New GENERAL APTITUDE · 2019. 4. 16. · GATE-2019 ORIGINAL PAPER – CE/1 GENERAL APTITUDE Q. 1 – Q. 5 carry one mark each. 1. If E = 10; J = 20; O = 30; and T = 40, what will

GATE-2019 ORIGINAL PAPER – CE/1

GENERAL APTITUDE Q. 1 – Q. 5 carry one mark each. 1. If E = 10; J = 20; O = 30; and T = 40, what will be P + E + S + T? (A) 51 (B) 120 (C) 82 (D) 164 Sol. Alphabetical position of E is 5. Given E=10. Similarly, J is given 20 for 10th position. O = 30 for 15, T=40 for 20. P’s position is 16 and S’ position is 19. Therefore, P=32. S=38. P+E+S+T = 32+10+38+40 = 120. Choice (B) 2. On a horizontal ground, the base of a straight ladder is 6 m away from the base of a vertical

pole. The ladder makes an angle of 45° to the horizontal. If the ladder is resting at a point located at one-fifth of the height of the pole from the bottom, the height of the pole is ______ meters. (A) 15 (B) 25 (C) 35 (D) 30

Sol. Let AB be the pole with base at A. C is the end of the ladder which lies on the ground and D is the other end of the ladder which

touches the pole. Given, AC = 6 If AD = x, then AB = 5x Given ∠ACD = 45°

In ΔADC, tan45°= 𝐴𝐷

𝐴𝐶

⇒ AD = AC = 6 and AB = 5(6) = 30 Choice (D)


3. They have come a long way in _________ trust among the users.

(A) creating (B) created (C) creation (D) create

Sol The appropriate word to be filled in the blank would be a present participle verb. Choice (A) 4. The lecture was attended by quite ______ students, so the hall was not very ______.

(A) few, quite (B) a few, quite (C) few, quiet (D) a few, quiet

Sol. The appropriate phrase to fit in the first blank would be ‘a few’, which will suggest that some

students were indeed present in the class as against ‘few’. ‘Quiet’ is the correct spelling of the word meaning silent. Hence, Option (D) is the correct answer. Choice (D)

5. The CEO’s decision to quit was as shocking to the Board as it was to _____.

(A) myself (B) I (C) me (D) my

Sol. The correct word to be filled in the blank would be ‘me’ as it the object form of the pronoun ‘I’

and the pronoun has to be used in the capacity of an object, alongside the Board. Choice (C) Q. 6 – Q. 10 carry two marks each. 6. A square has sides 5 cm smaller than the sides of a second square. The area of the larger

square is four times the area of the smaller square. The side of the larger square is _____ (A) 15.10 (B) 8.50 (C) 18.50 (D) 10.00

Sol. Let x be the side of the larger square.

x – 5 be the side of the smaller square. Given,

𝑥2 = 4(𝑥 − 5)2

𝑥2 = 4𝑥2 – 40𝑥 + 100 3𝑥2 − 40𝑥 + 100 = 0

3𝑥2 − 30𝑥 − 10𝑥 + 100 = 0 3𝑥(𝑥 – 10)– 10(𝑥 – 10) = 0 (𝑥 – 10)(3𝑥 – 10) = 0

𝑥 = 10 or 10

3

But x has to be greater than 5, ∴ x = 10. Ans: 10


7. “The increasing interest in tribal characters might be a mere coincidence, but the timing is of

interest. None of this, though, is to say that the tribal hero has arrived in Hindi cinema, or that the new crop of characters represents the acceptance of the tribal character in the industry. The films and characters are too few to be described as a pattern."

What does the word ‘arrived’ means in the paragraph above? (A) came to conclusion (B) reached a terminus (C) went to a place (D) attained a status

Sol. The phrase ‘hero has arrived’ can be translated as ‘gained visibility’ or ‘is finally represented’.

Option (D) is the closest. Choice (D) 8. The new cotton technology, Bollgard-II, with herbicide tolerant traits has developed into a

thriving business in India. However, the commercial use of this technology is not legal in India. Notwithstanding that, reports indicate that the herbicide tolerant Bt cotton had been purchased by farmers at an average of Rs.200 more than the control price of ordinary cotton, planted in 15% of the cotton growing area in the 2017 Kharif season.

Which one of the following statements can be inferred from the given passage? (A) Farmers want to access the new technology for experimental purposes (B) Farmers want to access the new technology if India benefits from it (C) Farmers want to access the new technology even if it is not legal (D) Farmers want to access the new technology by paying high price

Sol. Option (A) is the only inference, among the given options, one can arrive at. Option (B) does not follow from the information mentioned in the passage as nothing is mentioned about India benefitting from it. Option (B) and (C) paraphrase the ideas in the passage. Choice (C)

9. P, Q, R, S and T are related and belong to the same family. P is the brother of S. Q is the wife of P. R and T are the children of the siblings P and S respectively. Which one of the following statements is necessarily FALSE? (A) S is the sister-in-law of Q (B) S is the aunt of R (C) S is the brother of P (D) S is the aunt of T

Sol. The family tree of the given data is as follows:

If S is considered female, then Choice (A) and (B) are true. If S is male, then (C) is true. But Choice (D) is necessarily false. Choice (D) 10. In a sports academy of 300 people, 105 play only cricket, 70 play only hockey, 50 play only

football, 25 play both cricket and hockey, 15 play both hockey and football and 30 play both cricket and football. The rest of them play all three sports. What is the percentage of people who play at least two sports? (A) 25.00 (B) 28.00 (C) 23.30 (D) 50.00

P S

T

Q

R


Sol. In the Venn diagram, people who play at least two sports are those represented in the

intersection region. Total number of people who play at least two sports = 75

Percentage of the people who play at least two sports = 75

300 × 100 = 25%

Note: The question treated each of the expression, ‘both cricket and hockey’ cricket and hockey but not football. Same is the case with ‘both hockey and football’ and ‘both cricket and football’. Ans: 25

CIVIL ENGINEERING Q. 1 – Q. 25 carry one mark each.

1. A soil has specific gravity of its solids equal to 2.65. The mass density of water is 1000 kg/m3.

Considering zero air voids and 10% moisture content of the soil sample, the dry density (in kg/m3, round off to 1 decimal place) would be ______

Sol. G = 2.65

w = 1000 kg/m3 na = 0 w = 10%

d = ? since na = n ac ; ac = 0 sr = 100% or 1

e = wG

s𝑟=

0.1 ×2.65

1= 0.265

d = G𝑤

1 + 𝑒 =

2.65 ×1000

1.235

= 2094.8 kg/m3

Ans: 2086.6 to 2095.0

H



2. A circular duct carrying water gradually contracts from a diameter of 30 cm to 15 cm. The

figure (not drawn to scale) shows the arrangement of differential manometer attached to the duct.

When the water flows, the differential manometer shows a deflection of 8 cm of mercury (Hg).

The values of specific gravity of mercury and water are 13.6 and 1.0, respectively. Consider the acceleration due to gravity, g = 9.81 m/s2. Assuming frictionless flow, the flow rate (in m3/s. round off to 3 decimal places) through the duct is _______

Sol. Applying Bernoulli’s theorem between 1 and 2

𝑃1

𝜌𝑔 +

𝑉12

2𝑔 =

𝑃2

𝜌𝑔 +

𝑉22

2𝑔 [ Z1 = Z2]

𝑃1− 𝑃2

𝜌𝑔 =

𝑉22− 𝑉1

2

2𝑔

h = 𝑉2

2− 𝑉12

2𝑔 → (1)

Also h = x(𝑆𝑚

𝑆− 1)

= 0.08 (13.6

1− 1)

h = 1.008 m and A1V1 = A2V2

4 𝐷1

2 V1 =

4 𝐷2

2 V2

V2 = 4V1 → (2) Put the value h = 1.008 m and equation (2) in equation (1) we get

1.008 = 16𝑉1

2− 𝑉12

2𝑔

V1 = 1.148 m/s

Flow rate Q =

4 𝐷1

2 V1

=

4 (0.3)2 1.148

= 0.081 m3/s Ans: 0.078 to 0.085

15 cm V2

30 cm

V1

Mercury

8 cm


3. The coefficient of average rolling friction of a road is fr and its grade is +G%. If the grade of

this road is doubled, what will be the percentage change in the braking distance (for the design vehicle to come to a stop) measured along the horizontal (assume all other parameters are kept unchanged)?

(A) 0.02 G

f𝑟 + 0.01 G × 100

(B) 2f𝑟

f𝑟 + 0.01 G × 100

(C) 0.01 G

f𝑟 + 0.02 G × 100

(D) f𝑟

f𝑟 + 0.02 G × 100

Sol. Braking distance is given by

S = )

100

Gf(g2

V2

For positive gradient

S1 = )G01.0f(g2

V2

+

For G2 = 2G

S2 = )G02.0f(g2

V2

+

Percentage change = 100s

ss

1

21 −

= 100

)G01.0f(g2

V

)G02.0f(g2

V

)G01.0f(g2

V

2

22

+

+−

+

= 100

)G01.0f(

1

)G02.0f(

1

)G01.0f(

1

+

+−

+

= 100)G01.0f()G02.0f()G01.0f(

G01.0fG02.0f+

++

−−+

= 100)G02.0f(

G01.0

+ Choice (C)

4. The probability that the annual maximum flood discharge will exceed 25000 m3/s, at least once in next 5 years is found to be 0.25. The return period of this flood event (in years, round off to 1 decimal place) is ________

Sol. Probability of an event accuring atleast once in ‘n’ successive years is Risk (R) R = 1 – qn

0.25 = 1 – q5 q = 0.944 1 – p = 0.944

1

T = P = 0.0559

T = 17.89 years Ans: 17.8 to 18.0


5. A simple mass-spring oscillatory system consists of a mass m, suspended from a spring of

stiffness k. Considering z as the displacement of the system at any time t, the equation of motion for the free vibration of the system is mz̈ + kz = 0. The natural frequency of the system is

(A) m

k

(B) k

m

(C) √m

k

(D) √k

m

Sol. For simple harmonic motion m z̈ + kz = 0

z̈ + k

m z = 0 → (1)

Standard equation is d2𝑥

dt2 + 𝑤2𝑥 = 0 → (2)

By comparing (1) and (2)

= √k

m Choice (D)

6. The interior angle of four triangles are given below

Triangle Interior Angles

P 85°, 50°, 45°

Q 100°, 55°, 25°

R 100°, 45°, 35°

S 130°, 30°, 20°

Which of the triangles are ill-conditioned and should be avoided in Triangulation surveys?

(A) Both Q and S (B) Both P and S (C) Both Q and R (D) Both P and R

Sol. For a triangle to be well conditioned, none of the angles should be less than 30° nor greater

than 120°. Out of the given triangles, triangle Q and S are ill-conditioned Choice (A)

7. If the path of an irrigation canal is below the bed level of a natural stream, the type of cross-drainage structure provided is (A) Level crossing (B) Super passage (C) Aqueduct (D) Sluice gate

Sol. If the path of an Irrigation canal is below the bed level of a natural stream, it is called as a

super passage Choice (B)


8. An isolated concrete pavement slab of length L is resting on a frictionless base. The

temperature of the top and bottom fibre of the slab are Tt and Tb, respectively. Given: the

coefficient of thermal expansion = and the elastic modulus = E. Assuming Tt > Tb and the unit weight of concrete as zero, the maximum thermal stress is calculated as

(A) Eα(Tt−Tb)

2

(B) L(Tt – Tb)

(C) E(Tt – Tb) (D) zero

Sol. It is given that the pavement is resting on Friction less base. Hence until a load is acting on

the pavement, it will not be subjected to any thermal stresses Choice (D) 9. A plane truss is shown in the figure (not drawn to scale). Which one of the options contains ONLY zero force members in the truss?

(A) FG, FI, HI, RS (B) FI, FG, RS, PR (C) FG, FH, HI, RS (D) FI, HI, PR, RS

Sol. If two member forces are collinear, the force in the other members at that joint will be zero EF = FG collinear EG = GI

GF = 0, FI = 0 Similarly SR = 0, PR = 0. Choice (B) 10. Which one of the following is a secondary pollutant?

(A) Carbon Monoxide (B) Hydrocarbon (C) Ozone (D) Volatile Organic Carbon (VOC)

Sol. Ozone is a secondary pollutant. It will form when Hydrocarbons and nitrogen oxides combine

in the presence of sunlight. Choice (C)

20 kN

20 kN 20 kN

20 kN

L

G

E

S

T

P

N K

I

F H J M O R

2 m 2 m 2 m 2 m 2 m 2 m 2 m

1 m

1 m

1 m

1 m


11. A concentrated load of 500 kN is applied on an elastic half space. The ratio of the increase in

vertical normal stress at depths of 2 m and 4 m along the point of the loading, as per Boussinesq’s theory, would be ______

Sol. As per Boussinesq’s equation,

z = 3𝑄

2𝛱𝑧2 [1

1+(𝑟2⁄ )

2]

52⁄

r = 0

z 2z

1

2

2

4

2

2

4=

4

2

= 4 Ans: 4

12. Consider a two-dimensional flow through isotropic soil along x direction and z direction. If h

is the hydraulic head, the Laplace’s equation of continuity is expressed as

(A) ∂h

∂x+

∂h

∂x ∂h

∂z+

∂h

∂z= 0

(B) ∂h

∂x+

∂h

∂z= 0

(C) ∂2h

∂x2 + ∂2h

∂x ∂z+

∂2h

∂z2 = 0

(D) ∂2h

∂x2 +∂2h

∂z2 = 0

Sol. For a two dimensional flow through an isotropic soil along x direction and z direction Laplace’s

equation of continuity is given by ∂2 h

∂ x2 +

∂2 h

∂ z2 = 0

h – hydrantic head Choice (D) 13. For a given loading on a rectangular plain concrete beam with an overall depth of 500 mm,

the compressive strain and tensile strain developed at the extreme fibers are of the same

magnitude of 2.5 10–4. The curvature in the beam cross-section (in m–1, round off to 3 decimal places), is ________

Sol. D = 500mm

given, c = t = 2.5 10-4

we have, f

y=

E

R

curvature = 1

R =

f

yE =

E

y

= 2.5 ×10−4

5002⁄

= 1 10–6 mm = 0.001m Ans: 0.001 m



14. A retaining wall of height H with smooth vertical backface supports a backfill inclined at an

angle with the horizontal. The backfill consists of cohesionless soil having angle of internal

friction . If the active lateral thrust acting on the wall is P, which one of the following statements is TRUE?

(A) Pa acts at a height H/3 from the base of the wall and at an angle with the horizontal

(B) Pa acts at a height H/2 from the base of the wall and at an angle with the horizontal

(C) Pa acts at a height H/3 from the base of the wall and at an angle with the horizontal

(D) Pa acts at a height H/2 from the base of the wall and at an angle with the horizontal Sol.

Pa will act at a distance of H 3⁄ from the bottom at an angle of β. Choice (C)

15. A completely mixed dilute suspension of sand particles having diameters 0.25, 0.35, 0.40, 0.45 and 0.50 mm are filled in a transparent glass column of diameter 10 cm and height 2.50 m. The suspension is allowed to settle without any disturbance. It is observed that all particles of diameter 0.35 mm settle to the bottom of the column in 30 s, For the same period of 30 s, the percentage removal (round off to integer value) of particles of diameters 0.45 and 0.50 mm from the suspension is _______

Sol. d = 10 cm h = 2.5 m particles of size 0.35 mm ; 100% removal in 30 seconds So in same 30s, particles of dia 0.45 and 0.50 mm

(size = 0.35 mm) s d

Removal : 100%. Ans: 100

16. In the reinforced beam section shown in the figure (not drawn to scale), the nominal cover provided at the bottom of the beam as per IS 456-2000, is

(A) 30 mm

(B) 50 mm (C) 42 mm (D) 36 mm

2 – 28 φ

2-legged, 12φ Stirrups @ 90 c/c

50

50

350

3 – 16 φ

All dimensions are in mm


Sol. Nominal cover at bottom = effective cover – stirrup diameter – dia of main bar

2

= 50 – 12 – 16

2

= 30 mm Choice (A) 17. Which one of the following is correct?

(A) lim𝑥 →0

(sin 4𝑥

sin 2𝑥) = 1 and lim

𝑥 →0(

tan 𝑥

𝑥) = 1

(B) lim𝑥 →0

(sin 4𝑥

sin 2𝑥) = ∞ and lim

𝑥 →0(

tan 𝑥

𝑥) = 1

(C) lim𝑥 →0

(sin 4𝑥


𝑥 →0(

tan 𝑥

𝑥) = ∞

(D) lim𝑥 →0

(sin 4𝑥


𝑥 →0(

tan 𝑥

𝑥) = 1

Sol. limx→ 0

(sin 4x

sin 2x) = lim

x→ 0(

4 cos 4x

2 cos 2x)

(By L’ Hospital’s Rule)

= 4

2

= 2

Limx→ 0

(tan x

x) = 1 Choice (D)

18. The maximum number of vehicles observed in any five minute period during the peak hour is 160. If the total flow in the peak hour is 1000 vehicles, the five minute peak hour factor (round off to 2 decimal places) is _______

Sol. q = 160 × 60

5

= 1920 vehicles / hour

PHF = Hourly volume

Max.rate of flow

= 1000

1920

= 0.52 Ans: 0.51 to 0.53

19. Consider the pin-jointed plane truss shown in the figure (not drawn to scale). The RP, RQ and RR denote the vertical reactions (upward positive) applied by the supports at P, Q, and R, respectively, on the truss. The correct combination of (RP, RQ, RR) is represented by

(A) (10, 30, –10) kN

(B) (30, –30, 30) kN (C) (20, 0, 10) kN (D) (0, 60, –30) kN

3 m

3 m

3 m 3 m

2 m

Q R P

30 kN


Sol.

Since truss is carrying vertical load, horizontal reaction at P would be equal to zero

V = 0

RP + RQ + RR = 30 → (1)

MR = 0

(RP × 9 – 30) × (6 + RQ × 3) = 0 → (2) Applying method of section (1 – 1) Consider LHS

V = 0, RP – 30 = 0

RP = 30 kN [] Now substitute RP in equation–––– (2) 30 × 9 – 30 × 6 + RQ × 3 = 0 RQ = –30 kN = 30 kN [↓] Substitute RQ and RP in equation (1) 30 – 30 + RR = 30

RR = 30 kN [] Choice (B)

20. For a small value of h, the Taylor series expansion f(x + h) is

(A) f (x) + ℎ𝑓′ (𝑥) + ℎ2

2 𝑓" (𝑥) +

ℎ3

3 𝑓′′′ (𝑥) + ⋯ ∞

(B) f (x) – ℎ𝑓′ (𝑥) + ℎ2

2! 𝑓" (𝑥) −

ℎ3

3! 𝑓′′′ (𝑥) + ⋯ ∞

(C) f (x) + ℎ𝑓′(𝑥) + ℎ2

2! 𝑓" (𝑥) +

ℎ3

3! 𝑓′′′ (𝑥) + ⋯ ∞

(D) f (x) – ℎ𝑓′ (𝑥) + ℎ2

2 𝑓" (𝑥) −

ℎ3

3 𝑓′′′ (𝑥) + ⋯ ∞

Sol. The Taylor series expansion for f(x + h) is

f(x + h) = f(x) + hf’(x) + h2

2! f”(x) +

h3

3! f”’(x) + …. Choice (C)


21. Assuming that there is no possibility of shear buckling in the web, the maximum reduction

permitted by IS 800-2007 in the (low-shear) design bending strength of a semi-compact steel section due to high shear is (A) zero (B) governed by the area of the flange (C) 50% (D) 25%

Sol. As per is : 800 2007 For semi – compact section,

a) In the case of low shear [V 0.6 Vd]

Md = Ze fy

mo

→ (i)

b) In the case of high shear [V > 0.6 Vd]

Md = Ze fy

mo

→ (ii)

So, from equation (i) and (ii) reduction is zero Ans: 0

22. An element is subjected to biaxial normal tensile strains of 0.0030 and 0.0020. The normal strain in the plane of maximum shear strain is (A) Zero (B) 0.0050 (C) 0.0010 (D) 0.0025

Sol. x = 0.0030

y = 0.0020

n = ∈𝑥+ ∈𝑦

2

= 0.0030+0.0020

2 = 0.0025 Choice (D)

23. In a rectangular channel, the ratio of the velocity head to the flow depth for critical flow condition, is

(A) 3

2

(B) 1

2

(C) 2

3

(D) 2

Sol. At critical condition Fr = 1

V

√gyc = 1

Square both sides V2

gyc = 1

V2

g = yc

V2

2g =

yc

2

So, velocity head

critical depth =

V2

2g⁄

yc = 1 2⁄ . Choice (B)


24. In a soil specimen, the total stress, effective stress, hydraulic gradient and critical and

hydraulic gradient are , ', i and ic, respectively. For initiation of quicksand condition, which one of the following statements is TRUE?

(A) ' = 0 and i = ic

(B) ' 0 and i = ic

(C) ' 0 and i ic

(D) = 0 and i = ic

Sol. For initiation of quick sand condition, effective stress, = 0 Critical hydraulic gradient (ic) = exit grisliest (ie) Choice (A)

25. A catchment may be idealised as a rectangle. The are three rain gauges located inside the catchment at arbitrary locations. The average precipitation over the catchment is estimated by two methods: (i) Arithmetic mean (PA), and (ii) Thiessen polygon (PT). Which one of the following statements is correct? (A) PA is always smaller than PT (B) PA is always equal to PT (C) There is no definite relationship between PA and PT (D) PA is always greater than PT

Sol. There is no any definite relationship between precipitation (P) values calculated by Arithmetic mean and Theissen polygon method. Choice (C)

Q. 26 – Q. 55 carry two marks each.

26. A one-dimensional domain is discretized into N sub-domains of width x with node numbers

i = 0,1,2,3,……N. If the time scale is discretized in step of t the forward-time and centered-space finite difference approximation at ith node and nth time step, for the partial differential

equation 𝜕𝑣

𝜕𝑡= 𝛽

𝜕2𝑣

𝜕2𝑥2 is

(A) 𝑣𝑖+1

(𝑛+1)− 𝑣𝑖(𝑛)

∆𝑡 = 𝛽 [

𝑣𝑖+1(𝑛)− 2𝑣𝑖

(𝑛) + 𝑣𝑖−1(𝑛)

2∆𝑥]

(B) 𝑣𝑖

(𝑛+1)− 𝑣𝑖(𝑛)

∆𝑡 = 𝛽 [


(𝑛) + 𝑣𝑖−1(𝑛)

(∆𝑥)2 ]

(C) 𝑣𝑖

(𝑛)− 𝑣𝑖(𝑛−1)

∆𝑡 = 𝛽 [


(𝑛) + 𝑣𝑖−1(𝑛)

(∆𝑥)2 ]

(D) 𝑣𝑖

(𝑛)− 𝑣𝑖(𝑛−1)

2∆𝑡 = 𝛽 [


(𝑛) + 𝑣𝑖−1(𝑛)

2∆𝑥]

Sol. At ith node and nth time step,

∂V

∂t =

Vi(n+1)

–Vi(n)

△t → (1)

and ∂2V

∂x2 = Vi+1

(n)–2Vi

(n)+Vi–1

(n)

(△x)2 → (2)

Given PDE is

∂V

∂t =

∂2V

∂x2

Vi

(n+1)–Vi

(n)

△t = [

Vi+1(n)

–2Vi(n)

+Vi–1(n)

(△x)2 ] (From (1) & (2)) Choice (B)



27. A staff is placed on a benchmark (BM) of reduced level (RL) 100,00 m and a theodolite is

placed at a horizontal distance 0f 50 m from the BM to measure the vertical angles. The measured vertical angles from the horizontal at the staff readings of 0.400m and 2.400 m are found to be the same. Taking the height of the instrument as 1.400 m, the RL (in m) of the theodolite station is ______

Sol.

1 = 2

d = S

tan ∝1+ tan ∝1

50 = 2.4 −0.4

2tan∝

= 1.1457

tan2 = 𝑥

50

x = 1 m s = 1.400 RL of instrument station = 100 + 1.400 – 1.400 = 100.00m Ans: 100.00 m

28. The hyetograph of a storm of event of duration 140 minutes is shown in the figure. The infiltration capacity at the start of this event (t = 0) is 17 mm/hour, which linearly decreases

to 10 mm/hour after 40 minutes duration. As the event progresses, the infiltration rate further drops down linearly to attain a value of 4 mm/hour at t = 100 minutes and remains constant

thereafter till the end of the storm event. The value of the infiltration index, (in mm/hour, round off to 2 decimal places), is ______

Infiltration Capacity

Rainfall Intensity 17

15

10

8

3 1

8

4

0 20 40 60 80 100 120 140

Time, t (minutes)

Infiltra

tion C

ap

acity (

mm

/hour)

Rain

fall

Inte

nsity (

mm

/hour)


Sol. Total precipitation,

P = 20

60 × [4 + 8 + 15 + 10 + 8 + 3 + 1] = 16.3 mm

Infiltrated I = Area under curve

= 20

60 x ( 4 + 8 + 3 + 1) +

1

2 x (10 – 4) x

60

60 + 4 x

60

60

= 12.33 mm Runoff = R = P – I = 16.33 – 12.33 = 4 mm For te = 140 minutes Pe = 16.33 mm R = 4 mm

index = Pe –R

te=

16.33 –4

140/60= 5.29 mm/hr

but for time interval, 0 – 20, i = 4 mm/hr 100 – 120 i = 3 mm/hr 120 – 140 i = 1 mm/hr So repeat for te = 140 – 60 = 80 minutes

index = (8+15+10+8) ×

20

60 – 4

80 = 7.25 mm/hr Ans: 7.00 to 7.30

29. Average free flow speed and the jam density observed on a road stretch are 60 km/h and 120 vehicles/km, respectively. For a linear spead-density relationship, the maximum flow on the road stretch (in vehicles/h) is _______

Sol. qmax = V𝐾𝑗

4

= 60 ×120

4

= 1800 vehhr⁄ Ans: 1800

30. Traffic on a highway is moving at a rate of 360 vehicles per hour at a location. If the number of vehicles arriving on this highway follows Poisson distribution, the probability (round off to 2 decimal places) that the headway between successive vehicles lies between 6 and 10 seconds is _______

Sol. Prob of time headway between 6 to 10 sec = p(0 < t < 10) – p(0 < t < 6)

= ∫ f (t)dt10

0 – ∫ f (t)dt

6

0

(1 – e –µ(10)) – (1 – e –µ(6))

1 hr → 360 Vehicles

60 min → 360 Vehicles

1 min → 6 Vehicles

60 sec → 6 Vehicles

sec → (1

10) Vehicles

µ = 0.1

p (6 → 10 sec)

(1 – e –0.1 10) – (1 – e –0.1 0.6) = 0.18 (OR)

Poison distribution function, f(t) = (t)n e–t

n!

Here n = 0 in both cases.

p(Vehicles b/w 6 and 10) = e–0.1 ×6 – e–0.1 × 10 = 0.18 Ans: 0.17 to 0.19


31. Sedimentation basin in a water treatment plant is designed for a flow rate of 0.2 m3/s. The

basin rectangular with a length of 32 m, width of 8 m, and depth of 4 m. Assume that the settling velocity of these particles is governed by the stokes’ law. Given: density of the particles = 2.5 g/cm3; density of water = 1 g/cm3; dynamic viscosity of water = 0.01 g/(cm.s): gravitational acceleration = 980 cm/s2. If the incoming water contains particles of diameter

25m (spherical and uniform), the removal efficiency of these particles is (A) 100% (B) 51% (C) 78% (D) 65%

Sol. flow rate = 0.2 m3/s L= 32m B = 8m H = 4m

= 2.5g/cm3

= 1g/cm3

= 0.01g/cm.s g = 980 cm/s2

d = 25m = 25 10-4cm As per stokes law,

vs = 1

18 gd2(ρ− 𝜌𝜔)

μ

= 1

18 ×

980 ×(25×10−4)2 (2.5−1)

0.01

= 0.0510 cm/s

Overflow rate = Q

L ×B=

0.2

32 ×8

= 7.812 × 10–4 m3

m2s⁄⁄

Vs = overflow rate

= 0.0510 ×10−2

7.812 × 10−4 × 100

= 65.3% Choice (D)

32. A 0.80 m deep bed of sand filter (length 4 m and width 3 m) is made of uniform particles

(diameter = 0.40 mm, specific gravity = 2.65, shape factor = 0.85) with bed porosity of 0.4. The bed has to be backwashed at a flow rate of 3.60 m3/min. During backwashing, if the terminal settling velocity of sand particles is 0.05 m/s, the expanded bed depth (in m, round off to 2 decimal places) is ______

Sol. ze

z⁄ = 1–n

1–ne

z = 0.8 m L = 4 m B = 3 m d = 0.4 mm S = 2.65 𝜐𝑠 = 0.05 m/s 𝜙 = 0.85 n = 0.4 Q(backwash) = 3.6 m3/min

b = s . ne4.5

b = 0.05 ne4.5


b = Q(b)

A =

(3.6/60) m3/s

4 × 3 m2

= 5 10–3 m/s 0.005 = 0.05 . ne

4.5

ne = 0.602

ze = z(1–n)

1–ne =

(0.8)(1–4)

(1–0.602)

= 1.20603 m Ans: 1.15 to 1.25

33. Which one of the following is NOT a correct statement?

(A) The function √𝑋𝑥

, (x>0), has the global minima at x = e

(B) The function √𝑋𝑥

, (x>0), has the global maxima at x = e (C) The function x3 has neither global minima nor global maxima (D) The function |𝑥| has the global minima at x = 0

Sol. Let y = √xx

= x

x⁄

ln y = ln (x

x⁄ )

ln y = 1

x ln (x)

Differentiating w.r.t. x on both sides, 1

y dy

dx =

1

x (

1

x) + (

–1

x2) ln x

dy

dx = y [

1

x2(1– lnx)]

dy

dx = 0

y

x2(1 – lnx)

1 – lnx = 0

lnx = 1 x = e

Now dy

dx =

y (1– lnx)

x2

d2y

dx2 = d

dx(

y (1– lnx)

x2 )

= dy

dx(

1– lnx

x2 ) + y d

dx(

1– lnx

x2 )

= (y( 1– lnx)

x2 ) ( 1– lnx

x2 ) + y ( x2(

–1

x)– (1–lnx)2x

x4 )

= y ( 1– lnx

x2 )2 + y (

–3x+ 2xlnx

x4 )

At x = e, d2y

dx2

= e1

e⁄ ( 1– lne

e2 )2 + e

1e⁄ (

–3e+ 2elne

e4 )

= –𝑒1 𝑒⁄

e3 < 0

y = √xx

has a maximum at x = e Option (A) is NOT correct Choice (A)


34. The rigid-jointed plane frame QRS shown in the figure is subjected to a load P at the joint R.

Let the axial deformations in the frame be neglected. If the support S undergoes a settlement

of ∆ = PL3

βEI, the vertical reaction at the support S will become zero when is equal to

(A) 7.5 (B) 3.0 (C) 48.0 (D) 0.1 Sol. Distribution Factors

(D.F)RQ = (D.F)RS = 0.5

MFQR = −6𝐸Ι𝛿

ℓ2 = MFRQ

MFRS = MFSR = 0

QR RQ RS SR

0.5 0.5

F.E.M −6𝐸Ι𝛿

ℓ2

−6𝐸Ι𝛿

ℓ2

0 0

Blance

50% 3𝐸Ι𝛿

ℓ2 3𝐸Ι𝛿

ℓ2

50%

C.O.M 1.5𝐸Ι𝛿

ℓ2

1.5𝐸Ι𝛿

ℓ2

F.M. −4.5𝐸Ι𝛿

ℓ2

−3𝐸Ι𝛿

ℓ2

−3𝐸Ι𝛿

ℓ2

1.5𝐸Ι𝛿

ℓ2

Take QR member

Force due to settlement at R is 7.5𝐸Ι𝛿

ℓ3 = P

= 𝑃ℓ3

7.5𝐸Ι, So, = 7.5 Choice (A)

R Q

EI

P

EI L

S

L



35. A reinforced concrete circular pile of 12 m length and 0.6 m diameter is embedded in stiff clay

which has an undrained unit cohesion of 110 kN/m2. The adhesion factor is 0.5. The Net Ultimate Pullout (uplift) Load for the pile (in kN, round off to 1 decimal place) is _____

Sol. L = 12m

Ø = 0.6m

= 110 kN/m2

= 0.5 Net ultimate pullout load = skin friction force = As Fs

= Ø L C

= ×0.6 × 12 × 0.5 × 110 = 1244 kN

Ans: 1240 to 1250

36. For the following statements: P – The lateral stress in the soil while being tested in an oedometer is always at-rest. Q – For a perfectly rigid strip footing at deeper depths in a sand deposit, the vertical

normal contact stress at the footing edge is greater than that at its centre. R – The corrections for overburden pressure and dilatancy are not applied to measured

SPT- N values in case of clay deposits.

The correct combination of the statements is (A) P – TRUE; Q – TRUE; R – FALSE (B) P – FALSE; Q – FALSE; R – TRUE (C) P – FALSE; Q – FALSE; R – FALSE (D) P – TRUE; Q – TRUE; R – TRUE Sol. All of the above given statements are true. Choice (D) 37. A rectangular open channel has a width of 5 m and a bed slope of 0.001. For a uniform flow

of depth 2 m, the velocity is 2 m/s. The Manning’s roughness coefficient for the channel is (A) 0.050 (B) 0.002 (C) 0.033 (D) 0.017

Sol. B= 5m

S = 0.001 y = 2m v = 2m/s

v = 1

n R2/3 S1/2

2 = 1

n (

𝐴

p)

2/3 S1/2

2 = 1

n (

5 ×2

5+2 ×2)

2/3 (0.001)1/2

n = 0.017 Choice (D)


500mm

100mm

350kN

500mm

38. A 16 mm thick gusset plate is connected to the 12 mm thick flange plate of an I-section using

fillet welds on both sides as shown in the figure (not drawn to scale). The gusset plate is subjected to a point load of 350 kN acting at a distance of 100 mm from the flange plate. Size of fillet weld is 10 mm.

The maximum resultant stress (in MPa, round off to 1 decimal place) on the fillet weld along

the vertical plane would be _______ Sol.

Shear stress due to P, is = force

Area

T = 350 × 103

2 ×500 ×0.7 × 10

= 50 MPa

Bending moment = 350 100 = 35 kNm

Bending stress = M

I y

350 ×100 × 103

2 × (0.7 ×10) (500)2

6

60 MPa

Resultant stress = √𝑇2 + 3𝜎2

√502 + 3(60)2

105.35 MPa Ans: 78.0 to 78.2 or 105.3 to 105.5

16 mm thick gusset plate

30 kN 100 mm

500 mm

Fillet weld I-section

Flange (12 mm thick)

(Front view) (Side view)

Fillet weld

500 mm


F = 240 N

39. The cross-section of a built-up wooden beam as shown in the figure (not drawn to scale) is

subjected to a vertical shear force of 8 kN. The beam is symmetrical about the neutral axis

(N.A.) shown, and the moment of inertia about N.A. is 1.5 109 mm4. Considering that the nails at the location P are spaced longitudinally (along the length of the beam) at 60 mm, each of the nails at P will be subjected to the shear force of

(A) 240 N (B) 480 N (C) 60 N (D) 120 N

Sol. Shear stress acting on the Joint at which the nail ‘P’ is resisting the shear

= max = 𝐹. 𝐴�̅�

Ι𝑏

= 8 × 103 ×(50 ×100) × 150

1.5 × 102 ×50 = 0.08 MPa

The shear force F = Area

F = 0.08 60 50

Ans: 240 N

40. Tie bars of 12 mm diameter are to be provided in a concrete pavement slab. The working

tensile stress of the tie bars is 230 MPa. the average bond strength between a tie bar and concrete is 2 MPa, and the joint gap between the slabs is 10 mm. Ignoring the loss of bond and the tolerance factor, the design length of the tie bars (in mm, round off to the nearest integer) is _____

Sol. Length of tie bar, LT

2=

∅σS

τbd

LT = 2 × ∅σS

τbd =

2 ×12 ×230

4 × 2 = 690mm

Total length = 690 + 10 = 700 mm Ans: 700 mm

P

50 300 50

100 100

400 N.A.

50

All dimensions are in mm


41. Consider two functions: x = ln and y = ln. Which one of the following is the correct

expression for ∂

∂x ?

(A) ln

ln ln –1

(B) xln

ln ln –1

(C) ln

ln ln –1

(D) xln

ln ln –1

Sol. Given x = ln → (1)

ln = x

𝛹 → (2)

And y = ln→ (3)

ln = y

∅ → (4)

From (1) ; = x

ln

∂𝛹

∂x =

ln – x∂

∂x(ln )

(ln )2

= ln – x(

1

.

∂

∂x)

(ln )2

∂𝛹

∂x =

ln – x

.

∂

∂x

(ln )2 → (5)

From (3), = y

ln𝛹

∂

∂x =

(ln 𝛹)0 – y∂

∂x(ln 𝛹)

(ln 𝛹)2

= 0 – y(

1

𝛹 .

∂𝛹

∂x)

(ln 𝛹)2

𝜕

𝜕𝑥 =

–(y

𝛹⁄ )∂𝛹

∂x

(ln 𝛹)2 → (6)

Substituting (6) in (5), we get

∂𝛹

∂x =

ln – (x

)(

– (y

𝛹⁄ )∂𝛹∂x

(ln 𝛹)2 )

(ln )2

=

ln + (

x𝛹

) (y

)∂𝛹∂x

(ln 𝛹)2

(ln )2

= ln +

(ln 𝛹) (ln )∂𝛹∂x

(ln 𝛹)2

(ln )2 (From (2) and (4)


= ln +

ln

ln 𝛹 ∂𝛹

∂x

(ln )2

= 1 +

1

ln 𝛹 ∂𝛹

∂x

ln

∂𝛹

∂x =

ln 𝛹+ ∂𝛹

∂x

ln .ln 𝛹

ln . ln ∂𝛹

∂x = ln +

∂𝛹

∂x

ln . ln ∂𝛹

∂x –

∂𝛹

∂x = ln

(ln . ln – 1) ∂𝛹

∂x = ln

∂𝛹

∂x =

ln 𝛹

ln .ln 𝛹 –1 Choice (A)

42. A parabolic vertical curve is being designed to join a road of grade +5% with a road of grade –3%. The length of the vertical curve is 400 m measured along the horizontal. The vertical

point of curvature (VPC) is located on the road of grade +5%. The difference in height between VPC and vertical point of intersection (VPI) (in m, round off to the nearest integer) is _______

Sol.

Vertical distance between point of curve and vertical point = 5

100 ×

400

2= 10 𝑚

Ans: 10 m

43. A sample of air analysed at 0°C and 1 atm pressure is reported to contain 0.02ppm (parts per

million) of NO2. Assume the gram molecular mass of NO2 as 46 and its volume at 0°C and 1 atm pressure as 22.4 liters per mole. The equivalent NO2 concentration (in microgram per cubic meter, round off to 2 decimal places) would be________

Sol. At S.T.P i.e 0o and 1 atmp Vol. of 1 mole of NO2 = 22.2 lt @ S.T.P and 1 atmp and 0o, 22.4 lt of NO2 has mass of 46 gm

Mass of 1 litre of NO2 : 46

22.4

= 2.053 gm


Mass of 1 m3 or 1000 lt

= 2.053 103 gm

1ppm of NO2 = 2.053 × 103

106 gm

m3

= 2.053 × 109μg

106 m3

1 ppm = 2053.57 1g/m3

0.02 ppm NO2 = 0.02 2053.57

= 41.07 1g/m3

(∵ 1 ppm NO2 : 1 part NO2

106 parts air

= 1 m3 NO2

106 m3 air) Ans: 41.00 to 41.10

44. A granular soil has a saturated unit weight of 20 kN/m3 and an effective angle of shearing

resistance of 30°. The unit weight of water is 9.81 kN/m3. A slope is to be made on this soil deposit in which the seepage occurs parallel to the slope up to the free surface. Under this seepage condition for a factor of safety of 1.5, the safe slope angle (in degree, round off to 1 decimal place) would be_______

Sol. sat = 20 kN /m3

Ø = 30

w = 9.81 kN/m3 FOS = 1.5 Given water table is at ground level

FOS = sub

sat

tan ∅ 1

tan β

1.5 = 20 –9.81

20 tan 30

tan β

= 11.09 Ans: 10.8 to 11.3

45. A portal frame shown in figure (not drawn to scale) has a hinge support at joint P and a roller

support at joint R. A point load of 50 kN is acting at joint R in the horizontal direction. The flexural rigidity. EI. of each member is 106 kNm2. Under the applied load, the horizontal displacement (in mm, round off to 1 decimal place) of joint R would be _______

P 5 m

EI

Q

EI 10 m

R 50 kN


(RR)v = 100(kN)

(RP)v = 100(kN)

m = M

50

Sol.

Fx = 0 (RP)H – 50 = 0 (RP)H = 50 (kN)

Fy = 0 (RP)v – (RR)v = 0 –––- (1)

(M)p = 0

50 10 = (RR)v 5 So, from Ist Equation

By Unit Load method,

∆net = ∫Mm

EI dx Where

Span Origin Length Moment m = M

50

PQ P 5 m 100x 2x

QR R 10 m 50x x

∆net = ∫Mm

EI dx

= ∫ (100x ×2x dx

EI)

5

0 + ∫ (

50x (x) dx

EI)

10

0

= 200

EI [

x3

3]

0

5

+ 50

EI [

x3

3]

0

10

= 200

EI [

125

3] +

50

EI [

1000

3]

= 25000 + 50000

3EI =

75000

3EI

= 25000

EI =

25000

106 103 = 25 mm Ans: 24.9 to 25.1

5 m

EI

Q

(RP)H

(RP)V

EI

(RR)V

R 50 kN

10 m

P

x

x


46. If the section shown in the figure turns from fully-elastic to fully-plastic, the depth of neutral

axis (N.A.). �̅�, decreases by

(A) 12.25 mm (B) 15.25 mm (C) 10.75 mm (D) 13.75 mm

Sol.

Plastic state means compression Area = Tensile Area.

N.A (elastic state) Aixi

Ai

[60 × 5 ×2.5 ]+[60 ×5 ×(5+30)

(60 ×5)+60 ×5

18.75 mm. N.A @ plastic state

18.75 – 5

13.75 mm. Choice (D) 47. Consider a laminar flow in the x-direction between two infinite parallel plates (Couette flow).

The lower plate is stationary and the upper plate is moving with a velocity of 1cm/s in the x-direction. The distance between the plates is 5 mm and the dynamic viscosity of the fluid is

0.01 N-s/m2. If the shear stress on the lower plate is zero, the pressure gradient, 𝜕𝑝

𝜕𝑥, (in N/m2

per m, round off to 1 decimal place) is ________

NA 60 mm

5 mm

60 mm

5 mm

y

60

N.A.

5

60

5

Figure not to scale All dimensions are in mm

y̅


Sol. 𝜏𝐿𝑜𝑤𝑒𝑟𝑝𝑙𝑎𝑡𝑒 = 0

Velocity distribution equation for couette flow

u = – 1

2𝜇(

𝜕𝑝

𝜕𝑥)(By – y2) +

𝑉

𝐵 y

𝑑𝑢

𝑑𝑦 = –

1

2𝜇(

𝜕𝑝

𝜕𝑥)(B – 2y) +

𝑉

𝐵

𝑑𝑢

𝑑𝑦]

𝑦=0= –

1

2𝜇(

𝜕𝑝

𝜕𝑥)B +

𝑉

𝐵

w = 𝑑𝑢

𝑑𝑦]

𝑦=0

0 = [−1

2𝜇(

𝜕𝑝

𝜕𝑥) B +

𝑉

𝐵]

𝐵

2𝜇 (

𝜕𝑝

𝜕𝑥) =

𝑉

𝐵

𝜕𝑝

𝜕𝑥 =

2𝜇𝑉

𝐵2 = 2 ×0.01 ×0.01

(0.005)2 = 8N/m2 per m Ans: 7.9 to 8.1

48. A square footing of 4 m side is placed at 1 m depth in a sand deposit. The dry unit weight () of sand is 15 kN/m3. This footing has an ultimate bearing capacity of 600 kPa. Consider the

depth factors: dq = d = 1.0 and the bearing capacity factor: N = 18.75. This footing is placed at a depth of 2 m in the same soil deposit. For a factor of safety of 3.0 as per Terzaghi’s theory. The safe bearing capacity (in kPa) of this footing would be _____

Sol. Square footing, B = 4m D1 = 1m D2 = 2m

d = 15 kN/m3

qu = 600 kpa

N = 18.75 For D1 = 1m

qu = D1 Nq + 0.4 B N

qu = D1 Nq + 0.4 × 15 × 4 × 18.75

600 = (15 1 Nq) + (0.4 15 4 18.75) Nq = 10 For D2 = 2m

qu = D2 Nq + 0.4BN = (15 × 2 × 10) + (0.4 × 15 × 4 × 18.75) = 750 kPa


Net ultimate

qnu = qu – D2 = 750 – 15 × 2 = 720 kPa Net Safe

qns = 𝑞𝑛𝑢

FOS +

720

3 = 240 kPa

Gross safe bearing capacity

qs = qns + Df = 240 + 15 × 2 = 270 kPa Ans: 240 or 250 or 270

49. A box is measuring 50 cm 50 cm 50 cm is filled to the top with dry coarse aggregate of mass 187.5 kg. The water absorption and specific gravity of the aggregate are 0.5% and 2.5, respectively. The maximum quantity of water (in kg, round off to 2 decimal places) required to fill the box completely is________

Sol. Volume of box = (0.5)3 = 0.125m3

Mass of aggregate = 187.5 kg Gagg = 2.5

Volume of aggregate = Mass of aggregate

density of aggregate

= 187.5

2.5 ×1000 = 0.075m3

Volume of empty space = 0.125 – 0.075 = 0.05m3 water absorption = 0.5%

Volume of water absorbed = 0.5

100 × 0.075 = 3.75 × 10–4𝑚3

Total volume of water that can be filled = (3.75 × 10–4) + 0.05 = 0.050375m3

Density of water = Mass of water

Volume of water

Mass of water = 1000 × 0.050375 = 50.375 kg Ans: 50.5 to 51.2

50. A wastewater is to be disinfected with 35 mg/L of chlorine to obtain 99% kill of micro-organisms. The number of micro-organisms remaining alive (Nt) at time t, is modelled by

Nt = N0 𝑒−𝑘𝑡, where N0 is number of micro-organisms at t=0, and k is the rate of kill. The waste water flow rate is 36 m3/h. and k = 0.23 min–1. If the depth and width of the chlorination tank are 1.5 m and 1.0 m, respectively, the length of the tank (in m, round off to 2 decimal places) is _______

Sol. = (No–Nt

No) 100

= (No–Noe

–kt

No) 100

99 = (1 – e–kt) 100 (1 – e–0.23kt) = 0.99 t = 20.022 min V = Q.t

= (36

60) m3

min⁄ 20.022 (min) = 12.013 m3

L = V

B.H =

12.0.13

1 × 1.5 = 8.01 m Ans: 7.95 to 8.15


51. A survey line was measured to be 285.5 m with a tape having a nominal length of 30 m. On

checking, the true length of the tape was found to be 0.05 m too short. If the line lay on a slope of 1 in 10, the reduced length (horizontal length) of the line for plotting of survey work would be (A) 285.6 m (B) 284.5 m (C) 283.6 m (D) 285.0 m

Sol. Measured length = 285.5m Nominal length of tape = 30m True length of tape = 30 – 0.05 = 29.95m

True length of line = 285 .5 ×29.95

30

= 285.024 m

tan = 1

10

= 5.71

Cos (5.71) = x

285.024

horizontal length = 283.61m Choice (C)

52. Consider the ordinary differential equation 𝑥2 𝑑2𝑦

𝑑𝑥2 – 2𝑥𝑑𝑦

𝑑𝑥+ 2𝑦 = 0. Given values of y(1) = 0

and y(2) = 2, the value of y(3) (round off to 1 decimal place), is ______

Sol. Given x2 d2y

dx2 – 2x dy

dx + 2y = 0 → (1)

where y(1) = 0 and y(2) = 2 → (2)

Put x = ez (OR) z = ln x

x dy

dx = y and x2d2y

dx2 = (–1)y,

where = d

dz

(1) becomes,

( – 1)y – 2y + 2y = 0

[( – 1) – 2 + 2]y = 0

(2 – 3 + 2)y = 0 → (3) Its auxiliary equation is m2 – 3m + 2 = 0

(m – 1)(m –2) = 0

m = 1; m = 2

The general solution of (3) is y = c1e2z + c2ez = c1(ez)2 + c2(ez)

The general solution of (1) is

y = c1x2 + c2x → (4) From (2), y(1) = 0

0 = c1 + c2

c1 + c2 = 0 → (5)


y(2) = 2 2 = c1(2)2 + c2(2)

2c1 + c2 = 1 → (6) Solving (5) and (6), we get c1 = 1 and c2 = –1

(4) becomes, y = x2 – x

y(3) = 32 – 3 = 6 Ans : 5.9 to 6.1 53. The network of a small construction project awarded to a contractor is shown in the following

figure. The normal duration, crash duration, normal cost, and crash cost of all the activities are shown in the table. The indirect cost incurred by the contractor is INR 5000 per day.

Activity Normal

Duration (days)

Crash Duration (days)

Normal cost (INR)

Crash Cost (INR)

P 6 4 15000 25000

Q 5 2 6000 12000

R 5 3 8000 9500

S 6 3 7000 10000

T 3 2 6000 9000

U 2 1 4000 6000

V 4 2 20000 28000

If the project is targeted for completion in 16 days, the total cost (in INR) to be incurred by the

contractor would be ______ Sol.

Activity tnormal

(days) tcrash

(days)

Crash cost (cc)

Normal cost (cn)

Cost Slope 𝑪𝑪 – 𝑪𝑵

𝐭𝐧𝐨𝐫𝐦𝐚𝐥– 𝐭𝐜𝐫𝐚𝐬𝐡

P 6 4 15,000 25,000 5000

Q 5 2 6,000 12,000 2000

R 5 3 8,000 9,500 750

S 6 3 7,000 10,000 1000

T 3 2 6,000 9,000 3000

U 2 1 4,000 6,000 2000

V 4 2 20,000 28,000 4000

P

Q

S

R T

U

V


For 18 days, Direct cost = 60,000 Indirect cost = 90,000 Total project cost = 1,56,000 1st stage crashing, Crash activity by 1 day New project duration = 17 days Total project cost = 156000 + 1 x 750 – 1 x 5000 = 151750 2nd stage crashing Crash activity Q and R simultaneously by 1 day New project duration = 16 days Total project cost = 151750 + 1 x (750 + 2000) = 149500 Ans: 149500

54. A 3 m 3 m square precast reinforced concrete segments to be installed by pushing them through an existing railway embankment for making an underpass as shown in the figure. A reaction arrangement using precast PCC blocks placed on the ground is to be made for the jacks.

At each stage, the jacks required to apply a force of 1875 kN to push the segment. The jacks

will react against the rigid steel plate placed against the reaction arrangement. The footprint area of reaction arrangement on natural ground is 37.5 m2. The unit weight of PCC block is

24 kN/m3. The properties of the natural ground are : c = 17 kPa: = 25° and = 18 kN/m3. Assuming that the reaction arrangement has rough interface and has the same properties that of soil, the factor of safety (round off to 1 decimal place) against shear failure is ______

Sol. Applied force P = 1875 kN

Strength = (C + tan)A Factor of safety FOS

(𝐶 + 𝜎 𝑡𝑎𝑛∅) 𝐴

𝑃

Where = 𝐹𝑜𝑟𝑐𝑒

𝐴𝑟𝑒𝑎 = 24 7.5 kPa

FOS = [(17 +(24 × 7.5 × 𝑡𝑎𝑛25°)]37.5

1875

= 2.0187 = 2 Ans: 1.8 to 2.1

Railway embankment

Jacks

Natural Ground Not to scale

Precast RCC

segment

Precast RCC Blocks

Ste

el pla

te

7.5 m


55. Two water reservoirs are connected by a siphon (running full) of total length 5000 m and

diameter of 0.10 m, as shown below (figure not drawn to scale). The inlet leg length of the siphon to its summit is 2000 m. The difference in the water surface

levels of the two reservoirs is 5 m. Assume the permissible minimum absolute pressure at the summit of siphon to be 2.5 m of water when running full. Given: friction factor f = 0.02 through out, atmospheric pressure = 10.3 m of water, and acceleration due to gravity g = 9.81 m/s2. Considering only major loss using Darcy-Weisbach equation, the maximum height of the summit of siphon from the water level of upper reservoir, h (in m, round off to 1 decimal place) is_____

Sol. Applying Bernoulli’s theorem between 1 and 2 including losses (major)

𝑃1

𝜌𝑔 +

𝑉12

2𝑔 + Z1 =

𝑃2

𝜌𝑔 +

𝑉22

2𝑔 + Z2 + hL

0 + 0 + 5 = 0 + 0 + 0 + 𝑓𝐿𝑉2

2𝑔𝐷

5 = 0.02 × 5000 × 𝑉2

2𝑔 × 0.1

V2 = 0.0981 V = 0.313 m/s Now applying Bernoulli’s theorem between 1 and S including losses (major)

𝑃1

𝜌𝑔 +

𝑉12

2𝑔 + Z1 =

𝑃𝑠

𝜌𝑔 +

𝑉𝑠2

2𝑔 + Z2 + hL

10.3 + 0 + 0 = 2.5 + 0.0981

2𝑔 + h +

𝑓𝐿′𝑉2

2𝑔𝐷

10.3 = 2.5 + 0.0981

2𝑔 + h +

0.02 × 2000 × 0.0981

2 × 9.81 × 0.1

h = 5.795 m = 5.8 m Ans: 5.7 to 5.9

Summit

h = ?

5 m


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New GENERAL APTITUDE · 2019. 4. 16. · GATE-2019 ORIGINAL PAPER – CE/1 GENERAL APTITUDE Q. 1 – Q. 5 carry one mark each. 1. If E = 10; J = 20; O = 30; and T = 40, what will