R



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Page22

uPR sin θ =12

uPQ + uPR cos θ = 0

uPQ = –uPR cos θ = − θθ

1 .cos2sin

uP Q = − θ1 cot2 =

− × −=

1 4 / 3 22 3

ΔR = ( )−× λ = − =PQ PQ

2u 3 2mm upwards3

Q.40 The smallest angle of a triangle is equal to two third of the smallest angleof a quadrilateral. The ratio between the angle of the quadrilateral is3 : 4 : 5 : 6. The largest angle of the triangle is twice its smallest angle, whatis the sum, in degrees of the second largest angle of the triangle and thelargest angle of the quadrilateral.

Sol.

θ2

θ3 θ1

3α 4α

5α 6α 18 = 360 = 20°α

α

Largest angle of quadrilateral = 120°Smallest angle of quadrilateral = 60°

⇒ Smallest angle of triangle = × × = °2 (2 20) 403

Largest angle of triangle = 2 × 40 = 80°⇒ Third angle of triangle = 60°⇒ Sum of largest angle of quadrilateral and second largest angle of triangle

= 120° + 60° = 180°

Q.41 A straight 100 m long raw water gravity main is to carry water from intaketo the jackwell of a water treatment plant. The required flow of water is0.25 m3/s. Allowable velocity through main is 0.75 m/s. Assume f = 0.01,g = 9.81. The minimum gradient (in cm/100 length) required to be given tothis main so that water flow without any difficulty should be ________ .



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Page23

Sol.Q = 0.25 m3/s

Allowable velocity = 0.75 m/sf = 0.01g = 9.81

π 2d4 = = = 2Q 0.25 1 m

V 0.75 3

⇒ d = 0.6515 m

⇒2flv

2gd = ( )× ×=

× ×

2

f0.01 100 0.75h m

2 9.81 0.6515

= 0.044 m = 4.4 cm

⇒ Minimum gradient = =lfh 4.4 cm

100m

Hence answer is 4.4.

Q.42 For a beam cross section W = 230 mm, effective depth = 500 mm, the numberof reinforcement bars of 12 mm diameter required to satisfy minimumtension reinforcement requirement specified by IS-456-2000 (assume gradeof steel is Fe500) is __________.

Sol.

st minAbd =

y

0.85f

Ast = × × 20.85 230 500 mm500

n =( )

×= ππst2 2

A 0.85 230d 12

44= 1.729 = 2 bars

Q.43 The perception - reaction time for a vehicle travelling at 90 km/h, given thecoefficient of longitudinal friction of 0.35 and the stopping sight distance of170 m (assume g = 9.81 m/s2) is ________ seconds.



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Sol.

SSD = ( )+2

r0.278v0.278vt

2gf

⇒ 170 = ( )×× × +

× ×

2

r0.278 900.278 90 t

2 9.81 0.35

⇒ tr = 3.1510 sec.

Q.44 A traffic office impose on an average 5 number of penalties daily on trafficviolators. Assume that the number of penalties on different day is independentand follows a Poisson distribution. The probability that there will be less than4 penalties in a day is _________.

Sol.Mean λ = 5

P (x < 4) = p (x = 0) + p (x = 1) + p (x = 2) + p (x = 3)

=− − − −

+ + +5 0 5 1 5 2 5 3e s e 5 e 5 e 50! 1! 1! 3!

= − −⎡ ⎤ ⎛ ⎞+ + + = =⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦5 525 125 118e 1 5 e

2 6 30.265

Q.45 The speed-density (v-k) relationship on a single lane road with unidirectionalflow is v = 70 – 0.7 K, where v is in km/hr and k is in veh/km. The capacityof the road (veh/hr) is

Ans. (a)Capacity = Velocity × Density

⇒ C = V × K= 70 K – 0.7 K2

Now, dCdK = 70 – 1.4 K = 0

⇒ K = 50⇒ Capacity, C = 70 × 50 – 0.7(50)2

= 1750 veh/hr



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Q.46 A particle moves along a curve whose parametric equation are x = t3 + 2t,y = –3e-2t and z = 2 sin (5t), where x, y and z show variation of the distancecovered by the particles in (cm) with time (t) (in second). The magnitudeof the acceleration of the particle (in cm/s2) at t = 0 is __________.

Sol.x = t3 + 2ty = –3 e–2t

z = 2 sin (5t)dxdt = 3t2 + 2

⇒ ax = =2

2d x 6 tdt

dydt = –3e–2t × (–2) = 6e–2t

⇒ ay = −= −2

2t2

d y 12 edt

⇒dzdt = –10 × 5 sin (5t) = –50 sin5t

⇒ az = = −2

2d z 50sin 5tdt

a� = + +x y zˆˆ ˆa i a j a k

a� at t = 0 = − + ˆˆ ˆ0 i 12 j 0 k

a� = − ˆ12 j⇒ Magnitude of acceleration at

t = 0 = 12 cm/s2

Q.47 For a cantilever beam of a span 3m as shown a concentrated load of 20 kNapplied to the free end causes a vertical displacement of 2mm at a sectionlocated at a distance of 1m from the fixed end (with no other load on thebeam) the maximum vertical displacement in the same (in mm) is ___________.

1 m 2 m

2 mm

20 kN



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Sol.

2 m

20 kN

1 m Δ12

2 m1 m

10 kN

Δ21

From Betti’s law P1 × Δ12 = P2 × Δ21

⇒ 10 × 2 = 20 × Δ21

⇒ Δ21 = 1 mm

Q.48 An isolated three-phase traffic signal is designed by webster’s method. Thecritical flow ratio for three phase are 0.2, 0.3 and 0.25 respectively and losttime per phase is 4 second. The optimum cycle length (in sec.) is ________.

Sol.Sum of the flow, y = y1 + y2 + y3

= 0.2 + 0.3 + 0.25 = 0.75Total lost time in a cycle L = 4 × 3 = 12 sec.

Optimum cycle length, C0 = +−

1.5L 51 y

× +−

1.5 12 51 0.75 = 92 sec.

Q.49 Mathematical idealization of a crane has three bar with their verticesarranged as shown with load of 80 kN hanging vertically. The coordinate ofthe vertices are given in parenthesis. The force in member QR is ______.



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Page27

P(0,4)

22.84°

104.03°R(3,0)55.13°

80 kN

Q(1,0)

y

x

Sol.

P(0,4)

22.84°

104.03°R(3,0)55.13°

80 kN

Q(1,0)

4m

θφ

1m

3m

VQ VR

VQ + VR = 80 ...(i)ΣMR = 0

⇒ 80 × 3 = VQ × 2VQ = 120 kN

⇒ VR = –40 kN From eq. (i)

tanφ = 14

⇒ sinφ = φ =1 4, cos17 17

tanθ = 34

⇒ cosθ = 45



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Consider joint Q

FQR

FPQ

α

VQ

α = 104.03° – 90° = 14.03°ΣFy = 0

⇒ FPQ cosα = +VQ = 0⇒ FPQcos14.03° + 120 = 0⇒ FPQ = –123.6897 kN

ΣFx = 0⇒ FPQ sinα = FQR

FQR = –29.986 kN ≈ 30 kNm

Q.50 The flow net constructed for a dam is shown in the figure below. Takingcoefficient of permeability as 3.8 × 10–6 m/s, the quantity of flow (in cm3/sec)under the dam per m is _________.

50m

6.3 m

1.6 m

9.4 m

17.2

m

Sol.

Quantity of flow, Q = f

d

NKHN



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Here, Nf = No. of flow channels = 3Nd = No. of equipotential drops = 10

Given, K = 3.8 × 10–6 m/sand H = 6.3 m

∴ Q = −× × ×6 33.8 10 6.310

= 7.182 × 10–6 m3/s/m= 7.182 × 10–6 × 106 cm3/s/m

Q = 7.182 cm3/s/m

Q.51 The full data are given for laboratory sample σ′0 = 175 kPa, e0 = 1.1 σ′0 +Δσ′0 = 300 kPa, e = 0.9. If thickness of the clay specimen is 25 mm, the valueof coefficient of volume compressibility is ________ × 10–4 m2/kN.

Sol.

mv =ΔΔσ=

+ +v

0 0

ea

1 e 1 e

= ( )−×

1.1 0.9125 2.1

= 7.619 × 10–4 m2/kN

Q.52 The reinforced concrete section, the stress at extreme fibre in compressionis 5.8 MPa. The depth of Neutral Axis in the section is 58 mm and gradeof concrete is M25. Assuming Linear elastic behavior of the concrete, theeffective curvature of the section (in per mm) is(a) 2 × 10–6 (b) 3 × 10–6

(c) 4 × 10–6 (d) 5 × 10–6

Ans. (c)

M25 conrature

X = 58 mmu

5.8 MPaB

d



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Page30

Modulus of elasticity of concrete

E = = 25000 25 25000 N/mm

∴MI =

σ = Ey R

⇒1R =

σ = =×u

5.8 5.8Ey EX 58 25000

= 4 × 10–6 Per mm⇒ Curvature = 4 × 10–6 per mm

Q.53 A venturimeter having diameter of 7.5 cm at the throat and 15 cm at theenlarged end is installed in a horizontal pipeline of 15 cm diameter. The pipecarries incompressible fluid at steady rate of 30 l/s. The difference of pressurehead measured in terms of the moving fluid in between the enlarged andthe throat of the vent is observed to be 2.45 m. Taking the g = 9.8 m/s2,the coefficient of discharge of venturimeter (correct upto 2 decimal) is_________.

Sol.

2

21

1

7.5 cm φ15 cm φ

Q = 30 l/s

⎛ ⎞ ⎛ ⎞+ − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠1 2

1 2P PZ Zw w = 2.45 m = h

g = 9.81

Q =−

1 2d 2 2

1 2

A AC 2ghA A

Cd =

−1 2

2 21 2

QA A 2gh

A A



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=( ) ( )

( ) ( )

−×π ×

× × ×−

3 3

2 2

2 2

30 30 m /s

0.15 0.0754 2 9.81 2.45

0.15 0.075

Cd = 0.95

Q.54 A traffic surveying conducted on a road yield an average daily traffic countof 5000 vehicle. The axle load distribution on the same road is given in thefollowing table.

18 1014 2010 358 156 20

Axle load (tones) Frequency of traffic (f)

The design period of the road is 15 years. The yearly Traffic growth rateis 7.5% the load safety factor (LSF) is 1.3. If the vehicle damage factor (VDF)is calculated from the above data, the design traffic (In million standard axleload MSA) is _______ .

Sol.Calculation of vehicle damage factor

VDF =⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

+ + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠+ + + +

4 y 4 4 43 51 2 4

1 2 3 4 5s 3 s s s

1 2 3 4 5

W WW W WV V V V VW W W W WV V V V V

where Ws = standard axle load = 80 kN = 8.2 tonn

⇒ VDF =⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + × + ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

+ + + +

4 4 4 4 418 14 10 8 610 20 35 15 208.2 8.2 8.2 8.2 8.210 20 35 15 20

= 4.989

⇒ N =⎡ ⎤× −⎣ ⎦ ×

15365 5000 (1.075) 14.989

0.075

= 237.806 MSA



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Page32

Q.55 An incompressible fluid is flown at steady rate in horizontal pipe. From asection the pipe divides into two horizontal parallel pipes of diameter(d1 and d2) that run, for a distance of L each and then again join back toa pipe of the original size. For both the pipes, assume the head loss dueto friction only and the Darcy-weisbach friction factor to be same. Thevelocity ratio between bigger and smaller branched pipe is _________.

Sol.

d1Q1

C

D

Q2d2

A B

Ld1 = 4d221

1

flv2gd =

22

2

flv2gd

⇒21

1

Vd =

22

2

Vd

2122

VV

= =1

2

d 4d

1

2

vv = 2

Section - II (General Aptitude)One Mark Questions

Q.56 Rajan was not happy that Sajan decided to do the project on his own onobserving his unhappiness, sajan explained to Rajan that he preferred towork independently.Which one of the statement below is logically valid and can be inferred fromthe above sentences?(a) Rajan has decided to work only in group.(b) Rajan and Sajan were formed into a group against their wishes.(c) Sajan decided to give into Rajan’s request to work with him.(d) Rajan had believed that Sajan and he would be working together.

Ans. (d)



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Page33

Q.57 A boundary has a fixed daily cost of Rs 50,000 whenever it operates andvariable cost of Rs. 8000 Q, where Q is the daily production in tonnes. Whatis the cost of production in Rs. per tonne for a daily production of 100 tonnes.

Sol.Total cost for 100 tonne

= 8000 × 100 + 50000 = 850000

Cost per tonne = =850000 Rs.100

8500/tonne.

Q.58 Choose the most appropriate word from the option given below to completethe following sentences, one of his biggest _______ was his ability to forgive.(a) Vice (b) Virtues(c) Choices (d) Strength

Ans. (b)

Q.59 A student is required to demonstrate a high level of comprehension for thesubject, especially in the social sciences.The word closes in meaning to comprehension is(a) Understanding (b) Meaning(c) Concentration (d) Stability

Ans. (a)

Q.60 Find the odd one in the following group ALRVX, EPVZB, ITZDF, OYEIX(a) ALRVX (b) EPVZB(c) ITZDF (d) OYEIX

Ans. (d)

Two Marks Questions

Q.61 One percent of the people of country X are taller than 6 ft, 2 percent ofthe people of country Y are taller than 6 ft. There are thrice as many peoplein country X as in country Y. Taking both countries together. What % ofpeople are taller than 6 ft?(a) 3 (b) 2.5(c) 1.5 (d) 1.25



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Page34

Ans. (d)Let the population of county Y is P∴ Population of country X is 3P

% of people taken than 6 ft =

× ×+

+

3P 1 P 2100 100

3P P

= 1.25

Q.62 With reference to the conventional certesion (X, Y) the vertices of a trianglehave x1, y1 = 1, 0, x2, y2 = 2, 2, and x3, y3 = 4, 3, the area of triangle is

(a)32 (b)

34

(c)45 (d)

52

Ans. (a)

x

y

A(1,0)

B(2,2)

C(4,3)a

bc

Area of triangle is

A = ( ) ( ) ( )− − −p p a p b p c

when P =+ +a b c

2

a = ( ) ( )− + − =2 24 2 2 1 5

b = ( ) ( )− + =2 24 1 3 3 2

c = ( ) ( )− + =2 22 1 2 5

p =+ + = +5 5 3 2 35

2 2

A =⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

3 3 3 35 5 3 2 5 3 2 5 52 2 2 2

= 32

www.madeeasy.inWrite us at [email protected] | Phone: 011-45124612, 9958995830

GATE-2014

Detailed Solutions

Civil Engineeringof

Evening Session

Expert Opinion

Dear Students,

The Questions of GATE 2014 are based on fundamental

and basic concepts of the syllabus. There is no ambiguity

and misprint noticed till now, however, it is an observation based

on students feedback.

The level and standard of GATE 2014 questions are relatively easier

than the exam of GATE 2013. There are 3 important observations

made by me about GATE 2014 exam.

1. The GATE 2014 exam is conducted in two seating i.e. morning session and afternoon

session. The question papers of both seatings are different. The difficulty level of questions

are nearly same and due care has been taken to balance both the papers, however small

differences are certainly there. The morning session paper seems to be little easier by 2

to 5%, however, it varies on the perception of person to person also.

The average marks of both the papers should be equated and necessary scaling criteria

should be adopted for this purpose.

2. The GATE 2014 cut-off will be nearly same as that of last year, perhaps it may be little

lesser than that of GATE 2013. GATE-2013 cutoff was 33 marks. Though the paper of GATE

2013 was tougher and number of students were less, 6 marks questions were wrongly

framed and hence, these 6 marks were awarded to all the candidates, which was certainly

a kind of bonus.

Therefore expected cut-off for GATE 2014 may be between 30 to 34 marks (General category).

It may be noted that the following formulae is used to evaluate GATE cut-off marks.

GATE Cutoff =

Total Marks obtained by all the candidates

Total number of candidates

3. The topper’s marks in GATE 2013 was nearly 83 in which 6 marks of bonus to all are

included. In my opinion topper’s marks in GATE 2014 will be between 80 to 85.

GATE cutoff <| 25 Marks

DisclaimerDear Students, MADE EASY has taken due care in collecting the data and questions. Since ques-tions are submitted by students and are memory based, therefore the chances of error can not be ruled out. Therefore MADE EASY takes no responsibility for the errors which might have incurred.

If any error or discrepancy is recorded then students are requested to inform us at: [email protected]

B. Singh (Ex. IES)CMD, MADE EASY Group

Super Talent Batches

announcing

Super Talent Batchesat Kalu Sarai premise of Delhi

st Batch : Commencing from May 20th Morning Batch

nd Batch : Commencing from June 15th Evening Batch

Eligibility

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• Better Teaching Environment

• Extra teaching hours

• In-depth coverage of subjects

(Any of the following)

Civil Engineering


GATE-2014 Exam SolutionsCivil Engineering (Evening Session)

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Page1

Section - I (Civil Engineering)One Mark Questions

Q.1 If Gs = 2.7%, n = 40%, w = 20% then the degree of saturation is __________.

Sol.Se = wG

and e = =−n 0.67

1 n

⇒ S = ×2.71 200.67 = 81.3%

Q.2 The determinant of matrix is __________.0 1 2 31 0 3 02 3 0 13 0 1 2

Sol.

Δ =

0 1 2 31 0 3 02 3 0 13 0 1 2

R4 → R4 – R2 – R3

Δ =− −

0 1 2 31 0 3 02 3 0 10 3 2 1

R4 → R4 + 3R1

Δ =

0 1 2 31 0 3 02 3 0 10 0 4 10

R3 → R3 – 3R1

Δ = − −

0 1 2 31 0 3 02 0 6 80 0 4 10



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Page2

Interchanging column 1 and column 2 and taking transpose

Δ =−

−−

1 0 0 00 1 2 02 3 6 43 0 8 10

=− × −

−

1 2 01 3 6 4

0 8 10

= –1 × {1(–60 + 32) + 2(0 – 30)}= –(–28 – 60) = 88

Q.3 The static indeterminacy of two span continuous beam with internal hingeis __________.

Internal Hinge

Sol.

Internal Hinge

Number of member, m = 4Number of external reaction, re = 4Number of joint, j = 5Number of reaction released, rr = 1Degree of static indeterminacy,

Ds = 3 m + re – 3j – rr= 3 × 4 + 4 – 3 × 5 – 1= 0

Q.4 A plane blow leak velocity component = = −1 2

x yu , vT T and w = 0 along x, y

and z direction respectively where T1(≠ 0), T2(≠ 0) are constants havingdimension of time. The given blow is incompressible if



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Page3

(a) T1 = –T2 (b) = − 21

TT2

(c) = 21

TT2

(d) T1 = T2

Ans. (d)For a flow to exist

⇒∂ ∂+∂ ∂u vx y = 0

⇒ −1 2

1 1T T = 0

⇒ T1 = T2

Q.5 Groups-I contains representative stress-strain curves as shown in figure.While Group-II gives the list of materials. Match the stress-strain curveswith corresponding materials.

Stress

Strain

JK

L

Group-I Group-IIP. Curve J 1. Cement pasteQ. Curve K 2. Coarse aggregateR. Curve L 3. Concrete(a) P-1, Q-3, R-2 (b) P-3, Q-1, R-2(c) P-2, Q-3, R-1 (d) P-3, Q-2, R-1

Ans. (d)

Q.6 As per IS : 456-2000 in design of concrete target mean strength is takenas(a) fck + 0.825σ (b) fck + 1.64σ

(c) fck + 0.50σ (d) fck + 0.725σ

Ans. (b)



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Page4

Q.7 Modulus of elasticity of concrete is calculated (as per IS 456 : 2000) by(a) Secent modulus (b) Tangent modulus(c) Initial tangent modulus (d) None of these

Ans. (a)

Q.8 The flexural tensile strength of M 25 grade of concrete in N/mm2, as perIS:456-2000, is ________.

Sol.

Flexural strength = ck0.7 f

= 3.5 N/mm2

Q.9 Survey which is conducted for geological features like river, natural resources,building, cities etc. is denoted as(a) Land survey (b) Geological survey(c) Engineering survey (d) Topographical survey

Ans. (a)

Q.10 The integrating factor for the differential equation −+ = tk t2 1 0

dP k P k L edt

is

(a) − 1k te (b) − 2k te

(c) 1k te (d) 2k teAns. (d)

I.P. = ∫ 2k dte

Q.11 Polar moment of inertia (Ip) in cm4 at a rectangular section having widthb = 2 cm and depth d = 6 cm is ___________ .

Sol.

Polar moment of inertia, Ip = Ix + Iy = +3 3bd db

12 12

= ( )+2 2bd b d12 = ( )× +2 22 6 2 6

12 = 40 cm4



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Q.12 The average spacing between vehicles in a traffic stream is 50 m, then thedensity (in veh/km) of stream is ___________.

Sol.

Capacity = × = ×1000 V V densityS

⇒ Density = 1000S = 20 veh/km

Q.13 A fair (unbiased) coin was tossed 4-times in a succession and resulted inthe following outcomes (I) H (II) H (III) H (IV) H. The probability of obtaininga “TAIL’ when the coin is tossed again is

(a) 0 (b) 12

(c) 45

(d) 15

Ans. (b)

P(E) = ( )( )

n En S

n(s) = [{H}, {T}] = 2n(E) = {(T)} = 1

∴ P(E) = 12

Q.14 As per ISSCS (IS : 1498 -1970) an expression of A-line is(a) Ip = 0.73 (WL – 20) (b) Ip = 0.70 (WL – 20)(c) Ip = 0.73 (WL – 10) (d) Ip = 0.70 (WL – 10)

Ans. (a)

Q.15 The contact pressure for a rigid footing resting on clay at the centre andthe edges are respectively(a) maximum and zero (b) maximum and minimum(c) zero and maximum (d) minimum and maximum



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Ans. (d)

Stress distribution for cohesive soilMinimum at centre and maximum at edge.

Q.16 The clay mineral primarly governing the swelling behaviour of black cottonsoil is(a) Halloysite (b) Illite(c) Kaolinite (d) Montmorillonite

Ans. (d)

Q.17 Dominating micro-organisms in Active Sludge Reactor process.(a) Aerobic heterotrops (b) Anaerobic heterotrops(c) Autotrops (d) Phototrops

Ans. (a)

Two Marks Questions

Q.18 A rectangular channel of 2.5 m width is carrying a discharge of 4 m3/s.Considering that acceleration due to gravity as 9.81 m/s2, then velocity offlow (in m/s) corresponding to the critical depth (at which the specific energyis minimum) is ________ .

Sol.Q = 4 m3/s, B = 2.5 m

⇒ q = 4/2.5 m3/s/m = 1.6 m3/s/m

yc3 =

2qg

⇒ yc = 0.639 m



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At critical depth velocity head = cy2

⇒2V

2g = 0.6392

⇒ V = ×0.639 g

= 2.504 m/s

Q.19 A waste water stream (flow = 2 m3/s, ultimate BOD = 90 mg/l) is joininga small river (flow = 12 m3/s, ultimate BOD = 5 mg/l). Both water streamsget mixed up instantaneously. Cross-section area of the river is 50 m2.Assuming the de-oxygenation rate constant, k′ = 0.25/day. The BOD (in mg/l)of the river water 10 km downstream of the mixing point is(a) 1.68 (b) 2.63(c) 15.46 (d) 1.37

Ans. (c)Flow of waste water stream, Qw = 2 m3/secUltimate BOD, Yw = 90 mg/l = 90 gm/m3

Flow of river, QR = 12 m3/secUltimate BOD of river,

YR = 0.5 mg/l = 5 gm/m3

BOD of mixture, Y0 = ( ) ( )× + ×+

2 90 12 52 12

= + = = 3180 60 240 17.143 gm/m14 14

kD = 0.434 K = 0.434 ×0.25 = 0.1085

Yt = ( )−⎡ ⎤−⎣ ⎦Dk t

0Y 1 10 ...(i)

Area of river = 50 m2

Flow of river = 12 m3/sec

Stream velocity = =14 0.28 m /sec50

Time taken, t = =10km 9.921 days0.28m /s

∴ From eq. (i) Yt = ( )− ×⎡ ⎤−⎣ ⎦0.1085 9.92117.143 1 10

= 15.70 gm/m3 = 15.70 mg/lit



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Q.20 The beam of an overall depth 250 mm (shown below) is used in a buildingsubjected to two different thermal environments. The temperature at thetop and the bottom surfaces of the beam are 36°C and 72°C respectively.Considering coefficient of thermal expansion (α) as 1.50 × 10–5/°C, thevertical deflection of the beam (in mm) at its mid span due to temperaturegradient is _________.

1.5 m 1.5 m

36°C

72°C

250

mm

Sol.

L/2 L/2

σ

R =αhT

From properties of circle

(2R – δ)δ = ×L L2 2

(Considering ‘δ’ very small so neglect δ2)

⇒ δ =2 α=

2L L T8R 8h

1.5 m 1.5 m

36°C

72°C

250

mm



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Given, α = 1.50 × 10–5/°C

⇒ δ = α 2TL8h

=( )

( )−

−

× × ° − ° ×

× ×

5 2

31.50 10 72 36 3

8 250 10

= 2.43 mm

Q.21 The expression α

α→

−α0

x 1lim is equal to

(a) log x (b) 0(c) x log x (d) ∞

Ans. (a)α

α→

−α0

x 1lim ⎡ ⎤⎢ ⎥⎣ ⎦0 form0

Use L-Hospital Rule

α→0lim

α −α

lnxe 1

α→0lim

α lnxe ln x1= ln x

Q.22 Match the List-I (Soil exploration) with List-II (Parameters of subsoilstrength characteristic) and select the correct answer from the codes givenbelow:

List-I List-IIP. Pressure meter test (PMT) 1. Menard’s method (Em)Q. Static cone penetration (SCPT) 2. Number of blows (N)R. Standard penetration (SPT) 3. Skin resistance (fc)S. Vane shear test (VST) 4. Undrained cohesion (Cu)(a) P-1, Q-3, R-2, S-4 (b) P-1, Q-2, R-3, S-4(c) P-2, Q-3, R-4, S-1 (d) P-4, Q-1, R-2, S-3

Ans. (a)



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Q.23 The suspension of sand like particles in water with particles of diameter0.10 mm and below is flowing into a settling tank at 0.10 m3/s. Assumeg = 9.81 m/s2, specific gravity of particles = 2.65, and kinematic viscosity ofwater = 1.0105 × 10–2 cm2/s. The minimum surface area (in m2) requiredfor this settling tank to remove particles of size 0.06 mm and above with100% efficiency is ________.

Sol.

Vs = ( )−ν

2sG 1 gd18

=( ) ( )−

−

− × × ×

× ×

22

2

2.65 1 9.81 6 1018 1.0105 10

=−

−× × ×

× × ×

4

2 21.65 9.81 36 10

18 1.0105 10 10= 3.204 × 10–3 m/s

Vs = QBL

⇒ BL = −× 30.1

3.204 10= 31.214 m2

Q.24 The values of axial stress (σ) in kN/m2, bending moment (M) in kNm andshear force (V) in kN acting at point P for the arrangement shown in figureare respectively

50 kN

3 m

FrictionlessPulley

Cable

Beam(0.2 m × 0.2 m) Q

P

(a) 1000, 75 and 25 (b) 1250, 150 and 50(c) 1500, 225 and 75 (d) 1750, 300 and 100



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Ans. (b)Loading after removing the cable

50 kN50 kN

50 kN

Axial stress =×50

0.2 0.2 = 1250 kN/m2

Bending moment = 50 × 3 = 150 kNm

Q.25 Considering the symmetry of a rigid frame as shown, the magnitude of thebending moment (in kNm) at P (Preferably using the moment distributionmethod) is

8 m 8 m

6 m

4Ic 4IcP

IcIc Ic

24 kN/m

A

B C

DE

(a) 170 (b) 172(c) 176 (d) 178

Ans. (c)Distribution Factor

Joint Member RS TRS D.F.

BBA

BP

P

PB

PC

PE

CCP

CD

I6

4I8

I6

4I8

4I8

I6

4I8

2 I3

7 I6

2 I3

1434371737

3414



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Fixed End Moment

ABM = BAM = PEM = EPM = CDM = DCM = 0

BPM = ( )×− = −

224 8 128 kNm12

PBM = +128 kNm

PCM = –128 kNm

CPM = 128 kNmDistribution Table

1 34 4

3 37 7

3 14 4

17

A

0 0

B P C D P E

32 96 0

–128 128 –128 128 0 0 0 0

0 –96 –32 0

016 32 –32 48 0–48 –16

16 32 –32 176 32–176 –16 0 0Final

32 kNm

32 kNm

88 kNm 176 kNm 88 kNm

32 kNm32 kNm

16 kNm 16 kNm

P

+ +

–

–

+– – +

–

Q.26 An effluent at a flow rate of 2670 m3/d from a sewage treatment plant isto be disinfected. The laboratory data at disinfected studies with a chlorinedosage of 15 mg/l yield the model Nt = N0e–0.145t where Nt = number of microorganism surviving at time t (in min) and N0 = number of microorganism,



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present initially (at t = 0). The volume of disinfection unit (in m3) requiredto achieve a 98%. Kill of M.O. is _______ .

Sol. (50 m3)

Q.27 A horizontal nozzle of 30 mm diameter discharges a study Jet of water intothe atmosphere at a rate of 15 litres/second. The diameter of inlet to thenozzle is 100 mm. The Jet impinges normal to a flat stationary plate heldclose to the nozzle end. Neglecting air friction and considering density ofwater as 1000 kg/m3, the force exerted by the Jet (in N) on the plateis_________ .

Sol.Force = ρaV2

Q = aV = 15 litre/sec (given)

⇒ Force = ρ2Q

a

= −−× × ×π ×

2 66 2

3 2 62

15 kg m 1100 10 mm s 1030

4

= ×π1100 N = 31.83 N

Q.28 On a section of a highway the speed density relations is linear and is given

by ⎡ ⎤ν = −⎢ ⎥⎣ ⎦280 k3

; where ν is in km/hr and k in vehicle/km. The capacity (in

veh/hr) of this section of the highway would be(a) 1200 (b) 2400(c) 4800 (d) 9600

Ans. (b)

v = − 2k803

Capacity, q = v × k

= −22k80k

3

For q to be maximum dqdk = 0

⇒dqdk = − =4k80 0

3



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⇒ k = 60

Maximum capacity, q = ( )× − 2280 60 603

= − ×24800 36003

= 2400

Q.29 Irrigation water is to be provided to a crop in a field to bring the moisturecontent of the soil from the existing 18% to the field capacity of the soilat 28%. The effective root zone of the crop is 70 cm. If the densities of thesoil and water are 1.3 g/cm3 and 1 gm/cm3 respectively, the depth of theirrigation water (in mm) required for irrigating the crop is __________ .

Sol.Given,Root zone depth, d = 70 cmField capacity, FC = 28%Existing moisture content, w = 18%Density of soil, γ = 1.3 gm/cm3

Density of water, γw = 1.0 gm/cm3

Depth of irrigation water required,

dw = ( )γ −γ c

wd F w

= ( )( )× × −1.3 70 10 28% 18%1.0

= ( )× × × 101.3 70 10100

= 91 mm

Q.30 The rank of the matrix ⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− −⎣ ⎦

6 0 4 42 14 8 18

14 14 0 10 is _______

Sol.

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− −⎣ ⎦

6 0 4 42 14 8 18

14 14 0 10R3 → R3 – 2R1 + R2



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

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− + − − − + − + − − +⎣ ⎦

6 0 4 42 14 8 18

14 2 6 2 14 2 0 14 0 2 4 8 10 2 4 18

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

6 0 4 42 14 8 18

0 0 0 0

Determinant of matrix ⎡ ⎤⎢ ⎥−⎣ ⎦

6 02 14

is not zero.

∴ Rank is 2.

Q.31 An infinitely long slope is made up of a C-φ soil having the propertiescohesion (C) = 20 kPa and dry unit weight (γd) = 16 kN/m3. The angle ofinclination and critical length of slope are 40° and 5 m respectively. Tomaintain the limiting equilibrium, the angle of internal friction of soil (indegree) is ________ .

Sol.As the given slope is in dry condition, therefore, factory of safety shouldbe more than 2.

Fs = φ ≥tan 2tan b and β = 40°

⇒ tanφ ≥ 2tan (40°)⇒ tanφ ≥ 1.6782⇒ φ ≥ 59.21°

Q.32 A student riding a bicycle on a 5 km one way street takes 40 minutes toreach home. The student stopped for 15 minutes during this ride. 60 vehiclesover took the student (Assume the number of vehicle overtaken by thestudent is zero) during the ride and 45 vehicles while the student stopped.The speed of vehicle stream on that road (in km/hr) is(a) 7.5 (b) 12(c) 40 (d) 60

Ans. (d)

Velocity of student = ( ) =−5km 12km/hr

40 15 min



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⎛ ⎞⎜ ⎟⎝ ⎠ moving

Vehicle/minRelative speed of vehicle w.r.t. student =

⎛ ⎞⎜ ⎟⎝ ⎠ standing

Vehicle/minRelative speed of vehicle w.r.t. student

⇒−

60 / 25x 12 =

−

54 /15x 0

⇒ x = 60 km/hr

Q.33 In a Marshall sample, the bulk specific gravity of mix and aggregates are2.324 and 2.546 respectively. The sample includes 5% of bitumen (by totalwt. of mix) of specific gravity 1.10. The theoretical maximum specific gravityof mix is 2.441. The void filled with bitumen (VFB) in the Marshall sample(in %) is ________.

Sol.

Vv = − ×t m

m

G G 100G

= − × =2.441 2.324 100 5.03%2.324

Vb = = × =bm

b

w 5G 2.324 10.564G 1.1

VMB = Vv + Vb = 15.594%

VFB = × = ×bV 100 10.564 100VMB 15.594

= 67.74%

Q.34 With reference to a standard Cartesian (x, y) plane, the parabolic velocitydistribution profile of fully developed laminar flow in x-direction betweentwo parallel plates. Stationary and identical plates that are separated bydistance, h, is given by the expression

⎡ ⎤⎛ ⎞= − −⎢ ⎥⎜ ⎟⎝ ⎠μ ⎢ ⎥⎣ ⎦

22h dP yu 1 48 dx h

In this equation the y = 0 axis lies equidistant between the plates at adistance h/2 from the two plates, P is the pressure variable and μ is thedynamic viscosity term, the maximum and average velocities are respectively



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(a) = −μ

2

maxh dPU8 dx and =avg max

2U U3

(b) =μ

2


2U U3

(c) = −μ

2


3U U8

(d) =μ

2


3U U8

Ans. (a)

h

h/2

y

x

Velocity expression for a laminar flow between two parallel plates is

U =⎡ ⎤⎛ ⎞ ⎛ ⎞− −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠μ ⎢ ⎥⎣ ⎦

22h dP y1 48 dx h

End condition, U = Umax at y = 0

⇒ Umax =⎛ ⎞− ⎜ ⎟⎝ ⎠μ

2h dP8 dx

Discharge, dQ = Area × Velocity

⇒ dQ = ( )⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞⎢ ⎥− − ×⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠μ⎢ ⎥⎝ ⎠⎣ ⎦

22h dP y1 4 dy 18 dx h

⇒ Q = −

⎛ ⎞⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠μ ⎝ ⎠∫2 2h/2

2h/2h dP 4y1 dy8 dx h

=−

⎡ ⎤⎛ ⎞− −⎢ ⎥⎜ ⎟⎝ ⎠μ ⎢ ⎥⎣ ⎦

h/22 3

2h/2

h dP 4yy8 dx 3h

=⎡ ⎤⎧ ⎫⎛ ⎞⎛ ⎞⎧ ⎫ ⎪ ⎪⎛ ⎞ ⎛ ⎞⎢ ⎥− − − − − −⎨ ⎬ ⎨ ⎬⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠μ ⎢ ⎥⎝ ⎠⎩ ⎭ ⎝ ⎠⎪ ⎪⎩ ⎭⎣ ⎦

2 3 3

2h dP h h 4 h h8 dx 2 2 8 83h



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= ⎛ ⎞− ⎜ ⎟⎝ ⎠μ

3h dP12 dx

∵ Q = AV

⎛ ⎞− ⎜ ⎟⎝ ⎠μ

3h dP12 dx = (h × 1) × Uavg

⇒ Uavg =⎛ ⎞− ⎜ ⎟⎝ ⎠μ

2h dP12 dx

avg

max

UU =

⎛ ⎞− ⎜ ⎟⎝ ⎠μ = =⎛ ⎞− ⎜ ⎟⎝ ⎠μ

2

2

h dP8 212 dx

12 3h dP8 dx

∴ Uavg = max2 U3

Q.35 A venturimeter having a throat diameter of 0.1 m is used to estimate theflow rate of a horizontal pipe having a dia of 0.2 m for an observed pressuredifference of 2 m of water head and coefficient of discharge equal to unity,assuming that the energy losses are negligible. The flow rate (in m3/s)through the pipe is approximately equal to(a) 0.500 (b) 0.150(c) 0.050 (d) 0.015

Ans. (c)Diameter of throat, d = 0.1 mDiameter of pipe, D = 0.2 m

Pressure difference, −1 2P Pw = 2 m = h

d = 0.01 m

D =

0.2

m

1

1

2

2

Coefficient of discharge, CD = 1

Discharge, Q =−

D 1 22 21 2

C .A A rghA A



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A1 = ( )π 20.24

A2 = ( )π 20.14

⇒ Q =( ) ( )

( ) ( )

π⎛ ⎞× × × ×⎜ ⎟⎝ ⎠π⎛ ⎞ −⎜ ⎟⎝ ⎠

22 2

4 4

1 0.2 0.1 2 9.81 24

0.2 0.14

= 0.0508 ≈ 0.050 m3/sec

Q.36 A prismatic beam (shown) has plastic moment capacity of Mp, then thecollapse lead P of the beam is

L/2 L/2 L/3

P P/2

(a) p2ML (b) p4M

L

(c) p6ML (d) p8M

L

Ans. (c)Here degree of static indeterminacy = 0∴ Number of plastic hinges required for mechanical

= Ds + 1 = 0 + 1 = 1

θθ

Mp Mp

θ

P/2

P



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From principal of virtual work

− θ − θ + θ − × θp pL P LM M P2 2 3 = 0

⇒ − θ + θ − θpPL PL2M2 6

= 0

⇒ 2Mp = ( )−× = =

3 1 PLPL PL 1 PL2 6 6 3

⇒ P = 6MpL

Q.37 The axial load (in kN) in the member PQ for the assembly/arrangementshown in figure given below is ________.

2 m

2 m 2 m

160 kN

S RQ

P

Sol.Free body diagramFor principle of superposition

2 m 2 m

160 kN

SQ

VQ

VQ

× ×+ × −33 2

QV L160 2 160 2 23EI 2EI 3EI = QV L

AE ...(i)



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Deflections due to axial forces will be very less as compared to bendingforces.So we can neglect the axial deformation.∴ From equation (i)

× ×+ × −33 2

QV 4160 2 160 2 23EI 2EI 3EI = 0

⇒× × ×+

3 2160 2 160 2 23 2 =

× 3QV 4

3

⇒ VQ = 50 kN

Q.38 Water is flowing at a steady rate through a homogeneous and saturatedhorizontal soil strip at 10 meter length. The strip is being subjected to aconstant water head (H) of 5 m at the beginning and 1 m at the end. If the

governing equation of the flow in the soil strip is =2

2d H 0dx

(where x is the

distance along the soil strip), the value of H (in m) at the middle of thestrip is __________ .

Sol.Given, equation of the flow of soil strip is

2

2d Hdx = 0

⇒dHdx = C1

⇒ H = C1x + C2at x = 0, H = 5 m⇒ C2 = 5⇒ H = C1x + 5at x = 10, H = 1 m⇒ 1 = C1 × 10 + 5⇒ 10C1 = –4

⇒ C1 = − 25

⇒ H = − +2 x 55



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at x = 5 m

H = − × +2 5 55

H = 3 m

Q.39 An observer counts 240 veh/hr at a specific highway location. Assume thatthe vehicle arrival at the location is Poisson distributed, the probability ofhaving one vehicle arriving over 30 sec time interval is _______.

Sol.

P(n, t) = ( )−λ λ nte . tn!

Here, λ = no. of vehicles = 240 vehicle/km

P(1, 30) =

− ×⎛ ⎞×⎜ ⎟⎝ ⎠

240 30 13600 240e 303600

1!

= 2.e–2

= 0.2707

Q.40 A tachometer was placed at point P to estimate the horizontal distance PQand PR. The corresponding stadia intercept with the telescope kept horizontalare 0.320 and 0.210 m respectively. The ∠QPR is measured to be 61°30′30′′.If the stadia multiplication constant = 100 and stadia addition constant =0.10 m, the horizontal distance (in m) between the point Q and R is ________.

P

Q

R

Sol.PQ = ks + C

= 100(0.32) + 0.1 = 32.1 mPR = ks + C

= 100(0.21) + 0.1 = 21.1 m



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Applying the cosine rule

QR = ( )( )+ + θ2 2PQ PR 2 PQ PR cos

Where, θ = 61°30′30′′ = 61.508°

= ( )( )+ + °1030.41 445.21 2 32.1 21.1 cos61.508

= + +1030.41 445.21 696.2= 46.06 m

Q.41 For the state of stresses (in MPa) shown in the figure below, the maximumshear stress (in MPa) is

2

4

2

4

4

4

Sol.

τmax =σ − σ⎛ ⎞σ − σ = + τ⎜ ⎟⎝ ⎠

x y 21 2xy2 2

= 5.0 MPa

Q.42 The tension (in kN) in a 10 m long cable shown in figure neglecting its selfweight is

3 m 3 m

120 kN

P QS

yCable Cable

R



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(a) 120 (b) 75(c) 60 (d) 45

Ans. (b)

3 m 3 m

120 kN

P QS

R

θθ

4 m5 m5 m

Freebody diagram

120 kN

T TS

θθ

4 m5 m5 m

ΣFy = 0⇒ 2Tcosθ = 120 ...(i)

Here, cosθ = 45

× 42T5 = 120

⇒ T = × =×

120 5 75 kN2 4

Q.43 A pre-timed four phase signal has critical lane flow rate for the first threephases as 200, 187 and 210 veh/hr with saturation flow rate of 1800 veh/hr/lane for all phases. The lost time is given as 4 seconds for each phase.If the cycle length is 60 seconds. The efficiency gree time of 4th phase is____________ (in seconds).



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Sol.Flow rates for the first three phase are given as

q1 = 200 veh/hrq2 = 187 veh/hr

and q3 = 210 veh/hrSaturation flow rate is 1800 veh/hr/laneLost time, L = 4 × 4 = 16 seclength of the cycle, C0 = 60 sec

Now, y1 = =1

1

q 200s 1800

y2 = =2

2

q 187s 1800

y3 = =3

3

q 210s 1800

C0 = +−

1.5L 51 y

⇒ 60 = × +−

1.5 16 51 y

⇒ 60 = +−

24 51 y

⇒ 1 – y = 2960

⇒ y = − −=1 29 60 2960 60

= 0.517And y = y1 + y2 + y3 + y4

⇒ 0.517 = + 4597 y

1800⇒ y4 = 0.185

G = ( )−40

y C Ly

= ( )−0.185 60 160.517

= 15.745 sec



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Q.44 A single vertical friction pile of dia 500 mm and length 20 m is subjectedto a vertical compressive load. The pile is embedded in a homogeneous sandystratum where; angle of internal friction (φ) = 30°, dry unit weight(γd) = 20 kN/m3 and angle of wall friction (δ) = 2φ/3. Considering the coefficientof lateral earth pressure (k) = 2.7 and the bearing capacity factor (Nq) = 25,the ultimate bearing capacity of the pile is ________

Sol.

20 m

500

Homogeneous sandy stratumφ = 30°

γd = 20 kN/m3

Wall friction angle, δ = φ = × = °2 2 30 203 3

Lateral earth pressure coefficient,K = 2.7

Bearing capacity factor,Nq = 25

Ultimate load capacity,Qu = ?

Vertical effective stress at 20 m,

σv = 20 × 20 = 400 kN/m2

From 0 to 20 m, unit point bearingresistance and skin friction resistance remain constant at

σv = 400 kN/m2

The ultimate load capacity is given byQu = qup⋅Ap + qs⋅As

= Qup + fs

where qup = σv qN= 400 × 25= 10000 kN/m2

and qs = ⋅ σ ⋅ ⋅ δv1 K tan2



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= × × °1 400 2.7tan202

= 196.54 kN/m2

Skin friction resistance,fs = 196.54 × π × 0.5 × 20

= 6174.49 kN

Qup =π×

2

up(0.5)q

4

= π×2(0.5)10,000

4= 1963.49 kN

∴ Ultimate load on the pile= 1963.49 + 6179.49= 8137.98 kN

Q.45 The chainage of the intersection point of two straights is 1585.60 meter andthe angle of inter section is 140°. If the radius of a circular curves is 600meter, the tangent distance (in m) and length of the curve (in m) respectivelyare(a) 418.88 and 1466.08 (b) 218.38 and 1648.49(c) 218.38 and 418.88 (d) 418.88 and 218.38

Ans. (c)

V¢

(P.I)V

T (P.C)1 T (P.T)2

R

D

C

Δ/2 Δ/2

Δ = 40°

R

O

Δ = 180° – 140° = 40°



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Length of the curve = π Δ°

R180

= π × × =600 40 418.82m180

Tangent distance (T) is the distance between P-C to P.I (also the distancefrom P.I to P.T)

⇒ T = T1V = T2V = Δ Δ=1OT tan R tan2 2

= 600 tan 20°= 218.88 m

Q.46 A surface water treatment plant operates round the clock with a flaw rateof 35 m3/min. The water temperature is 15°C and Jar testing indicated andalum dosage of 25 mg/l with flocculation at a Gt value of 4 × 104 producingoptimal results. The alum quantity required for 30 days (in kg) of operationof the plant is _________ .

Sol.Given dataFlow rate, Q = 35 m3/min

Gt = 4 × 104

Alum dosage = 25 mg/lit.Alum quantity (kg) required for 30 days

= 35 × 103 × 60 × 24 × 30 × 25 × 10–6

= 37,800 kg

Section - II (General Aptitude)One Mark Questions

Q.47 The population of a new city is 5 million and is growing at a rate of 20%annually. How many year would it take to double at this growth rate.(a) 3-4 year (b) 4-5 year(c) 5-6 year (d) 6-7 year



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Ans. (a)

Pnew = ⎛ ⎞+⎜ ⎟⎝ ⎠

n

oldrP 1

100

⇒ 10 = ⎛ ⎞+⎜ ⎟⎝ ⎠

n205 1100

⇒ 2 = (1.2)n

⇒log2

log1.2 = n

⇒ n = 3.8 year

Q.48 A person affected by Alzheimers disease _______ short term memory loss.(a) experiences (b) has experienced(c) is experiencing (d) experienced

Ans. (a)

Q.49 Select the closest in meaning“As a women, I have no country”(a) Women have no country.(b) Women are not citizens of any country.(c) Women solidarity knows no national boundary.(d) Women of all country have equal legal rights.

Ans. (c)

Two Marks Questions

Q.50 If X is 1 km North east of Y, Y is 1 km South east of Z and W is 1 kmWest of Z. P is 1 km South of W, Q is 1 km East of P. Then find the distancebetween X and Q.

(a) 1 (b) 2

(c) 3 (d) 2



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Ans. (c)

Y(0, 0)

⎛ ⎞−⎜ ⎟⎝ ⎠1 1Z ,2 2

⎛ ⎞− −⎜ ⎟⎝ ⎠1 1W 1 ,2 2

⎛ ⎞− − − +⎜ ⎟⎝ ⎠

1 1P 1 , 12 2

⎛ ⎞− − +⎜ ⎟⎝ ⎠1 1Q , 12 2

⎛ ⎞⎜ ⎟⎝ ⎠

1 1X ,2 2

XQ = ( )⎛ ⎞+ + − =⎜ ⎟⎝ ⎠

221 1 1 3

2 2

Q.51 In the group of four children. If Som is younger than Riyaz and Shiv is elderthan Anshu. Anshu is youngest in the group, so the eldest in the group1. Shiv is younger than Riyaz.2. Som is younger than Anshu.(a) Statement (1) is sufficient to recognize.(b) Statement (2) is sufficient to recognize.(c) Statement (1) and statement (2) both are required.(d) Statement (1) and (2) both are insufficient to find.

Ans. (a)

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