Detailed Solution - IES Master · 2018. 2. 19. · Detailed Solution 11-02-2018 | AFTERNOON SESSION CE F-126, Katwaria Sarai, New Delhi - 110 016 Ph: 011-41013406, Mobile: 8130909220,

SolutionDetailed 11-02-2018 | AFTERNOON SESSION

CE

F-126, Katwaria Sarai, New Delhi - 110 016 011-41013406, Mobile: 8130909220, 9711853908Ph:

Web: www.iesmaster.org | E-mail: [email protected]

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Section : General Aptitude1. For non-negative intergers, a, b, c what would

be the value of a+ b + c if

log a + log b + logc = 0?(a) 3 (b) 1(c) 0 (d) –1

Sol–1 : (a)

log a + log b + log c = 0 log(abc) = 0

abc = e0 = 1

a, b, c I and abc = 1

which is possible only when a = b = c = 1.

So, a + b + c = 1 + 1 + 1 = 3

2. 2

n timesa a a ... a a b &

2

n timesb b b ...b ab ,

where a, b, n and m are natural numbers.What is the value of

n times n timesm m m ... m n n n ... n ?

(a) 2a2b2 (b) a4b4

(c) ab (a+ b) (d) a2 + b2

Sol–2 : (b)

na = a2b n = ab

mb = ab2 m = ab

Now, n times m times

(m m m .....m)(n n n .....n) =

(nm).(mn) = (mn)2 = (ab.ab)2 = a4b4

3. “Although it does contain some pioneeringideas, one would hardly characterize the workas______”.

The word the best fills the blank in the abovesentence is

(a) Innovative (b) Simple(c) Dull (d) Boring

Sol–3 : (a)

The word “Although” is a crucial signpost here.The work contains some pioneering ideas, butapparently it is not overall a pioneering work.Hence innovative best fits the sentance.

4. “His face____ with joy when the solution ofthe puzzle was ____ to him.”

The words that best fill the blanks in the abovesentence are

(a) shone, sown (b) shone, shone

(c) shown, shone (d) shown, shown

Sol–4 : (a)

5. A three-member committee has to be formedfrom a group of 9 people. How may suchdistinct committees can be formed?

(a) 27 (b) 72

(c) 81 (d) 84

Sol–5 : (d)

Three member committee from 9 members =9C3 = 84.

6. In manufacturing industries, loss is usuallytaken to be proportional to the square of thedeviation from a target. If the loss is Rs. 4900for a deviation of 7 units, what would be theloss in Rupees for a deviation of 4 units fromthe target?

(a) 400 (b) 1200

(c) 1600 (d) 2800

Sol–6 : (c)

Loss (Deviation)2

i.e., Loss = K.D2

Now, 4900 = K(7)2 K = 100

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So, loss is 4 units = K(4)2

= 100 × 16 = 1600

7. Given the

log P log Q log R 10y z z x x y for x

y z, what is the value of the product PQR?(a) 0 (b) 1(c) xyz (d) 10xyz

Sol–7 : (b)

logPy z =

logQz x

= logRx y = 10

log P = 10(y – z), log Q = (z – x)10, logR = 10(x – y)

So, log P + log Q + log R = 10[(y – z) + (z– x) + (x – y)]

log PQR = 0

PQR = e0 = 1

8. A faulty wall clock is known to gain 15 minutesevery 24 hours. It is synchronized to the correcttime at 9 AM on 11th July. What will be thecorrect time to the nearest minute when theclock shows 2 PM on 15th July of the sameyear?(a) 12:45 PM (b) 12:58 PM(c) 1:00 PM (d) 2:00 PM

Sol–8 : (b)From 9 am of 11th July to 2 pm of 15th July

No. of hours = 4 days and 5 hours

Now in 24 hours (1 day) clock gains = 15minutes

So in 4 days clock will gain = 15 × 4 = 60mins.

Again in 5 hours clock will gain

= 15 5 min

24

= 13 min8

So, overall clock will gain

= 160 38

= 1638 min.

So, correct time = 2:00 pm – 1638 min

12:57 PM

9. Each of the letters in the figure represents aunique integer from 1 to 9. The letters arepositioned in the figure such that each of(A+B+C). (C+D+E), (C+D+G) and (G+H+K)is equal to 13. Which integer does Erepresent?

A B CDE F G

HK

(a) 1 (b) 4(c) 6 (d) 7

Sol–9 : (b)

A=6 B=5 C=2

D=7

E=4 F=8 G=1

H=9

K=3

10. The annual average rainfall in a tropical city is1000 mm. On a particular rainy day (24-hourperiod), the cumulative rainfall experienced bythe city is shown in the graph. Over the 24hours period, 50% of the rainfall falling on arooftop, which had an obstruction free area of50 m2, was harvested into a tank. What is thetotal volume of water collected in the tank inliters?


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400

350]

250]

200

150

100

50

0 3 6 9 12 15 21 24

Cum

ulat

ive

rain

fall

(mm

)

Hours

300

18

(a) 25,000

(b) 18,750

(c) 7,500

(d) 3,125

Sol–10 : (c)

Hours Rainfall (in mm)0 3 253 6 506 9 259 12 100

12 1515 18 10018 2121 24

Hence, total rainfall in 24 hours = 300 mm

So, total rainfall on rooftop = 50% of (300)

= 150 mm = 15 cm

Area of rooftop = 50 m2

= (50 × 100 × 100) cm2

So, volume of water collected in a tank

= 15 cm × (50 × 100 × 100) cm2

= 7500 × 1000 cm3

= 7500 litres.

Civil Engineering

1. Dupit’s assumptions are valid for(a) artesian aquifer (b) confined aquifer(c) leaky aquifer (d) unconfined aquifer

Sol–1 : (d)The Dupit-Forchheimer assumption holds thatthe groundwater flows horizontally in an uncon-fined aquifer and that groundwater discharge isproportional to the saturated aquifer thickness.

2. A culvert is designed for a flood frequency of100 years and a useful life of 20 years. Therisk involved in the design of the culvert (inpercentage, up to two decimal places)is________

Sol–2 : 18.19%T = 100 years

P = 1

100 = 0.01

q = 0.99Risk = 1 – (0.99)20

Risk 18.19%

3. Which one of the following statements is NOTcorrect?

(a) When the water content of soil lies betweenits liquid limit and plastic limit, the soil issaid to be in plastic state.

(b) Boussinesq’s theory is used for theanalysis of stratified soil.

(c) The inclination of stable slope in cohesivesoil can be greater than its angle of internalfriction.

(d) For saturated dense fine sand after applyingoverburden correction, if the standardpenetration test value exceeds 15,dilatancy correction is to be applied.

Sol–3 : (b)Westergaard’s theory is used for the analysisof stratified soil not Boussinesq’s theory.

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4. For a given discharge in an open channel,there are two depths which have the samespecific energy. These two depths are knownas(a) alternate depths(b) critical depths(c) normal depths(d) sequent depths

Sol–4 : (a)For a given discharge in an open channel, al-ternate depths have the same specific energy.

Depth

y1

y2

Specific EnergyQ1

Q2

Q3

Q < Q < Q1 2 3

E

y1 and y2 are alternate depths.5. All the members of the planar truss (see

figure), have the same properties in terms ofarea of cross-section (A) and modulus ofelasticity (E).

P

L

L

P

For the loads shown on the truss, thestatement that correctly represents the natureof forces in the members of the truss is:

(a) There are 3 members in tension, and 2members in compression

(b) There are 2 members in tension 2members in compression and 1 zero-forcemember

(c) There are 2 members in tension, 1 memberin compression, and 2 zero-force members

(d) There are 2 members in tension, and 3zero-force members.

Sol–5 : (d)P B P C P

PDPAP

L

L

AB, CD and AC are zero force membersForce in member AC is zero because themember in between support at A and C whichhas no tendency to extend due to applied force.Had there been temperature change or lack offit in AC then only it could have carried any force.

6. The graph of a function f(x) is shown in thefigure.

3h

2h

h

0 1 2 3

f(x)

x


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For f(x) to be a valid probability density function,the value of h is

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

Sol–6 : (a)From given graph

f(x) = hx, 0 x 1

2h(x 1) 1 x 23h(x 2) 2 x 3

f(x) is valid p.d.f. so f(x)dx

= 1

1 2

0 1

hx dx 2h(x 1)dx + 3

2

3h(x 2)dx = 1

h 1 12h 3h2 2 2

= 1 h = 13

7. The setting time of cement is determined using(a) Le Chatelier apparatus(b) Briquette testing apparatus(c) Vicat apparatus(d) Casagrande’s apparatus

Sol–7 : (c)

8. A fillet weld is simultaneously subjected tofactored normal and shear stress of 120 MPaand 50 MPa, respectively. As per IS 800:2007,the equivalent stress (in MPa, up to twodecimal places) is_________

Sol–8 : 147.99

Equivalent stress = 2 23 = 147.99 MPa

9. The intensity of irrigation for the Kharif seasonis 50% for an irrigation project with culturablecommand area of 50,000 hectares. The dutyfor the Kharif season is 1000 hectare/cumec.Assuming transmission loss of 10%, therequired discharge (in cumec, up to twodecimal places) at the head of the canalis________

Sol–9 : (27.78)Area = 50,000 haArea under cultivation = 50% of 50,000 ha= 25000 haGiven :

Duty = Area

Discharge

Discharge = 325000 m / s1000

Required discharge = 27.78 m3/s10. The initial concavity in the load-penetration

curve of a CBR test is NOT due to(a) uneven top surface(b) high impact at start of loading(c) inclined penetration plunger(d) Soft top layer of soaked soil

Sol–10 : (b)11. A probability distribution with right skew is

shown in the figure.f(x)

xThe correct statement for the probability distri-bution is

(a) Mean is equal to mode(b) Mean is greater than median but less than

mode(c) Mean is greater than median and mode(d) Mode is greater than median

Sol–11 : (c)

O

Mode

Median

Mean

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12. A vertical load of 10 kN acts on a hinge locatedat a distance of L/4 from the roller support Qof a beam of length L (see figure).

10 kNP

3L/4 L/4

Q

The vertical reaction at support Q is

(a) 0.0 kN (b) 2.5 kN(c) 7.5 kN (d) 10.0 kN

Sol–12 : (a)10 kN

H

3L/4 L/4

P Q

R2R1 R

M +H = 0

LR4

= 0

R = 0

13. A structural member subjected tocompression, has both translation and rotationrestrained at one end, while only translationis restrained at the other end. As per IS 456:2000, the effective length factor recommendedfor design is(a) 0.50 (b) 0.65(c) 0.70 (d) 0.80

Sol–13 : (d)As per IS 456 : 2000, Table 28(Clause E–3)Recommended value of effective length = 0.80

Theoretical value = 0.70Recommended value = 0.80

14. Probability (up to one decimal place) ofconsecutively picking 3 red balls withoutreplacement from a box containing 5 red ballsand 1 white ball is_______

Sol–14 : (0.5)Required Probability

= 1 1 1

1 1 1

5 4 3

6 5 4

C C C

C C C

= 5 4 36 5 4 =

12

= 0.5

15. The contact pressure and settlementdistribution for a footing are shown in the figure.

The figure corresponds to a(a) rigid footing on granular soil(b) flexible footing on granular soil(c) flexible footing on saturated clay(d) rigid footing on cohesive soil.

Sol–15 : (a)For rigid footing on granular soil


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16. The solution of the equation dyx y 0dx

passing through the point (1, 1) is(a) x (b) x2

(c) x–1 (d) x–2

Sol–16 : (c)

xdy ydx

= 0; y(1) = 1

dyy =

dxx

On integrating log y = –log x + log c

xy C

Using y(1) = 1 C = 1 so xy 1

i.e., y = 1x

= x–1

17. As per IS 456:2000, the minimum percentageof tension reinforcement (up to two-decimalplaces) required in reinforced-concrete beamsof rectangular cross-section (consideringeffective depth in the calculation of area) usingFe500 grade steel is_____

Sol–17 : (0.17)

Astbd =

0.85fy

Astbd =

0.85500

Percentage = stA100

bd

= 0.85 100500

= 0.1718. Peak hour factor (PHF) is used to represent

the proportion of peak sub-hourly traffic flowwithin the peak hour. If 15-minute sub-hoursare considered, the theoretically possible rangeof PHF will be(a) 0 to 1.0 (b) 0.25 to 0.75(c) 0.25 to 1.0 (d) 0.5 to 1.0

Sol–18 : (c)

Peak hourly factor = 60

15

V4 V

Peak hourly factor varies from 0.25 to 119. The clay mineral, whose structural units are

held together by potassium bond is(a) Halloysite (b) Illite(c) Kaolinte (d) Smectite

Sol–19 : (b)Properties of Illite mineral1. 2 : 1 clay layer2. By Potassium bond, the two negative sur-faces of silica sheets are held together.

20. As per IS 10500:2012, for drinking water inthe absence of alternate source of water, thepermissible limits for chloride and sulphate, inmg/L, respectively are(a) 250 and 200 (b) 1000 and 400(c) 200 and 250 (d) 500 and 1000

Sol–20: (b)As per 10500 : 2012, table 2Permissible Limit of :Chlorides – 1000 mg/LSulphates – 400 mg/L

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21. As per IRC:37-2012, in order to controlsubgrade rutting in flexible pavements, theparameter to be considered is

(a) horizontal tensile strain at the bottom ofbituminous layer

(b) vertical compressive strain on top ofsubgrade

(c) vertical compressive stress on top ofgranular layer

(d) Vertical deflection at the surface of thepavement.

Sol–21 : (b)As per clause 6.3.2 of IRC 37:2012

N = 4.5337

8

v

14.1656 10

N = 4.5337

8

v

11.41 10

where,N = Number of cumulative standard axle, andEv = Vertical strain in the subgradeAs can be seen, the model considers the verti-cal strain in subgrade as the only variable forrutting, which is a measure of bearing capacityof the subgrade.

22. A flow net below a dam consists of 24 equi-potential drops and 7 flow channels. Thedifference between the upstream and down-stream water levels is 6m. the length of theflow line adjacent to the toe of the dam at exitis 1 m. The specific gravity and void ratio ofthe soil below the dam are 2.70 and 0.70,respectively. The factor of safety against pipingis(a) 1.67 (b) 2.5(c) 3.4 (d) 4

Sol–22 : (d)Nd = 24Nf = 7

H = 6m

G = 2.70e = 0.70Factor of safety against piping

FOS = c

e

ii

ic = G 11 e

= 2.7 11 0.7

= 1

ie = h

L

= d

HN

=

6m24 1m =

14

FOS = 114

= 4

23. In the figures, Group I represents theatmospheric temperature profiles (P, Q, R andS) and Group II represents dispersion ofpollutants from a smoke stack (1, 2, 3 and 4).In the figures of group I, the dashed linerepresents the dry adiabatic lapse rate, whereas the horizontal axis represents temperatureand the vertical axis represents the altitude.

P

Q

R

S

1

2

3

4

Super adiabatic

Adiabatic

Inversion

Inversion over super adiabatic

Fumigation

Fanning plume

Coning plume

Looping plume

Group I Group II

The correct match is


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(a) P–1, Q–2, R–3, S–4(b) P–1, Q–2, R–4, S–3(c) P–1, Q–4, R–3, S–2(d) P–3, Q–1, R–2, S–4

Sol–23 : (a)

SLAR

LARP Looping Plume

Q Coning Plume

ALR

ELR

R Fanning Plume

S Fumigating

24. The quadratic equation 22x 3x 3 0 is tobe solved numerically starting with an initialguess as x0 = 2. The new estimate of x afterthe first iteration using Newton-Raphsonmethod is_______

Sol–24 : (1)Consider f(x) = 2x2 – 3x + 3

then f (x) = 4x – 3 and x0 = 2

Iteration 1 : x1 = 0

00

f(x )x

f (x )

= f(2)2f (2)

=

525

= 1

25. A reinforced-concrete slab with effective depthof 80 mm is simply supported at two oppositeends on 230 mm thick masonry walls. Thecentre-to-centre distance between the wallsis 3.3 m. As per IS 456:2000, the effectivespan of the slab (in m, up to two decimalplaces) is________

Sol–25 : (3.15)

W W

Lo

leff = oW Wl2 2

lo = w w3.32 2

= 3.3 – 0.23 = 3.07 m

leff = o

o

l wmin

l d

= 3.07 0.23 3.30 m

min3.07 0.08 3.15 m

= 3.15 m

26. In a 5 m wide rectangular channel, the velocityu distribution in the vertical direction y is givenby 1/6u 1.25y . The distance y is measuredfrom the channel bed. If the flow depth is 2m,the discharge per unit width of the channel is(a) 2.40 m3/s/m (b) 2.80 m3/s/m(c) 3.27 m3/s/m (d) 12.02 m3/s/m

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Sol–26 : (a)u = 1.25y1/6

y

dy

dq = (1.25y1/6)dy

q = 2

1/6

0

(1.25y )dy

q = 2.405 m3/s/m.27. The prismatic propped cantilever beam of span

L and plastic moment capacity Mp is subjectedto a concentrated load at its mid-span. If the

collapse load of the beam is pML

the value

of is____________

Sol–27 : (6)

L/2 L/2

MP MP

MP

P3M = LW2

P6ML

= W

= 6 Correct answer is 6.

28. The total rainfall in a catchment of area 1000km2, during a 6 h storm, is 19 cm. The surfacerunoff due to this storm computed from

triangular direct runoff hydrograph is 8 31 10 m .

The index for this storm (in cm/h, up to onedecimal place) is_____

Sol–28 : (1.5)Area = 1000 km2

Depth = 19 cmRunoff = 1 × 108 m3

= Total Rain fall – Runoff

Time

Runoff = 8 3

6 21 10 m

1000 10 m

= 0.1 m

= 10 cm

= Rainfall – Runoff

Time

= 19 10

6

= 96 = 1.5 cm/hr..

29. A rough pipe of 0.5 m diameter, 300 m lengthand roughness height of 0.25 mm, carries water(kinematic viscosity = 0.9 × 10–6 m2/s) withvelocity of 3 m/s. Friction factor (f) for

10

1 r2 log 1.74,kf

where, Re = Reynolds

number, r = radius of pipe, k = roughnessheight and g = 9.81 m/s2. The head loss (inm, up to three decimal places) in the pipedue to friction is_________

Sol–29 : (4.599)Given, D = 0.5 mL = 300 m = 0.9 × 10–6 m2/secv = 3 m/secK = 0.25 mm

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= 63 0.5

9 10

= 1.667 × 106 > 2000

Hence flow is turbulent.

1f

= 10r2log 1.74K

= 10 30.252log 1.74

0.25 10

f = 0.0167

head loss, hf = 2fLv

2gD

= 20.0167 300 3

2 9.8 0.5

= 4.599 m

30. The value (up to two decimal places) of the

l ine integral 2 2

cF(r). dx. for F(r) x i y j

along C which is a straight line joining (0, 0)to (1, 1) is_____

Sol–30 : (0.67)

f = 2 2ˆ ˆx i y j and dr

= ˆ ˆdxi dyjSo, f.dr

= (x2 dx + y2 dy)Now, along (0, 0) to (1, 1) equation of straightline

y = x

dy = dx

0 x 1

So, LI = C

f.dr

= 2 2

C

(x dx y dy)

= 1

2

0

(2x dx) = 23 = 0.67

31. A prismatic beam P-Q-R of flexural rigidity EI= 1×104 kNm2 is subjected to a moment of180 kNm at Q as shown in the figure.

180 kNmP

5 m 4 m

R

The rotation at Q (in rad, up to two decimalplaces) is______

Sol–31 : (0.01)

5m 4m

P R

180 kN.m

Q

Fixed end momentMFQP = MFQR = 0Writing slope deflection equation of QP and QR

MQP = Q2EI0 (2 )5

= Q4 EI5

MQR = Q2EI (2 )4

= QEI

As MQP + MQR = –180 kN.m

Q Q4 EI EI5

= –180

Q = 4180 59 10

= –0.01 rad

Hence, Q = 0.01 radAlternative Solution :

EI5m

EI4m

M1 M2

M1 M2

180

4EI5

= M1

4EI4

= M2

M1 + M2 = 180 1.8EI = 180

= 100EI

= 410010

= 0.01 rad.

32. The 3 m high vertical earth retaining wallretains a dry grannular backfill with angle ofinternal friction of 30° and unit weight 20 kN/m3. If the wall is prevented from yielding (nomovement) the total horizontal thrust (in kNper unit length) on the wall is

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(a) 0 (b) 30(c) 45 (d) 270

Sol–32 : (c)

= 30° = 20 kN/m3

As wall does not move, so wall is at rest.Coefficient of earth pressure at rest k0 =1 sin

= 1 sin30 = 0.5Earth Pressure Distribution :

2o

1k H2

k H = 30 kN/mo2

Resultant Thrust (Po) = o1 (k H)(H)2

= 1 30 32

= 45 kN/m33. A singly reinforced rectangular concrete beam

of width 300 mm and effective depth 400 mm

is to be designed using M25 grade concreteand Fe500 grade reinforcing steel. For thebeam to be under-reinforced, the maximumnumber of 16 mm diameter reinforcing barsthat can be provided is(a) 3 (b) 4(c) 5 (d) 6

Sol–33 : (c)xu,lim for Fe 500 = 0.46 × 400 = 184 mmFor section to be under reinforced Ast AAst,lim

Ast,lim = ck ulim

y

0.36f bx0.87f =

0.36 25 300 1.840.87 500

= 1142.06 mm2

No. of bars for limited area of steel

= 2

1142.06

164 = 5.68

So, number of bars will be 5.34. A level instrument at a height of 1.320 m has

been placed at a station having a reducedlevel (RL) of 112.565 m. The instrument reads–2.835 m on a levelling staff held at the bottomof a bridge deck. The RL (in m) of the bottomof the bridge deck is(a) 116.720 (b) 116.080(c) 114.080 (d) 111.050

Sol–34 : (a)Height of instrument = 112.565 + 1.320= 113.885 mRL of bottom of bridge = 113.885 + 2.835= 116.720 m

35. A flocculation tank contains 1800 m3 of water,which is mixed using paddles at an averagevelocity gradient G of 100/s. The watertemperature and the corresponding dynamicviscosity are 30°C and 0.798 ×10–3 Ns/m2,respectively. The theoretical power required toachieve the stated value of G (in kW, up totwo decimal places) is_________


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Sol–35 : (14.36)

Velocity Gradient, G = PV

Given G = 100 S–1

P = ?, µ = 0.798 × 10–3 Ns/m2

V = 1800 m3

P = 1002 × 0.798 × 10–3 × 1800

= 14.364 kW

36. Two rigid bodies of mass 5 kg and 4 kg areat rest on a frictionless surface until actedupon by a force of 36 N as shown in thefigure. The contact force generated betweenthe two bodies is

5 kg 4 kg36 N

(a) 4.0 N (b) 7.2 N(c) 9.0 N (d) 16.0 N

Sol–36 : (d)

5 kg 4 kg36 N

Net force = Net mass × acceleration

36 = 9 × a

a = 4 ms–2

Now, considering 5 kg weight only

force on 5 kg = 5 × 4 N = 20 N

36N F

36N – F = 20N

F = 16N

37. A 6 m long simply-supported beam isprestressed as shown in the figure.

Neutral axis

Prestressing cable (straight profile)

50 mm

The beam carries a uniformly distributed loadof 6 kN/m over its entire span. If the effectiveflexural rigidity EI = 2 × 104 kNm2 and theeffective prestressing force is 200 kN, the netincrease in length of the prestressing cable(in mm, up to two decimal places) is________

Sol–37 : (0.12)

Downward UDL, w = 6 kN/m

Eccentricity, e = 50 mm

Prestressing force, P = 200 kN

EI = 2 × 104 kN m2

L = 6 m

Rotation due to prestress = PeL2EI

= 3

4200 50 10 6

2 2 10

= 1.5 × 10–3

Rotation due to UDL = 3wL

24EI

= 3

46 (6)

24 2 10

= 2.7 × 10–3

Net rotation = 2.7 × 10–3 – 1.5 × 10–3

= 1.2 × 10–3 radian

Elongation of the cable = 2 × 50 × 1.2 × 10–3

mm

= 0.120 mm

38. A schematic flow diagram of a completelymixed biological reactor with provision forrecycling of solids is shown in the figure.

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aeration tankS = 15 mg/LX = 3000 mg/LDetention time = 2 h Secondary

settling tank

Q = 15000 m /d3

S = 100 mg/L0X ~ 0 mg/L0

Sludge returnQ = 50 m /dw

3

S = 15 mg/LX = 10000 mg/Lu

Q – Q w= 14950 m /d

3

S = 15 mg/LX ~ 0 mg/Le

Q = 5000 m /dS = 15 mg/LX = 10000 mg/L

r

u

3

So, S = readily biodegradable soluble BOD, mg/L

Q, Qr, Qw = flow rates, m3/dX0, X, Xe, Xu = microorganism concentrations

(mixed-l iquor volati lesuspended solids or MLVSS),mg/LThe mean cell residence time (in days, up toone decimal place) is____

Sol–38 : (7.5)Mean cell residence time is given by

c = o w e w u

VX(Q Q )X Q X

Detention time = 2hVolume of tank

= o dQ t = 215000

24 = 1250 m3

Since, Xe = 0, c will be calculated as follows:

c = 1250 300050 10000

= 7.5 days

39. The compression curve (void ratio, e vs,effective stress, v ) for a certain clayey soilis a straight line is a semi-logarithmic plotand it passes through the points (e = 1.2;

v 50 kPa ) and (e = 0.6; v 800 kPa ).The compression index (up to two decimalplaces) of the soil is________

Sol–39 : (0.5)Compression index

CC = 2

1

e

log

= 1.2 0.6

800log50

= 0.6

log(16) = 0.50

40. A three-fluid system (immiscible) is connectedto a vacuum pump. The specific gravity valuesof the fluids (S1, S2) are given in the figure.

Patm

0.5 m1.0 m

1.5 m

Unit weight of water, = 9.81 kN/mw3

Atmospheric pressure P = 95.43 kPaatm

To vaccum pump

S = 0.881Fluid I

S = 0.952

Fluid II

Fluid III: Water

P1

The gauge pressure value (in kN/m2. up totwo decimal places) of p1 is

Sol–40 : (–8.73)

Pressure at bottom of fluid II = atm wP 0.5

= 95.43 + 0.5 × 9.81= 100.335 kPaAlso,

1 w wP 0.88 0.5 0.95 = 100.335 KPaP1 + 13.63 = 100.335P1 = 86.69 KPa Gauge Pressure = P1 – Patm

–8.73 kN/m41. A car follows a slow moving truck (travelling at

a speed of 10 m/s) on a two-lane two-wayhighway. The car reduces its speed to 10 m/


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IES

MASTER

s and follows the truck maintaining a distanceof 16 m from the truck. On finding a clear gapin the opposing traffic stream, the caraccelerated at an average rate of 4 m/s2,overtakes the truck and returns to its originallane. When it returns to its original lane, thedistance between the car and the truck is16m. The total distance covered by the carduring this period (from the time it leaves itslane and subsequently returns to its lane afterovertaking) is(a) 64 m (b) 72 m(c) 128 m (d) 144 m

Sol–41 : (b)Option 2 is correct

A B B A

16m b 16m

Speed of car when he starts overtaking,v = 10 m/sec

(16 + b + 16) = 21v.t at2

32 + b = 10t + 2t2 ...(i)Distanced travelled by slow moving truck dur-ing overtaking time tb = v.tb = 10t ...(ii)From (i) & (ii)32 + 10t = 10t + 2t2

16 = t2

t = 4 sec distance covered by car during the periodfrom the time it leaves its lane after overtaking= 32 + b= 32 + (10 × 4)= 72 m

42. A coal containing 2% sulfur is burnedcompletely to ash in a brick kiln at a rate of30 kg/min. The sulfur content in the ash wasfound to be 6% of the initial amount of sulfurpresent in the coal fed to the brick kiln. Themolecular weights of S, H and O are 32, 1and 16 g/mole, respectively. The annual rateof sulfur dioxide (SO2) emission from the kiln(in tonnes/year, up to two decimal places)is________

Sol–42 : (592.88)Total coal burnt in the kiln in a year= 30 × 60 × 24 × 365 kg= 15768000 kgSulphur present in the coal = 0.02 × 15768000kg= 315360 kgSulphur content in the ash = 0.06 × 315360 kg= 18921.6 kgSulphur oxidized into SO2 = 194 × 315360 kg= 296438.4 kg

2 2S O SO32 64

32 gm sulphur produce = 64 gm SO2

296438.4 kg sulphur produce =

64 296438.432

= 592876.8 kg= 592.876 tonnes.

43. The Laplace transform F(s) on the exponentialfunction. f(t) = eat when t 0, where a is aconstant and (s– a) > 0, is

(a)1

s a(b)

1

s a

(c)1

a s(d)

Sol–43 : (b)

L{f(t)} = st

0

e .f(t)dt

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MASTER


CE



So, L{eat} = st at

0

e .e dt

= 1

s a

44. Four bolts P, Q, R and S of equal diameterare used for a bracket subjected to a load of130 kN as shown in the figure.

PQ

R S

Centre line

200 mm

130 kN

100 mm

240 mm

The force in bolt P is(a) 32.50 kN (b) 69.32 kN(c) 82.50 kN (d) 119.32 kN

Sol–44 : (b)

Option (2) is correct.

Direct shear in bolt P,

1F = 130 32.5KN

4

Shear due to twisting moment

= 2 2

2 2 2

130 200 120 150

4( 120 50 )

2F = 50 KN

= 1 120tan50

= 67.38°

100 mm

240 mm

P2P1

130 kN

Q

S R

P

Resultant force in bolt P

= 2 21 2 1 2F F 2FF cos

= 2 232.5 50 2 32.5 50cos67.38

= 69.32 kN

45. The total horizontal and vertical stresses at apoint X in a saturated sandy medium are 170kPa and 300 kPa, respectively. The static pore-water pressure is 30 kPa. At failure, theexcess pore-water pressure is measured tobe 94.50 kPa, and the shear stresses on thevertical and horizontal planes passing throughthe point X and zero. Effective cohesion is 0kPa and effective angle of internal friction is36°. The shear strength (kPa, up to twodecimal places) at point X is________

Sol–45 : (52.52)

Major effective principal stress 1( )

= 300 30 94.5MPa

= 175.5 MPa

Minor effective principal stress


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MASTER

= 170 30 94.5MPa

3 = 45.5 MPa

Shear strength = C tan

C = 0, 36 (given)

= effective stress at the plane of failurenormal.

36° D

65

65130

A

B 175.545.50

y

OBA 90 36 = 54°

BD = 65cos54 = 38.21 KPa

= (45.5 65 38.21) = 72.3 KPa

shear strength = 72.3 tan36

= 52.52 KPa

46. The rank of the following matrix is

1 1 0 22 0 2 24 1 3 1

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

Sol–46 : (b)

A = 1 1 0 22 0 2 24 1 3 1

2 2 13 3 1

R R 2RR R 4R

1 1 0 20 2 2 60 3 3 9

3 3 23R R R2

1 1 0 20 2 2 60 0 0 0

So (A) = 2.

47. An aerial photograph of a terrain having anaverage elevation of 1400 m is taken at ascale of 1:7500. The focal length of camera is15 cm. The altitude of the flight above meansea level (in m. up to one decimal place)is____

Sol–47 : (2525)

Given; scale, S = 1

7500

elevation of terrain > h = 1400 m

Focal length of the camera, f = 15 cm

altitude of flight, H = ?

S = f

H h

17500 =

215 10H 1400

H = 2525 m

48. An 8 m long simply-supported elastic beamof rectangular cross-section (100 mm × 200mm) is subjected to a uniformly distributedload of 10 kN/m over its entire span. Themaximum principal stress (in MPa, up to twodecimal places) at a point located at theextreme compression edge of a cross-sectionand at 2 m from the support is

Sol–48 : (90)

IES

MASTER


CE



8 m2 m

100 m

200

m

8 KN/m

40 KN 40 KNat 2 m from support,

Bending moment, M = 40 2 10 2 1

= 60 KN-m

Shear force, V = 40 (10 2) = 20KN

At external compression edge (top fibre)

Shear stress P = 0

Bending stress, = MY

I

= 6

34

(60 10 N mm) 100mm100 200 mm

12

= 290N / mm

= 90 MPa

Since this edge is free from shear stress,hence it is a principal plane.

Thus maximum principal stress = 90 MPa

49. A 7.5 m wide two-lane road on a plain terrainis to be laid along a horizontal curve of radius510 m. For a design speed of 100 kmph,super-elevation is provided as per IRC: 73-1980. Considered acceleration due to gravityas 9.81 m/s2. The level difference betweenthe inner outer edges of the road (in m, upto three decimal places) is___

Sol–49 : (0.525)Design speed,V = 100 KmpH

Radius of curve, R = 510 m

g = 9.81 m/sec2

Super elevation e = 2(0.75V)

gR

=

250.75 10018

9.81 510

= 0.08675 .07

e = 0.07

heB

.07 = h

7.5

h = 0.525 m

B

h

Outer edge

50. At a small water treatment plant which has 4filters, the rate of filtration and backwashingare 200 m3/d/m2 and 1000 m3/d/m2,respectively. Backwashing is done for 15 minper day. The maturation, which occurs initiallyas the filter is put back into service aftercleaning, takes 30 min. It is proposed to recoverthe water being wasted during backwashingand maturation. The percentage increase inthe filtered water produced (up to two decimalplaces) would be _____

Sol–50 : (7.53)

Rate of filteration = 3 2200m / d / m

Backwashing rate = 3 21000m / d / m


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IES

MASTER

Time for backwashing = 15 min/day

Time wasted in maturation = 30 min

Let the area of filter to be unity

Total water to be produced

= 200m3/d/m2 × 23.25

24

= 193.75

Water wasted in backwashing

1000 1524 60

= 10.4167 m3

Water wasted during maturation

= 200 3024 60

= 4.1667 m3

Total filter water to be produced

= 193.75 + 10.4167 + 4.1667

= 208.3334 m3

Hence percentage increase in filtered waterproduced

= 208.3334 193.75

193.75

= 7.53%

51. A group of nine piles in a 3 × 3 square patternin embedded in a soil strata comprising densesand underlying recently filled clay layer, asshown in the figure. The perimeter of anindividual pile is 126 cm. The size of pile groupis 240 cm × 240 cm. The recently filled clayhas undrained shear strength of 15 kPa andunit weight of 16 kN/m3.

Recently filled clay

Dense sand

2 m

5 m

240 cm

240 cm

The negative frictional load (in kN, up to twodecimal places) acting on the pile group is_____

Sol–51 : (472.32)Negative skin friction by considering individualpile

Pn1 = un C Perimeter of single pile × L

According to the Cu value of soil it is classifiedas soft clay. Adhesion factor for soft clay is 1.So,

Pn1 = 9 1 15 1.26 2 340.2 kN

Negative friction load of pile group

Pn2 = u gC P L pile groupA L

= 15 × 4 × 2.4 × 2 + 16 × [2.4 × 2.4] × 2

= 472.32 kN

So, negative skin friction on pile group will bemaximum of Pn1 and Pn2 and given by 472.32kN.

IES

MASTER


CE



52. The matrix

2 44 2

has

(a) real eigen values and eigen vectors(b) real eigen values but complex eigen

vectors(c) complex eigen values but real eigen

vectors(d) complex eigen values and eigen vectors

Sol–52 : (d)

A = 2 44 2

1 2 + = Trace (A) = 0 1 2. = |A| = 12

1 = 2 and 1 1( )( ) = 12

21 = –12 1 = 2 3 i

and 2 = 2 3 i

Hence, Eigen values are complex andconsequently Eigen Vectors will also complex.

53. A cable PQ of length 25 m is supported attwo ends at same level as shown in the figure.The horizontal distance between the supportsis 20 m. A point load of 150 kN is applied atpoint R which divides it into two equal parts.

20 m

PR

150 kN

Q

Neglecting the self-weight of the cable, thetension (in kN, in integer value) in the cabledue to applied load will be__

Sol–53 : (125)

12.5 m 12.5 m

H1

V1 V2

H2P Q

150 KN

20 m

0

R

OR = 2 2(12.5) (10) = 7.5 m

1V = 2V = 75 KN

MR = 0 (of left hand side)

1H OR = 1V 10

1H = 75 10

7.5

= 100 KN

Tension in the cable T = 2 21 1H V

= 2 2100 75

= 125 KN

54. Three soil specimens (Soil 1, and Soil 3),each 150 mm long and 100 mm diameter areplaced in series in a constant head flow set-up as shown in the figure. Suitable screensare provided at the boundaries of thespecimens to keep them intact. The values ofcoefficient of permeability of Soil 1, Soil 2 andSoil 3 are 0.01, 0.03 and 0.03 cm/s,

100 mm

150 mm 150 mm 150 mm

Soil 1 Soil 2 Soil 3

560 mm

h

The value of h in the set-up is


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MASTER

(a) 0 mm (b) 40 mm(c) 255 mm (d) 560 mm

Sol–54 : (b)

150 mm

150 mm

150 mm

560 mm

Soil1 Soil2 Soil3

h

1K = .01cm / sec , 2K = 0.003 cm/sec,

3K 0.03cm / sec

Total head loss, h = 560 mm

Since soil specimen are connected in series,hence discharge through each soil specimenwill be same.

1Q = 2 3Q Q

1 11K A i = 2 2 2K A i = 3 3 3K A i

11 1

1

hK AL =

22 2

2

hK AL =

33 3

3

hK AL

1 1K h = 2 2 3 3K h K h = x

1 2 3( L L L 150mm 1 2 3A A A )

1xh

.01 , 2

xh0.003

, 3xh

.03

Thus x x x

0.01 0.003 .03 = 560

x 1.2

1h = 1.2.01

= 120 mm,

2h = 1.2 400mm.003

,

3h = 1.2 40mm.03

Now h = 560mm – (head loss between soil 1

& soil 2)

= 1 2560 (h h )

= 560 (120 400)

= 40 mm

55. The space mean speed (kmph) and density(vehicles/km) of a traffic stream are linearlyrelated. The free flow speed and jam densityare 80 kmph and 100 vehicles/km respectively.The traffic flow (in vehicles/h up to one decimalplace) corresponding to a speed of 40 kmphis_____

Sol. (2000)

Free flow speed, fV = 80 Kmph

Jam density, Kj = 100 Veh/km

Flow, q = ff

j

VK V k

K

We know V = ff

j

V .KVK

at V = 40 kmph

40 = 80 – 80 K100

K = 50Veh / km

q = 8050 80 50

100

= 2000 Veh/hr.

ESE-2018 Conventional Test Schedule, Civil Engineering

Institute for Engineers (IES/GATE/PSUs)

IES MASTER

Conventional Test 10:00 A.M. to 1:00 P.M.

Test Type Timing Day

Sunday

Conventional Full Length Test Paper-1 10:00 A.M. to 1:00 P.M. SundayConventional Full Length Test Paper-2 02:00 P.M. to 5:00 P.M. Sunday

Date Topic

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

:

:

:

:

:

:

:

:

:

:

:

M-1, M-3, M-4, SM-1, SM-3, SM-8

:

:

:

:

:

:

:

:

:

:

:

Full Length-1 (Test Paper-1 + Test Paper-2)



SA-1, SA-2, SA-5, HY-1, HY-4, HY-5, M-5

SM-1, M-1

DSS-4, DSS-5, FM-1, FM-4, FM-6

M-3, SA-1, SA-2

SA-6, SA-4, SA-3, EE-6, EE-5, EE-4

FM-4, FM-6, M-1, M-4, M-3, HY-1

FM-7, RCC-1, RCC-2, RCC-3, HY-2

SA-1, SA-2, SM-3, FM-6, EE-6

SM-4, DSS-1, DSS-2, DSS-3, RCC-4, RCC-5, RCC-6

SM-1, SA-3, EE-5

SU-1, SU-2, SU-3, SM-2, SM-5, SM-6, SM-7, HY-3, SU-5

FM-7, RCC-1, RCC-2, RCC-3, HY-1, EE-6

TF-1, TF-2, TF-3, TF-4, FM-5, M-2

RCC-5, DSS-1, DSS-2, SM-4, M-1, M-3, M-4, FM-4, SA-1

IR-1, IR-2, IR-3, IR-4, EE-7

SM-5, SM-6, FM-1, EE-5, DSS-3, DSS-4, HY-3, HY-4, HY-5, SU-1, SU-2

CPM-1, CPM-2, EE-1, EE-2, EE-3, SU-4 (Railway & Airport)

SM-4, FM-5, TF-1, TF-2, FM-7, SA-3, SU-3, SU-5, RCC-5

FM-2, FM-3, FM-8, Building Material, Ports & Harbors/Tunneling

IR-1, IR-2, HY-2, DSS-4, DSS-2, SA-1, SA-2, SA-3, RCC-6, EE-2, FM-6

Note : The timing of the test may change on certain dates. Prior information will be given in this regard.*N.T. : New Topic. *R.T. : Revision Topic

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R.T.

For Any Query Regarding The ProgramCall us : 09711853908, 011-41013406 or Mail us : [email protected]

Subject Code Details

Truss,Cables

Slab-One way,(LS, WS)Staircase

Earthquakeresistant structures,Beams (LS, WS),

Lintels

Fooling (LS, WS),Retaining

Wall

Column,(LS, WS)

Water Tanks

Network analysis, Pert, CPM, Crashing, Resource allogaction, Levelling,Smoothing, Rate Analysis, Estimation and costing, Quality control, Productivity, Opration cost, Land Aquisition

Cement, Concrete, Stone, Lime, Glass, Steel Brick Mortar Timber, Plastics, FRP, Ceramics, Aluminium

Construction equipments, Engineering Economy,Tendering Process and Contract Management

SU-1 SU-2 SU-3 SU-4 SU-5

IR-1 IR-2 IR-3 IR-4

HY-1 HY-2 HY-3 HY-4 HY-5

(SU)

(IR)

(HY)

Exploration of Soil,Expansive Soil, Geosynthetics,

Ground Modification Techniques,

Triangulation & TraversingPlane, tabling, Geology

Photogrammetry, Field Astronomy, GPS, Remote Sensing, Curve setting

Railways / Airports / Ports & Harbours / Tunneling

Free and Forced Vibrations , Concepts and use of computer aided design

,

,Power House

Solid Waste Management,

Documents

Detailed Solution - IES Master · 2018. 2. 19. · Detailed Solution 11-02-2018 | AFTERNOON SESSION CE F-126, Katwaria Sarai, New Delhi - 110 016 Ph: 011-41013406, Mobile: 8130909220,