Civil Engineering Forenoon Session Date of Exam: 09/02/2020

GATE-2020 ORIGINAL PAPER – CVL/1

Civil Engineering – Forenoon Session

Date of Exam: 09/02/2020


GENERAL APTITUDE

1. The sum of two positive numbers is 100. After subtracting 5 from each number, the product of the resulting numbers is 0. One of the original numbers is _____.

(A) 90

(B) 80

(C) 95

(D) 85

Sol. Let a, b be the two numbers.

Given, a + b = 100 and (a – 5)(b – 5) = 0.

This is possible only if one of the numbers is 5.

If one number is 5, the other number will be 95. Choice (C)

2. Five friends P, Q, R, S and T went camping. At night, they had to sleep in a row inside the tent. P, Q and T refused to sleep next to R since he snored loudly. P and S wanted to avoid Q as he usually hugged people in sleep.

Assuming everyone was satisfied with the sleeping arrangements, what is the order in which they slept?

(A) QRSPT

(B) SPRTQ

(C) QTSPR

(D) RSPTQ

Sol. The statement that ‘P, Q and T refuse to sleep next to R. From this statement, choices (A), (B) and (C) get eliminated. Arrangement RSPTQ does not violate any of the given conditions. Choice (D)

3. It is a common criticism that most of the academicians live in their _____, so, they are not aware of the real life challenges.

(A) homes (B) ivory towers (C) big flats (D) glass palaces Sol. To live in an ivory tower is to live in a state of privileged seclusion or separation from the facts and

practicalities of the real world. Choice (B)

4. Select the word that fits the analogy:

Fuse : Fusion :: Use : _____

(A) Usion

(B) Usage

(C) User

(D) Uses

Sol. ‘Fuse-Fusion’ is a verb-noun pair, hence the answer is ‘Use-Usage’. Choice (B)


5%

10%

15%

15%

10%

20%

25%

Health(5%)

Transport(10%)

Household Item(15%)

Education(15%)

Leisure(10%)

House Rent(20%)

Other(25%)

5. The unit’s place in 26591749110016 is _____.

(A) 1

(B) 9

(C) 3

(D) 6

Sol. The unit’s place of an depends on the unit’s digit of a and n.

For 26591749110016, the unit’s digit is 9 and it is raised to an even power.

For any even power of 9, the unit’s digit is 1.

The unit’s digit of 26591749110016 is 1. Choice (A)

6. Insert seven numbers between 2 and 34, such that the resulting sequence including 2 and 34 is an arithmetic progression. The sum of these inserted seven numbers is _____.

(A) 124 (B) 126 (C) 130 (D) 120

Sol. Given, seven numbers are inserted between 2 and 34 such that the nine numbers are in AP.

Sum of the n terms in an AP=n

2(first term + last term)

Sum of the 9 terms = 9

2(2 + 34) = 162.

Sum of the 7 terms inserted = 162 – (2 + 34) = 126. Choice (B)

7. The total expenditure of a family, on different activities in a month, is shown in the pie-chart. The extra money spent on education as compared to transport (in percent) is _____.

(A) 100

(B) 50

(C) 5

(D) 33.3


Sol. The amount spent on transportation is 10 %. The amount spent on education is 15%.

The extra amount spent on education compared to transportation = 15 −10

10× 100

= 50% Choice (B)

8. The American psychologist Howard Gardner expounds that human intelligence can be sub-categorised into multiple kinds, in such a way that individuals differ with respect to their relative competence in each kind. Based on this theory, modern educationists insist on prescribing multi-dimensional curriculum and evaluation parameters that enable development and assessment of multiple intelligences.

Which of the following statements can be inferred from the given text?

(A) Howard Gardner wants to develop and assess the theory of multiple intelligences. (B) Modern educationists want to develop and assess the theory of multiple intelligences. (C) Howard Gardner insists that the teaching curriculum and evaluation needs to be multi-

dimensional. (D) Modern educationists insist that the teaching curriculum and evaluation needs to be multi-

dimensional.

Sol. Choice (A) is not correct, since the theory in question is already formulated.

Modern educationists have little to do with developing theories, hence (B) is out.

Gardner has little to do with teaching curriculums,

hence (C) is out. (D) is supported by ‘modern educationists insist on prescribing multi-dimensional

curriculum and evaluation’. Choice (D)

9. If 0, 1, 2, …, 7, 8, 9 are coded as O, P, Q, …, V, W, X, then 45 will be codes as _____.

(A) TS

(B) SS

(C) ST (D) SU Sol. If 0, 1, 2…8, 9 are O, P, Q….W, X, then 45 would be the place value of number 4 and 5 i.e. ST. Hence the answer is ST. Choice (C)

10. His hunger for reading is insatiable. He reads indiscriminately. He is most certainly a/an _____ reader.

(A) Precocious

(B) voracious

(C) all-round

(D) wise

Sol. A voracious reader is one who is insatiable, needs always to be reading, and who can never read enough

books. Choice (B)


P

CIVIL ENGINEERING

1. Velocity of flow is proportional to the first power of hydraulic gradient in Darcy’s law. This law is applicable to

(A) turbulent flow in porous media (B) laminar flow in porous media (C) laminar as well as turbulent flow in porous media (D) transitional flow in porous media Sol. Darcy’s law is valid in porous media of laminar flow. Choice (B)

2. In an urban area, a median is provided to separate the opposing streams of traffic. As per IRC:86-1983, the desirable minimum width (in m, expressed as integer) of the median, is ______.

Sol. Desirable minimum width of median in urban roads is 5 m Ans: 5 m

3. A reinforcing steel bar, partially embedded in concrete, is subjected to a tensile force P. The figure that appropriately represents the distribution of the magnitude of bond stress (represented as hatched region), along the embedded length of the bar, is

(A) (B) (C) (D)

Sol. Actual variation of bond stress would be parabolic.

In general bond stress depends on shear. Bond stress will be max., where shear is max. But in calculation bond stress is assumed uniform Choice (B)


4. The area of an ellipse represented by an equation 𝐱𝟐

𝐚𝟐 +

𝐲𝟐

𝐛𝟐 = 1 is

(A) 𝛑𝐚𝐛

𝟒

(B) 𝛑𝐚𝐛

𝟐

(C) ab

(D) 𝟒𝛑𝐚𝐛

𝟑

Sol. Area of the ellipse x2

a2 + y2

b2 = 1 = 4 (Area of OAB) = 4 ∫𝑏

𝑎

𝑎

𝑥=0 √𝑎2 – 𝑥2 dx

(∵ 𝑥2

𝑎2 + y2

b2 = 1 y = b

a √𝑎2 – 𝑥2)

= ab

Choice (C)

5. In a two-dimensional stress analysis, the state of stress at a point P is

[𝛔] = [𝛔𝐱𝐱 𝛕𝐱𝐲

𝛕𝐱𝐲 𝛔𝐲𝐲 ]

The necessary and sufficient condition for existence of the state of pure shear at the point P, is

(A) 𝛔𝐱𝐱 𝛔𝐲𝐲 − 𝛕𝐱𝐲𝟐 = 0

(B) (𝛔𝐱𝐱 − 𝛔𝐲𝐲)𝟐 + 𝟒𝛕𝐱𝐲𝟐

= 0

(C) xy = 0

(D) xx + yy = 0

Sol. = [𝑥𝑥 𝜏𝑥𝑦

𝜏𝑥𝑦 𝑦𝑦]

For pure shear stress

The centre of Mohr’s circle coincides with the origin of – τ co – ordinate system

𝜏𝑦𝑦

Choice (D)

O

Y

X

B

A

a

𝑥 𝑦

𝜏𝑦𝑥

𝜏𝑥𝑦

τ

∴ 𝑥𝑥 = – 𝑦𝑦

∴ 𝑥𝑥 + 𝑦𝑦 = 0


6. The data for an agricultural field for a specific month are given below: Pan Evaporation = 100 mm Effective Rainfall = 20 mm (after deducting losses due to runoff and deep percolation) Crop Coefficient = 0.4 Irrigation Efficiency = 0.5 The amount of irrigation water (in mm) to be applied to the field in that month, is (A) 0 (B) 20 (C) 80 (D) 40

Sol. Depth of water required by crop : pan evaporation crop coefficient

= 100 0.4 = 40 mm

– 20 mm → effective rainfall

additional water required 20 mm

Considering efficiency of irrigation, depth of water required = 20 mm 0.5 = 40 mm Ans: 40 mm

7. Consider the planar truss shown in the figure (not drawn to the scale)

Neglecting self-weight of the members, the number of zero-force members in the truss under the

action of the load P, is (A) 7 (B) 9 (C) 6 (D) 8

Sol.

Total no of zero force members = 7 Choice (A)

L L L

L

P


8. Uniform flow with velocity U makes an angle with the y-axis, as shown in the figure

The velocity potential (), is

(A) -±U(x sin + y cos)

(B) -±U(x sin – y cos)

(C) -±U(y sin + x cos)

(D) -±U(y sin – x cos)

Sol. ∂ϕ

∂x = Ux

= Uxy f(x) + c

= –(u sinθ)x + f(y) + c (1)

–ϕ

y = Uy

= Uyy + f(x) + c (2)

= –u (x.sinθ + y cosθ)

ϕ

x = Ux :

ϕ

y = Uy

= U(x.sinθ + y cosθ)

= ±U(x.sinθ + ycosθ) Choice (A)

9. The probability that a 50 year flood may NOT occur at all during 25 years life of a project (round off to two decimal places), is _____.

Sol. Prob. of non-occurrence

p = 1

𝑇

q = 1 – p

= 1 – 1

𝑇

= 1 – 1

50

= 0.98

= p(r, n) = p(0, n) = nCo. po. qn = qn = (0.98)25 = 0.603 Ans: 0.59 to 0.61

y

U

x

θ


10. The value of 𝐥𝐢𝐦𝐱→∞

𝐱𝟐 − 𝟓𝐱 + 𝟒

𝟒𝐱𝟐 + 𝟐𝐱 is

(A) 1 (B) 0

(C) 𝟏

𝟐

(D) 𝟏

𝟒

Sol. limx→∞

(x2 –5x + 4

4x2 +2x) = lim

x→∞

x2 (1 –5

x +

4

x2)

x2 (4 +

2x

) = lim

x→∞

(1 –5

x +

4

x2)

4+ 2

x

= 1 – 0 + 0

4 + 0 =

1

4 Choice (D)

11. In the following partial differential equation, is a function of t and z, and D and K are functions

of

𝑫(𝜽)𝝏𝟐𝜽

𝝏𝒛𝟐 + 𝝏𝑲(𝜽)

𝝏𝒛−

𝝏𝜽

𝝏𝒕 = 0

The above equation is (A) a second degree linear equation (B) a second order non-linear equation (C) a second degree non-linear equation (D) a second order linear equation

Sol. The partial differential equation

D () ∂2

∂Z2 + ∂ K ()

∂Z –

∂

∂t = 0

is of second order

As D() and ∂2

∂Z2 are multiplied together, it is non - linear Choice (B)

12. A planar elastic structure is subjected to uniformly distributed load, as shown in the figure (not drawn to the scale)

Neglecting self-weight. The maximum bending moment generated in the structure (in kN.m, round off to the nearest integer), is _____.

Sol.

V = 0VA + VB = 12 × 8 = 96

MA = 0– VB × 8 + 12 × 8 × 8/2 = 0 VB = 48 kN and VA = 48 kN

Mxx = VB x – 12 𝑥2

2 Mxx = 48x – 6x2

At max. BM, SF will be zero, (i.e. SF = 0)

SF = 𝑑𝑀𝑥𝑥

𝑑𝑥 = 48 – 12x = 0 48 – 12x = 0 x = 4m

Mmax = 48 × 4 – 6(4)2 = 96 kN –m Ans: 95 to 97

12 kN/m

2.4 m

8.0 m

Parabolic Profile


13. Which one of the following statements is NOT correct? (A) The cohesion of normally consolidated clay is zero when triaxial test is conducted under

consolidated undrained condition. (B) The ultimate bearing capacity of a strip foundation supported on the surface of sandy soil

increases in direct proportion to the width of footing. (C) A clay deposit with a liquidity index greater than unity is in a state of plastic consistency. (D) In case of a point load, Boussinesq’s equation predicts higher value of vertical stress at a

point directly beneath the load as compared to Westergaard’s equation. Sol. In the case of NC clay in CU test, the failure (coulomb’s) envelope is as follows

So, statements A is correct So, statement C is wrong Choice (C)

14. An amount of 35.67 mg HCl is added to distilled water and the total solution volume is made to one litre. The atomic weights of H and CL are 1 and 35.5, respectively. Neglecting the dissociation of water, the pH of the solution, is

(A) 3.50 (B) 3.01 (C) 2.01 (D) 2.50

Sol. Concentration of HCl = 35.67 mg/l

HCl → H+ + Cl–

36.5 gm HCl produces 1 gm H+ ions

1 part HCl produces 1

36.5 parts H+ ions

35.67 mg/l HCl 1

36.5 × 35.67

= 0.977 mg/l of H+ ions

[H+] = 0.977 mg/l

= 0.977

1000 mole/l

= 9.77 × 10–4 mole/l

pH = –log[H+]

= –log[9.77 × 10–4]

= 3.01 Choice (B)

15. The Los Angeles test for stone aggregates is used to examine (A) abrasion resistance (B) crushing strength (C) specific gravity (D) soundness

Sol. Los angeles test → abrasion resistance Choice (A)


16. A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _____.

Sol. When flow is parallel to slope

F = 1

𝑠𝑎𝑡 .

𝑡𝑎𝑛

𝑡𝑎𝑛 𝑖

F = (18 –10)

18 tan(38)°

tan(30)° = 0.60 Ans: 0.59 to 0.61

17. In a drained triaxial compression test, a sample of sand fails at deviator stress of 150 kPa under confining pressure of 50 kPa. The angle of internal friction (in degree, round off to the nearest integer) of the sample, is _____.

Sol. Given; 𝑑 = 150 kpa, 𝑐= 3 = 50 kpa, =?

For sand, c= 0

At failure, 1 = 3 𝑡𝑎𝑛2 (45 +/2) + 2c tan (45 + /2)

𝑑 = 1 – 3 1 = 𝑑 + 3

𝑑 + 3 = 3 tan2 (45 + /2)

150 + 50 = 50 tan2 (45 + /2)

= 36.86° Ans: 36 to 38

18. A body floating in a liquid is in a stable state of equilibrium if its

(A) metacentre coincides with its centre of gravity

(B) metacentre lies below its centre of gravity

(C) centre of gravity is below its centre of buoyancy

(D) metacentre lies above its centre of gravity

Sol. GM > 0 M-lies above G. Choice (B)

19. During the process of hydration of cement, due to increase in Dicalcium Silicate (C2S) content in cement clinker, the heat of hydration

(A) does not change (B) increases (C) initially decreases and then increases (D) decreases

Sol. Among all the bogues compounds, C2 S has the least heat of hydration. Hence as the C2S proportion

increases, the heat of hydration of cement decreases. Choice (D)


20. In a soil investigation work at a site, Standard Penetration Test (SPT) was conducted at every 1.5 m interval up to 30 m depth. At 3 m depth, the observed number of hammer blows for three successive 150 mm penetrations were 8, 6 and 9, respectively. The SPT N-value at 3 m depth, is

(A) 15 (B) 23 (C) 14 (D) 17 Sol. SPT value [N] means, the no of blows required for 30 cm penetration so, neglect first 15 cm

n1 = 8 → 1st 15 cm

n2 = 6 → 2nd 15 cm

n3 = 9 → 3rd 15 cm N = n2 +n3 = 6 + 9 = 15

∴ N = 15 Choice (A)

21. During chlorination process, aqueous (aq) chlorine reacts rapidly with water to form Cl–, HOCl, and H+ as shown below

Cl2(aq) + H2O ⇌ HOCl + Cl– + H+

The most active disinfectant in the chlorination process from amongst the following, is (A) HOCl (B) H2O (C) H+ (D) Cl–

Sol. The most active disinfectant in chlorination process is HOCl. Choice (A)

22. A river has a flow of 1000 million litres per day (MLD), BOD5 of 5 mg/litre and Dissolved Oxygen (DO) level of 8 mg/litre before receiving the wastewater discharge at a location. For the existing environmental conditions, the saturation DO level is 10 mg/litre in the river. Wastewater discharge of 100 MLD with the BOD5 of 200 mg/litre and DO level of 2 mg/litre falls at that location. Assuming complete mixing of wastewater and river water, the immediate DO deficit (in mg/litre, round off to two decimal places), is _____.

Sol. DOdef = DOsat – DOmix

DOmix = (DO𝑠 × Qs) + (DOR × QR)

Qs + Q𝑅

= (2 × 100)+(8 ×1000)

1100 = 7.455 mg/l

DOdef = 10 – 7.45 = 2.545 mg/l Ans: 2.45 to 2.55

23. The true value of ln(2) is 0.69. If the value of ln(2) is obtained by linear interpolation between ln(1) and ln(6), the percentage of absolute error (round off to the nearest integer), is

(A) 35 (B) 69 (C) 84 (D) 48

Sol. Given the true value of ln 2 = 0.69 = T (say)

x 1 6

y = ln x 0 1.7918


The divided difference [ x0, x1] = y1– y0

x1 – x0 =

1.7918 –0

6 –1= 0.3584

By Newton’s divided difference formula, we have

y = y0 + (x – x0)[ x0, x1 ] l𝑛 2 = y0 + (2 – x0)[ x0, x1 ] l𝑛 2 = 0 + (0.3584) = 0.3584 = L (say)

Absolute error = |T – L| = |0.69 – 0.3584| = 0.3316

Percentage of absolute error = |T –L|

T 100 =

0.3316

0.69 100

= 48.058% 48% Choice (D)

24. A 4 m wide rectangular channel carries 6 m3/s of water. The Manning’s ‘n’ of the open channel is 0.02. Considering g = 9.81 m/s2, the critical velocity of flow (in m/s, round off to two decimal places) in the channel, is _____.

Sol. yc = (𝑞2

𝑔)

1/3

= ((1.5)2

9.81)

1/3

= 0.612m

V = √gYc = √9.81 × 0.612 = 2.45 m/s Ans: 2.35 to 2.55

25. A road in a hilly terrain is to be laid at a gradient of 4.5%. A horizontal curve of radius 100 m is laid at a location on this road. Gradient needs to be eased due to combination of curved horizontal and vertical profiles of the road. As per IRC, the compensated gradient (in %, round off to one decimal place), is _____.

Sol. Compensated gradient,

CG = G – G.C G.C = 30+𝑅

𝑅 (or)

75

𝑅 } lesser

= 30+100

100 (or)

75

100 = 0.75

CG = 4.5 – 0.75 = 3.75 = 4%

CG = 4% Ans: 4%

26. Surface Overflow Rate (SOR) of a primary settling tank (discrete settling) is 20000 litre/m2 per day.

Kinematic viscosity of water in the tank is 1.01 10–2 cm2/s. Specific gravity of the settling

particles is 2.64. Acceleration due to gravity is 9.81 m/s2. The minimum diameter (in m, round off to one decimal place) of the particles that will be removed with 80% efficiency in the tank, is _____.

Sol. dmin = ?

Us = 𝑔

18 (𝑠 – 1)𝑑2

= 𝑈𝑠

𝑈𝑜 10 = 80

Us = 0.8(U0)= 0.8 (20000 10–3

24 60 60) m3/s/m2 = 1.85 10–4 m/s

1.85 10–4 m/s = 9.81

18 (2.64 – 1)

𝑑2

1.01 10–6

(∵ = 1.01 10–2 cm2/s

= 1.01 10–6 m2/s

d = 14.46 10–6 m

= 14.46 m Ans: 14 to 15


27. For the Ordinary Differential Equation 𝒅𝟐𝒙

𝒅𝒕𝟐 − 𝟓𝒅𝒙

𝒅𝒕 + 6x = 0, with initial conditions x(0) = 0 and

𝒅𝒙

𝒅𝒕(0) = 10, the solution is

(A) 5e2t + 6e3t (B) 10e2t + 10e3t (C) –5e2t + 6e3t (D) –10e2t + 10e3t

Sol. Given d2x

dt2 – 5 dx

dt + 6x = 0 → (1)

where x (0) = 0 and dx

dt (0) = 10

Applying the Laplace transform on both sides of (1), we have

L (d2x

dt2 ) – 5L (dx

dt) + 6L (x) = L (0)

s2 x – S x(0) – x1 (0) – 5(s x – x (0)) + 6x = 0

where x = L (x)

s2 x – S 0 – 10 – 5 sx + 5 0 + 6x = 0

(s2 – 5s + 6) x = 10

x = 10

S2 – 5S + 6 =

10

(S – 2)(S − 3) x =

10

S –3 –

10

S –2

Applying inverse Laplace transform on both sides,

L–1(x) = L–1 (10

S –3) – L–1 (

10

S –2) x = 10e3t – 10e2t Choice (D)

28. A circular water tank of 2 m diameter has a circular orifice of diameter 0.1 m at the bottom. Water enters the tank steadily at a flow rate of 20 litre/s and escapes through the orifice. The coefficient of discharge of the orifice is 0.8. Consider the acceleration due to gravity as 9.81 m/s2 and neglect frictional losses. The height of the water level (in m, round off to two decimal places) in the tank at the steady state, is _____.

Sol. Q = water entering in tank

Cd . a . √2𝑔𝐻 = 20 × 10–3

0.8 × 𝜋

4(0.1)2

√2𝑔𝐻 = 0.02

H = 0.5164 m. Ans: 0.5 to 0.54

d = 0.1 m

Cd = 0.8

H

Q = 20 Lt sec


29. A continuous function f(x) is defined. If the third derivative at xi is to be computed by using the fourth order central finite-divided-difference scheme (with step length = h) the correct formula is

(A) f(xi) = −𝒇(𝒙𝒊+𝟑)+𝟖𝒇(𝒙𝒊+𝟐)−𝟏𝟑𝒇(𝒙𝒊+𝟏)+𝟏𝟑𝒇(𝒙𝒊−𝟏)−𝟖𝒇(𝒙𝒊−𝟐)+𝒇(𝒙𝒊−𝟑)

𝟖𝒉𝟑

(B) f(xi) = 𝒇(𝒙𝒊+𝟑)+𝟖𝒇(𝒙𝒊+𝟐)−𝟏𝟑𝒇(𝒙𝒊+𝟏)+𝟏𝟑𝒇(𝒙𝒊−𝟏)+𝟖𝒇(𝒙𝒊−𝟐)+𝒇(𝒙𝒊−𝟑)

𝟖𝒉𝟑

(C) f(xi) = −𝒇(𝒙𝒊+𝟑)−𝟖𝒇(𝒙𝒊+𝟐)−𝟏𝟑𝒇(𝒙𝒊+𝟏)+𝟏𝟑𝒇(𝒙𝒊−𝟏)+𝟖𝒇(𝒙𝒊−𝟐)+𝒇(𝒙𝒊−𝟑)

𝟖𝒉𝟑

(D) f(xi) = 𝒇(𝒙𝒊+𝟑)−𝟖𝒇(𝒙𝒊+𝟐)+𝟏𝟑𝒇(𝒙𝒊+𝟏)+𝟏𝟑𝒇(𝒙𝒊−𝟏)−𝟖𝒇(𝒙𝒊−𝟐)−𝒇(𝒙𝒊−𝟑)

𝟖𝒉𝟑

Sol. From the fourth order central finite - divided - difference scheme with step size h, the third derivative can

be expressed as

(d3 y

dx3 ) at x = xi = –yi + 3 + 8𝑦i − 2− 13yi + 1 + 13yi – 1−8yi − 2 + yi−3

8h3

f (xi) = −f(xi+3)+8f(xi+2)−13f(xi+1)+13𝑓(xi−1)−8f(xi−2)+f(xi−3)

8h3

Choice (A)

30. A water supply scheme transports 10 MLD (Million Litres per Day) water through a 450 mm

diameter pipeline for a distance of 2.5 km. A chlorine dose of 3.50 mg/litre is applied at the starting

point of the pipeline to attain a certain level of disinfection at the downstream end. It is decided

to increase the flow rate from 10 MLD to 13 MLD in the pipeline. Assume exponent for

concentration, n = 0.86. With this increased flow, in order to attain the same level of disinfection,

the chlorine dose (in mg/litre) to be applied at the starting point should be

(A) 3.95

(B) 4.75

(C) 4.40

(D) 5.55

Sol. As per kinetics of disinfection,

Cnt = Constant

C : Concentration of disinfectant

t : time of disinfection

n : dilution factor

t = V/Q

∴ cn × 𝑉

𝑄 = Constant

𝐶𝑛

𝑄 = constant

𝑐1

𝑛

𝑄1 =

𝑐2𝑛

𝑄2

(3.5)0.86

10 =

(𝑐2)0.86

13

C2 = 4.75 mg/l Choice (B)


31. Traffic volume count has been collected on a 2-lane road section which needs upgradation due to severe traffic flow condition. Maximum service flow rate per lane is observed as 1280 veh/h at level of service ‘C’. The Peak Hour Factor is reported as 0.78125. Historical traffic volume count provides Annual Average Daily Traffic as 12270 veh/day. Directional split the traffic flow is observed to be 60:40. Assuming that traffic stream consists of ‘All Cars’ and all drivers are ‘Regular Commuters’, the number of extra lane(s) (round off to the next higher integer) to be provided, is _____.

Sol. No of lanes needed is given by

DDHV

PHF ×MSF ×HVF × fp

DDHV = Directional distribution hourly volume

PHF = peak hour factor

MSF = Max service rate flow

HVF = Heavy vehicle adjustment factor = 1

fp = road under familiarly adjustment factor = 1 assumed

DDHV = 1,22,270 (veh/day)× 0.6 = 7362 veh/day

(∵ 0.6 is max split of traffic flow = 60%)

No of lanes needed = 7362

0.78125 ×1280 ×1 ×1 = 8 lanes

existin lanes=2

additional lanes=8 –2=6 lanes Ans: 6

32. Consider the system of equations

[

𝟏 𝟑 𝟐𝟐 𝟐 −𝟑𝟒 𝟒 −𝟔𝟐 𝟓 𝟐

] [

𝒙𝟏

𝒙𝟐

𝒙𝟑

] = [

𝟏𝟏𝟐𝟏

]

The value of x3 (round off to the nearest integer), is _____. Sol. Given system of equations is

[

1 3 22 2 −34 4 −62 5 2

] [

x1

x2

x3

] = [

1121

]

which can be considered as AX = B

Where A =[

1 3 22 2 −34 4 −62 5 2

]; X = [

x1

x2

x3

] and B = [

1121

]

Consider the augmented matrix

[A/B] = [

1 3 22 2 −34 4 −62 5 2

. 1

. 1

. 2

. 1

]

R2→ R2–2R1, R3→ R3–4R1, R4→ R4–2R1

[

1 3 20 −4 −70 −8 −140 −1 −2

. 1. −1. −2. −1

]


R2R4

[

1 3 20 −1 −20 −8 −140 −4 −7

1−1 – 2−1

]

R2→ R4 –

1

2 R2

[

1 3 20 −1 −20 −8 −140 0 0

1−1 – 20

]

R3→ R3 – 8R2

[

1 3 20 −1 −20 0 20 0 0

1−1 60

]

R3→1

2 R3

[A/B} [

1 3 20 −1 −20 0 20 0 0

1−1 60

]

The solution of given system is same as the solution of

[

1 3 20 −1 −20 0 10 0 0

] [

x1

x2

x3

] = [

1−130

] x1 + 3x2 + 2x3 = 1

–x2 – 2x3 = –1 x3 = 3

The value of x3 = 3 Ans : (3)

33. A stream with a flow rate of 5 m3/s is having an ultimate BOD of 30 mg/litre. A wastewater discharge of 0.20 m3/s having BOD5 of 500 mg/litre joins the stream at a location and instantaneously gets mixed up completely. The cross-sectional area of the stream is 40 m2 which remains constant. BOD exertion rate constant is 0.3 per day (logarithm base to e). The BOD (in mg/litre, round off to two decimal places) remaining at 3 km downstream from the mixing location, is _____.

Sol. BOD remaining after time ‘t’ is given by Lt = Loe–kt

Lo refers to ultimate BOD of mix (Lo)mix = (Lo)RQR + (Lo)SQS

QR + QS

For sewage; 𝑦𝑡𝑇 = 𝐿𝑜(1 − 𝑒−𝑘𝑡)

(𝐿𝑜)𝑠 = y𝑡

(1−e−𝑘𝑡)

(𝐿𝑜)𝑠 = 500

(1−e−0.3×5) = 643.66 mg/l

(𝐿𝑜)𝑚𝑖𝑥 = (5 ×30)+(0.2 ×643.66)

5+0.2 = 53.6 mg/l

Lt = Loe–kt = 53.6 e–0.3 × t

t = d/v

v = (Q𝑆 + 𝑄𝑅)

𝐴 =

(0.2 + 5)

40 = 0.13 m/s

t = 3 × 103 𝑚

0.13 ×24 ×60 ×60 𝑚 /𝑑𝑎𝑦 = 0.26 days

Lt = (53.6)e-0.3 × 0.26 = 49.57 mg/l Ans: 49 to 50


34. A rigid, uniform, weightless, horizontal bar is connected to three vertical members P, Q and R as shown in the figure (not drawn to the scale). All three members have identical axial stiffness of 10 kN/mm. The lower ends of bars P and R rest on a rigid horizontal surface. When NO load is applied, a gap of 2 mm exists between the lower end of the bar Q and the rigid horizontal surface. When a vertical load W is placed on the horizontal bar in the downward direction, the bar still remains horizontal and gets displaced by 5 mm in the vertically downward direction.

The magnitude of the load W (in kN, round off to the nearest integer), is ______.

Sol. A Rigid

= gap = 2 mm

P

δ = K each member = 10 kN/mm

displacement = y = 5 mm

p = R = 5 mm K = P

Q = 5 – 2 = 3 mm

FP = Fr = 5 10 = 50 kN.

FR = k = 3 10 = 30 kN.

P, Q, R bars are parallel

Total load = FP + FQ + FR

= 50 + 30 + 50

= 130 kN Ans: 129 to 131

P Q R

Horizontal Rigid uniform bar

gap

Rigid Horizontal

Surface

P Q R

Rigid Uniform Weightless Horizontal Bar


35. Water flows at the rate of 12m3/s in a 6 m wide rectangle channel. A hydraulic jump is formed in the channel at a point where the upstream depth is 30 cm (just before the jump). Considering acceleration due to gravity as 9.81 m/s2 and density of water as 1000 kg/m3, the energy loss in the jump is (A) 141J/s (B) 114.2 kW (C) 114.2MW (D) 141.2 h.p

Sol.

q = 12

6 = 2m3\3\m

Fr = (𝑞2

gy3)1/2

= (22

9.81×0.33)1/2

= 3.88

y2

y1 =

1

2(–1 + √1 + 8 × 𝐹1

2) = 1

2(–1 + √1 + (8 × (3.88)2)

y2 = 5.013 × y1 = 1.505

Jump = h1 = 𝑦2–𝑦1

4𝑦1, 𝑦2 = 0.968 m

P = ρgQH = 1000 × 9.81 × 12 × 0.968 = 114.2 kW Choice (B)

36. A 10 m thick clay layer is resting over a 3 m thick sand layer and is submerged. A fill of 2 m thick sand with unit weight of 20 kN/m3 is placed above the clay layer to accelerate the rate of consolidation of the clay layer. Coefficient of consolidation of clay is 9 × 10–2 m2/year and coefficient of volume compressibility of clay is 2.2 × 10–4 m2/kN. Assume Taylor’s relations between time factor and average degree of consolidation.

Water table

Clay

Sand 3 m

10 m

Impervious Stratum


The settlement (in mm, round off to two decimal places) of the clay layer, 10 years after the

construction of the fill, is ______.

Sol.

H for 10 years = ?

H = mv Ho (completely)

= 10 kN / m2

mv = 2.2 10–4 m2 / kN

H = 88 mm

for 10 years hour how much consolidation has taken place

t = Tv(Z)2

Cv

10 years = 𝑇𝑣(5)2

𝑞 × 10–2 𝑚2/ 𝑦𝑟𝑠

lets assume V < 60%

Tv = 0.036

Tv =

4 (U)2

0.036 =

4 (U)2

U = 0.2141

Hence assumption is right

H = 0.2141 (88 mm)

H = 18.84 mm Ans: 18.5 to 19.5

37. A simply supported prismatic concrete beam of rectangular cross-section, having a span of 8 m,

is prestressed with an effective prestresing force of 600 kN. The eccentricity of the prestressing

tendon is zero at supports and varies linearly to a value of e at the mid-span. In order to balance

an external concentrated load of 12 kN applied at the mid-span, the required value of e (in mm,

round off to the nearest integer) of the tendon, is ______.

Sol. By load balancing concept

𝑚𝑜𝑚𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜

𝑝𝑟𝑒𝑠𝑡𝑟𝑒𝑠𝑠𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒} = moment due to applied load

P . e = 𝑤𝑙

4

600 × e = 12 ×8

4

e = 0.04 m

∴ e = 40 mm Ans: 39 to 41

clay

Sand


38. The total stress paths corresponding to different loading conditions, for a soil specimen under the isotropically consolidated stress state (O), are shown below

The correct match between the stress paths and the listed loading conditions, is (A) OP – I, OQ – II, OR – IV, OS – III (B) OP – IV, OQ – III, OR – I, OS – II (C) OP – I, OQ – III, OR – II, OS – IV

(D) OP – III, OQ – II, OR – I, OS – IV

Sol. OP: Extension loading → (IV)

OQ: Extension unloading → (III)

q = σ1 – σ3 R

S

O

P Q

p = 𝜎1+ 2𝜎3

3

σ1

σ3

σ1

σ3

Loading Condition Stress Path

OP

OQ

OR

OS

I – Compression loading (𝛔1 – increasing; 𝛔3 – Constant)

II – Compression unloading (𝛔1 – constant; 𝛔3 – decreasing)

III – Extension unloading (𝛔1 – decreasing; 𝛔3 – constant)

IV – Extension loading (𝛔1 – constant; 𝛔3 – increasing)


OR: Compression loading → (I)

OS: Compression unloading → (II):

OP : (IV) OQ : (III) OR : (I) OS : (II) Choice (B)

39. Distributed loads(s) of 50 kN/m may occupy any position(s) (either continuously or in patches) on the girder PQRST as shown in the figure (not drawn to the scale)

The maximum negative (hogging) bending moment (in kN.m) that occurs at point R, is (A) 93.75 (B) 22.50 (C) 56.25 (D) 150.00

Sol.

ILD for max hogging moment

P Q R S T

1.5 m 3 m 2 m 1.5 m


To get max hogging moment, place the load on either PQ (or) RS segment Hogging moment = load × ordinate

= 50 [1

2 × 1.5 × 1.5]

= 56.25 kN – m Alternative method

B.M = load × ordinate

= 50 × 1

2 × 1.5 × 0.6 + 50 ×

1

2 ×1.5 × 0.9

= 56.25 kN –m To get max hogging moment, place the load on segment PQ & RS simultaneously Choice (C)

40. The soil profile at a site up to a depth of 10 m is shown in the figure (not drawn to the scale). The

soil is preloaded with a uniform surcharge (q) of 70 kN/m2 at the ground level. The water table is

at a depth of 3 m below ground level. The soil unit weight of the respective layers is shown in the

figure. Consider unit weight of water as 9.81 kN/m3 and assume that the surcharge (q) is applied

instantaneously.

Immediately after preloading, the effective stress (in kPa) at points P and Q, respectively, are

(A) 36 and 126

(B) 54 and 95

(C) 124 and 204

(D) 36 and 90

q = 70 kN/m2

Ground level

Ground Water table

Sand (Bulk unit weight = 18 kN/m3)

Silty Clay(Saturated unit weight = 20 kN/m3)

Clay(Saturated unit weight = 21 kN/m3)

Boulder Clay

3 m

4 m

3 m

P

Q


Sol.

Surcharge load is applied instantaneously hence excess pore pressure is developed @ P & Q

∴ q = �̅� = Excess pore pressure

@ Point P

= q +3 = 70 + (3 × 18) = 124 kN /m2

Pore Pr = �̅� + ui = 70 + 0 = 70

1 = – u = 124 – 70 = 54 kN /m2

@ Point Q

= q + 3 + 4γsat = 70 + (3 × 18) + (4 × 20) = 204

u = �̅� + ui = 70 + 4 × w = 70

= 70 + 4 × 9.51 = 109.24

1 = – u = 94.76 = 95 kN/m2 Choice (B)

41. Three reservoirs P, Q and R are interconnected by pipes as shown in the figure (not drawn to the scale). Piezometric head at the junction S of the pipes is 100 m. Assume acceleration due to gravity as 9.81 m/s2 and density of water as 1000 kg/m3. The length of the pipe from junction S to the inlet of reservoir R is 180 m.

Considering head loss only due to friction (with friction factor of 0.03 for all the pipes), the height of water level in the lowermost reservoir R (in m, round off to one decimal place) with respect to the datum, is ______.

Sol. Q1 + Q2 = Q3

Q3 =𝝅

𝟒(0.3)2(0.256) +

𝝅

𝟒(0.3)2(1.98)

Q3 = 0.3209 m3/s

h = 𝑓𝐿𝑄2

12.1𝑑5 = 0.03 ×180 ×(0.3209)2

12.1 × (0.45)5 = 2.49 m

hs – hR = 2.49 m hR = hs – 2.49 hR = 100 – 2.49 = 97.51 m Ans: 97.2 to 97.8

Reservoir Q

106 m

Reservoir R

109 m

Reservoir P

Diameter = 30 cm Flow velocity = 2.56 m/s

Length = 180 cm Diameter = 45 cm

Datum

Diameter = 30 cm Flow velocity = 1.98 m/s

10 m

g

90 m

S


42. A cantilever beam PQ of uniform flexural rigidity (EI) is subjected to a concentrated moment M at

R as shown in the figure

The deflection at the free end Q is

(A) 𝐌𝐋𝟐

𝟔𝐄𝐥

(B) 𝟑𝐌𝐋𝟐

𝟒𝐄𝐥

(C) 𝟑𝐌𝐋𝟐

𝟖𝐄𝐥

(D) 𝐌𝐋𝟐

𝟒𝐄𝐥

Sol. Q = R + QR . LRQ

= 𝑀(𝐿/2)2

2𝐸𝐼 +

𝑀 (𝐿/2)

𝐸𝐼 ×

𝐿

2

𝑄 = 𝑀𝐿2

8𝐸𝐼 +

𝑀𝐿2

4𝐸𝐼

Q = 3𝑀𝐿2

8𝐸𝐼 Choice (C)

43. The lengths and bearings of a traverse PQRS are:

Segment Length(m) Bearing

PQ 40 80°

QR 50 10°

RS 30 210°

The length of line segment SP (in m, round off to two decimal places), is ______

Sol. SP = √(C)2 + (D)2

C = 40cos 80° + 50cos 10° + 30cos 210°

= 30.20 m

D = 40sin 80° + 50sin 10° + 30sin 210°

= 33.07 m

length SP = √(30.2)2 + (33.07)2 = 44.79 m Ans: 44 to 45

P Q R

L/2

M L


44. The appropriate design length of a clearway is calculated on the basis of ‘Normal Take-off’

condition. Which one of the following options correctly depicts the length of the clearway? (Note:

None of the options are drawn to scale)

Sol. For normal take – off Condition;

Clearway < ½((1.15 × Take off distance) – (1.15 × lift off distance))

Clearway < ½((1.15 × 1625) – (1.15 × 875))

< 431.25 m

Ans: 431.25

Runway

Lift Off Point

Flight Path

1625 m

875 m >244 m

10.5 m Clearway

Airplane

Runway

Lift Off Point

Flight Path

1625 m

875 m < 244 m

10.5 m Clearway

Airplane

Runway

Lift Off Point

Flight Path

1625 m

875 m < 432 m

10.5 m Clearway

Airplane

Runway

Lift Off Point

Flight Path

1625 m

875 m < 432 m

10.5 m Clearway

Airplane


45. The singly reinforced concrete beam section in the figure (not drawn to the scale) is made of M25 grade concrete and Fe500 grade reinforced steel. The total cross-sectional area of the tension steel is 942 mm2.

As per Limit State Design of IS 456:2000, the design moment capacity (in kN.m, round off to two decimal places) of the beam section, is ______.

Sol. Ast= 942 MM2 Fe – 500 & M25

xu = 0.87 𝑓𝑦𝐴𝑠𝑡

0.36 𝑓𝑐𝑘𝑏

xu = 0.87 ×500 × 942

0.36 ×25 ×300 = 151.76 mm

xumax = 0.46d for Fe – 5w

= 0.46 × 450 = 207 mm

Xu < xumax → U.R.S

Mu(lim) = 0.87 fyAst [d – .42 xu]

= 0.87 × 500 × 942 [450 – 0.42 ×1.51.76]

Mu(lim) = 158.27 kN – m Ans: 158 to 158.60

450 mm

300 mm


46. An open traverse PQRST is surveyed using theodolite and the consecutive coordinates obtained are given in the table

Line Consecutive Coordinates

Northing(m) Southing(m) Easting(m) Westing(m)

PQ 110.2 - 45.5 -

QR 80.6 - - 60.1

RS - 90.7 - 70.8

ST - 105.4 55.5 -

If the independent coordinates (Nothing, Easting) of station P are (400 m, 200 m), the independent coordinates (in m) of station T, are

(A) 405.3, 229.9 (B) 194.7, 370.1 (C) 205.3, 429.9 (D) 394.7, 170.1

Sol. L = 110.2 + 80.6 – 90.7 – 105.4 = – 5.3

D = 45.5 + 55.5 – 60.1 – 70.8 = – 29.9 Coordinator of station = [400 – 5.3] = 394.7m = [200 – 29.9] = 170.1m Choice (D)

47. A gaseous chemical has a concentration of 41.6 mol/m3 in air 1 atm pressure and temperature 293 K. The universal gas constant R is 82.05 × 10–6 (m3 atm)/(mol K). Assuming that ideal gas law is valid, the concentration of the gaseous chemical (in ppm, round off to one decimal place), is _______.

Sol. Let the gaseous chemical be ‘x’

In order to convert concentration from gm/m3 to ppm,

xgm/m3 = xppm

𝑀𝑥

𝑉 103

Mx = molecular weight of gas in gm

V = vol occupied by gas in litres per mole

Given x = 41.6 .mol/m3

= 41.6 Mx gm/m3

41.6 Mx = xppm 𝑀𝑥

𝑉 103 –––––– (1)

PV = nRT

V/n = 𝑅𝑇

𝑃 =

82.05 10−6 293

1

= 24.04 10–3 m3/mole

= 24.04 ℓ/mole

from eqn (1);

41.6 Mx = xppm 𝑀𝑥

24.04 103

xppm = 1ppm Ans: 0.9 to 1.1


48. A rigid weightless platform PQRS shown in the figure (not drawn to the scale) can slide freely in the vertical direction. The platform is held in position by the weightless member OJ and four weightless, frictionless rollers. Point O and J are pin connections. A block of 90 kN rests on the platform as shown in the figure

The magnitude of horizontal component of the reaction (in kN) at pin O, is (A) 150 (B) 180 (C) 90 (D) 120

Sol. According to the assembly, the weight W = 90 kN, must be equl to vertical component of oj = Foj sin = 90

Foj = 𝑤

𝑠𝑖𝑛 =

90

𝑠𝑖𝑛

H0 = 90 × 4

3

H0 = 120kN Horizontal component of reaction e pin ‘o’

H0 = Foj cos

H0 = 𝑤

𝑠𝑖𝑛 × cos =

90

𝑠𝑖𝑛 × cos

H0 = 90 × cot Choice (D)


49. A dowel bar is placed at a contraction joint. When contraction occurs, the concrete slab cracks at predetermined location(s). Identify the arrangements, which shows the correct placement of dowel bar and the place of occurrence of the contraction crack(s).

Sol. For contraction joints. Dowel bar is placed at mid depth of slab and filler till (d/3 to d/4) where d is depth of slab. Contraction crack is at predetermined location at centre below filler.

Choice (B)


50. The relationship between traffic flow rate (q) and density (D) is shown in the figure

The shock wave condition is depicted by (A) Flow with respect to point 1 (q1 = qmax) (B) Flow with respect to point 4 and point 5 (q4 = q5) (C) Flow changing from point 2 to point 6 (q2 > q6) (D) Flow changing from point 3 to point 7 (q3 < q7)

Sol.

Shock wave is decipted by point 2 to Point 6 since there is a sudden decrease in flow (∵as flow density

relation there should be no sudden change in flow) Choice (C)

51. If C represents a line segment between (0, 0, 0) and (1, 1, 1) in Cartesian coordinate system, the value (expressed as integer) of the line integral

∫ [(𝐲 + 𝐳)𝐝𝐱 + (𝐱 + 𝐳)𝐝𝐲 + (𝐱 + 𝐲)𝐝𝐳]

𝐂

is _______.

Sol. we have to evaluate

∫ [(y + z)dx + (x + z)dy + (x + y)dz]

c

where C is a line segment between (0, 0, 0) and (1, 1, 1)

∫ [(y + z)dx + (x + z)dy + (x + y)dz

c]

= ∫ d(xy + yz + zx)

c

= [𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥](0,0,0)(1,1,1)

= 1 + 1 + 1 = 3 Ans : (3)

K(x) (K

j/2)

4

3

2

1

7

6

5

Density’D’ 0

q(x)

Flow

rate ‘q’

4

3

2

1

7

6

5

Density, D 0

Flow

rate ‘q’


52. In a homogeneous unconfined aquifer of area 3.00 km2, the water table was at an elevation of 102.00 m. After a natural recharge of volume 0.90 million cubic meter (Mm3), the water table rose to 103.20 m. After this recharge, ground water pumping took place and the water table dropped down to 101.20 m. The volume of ground water pumped after the natural recharge, expressed (in Mm3 and round off to two decimal places), is ______.

Sol. Recharge Volume VR = 0.9 Mm3 V = 3 × (10.3.2 – 102) = 3 ×.1.2 = 3.6 Mm3

𝑌𝑅 = 𝑅𝑒𝑐ℎ𝑎𝑟𝑔𝑒 𝑉𝑜𝑙𝑢𝑚𝑒

𝑉

= 0.9

3.6

𝑌𝑅 = 𝐷𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑉𝑜𝑙𝑢𝑚𝑒

𝑉

= 0.9

3.6 [3 × (103.2 – 101.2)]

VG = 1.5 Mm3 Ans: 1.4 to 1.6

53. The flange and web plates of the doubly symmetric built-up section are connected by continuous 10 mm thick fillet welds as shown in the figure (not drawn to the scale). The moment of inertia of the section about its principal axis X – X is 7.73 × 106 mm4. The permissible shear stress in the fillet welds is 100 N/mm2. The design shear strength of the section is governed by the capacity of the fillet welds.

The maximum shear force (in kN, round off to one decimal place) that can be carried by the section, is ______.

Sol.

100 mm

10 mm

10 mm

fillet weld (typical)

12

0 m

m

X X


[τ = Shear stress at level of weld] ≤ Permissible shear stress of shear strengths of weld

= 100 N/mm2

τ = 𝑆𝑎�̅�

𝐼𝑏 S = Shear Force in section

a = area above the level at which Shear stress is to be found = 100 10 mm2

I = Moment of Inertia = 7.73 106 mm4

�̅� = distance of centroid of area a from central neutral axis = 55 mm

b = width of weld = 4 te = 4 7 = 28 mm

Throat thickness

te = 0.7 ‘S’

= 0.7 10

= 7mm

[S = site of weld]

assuming fusion angle 60° – 40°

τ = 𝑆 ×100 ×10 ×55

7.73 × 106× 28

S = 393527.27 N

= 393.53 KN

= 393.5 KN

�̅� = 60.5 = 55 mm Ans: 393.5 to 399.1

54. A vertical retaining wall of 5 m height has to support soil having unit weight of 18 kN/m3, effective

cohesion of 12 kN/m2, and effective friction angle of 30°. As per Rankine’s earth pressure theory and assuming that a tension crack has occurred, the lateral active thrust on the wall per meter length (in kN/m, round off to two decimal places), is ______.

Sol. Total Active pressure After formation of tensile crack

Pa = (1

2 . 𝑘𝑎 . 𝑟. 𝑧2) − 2𝑐√𝑘𝑐 +

2𝑐2

r

Ka = 1 –sin

1 +𝑠𝑖𝑛 =

1

3

Pa = 1

3 × 18 ×

52

2 – 2 × 12 √

1

3 × 5 +

2 (12)2

18

Pa = 21.714 kN/m Ans: 21 to 22

= 30°


55. Water flows in the upward direction in a tank through 2.5 m thick sand layer as shown in the figure. The void ratio and specific gravity of sand are 0.58 and 2.7, respectively. The sand is fully saturated. Unit weight of water is 10 kN/m3.

The effective stress (in kPa, round off to two decimal places) at point A, located 1 m above the base of tank, is ______.

Sol. During upward seepage, 1 = 1z –𝑤 ℎ𝑓

hydraulic gradient, I = ℎ𝑓

𝑧

hf = i.z= 1.2

2.5 × z

𝑠𝑎𝑡

= 𝑤 [𝐺+𝑒]

1+𝑒 =

10 [2.7+0.58]

1+0.58 = 20.75 kN/m3

1 (20.75 – 10) × 1.5 – 10 × 1.2

2.5 × 1.5

1 = 8.925 kN / m2 Ans: 8.7 to 9.3

1.2 m

Sand A

Water 1 m

Outward

flow

2.5 m

1 m

Filter media

Inward flow

Civil Engineering Forenoon Session Date of Exam: 09/02/2020

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Transcript of Civil Engineering Forenoon Session Date of Exam: 09/02/2020