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QUESTION: 1

**Identify the correct spelling of the word.**

Solution:

The correct spelling of the word is 'definitive' which means '(of a conclusion or agreement) done or reached decisively and with authority.' Thus option 2 is the correct answer.

QUESTION: 2

This is the place that _______

Solution:

The preposition 'about' is mandatory here thus option 1 and 4 are eliminated. Option 2 is correct as the tense present perfect continuous fits here. It conveys the meaning that the person usually talked about the place.

QUESTION: 3

She is brave. Her brother is more brave.

Select the most suitable sentence with respect to grammar and usage.

Solution:

Options 1 and 2 are incorrect as they change the meaning of what is mentioned. Option 3 is incorrect as with 'less' we use the adjective in positive form and not comparative form. The correct word here should be 'brave.' Option 4 is thus the correct answer.

*Answer can only contain numeric values

QUESTION: 4

When a four digit number is divided by 65, it leaves a remainder of 29. If the same number is divided by 13, the remainder would be______

Solution:

Let number is N

N = 65 k +29

Now

∴ Remainder = 3

QUESTION: 5

Complete the following sentence.

I was ___ ___ for the bus and then I ___ sight of Craig passing by.

Solution:

The word 'here' and 'there' both can be used here but should be followed with 'waiting' as no other word can fit here. The word 'caught' is correct here as 'catch a sight' means 'to see something.' Option 2 thus has the correct combination of words. The word 'cot' means 'a small bed with high barred sides for a baby or very young child.'

*Answer can only contain numeric values

QUESTION: 6

4 – digit number greater than 5000 are randomly formed from the digits 0, 2, 3, 5 and 7. The probability of forming a number divisible by 5 when the digits are repeated is ______

Solution:

For a number to be greater than 5000, d_{1} should be filled with either 5 or 7

∴ Total numbers formed when the digits are repeated = 2 × 5 × 5 × 5 = 250

total cases = 250 -1 = 249 ( case of 5000 is not included)

Now, For the number to be divisible by 5, unit digit d_{4} should be either 0 or 5.

∴ Total no. of ways = 2 × 5 × 5 × 2 = 100

favorable cases = 100 - 1=9 ( 5000 is not included))

∴Required Probability = =0.397

QUESTION: 7

It is theoretically possible that bacteria developed on Venus early in its history and that some were carried to Earth by a meteorite. However, strains of bacteria from different planets would probably have substantial differences in protein structure that would persist over time, and no two bacterial strains on Earth are different enough to have arisen on different planets. So, even if bacteria did arrive on Earth from Venus, they must have died out.

The argument is most vulnerable to which of the following criticisms?

Solution:

The question asks which of the statements given in the options can weaken the argument put by the author that all bacteria from Venus must have died out.

The passage states that since there is a single strain of bacteria which exists on the Earth, they all must be belonging to the Earth or let's say a single planet. But here the author does not take into consideration (as can be argued from his theory) the fact that may be all the bacteria came from Venus and there are none which originally belong to the Earth. So this criticism as mentioned in option 3 makes the argument of the author vulnerable.

Options 1, 2 and 4 are completely irrelevant criticisms as they do not address the main argument. The argument claims that if there were Venusian bacteria on Earth, then they must have died out by now. Whether there are bacteria originally from Earth that have also disappeared from Earth is irrelevant to the question and has no effect on the given argument.

QUESTION: 8

A man sells three articles A, B, C and gains 10% on A, 20% on B and loses 10% on C. He breaks even when combined selling prices of A and C are considered, whereas he gains 5% when combined selling prices of B and C are considered. What is his net loss or gain on the sale of all the articles?

Solution:

Let a, b and c be the cost prices of the three articles A, B and C.

SP = CP + Profit (or) SP = CP – Loss

⇒ SP of A = 1.1a; SP of B = 1.2b; SP of C = 0.9c

By question,

1.1a + 0.9c = a + c ⇒ 0.1a = 0.1c ⇒ a = c

1.2b + 0.9c = 1.05(b + c) ⇒ 0.15b = 0.15c ⇒ b = c = a

Gain% = {(SP – CP)/CP} × 100

⇒ Net gain on the sale of all the articles = =

∴ Net gain on the sale of all the articles = 6.66%

QUESTION: 9

Which of the following inferences can be drawn from the above graph?

Solution:

Option 1 is false as graph says there is decrease in students qualifying in Physics in 2015 compared to 2014.

Option 2

Let no. of students qualifying in Biology in 2013 be 100

⇒ No. of students qualifying in Biology in 2014 = 100 – 10% of 100 = 90

⇒ No. of students qualifying in Biology in 2015 = 90 + 10% of 100 = 99

∴ The number of students qualifying in Biology in 2015 is less than that in 2013

Option 3 and option 4 are incorrect since no detail is given regarding how many students qualified the subject in 2013.

QUESTION: 10

DRQP is a small square of side a in the corner of a big square ABCD of side A. What is the ratio of the area of the quadrilateral PBRQ to that of the square ABCD, given A/a = 3?

Solution:

Area of triangle PAB = Area of triangle RCB (By symmetry)

∴ Area of Δ PAB = ½ × PA × AB

= ½ × 2A/3 × A = A^{2}/3

Area of ΔRCB = A^{2}/3

Now, Area of ABCD = Area of DRQP + Area of PAB + Area of RCB + Area of PBRQ

A^{2} = a^{2} + A^{2}/3 + A^{2}/3 + Area of PBRQ

As, A/a = 3 ⇒ a = A/3

⇒ Area of PBRQ =

*Answer can only contain numeric values

QUESTION: 11

If the Laplace transform of y(t) is given by Y(s) = L(y(t)) = then y(0) + y'(0) = _____.

Solution:

Y(s) = L(y(t)) =

Apply inverse Laplace transform,

⇒ y(t) =

Differentiate with respect to ‘t’.

⇒ y(0) + y'(0) = 1

*Answer can only contain numeric values

QUESTION: 12

The value of will be _____

Solution:

= 1

QUESTION: 13

Consider the following statements P and Q:

(P):If M = then M is singular.

(Q): Let S be a diagonalizable matrix. If T is a matrix such that S + 5 T = I, then T is diagonalizable.

Which of the above statements hold TRUE?

Solution:

A matrix is said to be singular, if determinant of that matrix is zero.

= 1 (18 – 12) - 1 (9 – 4) + 1 (3 – 2)

= 6 – 5 + 1 = 2 ≠ 0

M is non singular

(Q) A matrix can be diagonalizable when it has distinct eigen values

S is a diagonalizable matrix. Hence, has distinct eigen values.

Let S be a 3 × 3 matrix and the eigen values of s are λ_{1}, λ_{2}, λ_{3}

Given that, S + 5T = I

From the properties of Eigen values,

(a) If λ_{1} is an eigen value of matrix A, then -λ_{1 }will be on eigen value of matrix -A.

(b) If λ_{1} is an eigen value of matrix A, then (λ_{1} + 1) will be an eigen value of matrix (A + I)

(c) If λ_{1} is an eigen value of matrix A, then will be an eigen value of matrix where K is a scalar.

From the above properties, eigen values of T are,

As λ_{1}, λ_{2}, λ_{3} are distinct values, λ′_{1},λ′_{2},λ′_{3} will be distinct.

Hence, matrix T is diagonalizable

So, only Q is true.

QUESTION: 14

= 0 .If x = 0 at t = 0 and x = 1 at t = 1, the value of x at t = 2 is

Solution:

A.E is, (D^{2} – 3D + 2) = 0

⇒ (D – 1) (D – 2) = 0

⇒ D = 1, 2

x(t) = C_{1}e^{t} + C_{2}e^{2t}

x = 0 at t = 0

⇒ 0 = C_{1} + C_{2} ⇒ C_{1} – C_{2}

x = 1 at t = 1

⇒ 1 = C_{1}e^{1} + C_{2}e^{2}

⇒ 1 = C_{1}e^{1} + (-C_{1}) e^{2}

⇒ C_{1}(e^{1} – e^{2}) = 1

x(2) = C_{1}e^{2} + C_{2}e^{4}

= e + e^{2}

*Answer can only contain numeric values

QUESTION: 15

If for two vectors and , sum is perpendicular to the difference The ratio of their magnitude is

Solution:

QUESTION: 16

Consider plane truss below and identify the correct option.

Solution:

**Static indeterminacy of the Planer structure:**

D_{S} = R – r

R → no. of unknowns = 3

r → no. of equilibrium equations = 3

D_{S }= 3 – 3 = 0

As all the three Reactions (1, 2, 3) are non-parallel, non-concurrent and non-trival, So structure is externally stable.

**For Internal stability of truss:**

m ≥ 2J -3

m → no. of member

J → no. of Joints

m = 6

J = 5

6 ≥ 2 × 5 – 3

6 ≥ 10 – 3 = 7

As no. of members is less than (2J – 3), therefore the truss is **internally unstable.**

QUESTION: 17

Which of the following given statement is not correct?

Solution:

Factor affecting concrete strength

a) concrete porosity

b) water/cement ratio

c) Soundness of aggregate

d) Aggregate paste bond

e) Cement related parameters.

Hence, The water cement ratio (w/c) is not only factor influencing the strength of concrete.

At water/cement ratio more than 0.6, the increase in volume of hydrated product will not be able to occupy the space already filled with water. Hence, porosity increase and strength decreases.

The addition of the surfactant in the concrete mix results in the decrease in the cement ratio.

QUESTION: 18

The permissible limit for chlorides in water used for mixing concrete for reinforced concrete work as per IS 456 : 2000 is

Solution:

__As per clause 5.4 of IS 456 : 2000__

Potable water is considered satisfactory for mixing Concrete and the permissible limits for solids is shown in table below:

QUESTION: 19

The ratio of shear modulus to bulk modulus of the material for poisson ratio of 0.21 will be

Solution:

** Concept**:

Using the relation,

E = 3k(1 – 2μ) ……(i)

E = 2G(1 + μ) ……(ii)

From equation (i) and (ii)

3k(1 – 2μ) = 2G (1 + μ)

where k = Bulk modulus of the material

G = shear modulus of the material

μ = poisson’s ratio

** Calculation**:

μ = 0.21

3k(1 – 2μ) = 2G (1 + μ)

QUESTION: 20

A partially saturated soil sample is taken from a borrow pit and tested for its specific gravity, which is found to be 2.64. If the air content and water content of this sample is 20% and 26.5% respectively. Calculate the void ratio of the partially saturated soil sample when water content is 26.5% and change in the air content of the soil sample if the water content increased to 30%

Assume void ratio to be constant.

Solution:

__Soil sample when water content is 26.5%:__

Using the relation Se = wG

a_{c} = 20%

a_{c} + S = 100%

S = 100% - a_{c}

S = 100% - 20%

S = 80%

w = 26.5%

0.80 × e = 0.265 × 2.65

e = 0.878

e = 87.8% (ans)

__Soil sample when water content is 30%:__

w_{I} = 30%

G = 2.65

e = 0.878

S_{I} e = W_{i} G

S_{I} × 0.878 = × 2.65

S_{I} = 0.905

S_{I} = 90.5%

a_{c1}+S_{I }= 100%

a_{c1} = (100 − 90.5)%

a_{c1}+S_{I }= 100%

a_{c1 }= (100 − 90.5)%

change in the air content = a_{c} − a_{cI}

Δa_{c} = (20 – 9.5)%

Δa_{c} = 10.5% (Ans)

*Answer can only contain numeric values

QUESTION: 21

A plate load test was conducted in clay on a 450 mm diameter plate and the settlement of the plate is recorded as 10mm. if the settlement of foundation is not to exceed 35 mm, the width of the foundation required will be _____mm.

Solution:

__Concept:__

Using the relation

Where

S_{f} = settlement of foundation

S_{p} = settlement of plate

B_{f} = width of footing/foundation

B_{p} = width of plate

** Calculation**:

S_{f} = 35 mm

S_{p} = 10 mm

B_{p} = 450 mm

B_{f} = 1575 mm

Therefore width of foundation required will be 1575 mm.

QUESTION: 22

Following results are obtained after the sieve analysis of the soil sample:

(i) % of particles passing through 0.075 mm IS sieve = 4%

(ii) % of particles having size greater than 0.075 mm retained over 4.75 mm IS sieve = 62%

(iii) Coefficient of uniformity = 5

(iv) Coefficient of curvature = 1.5

As per ISSC system, the soil is classified as:

Solution:

__From the test results:__

(i) % of particles retained over 0.075mm IS sieve = (100 – 4)% = 96%

So, soil is **coarse Grained soil**

(ii) % fineness = 4% < 5%

(iii) As 62% (more than 50%) coarse fraction is retained over 4.75 mm IS sieve, the soil is termed as** Gravel**

Also, C_{u} = 5 > 4

C_{c} = 1.5 which lies between 1 and 3

So, soil is termed as **well graded Gravel**

Hence, soil is** GW**

QUESTION: 23

Which among the following clay mineral has least base exchange capacity?

Solution:

**Base Exchange Capacity:** The ability of the clay particles to absorb ions on its surface or edges is called Base or cation exchange capacity. Base exchange capacity depends upon the size of particles and and mineral structure.

QUESTION: 24

The given network diagram shows the detailed necessary for the CPM network analysis:

All the numerical values are given in days. The free float for the activity (3) – (5) will be

Solution:

Free float (F_{F}) = Total float (F_{T}) – Head event slack (S_{j})

Total float is the maximum available time in excess to activity completion time.

FF for the activity [(3) – (5)] = F_{T} [(3) – (5)] – S_{(5)}

= (14 – 8) – 4 – 2

= 0 days

*Answer can only contain numeric values

QUESTION: 25

A rectangular channel of 3m width carrying a discharge of 6m^{3}/sec. If the specific Energy at depth (y_{1}) is 20% more than specific Energy at critical depth, Then the depth of flow (y_{1}) will be ________ m.

Take acceleration due to gravity to be 9.81 m/s^{2}

Solution:

** Concept**:

Critical depth of flow:

q = discharge per unit width

** Calculation**:

Specific Energy at critical depth =

= 1.1122m

Specific Energy at depth “y_{1}” = 1.20 E_{c}

E_{1} = 1.20 × 1.1122

E_{1} = 1.3347m

y_{1} = 1.191m

QUESTION: 26

As per IS 10500: 2012, the permissible limits in absence of alternate source of water for calcium and magnesium in the drinking water respectively will be (in ppm)

Solution:

As per IS 10500: 2012,

As per IS 1055: 2012, the permissible limits in absence of alternate source of water for calcium and magnesium in the drinking water respectively will be 200 ppm and 100 ppm.

*Answer can only contain numeric values

QUESTION: 27

A confined coastal aquifer has an effective depth of 25 m. During the experiment to check the salt concentration in the water with discharge 0.5 m^{3}/s, the head difference in the two observation well was 0.5 m, located at 271.8 m and 100 m distance form the center of the test well. The hydraulic conductivity of the aquifer (in mm/s) will be _______.

Solution:

Discharge form well in confined aquifer (Q)

where, k = hydraulic conductivity

d = depth of the aquifer

S_{1} = drawdown in the observation well 1

S_{2} = drawdown in the observation well 2

(S_{1} - S_{2}) = 0.5m given

r_{1} = distance of observation well 1 form test well

r_{2} = distance of observation well 2 from test well.

⇒ k = 0.00637 m/s = 6.37 mm/s

QUESTION: 28

As per IRC: 31-102, the Vehicle Damage Factor (VDF) is a multiplier to convert the number of commercial vehicles of different axle loads and axle configuration into the number of repetitions of standard axle load of magnitude.

Solution:

The Vehicle Damage Factor (VDF) is a multiplier to convert the number of commercial vehicles of different axle loads and axle configuration into the number of repetitions of standard axle load of magnitude 80 kN.

It may also define as equivalent number of axles per commercial vehicles. The VDF varies with the vehicle axle configuration and axle loading.

QUESTION: 29

The speed and the theoretical maximum capacity of a single lane highway are 5m/s and 900 vehicle per hour respectively. The space headway and time headway respectively are:

Solution:

Theoretical maximum capacity of single lane can be expressed as

where, S = Capacity of single lane in vehicle per hour.

V = Speed in kmph

S = Space headway

Theoretical capacity can also be written as

Where, H_{t} = time headway

Calculation:

Space Headway (S) =

Time Headway (H_{t}) = 3600/900 sec

= 4 sec

QUESTION: 30

Rapid curing bitumen are the bitumen cutback with commonly used distillate will be

Solution:

Cutback bitumen is a bitumen with less viscosity which, is achieved by addition of volatile diluent. Hence, to increase fluidity of the bitumen binder at low temperature the binder is blended with a volatile solvent.

The viscosity of the cut book and the rate of which hardness on the road depends on the characteristics and quantity of both bitumen and volatile oil used as diluent.

Cutback bitumen are available in there types:

1) Rapid curing → Naptha, gasoline

2) Medium curing → Kerosene or high diesel oil

3) Slow Curing → High boiling point gas oil.

QUESTION: 31

In Accordance with IS 1343: 1980, In the absence of data, the approximate value of shrinkage strain for design of post-tensioning concrete member if age of concrete at transfer is 3 days will be?

Solution:

**According to 1343: 1980, clause 5.2.4.1,**

The approximate value of shrinkage strain for design shall be assumed as follows:

(i) For Pre-tensioning = 0.0003

(ii) ForPost − tensioning =

For t = age of concrete at transfer (days)

At t = 3 days

Shrinkage strain will be:

*Answer can only contain numeric values

QUESTION: 32

A cantilever beam of rectangular cross-section having 20 m span is subjected to uniformly distributed load throughout. The minimum effective depth required as per IS 456: 2000, so that the beam will not fail in deflection will be _______m. (up to two decimal places)

Use Modification factor for tension and compression reinforcement to be 0.95 and 1.345 respectively.

Solution:

As per IS 456: 2000, clause 23.2.1

__For cantilever beam:__

For span ≤ 10m

For, span > 10m, the limit of deflection criteria is modified.

Where MF_{1} = Modification factor for tension = 0.95

MF_{2} = Modification factor for compression = 1.345

d ≥ 4.47m

QUESTION: 33

Which one of the following statements is not correct?

Solution:

In bolted connection, the minimum width of lacing bars shall be **three times** the nominal diameter of the end bolt.

For more detail refers (IS 800:2007 section 7.6.2, page)

*Answer can only contain numeric values

QUESTION: 34

The total volume of water applied in 100 hectares of land was 25 × 10^{7} liters. If the water stored in the root zone is 0.20 cm, then the field application efficiency will be_______%

Solution:

Total volume of water applied

= (25 × 10^{7}) × 10^{-3} m^{3}

= 25 × 10^{4} m^{3}

Depth of the applied water

= 0.25 m

= 25 m

Also, Depth of water stored in the root zone = 20 cm

Therefore, field application efficiency

= 80%

*Answer can only contain numeric values

QUESTION: 35

The coefficient of variation of the rainfall for 10 rain gauges stations in a catchment was found to be 15%. If the admissible error allowed in the estimation of the mean rainfall data is 6% then, the optimum number of rain gauge station will be________.

Solution:

Optimum Number of Rain gauge station (N)

where, C_{V} = Coefficient of variation

ϵ = admissible error

N = (2.5)^{2}

N = 6.25 @ 7

Hence, Optimum Number of Rain gauge station will be 7.

QUESTION: 36

Consider the matrix equation

The condition for existence of a non-trivial solution, and the corresponding normalised solution (up to a sign) is

Solution:

For non-trivial solution, the rank of the matrix should be less than the number of variables. i.e. r < n.

For this, |A| = 0

⇒ (4c – 3b) – (2c – 6) + (b – 4) = 0

⇒ 4c – 3b – 2c + 6 + b – 4 = 0

⇒ 2c – 2b + 2 = 0

⇒ b = c + 1

The vectors x_{1}, x_{2} ….. x_{n }are said to be linearly dependent, if there exist numbers λ_{1}, λ_{2} ……. λ_{n}, not all zero such that

λ_{1}x_{1} + λ_{2}x_{2} + ….. + λ_{n}x_{n} = 0

Here,

λ_{1} + λ_{2} + λ_{3 }= 0 ----(1)

λ_{1} + 2λ_{2} + 3λ_{3 }= 0 ----(2)

2λ_{1} + bλ_{2} + 2cλ_{3 }= 0

2λ_{1} + (c + 1) λ_{2} + 2cλ_{3 }= 0 ----(3)

From (1) & (3):

λ_{2} = -2λ_{3}

λ_{1} = λ_{3} = λ

λ_{2} = -2λ

so corresponding normalised solution:

*Answer can only contain numeric values

QUESTION: 37

Two points are chosen randomly on a line 9 cm long. Determine the probability that the distance between them is less than 3 cm

Solution:

x: The distance of first point from the start of the line

y: distance of second point from the start of the line segment

x,y ϵ[0,9]

So sample space is Area of region bounded by

x ≥ 0, y ≥ 0, x ≤ 9, y ≤ 9

This is square of side 9

Area = 81 cm^{2}

The region of our interest is

|x-y| < 3

0 ≤ x ≤ 9

0 ≤ y ≤ 9

Area of shaded region = 2 (area of triangle) + area of rectangle

= 45

Probability = 45/81 =0.55

QUESTION: 38

The area bounded by the curve y =x (3 – x)^{2}, the x-axis and the ordinates of the maximum and minimum points of the curve is

Solution:

y = x(3 – x)^{2}

=−x.2(3−x)+(3−x)^{2 }= 3(x^{2}−4x+3) = 0

(x - 3) (x – 1) = 0 ⇒ x = 3, x = 1

= 4sq.unit

QUESTION: 39

Given N > 0, the iterative equation for finding using Newton-Raphson method is:

Solution:

QUESTION: 40

If y = 3e^{2x} + e^{-2x} - αx is the solution of the initial value problem = 1. Where, α, β ϵ R, then

Solution:

y = 3e^{2x} + e^{-2x} – αx

Given that,

⇒ 1 = 6 – 2 - α

⇒ α = 3

Complementary solution

(D^{2} + β) = 0 ----(1)

The given is, y = 3e^{2x} + e^{-2x} – αx

It indicates, 2 and -2 are the roots of auxiliary equations.

⇒ (D + 2) (D – 2) = 0

⇒ D^{2} – 4 = 0

By comparing this equation with equation (1)

β = -4

QUESTION: 41

A 5 kg object with a speed of 20 m/s strikes a steel plate at an angle of 40° and rebound at the same speed at angle, change in linear momentum of the object will be

Solution:

Initial momentum (Pi) = M_{object }× V_{t}

= 5 × 20

= 100 kg m/s

Final momentum (P_{f}) = M_{object} × V_{final}

= 5 × 20

= 100 kg. m/s

Change of momentum in x-direction

= (P_{f})_{x} – (P_{i})_{x} = 100 cos 40° - 100 cos 40°

= 0

Change of momentum in y – direction

= (P_{f})_{y} – (P_{i})_{y}

= 100 sin 40° - (-100 sin 40°)

= 2 × 100 × sin 40°

= 2 × 100 × 0.643

= 128.6 kg. m/s

QUESTION: 42

Two angle section ISA 100 × 100 × 10 are connected to either side of 14 mm gusset plate and is subjected to a working pull of 400 kN. If the welding is done in the workshop then, the length of the weld l_{1 }and l_{2} as shown in figure respectively will be?

(Take Ultimate strength as 410 MPa)

Solution:

Maximum size of the weld = × thickness of angle with rounded toe.

= × 10 = 7.5mm

Hence, provided weld size (w) = 7.5 mm

Length of weld = l_{1} + l_{2} = L (say)

∴ Design the strength of the weld

= 994.2 L

Also, Force in each angle section will be =

(Because of two angle section)

∴ 300 × 10^{3 } = 994.2 × L

⇒ L = 301.75 mm

∴ L = l_{1} + l_{2}

To provide zero eccentricity

Also, l_{1 }× 28.4 = l_{2 }× (100−28.4)

⇒ l_{1} = 216.053 mm = 216 mm

l_{2} = 302 – 216

l_{2} = 86

Hence l_{1} = 216 mm and l_{2} = 86 mm

*Answer can only contain numeric values

QUESTION: 43

A reinforced Concrete column of size 450mm × 550mm (overall depth) is provided with area of steel in compression as 4 bars of 25mm diameter. If the effective length of the column is 3.5m, the ultimate load carrying capacity of the column will be _______kN.

Use M-20 concrete and Fe-415 steel.

Solution:

**Follow step by step method:**

**(i) check for short or long column**

λ = 7.77 < 12

it is a short column

**(ii) check for Minimum eccentricity**

**Minimum eccentricity (e _{x-x}) is greater of**

(ii) 20 mm

So, e_{x-x} = 25.33 mm

e_{x-x} < 0.05 × 550

25.33 < 27.5 mm

(O.K)

For e_{y−y }= = 22mm

e_{y-y} < 0.05 × 450

22 mm < 22.5 mm

(O.K)

So, Load carrying capacity of the column will be

P_{u} = 0.4 f_{ck} A_{c} + 0.67f_{y} A_{sc}

f_{ck} = 20 N/mm^{2}

= 1963.50mm^{2}

f_{y} = 415 N/mm^{2}

P_{u} = 0.4 × 20 × [450 × 550 – 1963.50] + 0.67 × 415 × 1963.50

P_{u} = 2510.24 kN

*Answer can only contain numeric values

QUESTION: 44

A saturated clay layer 5m thick is overlain by sand layer 3m deep. Consider the water table to be present at a depth of 2m below the ground level or surface. If 2m deep sand layer of Bulk unit weight 20 kN/m^{3} is placed immediately over the surface, the increase in the effective stress at centre of the clay layer at the instant will be ______ kpa.

Given:

(i) γ_{sat} of clay = 19 kN/m^{3}

(ii) γ_{sat} of sand = 20 kN/m^{3}

(iii) ) γ_{t} of sand above water table = 18 kN/m^{3}

(iv) γ_{w} = 10 kN/m^{3}

Solution:

__Case I: When there is no fill over the surface__

**Effective vertical stress at centre of clay layer:**

= 18 × 2+20 × 1+19 × 2.5 − 10 × 3.5

= 68.5kN/m^{2 }= 68.5kpa

__Case: II when 2m thick sand is immediately placed over the surface__

When 2m thick sand is immediately placed over the surface, all the stresses is taken up by pore water. Hence pore water pressure is increased by 2(γt)sand2(γt)sand

Increase in pore water pressure at centre of clay layer = 2 × 20 = 40 kN/m^{2}

Δ u = 40 kpa

Also, simultaneous increase in total vertical stress at centre of clay layer = 2 × 20 = 40 kN/m^{2}

Δ σ = 40 kN/m^{2}

__Effective stress at centre of clay layer:__

= 18 × 2 + 20×1 + 40 + 19×2.5 −10×3.5 −40

= 68.5kpa

*Answer can only contain numeric values

QUESTION: 45

A sand layer of 15 m is overlain by 7.5 m clay layer which is further overlain by 10 m deep pond. A standpipe is inserted in the sand layer at the bottom of clay stratum and rise of Elevation 7.5 m is observed in the standpipe due to artesian pressure in the sand stratum. Calculate the quantity of flow into the pond in litres per day from the clay layer per meter square of the pond bed.

Take γ_{sat} for clay and sand layer to be 20 kN/m^{3} and 19 kN/m^{3} respectively and permeability of clay layer is 1.5 × 10^{–5} cm/sec.

Use γ_{w} =10 kN/m^{3}

Solution:

** Concept**:

Using Darcy’s law

q = kiA

k → permeability of soil

i → hydraulic gradient

A → Area of Soil Sample

** Calculation**:

hydraulic gradient (i) is =

h = head causing flow in the clay layer

Take C_{1} as datum

Total head at C_{1} = Datum head + Pressure head

= 0 + (7.5 + 10 + 7.5) = 25 m

Total head at C_{2} = Datum head + Pressure head

= 7.5 + 10

= 17.5 m

(T.H)_{C1 }> (T.H)_{C2}

Flow is occurring from C_{1} to C_{2} (upward)

Head causing the flow = (T.H)_{c1 }− (T.H)_{c2}

= 25 – 17.5

H_{L} = 7.5m

L = 7.5m

A = 1 m^{2}

k = 1.5 × 10^{–5} cm/sec

k = 1.5 × 10^{–7} m/sec

q = 1.5 × 10^{–7} × 1 × 1

q = 1.5 × 10^{–7} × 60 × 60 × 24 × 1 × 1

q = 12. 96 × 10^{–3} m^{3}/day

q = 12.96 litres/day

*Answer can only contain numeric values

QUESTION: 46

If a hydraulic jump occurring in a rectangular channel 2.5 m wide such that the depth after the jump is 2.45m. If the discharge in the rectangular channel is 10 m^{3}/sec, then the ratio of energy loss (E_{L}) in the jump to the pre-jump depth will be?

Solution:

__Concept:__

Using the relation,

y_{1} = Pre-jump depth

y_{2} = Post-jump depth

F_{1} = Froude number before jump

F_{2} = Froude number after jump

**Energy loss is given by:**

__Calculation:__

y_{2} = 2.45 m

F_{2} = 0.333

y_{1} = 0.458 m

E_{L} = 1.761 m

*Answer can only contain numeric values

QUESTION: 47

Find the minimum sight distance in m to avoid head-on collision of two cars approaching at 80 kmph and 50 kmph.

Use reaction time for both the driver, t = 2.5 sec, co-efficient of longitudinal friction f = 0.4 and brake efficiency of 80% in either case?

Solution:

Given, V_{1} = 80 kmph

V_{2} = 50 kmph

f = 0.4

Brake efficiency, η = 80% = 0.80

SSD for slow speed car =

SSD_{50 }= 0.278×50×2.5 +

= 65.50 m

SSD for high speed car,

SSD_{80 }= 0.278×80×1.5 +

= 134.34 m

∴ stopping sight distance to avoid head-on collision of the two approaching cars,

SSD = SSD_{50} + SSD_{80}

= 65.50 + 134.34

= 199.84 m

*Answer can only contain numeric values

QUESTION: 48

An observer (P) is standing on the deck of a ship just sees the another observer (Q) whose eye is 12m above the sea level. If the height of eye of the observer ‘P’ is 8m above the seal level. Determine the distance between the two observers in kms.

Ignore the effect of Refraction.

Solution:

Let d_{1} is the distance of observer ‘P’ from the horizon

d_{1} = 10.097 kms

Let d_{2} is the distance of observer ‘Q’ from the horizon

h_{2} = 12m

d_{2} = 12.367 kms

Total distance between the two observers

= d_{1} + d_{2}

= 10.097 + 12.367

= 22.464 kms

QUESTION: 49

A cantilever beam of length ‘L’ and flexural rigidity EI Carries uniformly distributed load of w kN/m throughout the span and maximum deflection occur is ‘δ’. If the same beam of length ‘2L’and flexural rigidity “0.5 EI” is made simply supported, then the maximum deflection is observed to be

Solution:

Maximum Deflection of Cantilever beam of length L and flexural rigidity EI carries UDL of w kN/m is:

Maximum deflection of simply supported beam of length 2L and flexural rigidity 0.5 EI carries U.D.L. of w kN/m is:

*Answer can only contain numeric values

QUESTION: 50

For the Beam shown below, the value of unknown Reaction ‘X’ will be_____kN.

Solution:

**Using Moment distribution method:**

**Using principal of superposition:**

= 190.5kN − m

But moment at B must be zero, so applying a counter moment of 190.5 kN-m at B to make final moment at B equal to zero, Half of (-190.5 kN-m) is transferred to fixed end A.

So,final moment at A =

= 189.75kN − m(clockwise) = y

∑F_{V }= 0

x + Z = 150 kN

∑M_{B }= 0

x.10 – 50 × 5 – 100 × 2 + 189.75 = 0

10.x = 260.25

x = 26.025 kN

QUESTION: 51

Find the value of P for collapse of the beam shown in figure.

Solution:

Internal work = 3M_{P}(θ)+3M_{P}(θ)+3M_{P}(ϕ)+M_{P}.ϕ = 3M_{P}(θ)+3M_{P}(θ)+3M_{P}(ϕ)+M_{P}.ϕ

= M_{P}(6θ+4ϕ) = M_{P}(6θ+4ϕ)

= M_{P}(12ϕ+4ϕ) = 16M_{P}ϕ = M_{P}(12ϕ+4ϕ) = 16M_{P}ϕ

External work =

Internal work done = 3M_{P}(θ)+M_{P}(θ+θ)+M_{P}(θ) = 3M_{P}(θ)+M_{P}(θ+θ)+M_{P}(θ) = 6M_{P}θ

External work done =

Hence true collapse load is 9M_{p}/L.

*Answer can only contain numeric values

QUESTION: 52

A geometrically similar model is made to the scale ratio of 1:14 to study the characterstics of spillway. If the energy dissipated per second in prototype is 15000 N-m, what will be the Energy dissipated per sec in model ________ N – m.

Solution:

** Concept**: As it is the case of spillway, So Fraude model law is applicable

P_{m} = Power in model = ρQ_{m}gH_{m}

P_{p} = Power in Prototype = ρQ_{p}gH_{p}

** Calculation**:

P_{m} = 20.45 N – m per second

Hence power in model will be 20.45 N – m per second

*Answer can only contain numeric values

QUESTION: 53

An inverted U-tube differential manometer is connected to two pipes X and Y. Pipe X and Y is carrying the oil of specific gravity in the ratio = 0.96. If the difference in pressure between two pipes (p_{y} – p_{x}) is 1900 N/m^{2}, then the density of the fluid flowing in pipe is ________ kg/m^{3}.

Take the fluid in the manometer is oil of specific gravity 0.85.

Solution:

Ratio of specific gravity is given as:

= 0.96

= 0.96

ρ_{x} = 0.96 ρ_{y}

**Pressure in the left limb at A – A**

= p_{x} – ρ_{x}.g.0.45

= p_{x} – 0.96 ρ_{y}.g.0.45

**Pressure in the Right limb at A – A**

= p_{y} – ρ_{y}.g.0.35 – ρ_{m}.g.[0.45 + 0.25 – 0.35]

= p_{y} – ρ_{y}.g.0.35 – 0.85 × 1000 × g [0.35]

**Pressure in the left limb at A – A = Pressure in the right limb at A – A**

p_{x} – 0.96ρ_{y}.g.0.45 = p_{y} – ρ_{y}.g.0.35 – 297.5g

p_{x} – p_{y} = 0.432gρ_{y} – 0.35gρ_{y} – 297.5g

–1900 = 0.082gρ_{y} – 297.5g

ρ_{y} = 1266.09 kg/m^{3}

ρ_{x} = 1215.45 kg/m^{3}.

*Answer can only contain numeric values

QUESTION: 54

A transmission tower of height 14m was erected to top of a building. A theodolite is used to measure vertical angle to top and bottom of the tower and were measured as 9° and 6° respectively. Calculate the Reduced level (metres) of the top of the transmission tower if the staff reading of 1.425m was taken from the same point on a Benchmark of R.L 102.5m. Neglect curvature and refraction correction.

Solution:

In ΔPRS

In ΔPQS

V = Dtan6°

Dtan9° – Dtan6° = 14

D(tan9° – tan6°) = 14

D = 262.76 m

V = 262.76 tan6°

V = 27.62 m

R.L of top of the tower = (R.L)_{B.M} + Staff reading on BM + V + 14

R.L._{Q} = 102.5 + 1.425 + 27.62 + 14

R.L._{Q} = 145.545 m

QUESTION: 55

The number of photographs required to cover an area of 30 km × 25 km if the scale of an aerial photograph is 1 in 15000 is?

Longitudinal overlap = 58%

side overlap = 29%

Format of photograph taken = 24 cm × 24 cm

Solution:

__Conc____ept__**: **

No. of photograph in the direction of flight (N_{1})

No. of photographs perpendicular to the direction of flight (N_{2})

Where,

P_{l} = overlap in the direction of flight

P_{w} = overlap in the direction perpendicular to the direction of fligh

tl = length of photograph in the direction of flight

w = width of photograph normal to direction of flight

L_{1} = Length of ground to be covered

W_{1} = Width of ground to be covered

S = Scale of the photograph

Total number of photograph required: = N_{1} × N_{2}

__Calculation:__

L_{1} = 30000 m

W_{1} = 25000 m

P_{l} = 58%

P_{s} = 29%

l = 24 cm

N_{1} = 20.84

N_{2} = 10.78

Total no. of Photographs = N_{1} × N_{2}

N = 20.84 × 10.78

N = 224.65 ≅ 225

*Answer can only contain numeric values

QUESTION: 56

A pile is driven through 5 m of soft clay which is underlain by stiff clay. The average cohesion from 5 m to 15 m depth is 160 kN/m^{2} and the end bearing resistance is 190 kN/m^{2}. Calculate the Factor of safety of the pile against the safe working load of 500 kN. Take adhesion factor between stiff clay and pile to be 0.5. The diameter of the pile is 0.6 m. Neglect the skin friction from loose clay.

Take N_{c} = 9.

Solution:

Ultimate bearing capacity of pile is given by:

Q_{up} = Q_{eb} + Q_{sf}

Where,

Q_{eb} = End bearing resistance of the pile = CN_{c}A_{b}

Q_{sf} = Skin friction resistance of the pile =

Where,

A_{b} = base area of the pile

A_{sf} = Surface area of the pile

__Calculation:__

As no side resistance from the loose clay layer, so skin friction resistance is considered for stiff layer only for the length of (15-5) =10 m.

Q_{up} = 9×190× +0.5×160×π×0.6×10

Q_{up} = 483.49+1507.96

Q_{up} = 1991.45 kN

Which is greater than 2.5 (o.k).

*Answer can only contain numeric values

QUESTION: 57

A Pipeline “PQ” of diameter 500mm and length 300m carries water as shown in figure below. The flow is occurring from P to Q.

If the point Q is 20 metres above point P and the pressure at point ‘P’ is 20 N/cm^{2 }more than that of pressure at point Q. Then calculate the rate of flow (litres per second) through the pipe?

Take friction factor to be 0.034.

Solution:

__Applying Bernoulli’s equation at P and Q and take Point ‘P’ as datum:__

Z_{P} = 0[P as datum]

Z_{P} = 0[Pasdatum]

Z_{Q }= 20m

V_{P }= V_{Q }[FromContinuityequation]

P_{P}=P_{Q}+20×10^{4}

Q = 0.1198 m^{3}/sec = 119.8 liters/sec

*Answer can only contain numeric values

QUESTION: 58

A 8 hour unit hydrograph of a catchment of area 220 km^{2} is a triangular in shape with base width of 50 hours. A direct runoff hydrograph (DRH) of this catchment, due to 3 cm of effective rainfall for 4 hours will have a peak flow rate of _______m^{3}/s.

Solution:

Area of triangle OAB = Volume of runoff water

½ × Q_{P} × 50 × 60 × 60 = 220 × 10^{6} m^{2} × 1 cm

⇒Q_{P }= 220×10^{6 }×

Q_{p} = 24.44 m^{3}/s

For the Direct Runoff Hydrograph.

Q_{PDRH} = 3 × 24.44 m^{3}/s

Q_{PDRH} = 73.33 m^{3}/s

*Answer can only contain numeric values

QUESTION: 59

A rectangular concrete beam of cross-section 150mm wide and 400mm deep is prestressed by a cable with cross-sectional area of 350 mm^{2}. The effective span of the beam is 11m. The cable profile is parabolic with eccentricity of 75mm above the centroid of the cross-section at supports and 75mm below at midspan. If one end of the cable is tensioned, then percentage loss of prestress in the cable due to friction will be____?

Take wobble correction factor (k) and coefficient of friction (μ) between steel reinforcement and duct surface to be 0.15 per 100m and 0.55 respectively.

Solution:

** Concept**:

Loss of prestress due to friction is given by:

Loss = p_{0} (kx + μα)

Where,

p_{0} = initial prestress

k = wobble correction factor

x = distance from Jacking end

x = (when Jacking from both ends)

x = L (when Jacking from one end)

μ = Coefficient of friction between steel reinforcement and duct surface

α = Net change in gradient between two points of considerations.

**Equation of Cable:**

h = e_{1} + e_{2}

e_{1} = eccentricity at end

e_{2} = eccentricity at centre

For x = 0

** Calculation**:

e_{1} = 75 mm

e_{2} = 75 mm

h = e_{1} + e_{2} = 75 + 75 = 150 mm

= 0.10909radian

When Cable is tensioned at one end, the maximum loss of prestress occur at other end

Hence, at x = L = 11m

Loss of prestress = p_{0} (kx + μα)

x = 11m

μ = 0.55

α = 0.10909 radian

Loss = p_{0 }(0.0015 × 11 + 0.55 × 0.10909)

Loss = p_{0} (0.07649)

% loss = 7.65%

*Answer can only contain numeric values

QUESTION: 60

A PSC beam (simple supported over effective span of 9m) of size 300 × 500mm is prestressed by 700 mm^{2} area of cable stressed to 1350 mpa. If the straight cable is provided with eccentricity of 150mm above the Neutral axis, then the stress developed at top fibres at centre of the beam will be ________ mpa

The beam is subjected to 25 kN/m of load (inclusive of dead weight)

Solution:

Final stress diagram

B.M. at centre =

w = 25 kN/m

L = 9m

P = Prestressing force = A_{s} × p

P = 700 × 1350

P = 945 × 10^{3} N

Final stress developed at centre of beam at top fibres =

f_{t} = 6.3 + 11.34 + 20.25

f_{t} = 37.89 mpa

*Answer can only contain numeric values

QUESTION: 61

The theoretical volume of methane gas (in m^{3}/tonnes of waste) that would be expected from the anaerobic digestion of a tonne of waste having the composition C_{50}H_{100}O_{40}N will be ______.

Assuming, 1 mole of C_{50}H_{10}O_{40}N will produce 27.125 mole of methane (CH_{4}) and density of methane is 0.7167kg/m^{3}.

Solution:

Molecular mass of C_{50} H_{100} O_{40} N

= (12 × 50) + (1 × 100) + (16 × 40) + (14 × 1)

= 1354 gm

Molecular mass of CH_{4} = (1 × 12) + (4 × 1)

= 16 gm

∴ 1 mole of C_{50}H_{100}O_{40}N will produce 27.125 mole of methane

∴ 1354 gm of C_{50}H_{100}O_{40}N will produce 27.125 × 16 gm of methane

∴ Mass of methane =

= 320.5 kg/tonne

Volume of methane gas =

= 447.2 m^{3}/tonne of waste

*Answer can only contain numeric values

QUESTION: 62

The activated sludge process schematic flow diagram is shown in figure below:

where, Q, Q_{w} = flow rate, m^{3}/d

θ_{c} = sludge age, days

X, X_{r} = microorganism- concentration (mixed – liquor volatile suspended solids or MLVSS)

The volume of sludge that must be wasted each day if wastage is accomplished from point A will be ______m^{3}/day.

Solution:

According to definition

sludge age

Q_{w} X_{R} = 1000 kg/day

Q_{w} = 100 m^{3}/day

*Answer can only contain numeric values

QUESTION: 63

A filter baghouse is to process 15 m^{3}/s of waste gas. The baghouse is to be divided into eight section of equal cloth area so that one section can be shut down for cleaning and/or repair while the others continue operating. Laboratory analysis indicate an air to-cloth ratio of 9.0 m^{3}/min/m^{2}.cloth will provided sufficient treatment. The bags are 0.25 m in diameter and 7.0 m long. The number of bags for the above physical arrangement to meet the requirement will be _________.

Solution:

Flow of air = 15 m^{3}/s

Air to cloth ratio = 9 m^{3}/min/m^{2} cloth.

∴Area of colth rqurired =

= 100 m^{2}.

Also, Area of each bag = π DH

= 3.14 × 0.25 × 7

= 5.5 m^{2}

∴Number of bag requried =

= 18.19

Since baghouse is to be divided into eight equal section of equal cloth area so that one section can be shut down for cleaning or repair.

∴ Minimum number of bags required

= 3 × 8 = 24

*Answer can only contain numeric values

QUESTION: 64

In a Laboratory analysis to evaluate strength of the flexible pavement through Marshall method of mix design, the specific gravity of fine aggregate and bitumen was not measured accidently. The coarse aggregate, fine aggregate, finer and bitumen are mixed in the relative proportion (% by weight) of 50, 40, 6 and 4 respectively. The specific gravity of coarse aggregate and finer are 2.72 and 2.6 respectively. If the theoretical specific gravity of the mix and the effective specific gravity of the aggregate in the mix are respectively 2.525 and 2.7 then, the specific gravity of bitumen is______

Solution:

The effective specific gravity of aggregates (G_{e})

⇒ G_{FA} = 2.675

Where G_{FA} = specific gravity of fine aggregate.

Also Theoretical specific gravity

Where G_{b} = specific gravity of bitumen

⇒ G_{b} = 1.0099

⇒ G_{b} = 1.01

QUESTION: 65

A clayey soil sample in a consolidometer test showed decrease in void ratio from 1.15 to 1.05 when the pressure was increased from 0.30 kgf/cm^{2} to 0.60 kgf/cm^{2}. If the same soil shows 10% more decrease in void ratio when the pressure was increased from 0.30 kgf/cm^{2} to 0.80 kgf/cm^{2}, then change in coefficient of consolidation observed during the test will be?

Take initial coefficient of consolidation observed from test when the pressure was increased from 0.30 kgf/cm^{2} to 0.60 kgf/cm^{2} to be 12 m^{2}/year.

Solution:

Coefficient of compressibility (av)=

Δe = change in void ratio

increase in effective stress

Coefficient of volume compressibity

e_{0} = initial void ratio

Also,

k = c_{v}m_{v}γ_{w}

k = permeability of soil Sample

c_{v} = coefficient of consolidation

m_{v }= coefficient of volume compressibility

** Calculation**:

Δe_{1} = 1.15 – 1.05

Δe_{1} = 0.10

Δσ_{1} = 0.60 – 0.30

Δσ_{1} = 0.30 kgf/cm^{2}

= 3.80×10^{−3}cm^{2}/sec

k = 3.80 × 10^{–3} × 0.155 × γ_{w}

Now,

Δe_{2} = 1.10 × 0.10 = 0.11

Δσ_{2} = 0.80 – 0.30

Δσ_{2} = 0.50

As permeability of both the sample is Same_{s},

So,

3.80×10^{−3}×0.155=c_{v2}×0.122

c_{v2}=4.83×10^{−3}cm^{2}/sec

c_{v2}=15.22m^{2}/ year

Change in coefficient of consolidation:

Δc_{v}=c_{v2}−c_{v1 }

Δc_{v} = 15.22 – 12

Δc_{v} = 3.22 m^{2}/year

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