Previous Year Paper

UPPSC AE Civil 2013 Official Paper - II (Previous Year Paper)

Held in 2013

100 questions120 minutes
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Civil Engineering (100 questions)

Question 1

When a train passes on curves which have no super-elevation it will give thrust on

  1. AOption (a)

    The inner rail

  2. BOption (b)

    The outer rail

  3. COption (c)

    inner side of the inner rail

  4. DOption (d)

    inner side of the outer rail

View answer & explanation

Correct answer

Answer: (d)D.

inner side of the outer rail

Why this answer is correct

Concept:

Whenever the superelevation is not provided the inner side of the outer rail experiences thrust applied by the centrifugal force created due to movement on the curves.

The flanges of the rolling stock push against the inner side of the outer rail and this create extra pressure on the inner side of the outer wheel.

The pressure on outer rail is greater than that on the inner rail under these circumstances.

Whenever a train moves on curves it experiences a centrifugal force that acts outwards. Superelevation is provided to counteract this centrifugal force.

Other functions of superelevation are:

a) To ensure a better distribution of loads on both rails.

b) To reduce the wear and tear of rolling stock and rails

c) To provide comfort to passengers.

Tips:- In the presence of equilibrium superelevation, pressure on both the rails are equal.

Important Points

Superelevation or cant is the amount by which one rail is raised above the other rail. If the outer rail is raised above the inner rail it is called positive superelevation and vice-versa.

The value of equilibrium superelevation is given as:

e=GV2127R{\rm{e}} = \frac{{{\rm{G}}{{\rm{V}}^2}}}{{127{\rm{R}}}}

G = Gauge, V = Speed in km/h, and R = Radius of curve in V2

e=GV2127RWhe

Question 2

A scissor crossover between two parallel railway tracks contains

  1. AOption (a)

    after angle crossover

  2. BOption (b)

    a turn-table device

  3. COption (c)

    a diamond crossover

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

a diamond crossover

Why this answer is correct

Concept:

Scissor Cross Over or Double Cross Over

  • It enables the trains to change the track from either direction along the main track.
  • It consists 4 pairs of points, 6 acute angle crossings, 2 obtuse angle crossing, check rails and straight lengths.
  • It is used when space for 2 separate cross over is not available.
  • It is the combination of one cross over the other cross over in opposite direction

Turn- Table: Turn table or wheel house is a device for turning railway rolling stock, usually locomotives, so that they can be moved back in the direction from which they came.

Question 3

Runway length required under standard conditions is 1500 m then actual elevation of the site is 1100 m above M. S. L Runway length corrected for altitude will be

  1. AOption (a)

    1115 m

  2. BOption (b)

    1678 m

  3. COption (c)

    1885 m

  4. DOption (d)

    2050 m

View answer & explanation

Correct answer

Answer: (c)C.

1885 m

Why this answer is correct

Basic runway length is based on some assumptions:-

a) Derived at Mean Sea level

b) Derived for a standard temperature of 15°C

c) The gradient must be zero.

The actual runway need not follow all these guidelines and hence we must apply some corrections.

The sequential correction is applied in the basic runway length.

The first correction is altitude correction, the second correction is temperature correction and the last correction is gradient correction.

Altitude Correction:

Basic runway length is increased by 7% for every 300 m rise above mean sea level.

This is done due to the decrease in drag force on the aircraft owing to the decrease in density of air as the altitude increases.

Given:

Height above MSL = 1100 m, and Basic Runway length = 1500 m

After applying elevation correction,

Correction=1500×7100×1100300=385;m{\rm{Correction}} = 1500 \times \frac{7}{{100}} \times \frac{{1100}}{{300}} = 385{\rm{;m}}

Corrected Runway Length = 1500 + 385 = 1885 m

Tips and Tricks:-

Temperature correction and gradient corrections are applied on the subsequently corrected runway length.

Question 4

Based on spot speed data which speed is used for geometric design of roads?

  1. AOption (a)

    15 percentile

  2. BOption (b)

    50 percentile

  3. COption (c)

    85 percentile

  4. DOption (d)

    98 percentile

View answer & explanation

Correct answer

Answer: (d)D.

98 percentile

Why this answer is correct

If the spot speed data is plotted for a particular track as shown below:

The 98 percentile speed is adopted for the geometric design of the highway.

Geometrical design speed = 98th percentile speed

Safe speed limit = 85th percentile speed

Minimum speed limit = 15th percentile speed

Question 5

The rate of rise/fall of the road surface along its length, is called.

  1. AOption (a)

    Cant

  2. BOption (b)

    Super- elevation

  3. COption (c)

    Banking

  4. DOption (d)

    Gradient

View answer & explanation

Correct answer

Answer: (d)D.

Gradient

Why this answer is correct

Gradient: It is the rate of rise or falls along the length of the road with respect to horizontal.

Expressed as ‘1' vertical unit to 'N' horizontal units.

Types of Gradient:

i) Ruling gradient: The gradient is fixed in such a way that maximum power developed by the engine is equal to the power required to overcome the resistance to the motion on the grade at design speeds.

ii) Limiting gradient: A gradient steeper than the ruling gradient which may be used in restricted length, where the ruling gradient is not possible.

iii) Exceptional gradient: A gradient steeper than the limiting gradient which may be used in very short stretchers only in an exceptional case.

TerrainRulingLimitingsExceptional
Plain/Rolling3.35.06.7
Hilly5.06.07.0
Steep6.07.08.0

Question 6

As per I.R.C, it is considered appropriate that roads in rural areas should be designed for

  1. AOption (a)

    20 – 25 years

  2. BOption (b)

    15 – 20 years

  3. COption (c)

    10 – 15 years

  4. DOption (d)

    5 – 10 years

View answer & explanation

Correct answer

Answer: (c)C.

10 – 15 years

Why this answer is correct

Design life depends upon the environmental conditions, the type and strength of materials used, etc. For rural roads design life of 10 - 15 years is considered a standard value.

Important Points:-

Indian Roads Congress (IRC) was established in 1934 on the recommendations of the Jayakar committee. It is the governing technical body in India for the design and construction of roads.

Objectives of IRC are:

  • To promote scientific construction of roads.
  • To promote the use of standard specifications and practices.
  • To suggest legislation for the improvement, development and protection of roads.

Question 7

Relation between gradient (g) and camber (c) in case of roads, is

  1. AOption (a)

    g = c

  2. BOption (b)

    g=c2{\rm{g}} = \frac{{\rm{c}}}{2}

  3. COption (c)

    c=34g{\rm{c}} = \frac{3}{4}{\rm{g}}

  4. DOption (d)

    c=g2{\rm{c}} = \frac{{\rm{g}}}{2}

View answer & explanation

Correct answer

Answer: (d)D.

c=g2{\rm{c}} = \frac{{\rm{g}}}{2}

Why this answer is correct

Camber: It is the slope provided to the road surface in the transverse direction to drain off the rainwater from the road surface. It is also known as the cross slope of the road.

Camber can be written as 1 in n or x%.

There are generally three types of the cambers: (a) Straight Camber (b) Parabolic Camber (c) Mixed Camber.

Gradient: It is the slope provided to the surface of the road in the longitudinal direction for the vertical alignment of the road. There are three kinds of gradients:

(a) Ruling Gradient (b) Limiting Gradient (c) Exceptional Gradient

∴ The gradient and camber are interrelated to each other with the following relationship:

G = 2 × C

∴ C = G/2 is a correction option.

Question 8

Horizontal alignment in mountainous country is conditioned by

  1. AOption (a)

    Drainage structures

  2. BOption (b)

    Depth of cut and fill

  3. COption (c)

    Cost of right of way

  4. DOption (d)

    Maximum grade criteria

View answer & explanation

Correct answer

Answer: (a)A.

Drainage structures

Why this answer is correct

Concept:

As per IRC73-1980, Clause 9.1.1, uniformity of design standard is one of the essential requirements of a road alignment.

For a given section there must be consistent application of a design element to avoid unexpected situation being created for the drivers.

For mountainous terrain, the drainage of the road surfaces is very essential, so as to achieve consistency, any unnecessary break in horizontal alignment at cross-drainage structures should be avoided.

Thus the horizontal alignment in mountainous country is conditioned by drainage structures.

Question 9

The main function of fish plates is

  1. AOption (a)

    to join the two rails together

  2. BOption (b)

    to join rails with the sleeper

  3. COption (c)

    to allow rail to expand and contract freely

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

to join the two rails together

Why this answer is correct

Concept:

The different types of track fasteners used in railway engineering are as follows:

Track FastenerUse
Fish PlatesUsed at rail joints in order to maintain continuity It also allows thermal contraction and thermal expansion
SpikesUsed to hold rail on wooden sleepers For holding flat footed rails on wooden sleepers’ dog spikes are used.
ChairUsed to support bull headed rails on the sleepers Slide chairs are used to hold stock rail and tongue
BoltsUsed where sleepers directly rest over the steel girders Fish bolts are used to resist heavy transverse shear.
KeysUsed to fix rails to chairs on metal sleepers.
<br>

Confusion Point:

Since both options 1 and 3 are correct. But there is no such option, so we will go for option 1. As the fish plates are primarily used for connecting two rails together.

Question 10

In railways, most severe gradient is 

  1. AOption (a)

    Gradients of rack railways

  2. BOption (b)

    Pusher gradient

  3. COption (c)

    Momentum gradient 

  4. DOption (d)

    Exceptional gradient

View answer & explanation

Correct answer

Answer: (b)B.

Pusher gradient

Why this answer is correct

Order of different gradients in railways are as follows:

(Gradients in station yards) < (Ruling Gradient) < (Momentum Gradient) < (Pusher or helper Gradient)

Concept:

Type of gradientCharacteristics/Definitions
Ruling gradient (Gradient on railway tracks)1. It is the maximum gradient allowed on the railway tracks. Steep gradient requires powerful locomotives, hence steep gradients are avoided. 2. For plain terrain – (0.50% to 0.67%) 3. For hilly regions – (0.67% to 1%)
Momentum Gradient1. If rising gradient is followed by the falling gradient, the train while coming down in falling gradient acquires sufficient momentum, this momentum gives additional kinetic energy to the train. 2. This rising gradient is called momentum gradient. 3. It is steeper than the ruling gradient.
Pusher or Helper Gradient1. In mountainous regions where assistant engine or pusher engine or helper engine is required at the end of the train is known as pusher gradient. 2. In general for 1 in 75 and above pusher engine is used.
Gradients in station Yards1. It is provided to drain off the water used for cleaning rains. 2. It should be sufficiently low to prevent the movement of standing vehicles and also to prevent additional resistance at the start of vehicle. 3. Maximum gradient permitted on the station yards is 1 in 400 to 1 in 1000
<br>

Confusion:

Exceptional gradient is provided in highway engineering not in railway. And it is the steepest gradient which is provided in unavoidable circumstances.

Question 11

The transition curve used in the horizontal alignment of highways as per IRC recommendation is

  1. AOption (a)

    spiral

  2. BOption (b)

    lemniscate

  3. COption (c)

    cubic parabola

  4. DOption (d)

    any of the above

View answer & explanation

Correct answer

Answer: (a)A.

spiral

Why this answer is correct

Transition curve is a curve in plan which is provided to change the horizontal alignment from straight to circular curve gradually means the radius of transition curve varies between infinity to R or R to infinity.

IRC recommends Spiral or clothoid as the ideal transition curve due to following reasons:

i) It satisfies that rate of change of centrifugal acceleration is constant i.e., Ls.R = constant.

Where Ls = length of transition curve R = radius of curve.

ii) The calculation and field implementation of spiral curve is simple and easy.

iii) It enhances aesthetics also.

Question 12

For sandy soils the most common method of stabilization is

  1. AOption (a)

    soil-cement stabilization

  2. BOption (b)

    mechanical stabilization

  3. COption (c)

    soil-lime stabilization

  4. DOption (d)

    soil-bitumen stabilization

View answer & explanation

Correct answer

Answer: (a)A.

soil-cement stabilization

Why this answer is correct

Concept:

Sandy soils are generally stabilized with cement.

The amount of cement required to stabilize depends upon the quantity and quality of fines contained in sandy soils and final compacted density.

The cement required for sandy soils range between 5 and 12% by weight.

Stabilization of soil with cements consists of adding cement to pulverized soil and permitting the mixture to harden by hydration of cement.

The quantity of cement required for stabilization increases as soil plasticity increases.

Important Point:

Soil-lime stabilization: It is the process of adding lime to the soil to improve its properties like density, bearing capacity etc. The principle involved in lime-soil stabilization is the exchange of ions between lime and soil when added.

When clayey soil is treated with lime, cation exchange takes place between them which increases plastic limit and reduces plasticity index which finally results in increase in stability of soil. This method is used mostly for clayey flocculated soil.

Soil-bitumen stabilization: It is applied to highways and airfield construction. It is the method of designing and mixing local soil or aggregates with bituminous materials to form a stable and waterproof base course.

Bitumen stabilization is used for:-

a) For binding soil particles together for purpose of supplying cohesion for non-mechanical stabilized granular material.

b) For waterproofing, mechanical stabilized granular mixtures or for waterproofing cohesieve soils.

Question 13

The station which is selected closed to the main triangulation station, to avoid intervening obstruction, is known as

  1. AOption (a)

    satellite station

  2. BOption (b)

    Base station

  3. COption (c)

    supplementary station

  4. DOption (d)

    pivot station

View answer & explanation

Correct answer

Answer: (a)A.

satellite station

Why this answer is correct

Explanation

Triangulation surveying is the tracing and measurement of a series or network of triangles to determine distances and relative positions of points spread over an area, by measuring the length of one side of each triangle and deducing its angles and length of other two sides by observation from this baseline.

Pivot point:

The station where observations are not made, but the angles at the station are used in the triangulation series, is known as the Pivot station.

Satellite station:

A subsidiary station is established as near the true or principal station as possible, the station so established is called a satellite station or eccentric station.

Additional InformationTo secure well-conditioned triangles/ intervisibility, objects such as chimneys, towers, are selected as triangulation stations 

These can be sighted from other stations, but it is not possible to occupy these stations for taking observations 

In such cases, a satellite station is established and observations are taken and later reduced to what they would have been if the main station was occupied

Question 14

A tape of length l and weight w kg/m is suspended at its ends with a pull of p kg. the sag correction is:

  1. AOption (a)

    l2w224p2\frac{{{l^2}{w^2}}}{{24{p^2}}}

  2. BOption (b)

    l2w324p2\frac{{{l^2}{w^3}}}{{24{p^2}}}

  3. COption (c)

    l3w224p2\frac{{{l^3}{w^2}}}{{24{p^2}}}

  4. DOption (d)

    l2w224p3\frac{{{l^2}{w^2}}}{{24{p^3}}}

View answer & explanation

Correct answer

Answer: (c)C.

l3w224p2\frac{{{l^3}{w^2}}}{{24{p^2}}}

Why this answer is correct

Correction for sag is given by:

Cs=;−w2l324;P2=;−W2l24;P2{C_s} = ; - \frac{{{w^2}{l^3}}}{{24;{P^2}}} = ; - \frac{{{W^2}{l}}}{{24;{P^2}}}

W is the total weight of tape  = w × L

Where,

W = weight of tape per unit length (N/m)

l = length of tape suspended between supports (m)

p = pull applied (N)

Question 15

Θ1 and θ2 are the angels of elevation from ‘A’ to the top and bottom of a vertically held rod of length ‘S’ at B. The horizontal distance AB will be

  1. AOption (a)

    stan⁡θ1+tan⁡θ1\frac{{\rm{s}}}{{\tan {{\rm{\theta }}_1} + \tan {{\rm{\theta }}_1}}}

  2. BOption (b)

    stan⁡θ1−tan⁡θ2\frac{{\rm{s}}}{{\tan {{\rm{\theta }}_1} - \tan {{\rm{\theta }}_2}}}

  3. COption (c)

    stan⁡θ2−tan⁡θ1\frac{{\rm{s}}}{{\tan {{\rm{\theta }}_2} - \tan {{\rm{\theta }}_1}}}

  4. DOption (d)

    S(tanθ1 – tan θ2)

View answer & explanation

Correct answer

Answer: (b)B.

stan⁡θ1−tan⁡θ2\frac{{\rm{s}}}{{\tan {{\rm{\theta }}_1} - \tan {{\rm{\theta }}_2}}}

Why this answer is correct

Let the horizontal distancebetween A and B be ‘d’.

In ΔABD,

tanθ2=hd{\rm{tan}}{{\rm{\theta }}_2} = \frac{{\rm{h}}}{{\rm{d}}}

Again,

In ΔACD,

tanθ1=h+sd{\rm{tan}}{{\rm{\theta }}_1} = \frac{{{\rm{h}} + {\rm{s}}}}{{\rm{d}}}

Solving both the equation we get,

d=stan⁡θ1−tan⁡θ2{\bf{d}} = \frac{{\bf{s}}}{{\tan {{\bf{\theta }}_1} - \tan {{\bf{\theta }}_2}}}

Question 16

The radius of a simple circular curve is 300 m and length of its specified Chord is 30 m. The degree of the curve is

  1. AOption (a)

    5.73° 

  2. BOption (b)

    5.37°

  3. COption (c)

    3.57°

  4. DOption (d)

    3.75°

View answer & explanation

Correct answer

Answer: (a)A.

5.73° 

Why this answer is correct

Concept:

Given: Radius of the curve = 300 m

Formula:

Degree of curve is given by :

D=1720R;degree{\rm{D}} = \frac{{1720}}{{\rm{R}}}{\rm{;degree}},

Where,

D = Degree of the curve (degrees) and R = radius of the curve (m)

Calculation:

Degree of curve,

D=1720R=1720300=5.733∘{\rm{D}} = \frac{{1720}}{{\rm{R}}} = \frac{{1720}}{{300}} = 5.733^\circ

Note: Degree of curve is always calculated for 30m chain in general.

Question 17

The shift of a circular curve is given by __________

Where,

L = Length of transition curve and R = Radius of the circular curve

  1. AOption (a)

    L2 /6R

  2. BOption (b)

    L2/24R

  3. COption (c)

    L/24R

  4. DOption (d)

    L/6R

View answer & explanation

Correct answer

Answer: (b)B.

L2/24R

Why this answer is correct

To facilitate a smoother transition to the vehicle while traveling form transition curve to the circular curve. The circular curve is provided with some downward displacement to match the properties of the transition curve. The downward displacement is known as a shift of the curve. 

Empirically, the shift of the curve (s) is given by: S=L224R{\rm{S}} = \frac{{{{\rm{L}}^2}}}{{24{\rm{R}}}}

Where,

L = length of the curve and R = Radius of the curve.

Question 18

The multiplying constant of a tachometer is given by

When ‘f’ is the focal length of the objective, ‘i’ is the stadia interval and ‘d’ is the distance between the- objective and the vertical axis.

  1. AOption (a)

    fi\frac{f}{i}

  2. BOption (b)

    if\frac{i}{f}

  3. COption (c)

    f + d

  4. DOption (d)

    f+di\frac{{f + d}}{i}

View answer & explanation

Correct answer

Answer: (a)A.

fi\frac{f}{i}

Why this answer is correct

Concept:

Tachometer: 

It’s a transit theodolite fitted with a stadia diaphragm which gives both horizontal and vertical control with an accuracy of 1 in 1000.

Distance calculation involves the computation of two constants i.e. multiplying constant (k) and additive constant (c).

k=fi;and;c=f+d\rm{k = \frac{f}{i};{\rm{and;}}c = f + d}

Where

1f=1f1+1f2\rm{\frac{1}{f} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}}}

f = combined focal length

f1 = focal length of objective

f2 = focal length of the eyepiece

d = distance of the objective from the center of the instrument.

Question 19

Tachometric formula for horizontal distance using inclined sights through θ is obtained by multiplying

  1. AOption (a)

    the constants by sin2θ

  2. BOption (b)

    the constants by cos2 θ

  3. COption (c)

    the constants by cos θ

  4. DOption (d)

    the multiplying constant by cos2 θ and additive constant by cos θ

View answer & explanation

Correct answer

Answer: (d)D.

the multiplying constant by cos2 θ and additive constant by cos θ

Why this answer is correct

Tachometer: It’s a transit theodolite fitted with a stadia diaphragm which gives both horizontal and vertical control with an accuracy of 1 in 1000.

Horizontal Distance (D) ⇒

D=k×s×(cos⁡θ)2+c×cos⁡θ\rm{D = k \times s \times {\left( {\cos \theta } \right)^2} + c \times \cos \theta}

Vertical distance (V) ⇒

V=k×s×sin⁡θ×cos⁡θ+c×sin⁡θ\rm{V = k \times s \times \sin \theta \times \cos \theta + c \times \sin \theta}

Where,

k = multiplying constant, c = additive constant, s = staff intercept, and θ = angle of inclination of measurement

For given condition: (θ = 0°),

Assuming analltic lens i.e. k = 100 and c = 0

∴ Tachometric formula for horizontal distance using inclined sights through θ is obtained by multiplying the multiplying constant by cos2 θ and additive constant by cos θ.

Question 20

In aerial vertical photography, the longitudinal overlap is normally kept as

  1. AOption (a)

    50%

  2. BOption (b)

    60%

  3. COption (c)

    70%

  4. DOption (d)

    75%

View answer & explanation

Correct answer

Answer: (b)B.

60%

Why this answer is correct

Concept:

In aerial photography, two types of overlaps are provided:

a. Longitudinal Overlap: This overlapping of photographs along the flight strips is known as longitudinal overlap, end lap or forward overlap. Generally it is taken as 60%.

b. Side Overlap: Two adjacent flight strips are so photographed that there is a side lap or lateral overlap. Generally it is taken as 30%

Question 21

The main plate of a theodolite is divided into 1440 equal divisions. 60 divisions of the Vernier coincide exactly with 59 divisions of the main scale. The least count of the theodolite is

  1. AOption (a)

    5”

  2. BOption (b)

    10”

  3. COption (c)

    15”

  4. DOption (d)

    20”

View answer & explanation

Correct answer

Answer: (c)C.

15”

Why this answer is correct

The least count of a vernier is equal to the difference in length of one division of main scale and one division of vernier scale.

In a direct vernier (n-1) divisions of the main scale is equal to the ‘n’ divisions on the vernier scale.

Thus,

n 'v' = (n – 1) 's'

⇒v=(n−1)sn\Rightarrow {\rm{v}} = \frac{{\left( {{\rm{n}} - 1} \right){\rm{s}}}}{{\rm{n}}}

Thus the least count is given by,

L.C = s – v

⇒L.C=s−(n−1)sn\Rightarrow {\rm{L}}.{\rm{C}} = {\rm{s}} - \frac{{\left( {{\rm{n}} - 1} \right){\rm{s}}}}{{\rm{n}}}

⇒ L C = s/n

Given:

Total division on main scale = 1440

Thus,

Length of one division (s) = 360°/1440

n = 60

⇒L.C=360∘/144060×3600′′=15′′\Rightarrow {\rm{L}}.{\rm{C}} = \frac{{360^\circ /1440}}{{60}} \times {3600^{{\rm{''}}}} = {15^{{\rm{''}}}}

Important Points

A theodolite is an important instrument used to measure horizontal and vertical angles.

There are two types of theodolites based on the method of obtaining reading:

a) Vernier Theodolite:-

  • These theodolites use verniers to take readings and are most commonly used. The upper plate of a transit theodolite has two verniers which are diametrically opposite to each other. If this plate is unclamped the upper plate rotates and the reading of the vernier changes.

b) Precise optical theodolites:-

  • Mostly used for astronomical works.

Question 22

Perpendicular offset from a tangent to the junction of a transition curve and circular curve is equal to _____

Where ‘S’ is shift.

  1. AOption (a)

    S

  2. BOption (b)

    2S

  3. COption (c)

    3S

  4. DOption (d)

    4S

View answer & explanation

Correct answer

Answer: (d)D.

4S

Why this answer is correct

Concept:

The offset for the transition curve are found from-

offset;(y)=ℓ36RL{\bf{offset}};\left( {\bf{y}} \right) = \frac{{{\ell ^3}}}{{6{\bf{RL}}}}

Where,

ℓ = measured along the curve (m),

R = radius of the curve (m), and

L = length of the curve (m)

∴ For offset at junction point, ℓ = L

y=l36RL=L36RL=L26R{\bf{y}} = \frac{{{{\bf{l}}^3}}}{{6{\bf{RL}}}} = \frac{{{{\bf{L}}^3}}}{{6{\bf{RL}}}} = \frac{{{{\bf{L}}^2}}}{{6{\bf{R}}}}         ---(1)

Also, we know Shift of the curve,

S=L224R{\bf{S}} = \frac{{{{\bf{L}}^2}}}{{24{\bf{R}}}}         ----(2)

From (1) and (2)

yS=4;\frac{{\bf{y}}}{{\bf{S}}} = 4;

⟹ y = 4S

Question 23

In plane tabling the instrument used to measure horizontal and vertical distances directly is known as

  1. AOption (a)

    Simply alidade

  2. BOption (b)

    Telescopic alidade

  3. COption (c)

    Tachometer

  4. DOption (d)

    Clinometer

View answer & explanation

Correct answer

Answer: (b)B.

Telescopic alidade

Why this answer is correct

Concept:

Plane tabling is a graphical method of surveying in which the field surveying and plotting proceed simultaneously.

Instruments used in plane table survey are as follows:

  1. The plane table with levelling head having arrangements.
  2. Alidade for sighting.
  3. Plumbing fork and plumb bob.
  4. Spirit level.
  5. Compass.
  6. Drawing paper with a rainproof cover.

Tachometer: It is used for measuring horizontal and vertical distances. But it's not used in plane table surveying.

Clinometer: It is used for measuring slope of the ground as it measures the angular displacement.

Question 24

With usual notations, the expression v2gR\frac{{{{\rm{v}}^2}}}{{{\rm{gR}}}} represents

  1. AOption (a)

    centrifugal force

  2. BOption (b)

    centrifugal ratio

  3. COption (c)

    super-elevation

  4. DOption (d)

    radial acceleration

View answer & explanation

Correct answer

Answer: (b)B.

centrifugal ratio

Why this answer is correct

Concept:

Centrifugal Force (F=mv2R)\left( {{\bf{F}} = \frac{{{\bf{m}}{{\bf{v}}^2}}}{{\bf{R}}}} \right)

It is defined as the force arising from body’s inertia, which appears to act on a body moving in a circular path and is directed away from the centre around which the body is moving.

Centrifugal Ratio or Impact Factor

It is defined as the ratio of centrifugal force to the weight of the vehicle.

It is given by:

Impact;Factor=Centrifugal;forceWeight=;FW=mv2Rmg=v2Rg{\rm{Impact;Factor}} = \frac{{{\rm{Centrifugal;force}}}}{{{\rm{Weight}}}} = {\rm{;}}\frac{{\rm{F}}}{{\rm{W}}} = \frac{{{\rm{m}}{{\rm{v}}^2}}}{{{\rm{Rmg}}}} = \frac{{{{\rm{v}}^2}}}{{{\rm{Rg}}}}

Super-elevation (e=v2gR−f)\left( {{\bf{e}} = \frac{{{{\bf{v}}^2}}}{{{\bf{gR}}}} - {\bf{f}}} \right)

It is the amount by which the outer edge of the curve on a road should be raised above the inner edge in order to counter balance the centrifugal force.

Radial Acceleration (ar=v2R)\left( {{{\bf{a}}_{\bf{r}}} = \frac{{{{\bf{v}}^2}}}{{\bf{R}}}} \right)

It is the acceleration of the object along the radius directed towards the centre.

Symbols have their usual meaning.

Confusion:

Sometime super-elevation is also given by ;v2Rg;\frac{{{{\bf{v}}^2}}}{{{\bf{Rg}}}}. For this condition to be satisfied, the value of e and f should such that the lateral friction coefficient should be zero. Such state of super-elevation is known as equilibrium state.

Question 25

If the length of chain is 20 m, the degree of curve is given by

Where R radius of circular curve.

  1. AOption (a)

    1146/R

  2. BOption (b)

    1546/R

  3. COption (c)

    1519/R

  4. DOption (d)

    1119/R

View answer & explanation

Correct answer

Answer: (a)A.

1146/R

Why this answer is correct

Concept:

For 20 m chain length:

Degree of the curve (D) is given by:

D=1146R\rm{D = \frac{{1146}}{R}}

For 30 m chain length:

Degree of the curve (D) is given by:

D=1719R\rm{D = \frac{{1719}}{R}}

Where

R = radius of the curve

Question 26

If the image of a triangulation station of R.L. 500 m is 4 cm from the principal point of a vertical photo taken form an altitude of 2000 m above datum, the height of displacement will be

  1. AOption (a)

    6 mm

  2. BOption (b)

    8 mm

  3. COption (c)

    10 mm

  4. DOption (d)

    12 mm

View answer & explanation

Correct answer

Answer: (c)C.

10 mm

Why this answer is correct

Concept:

Relief displacement (d) is the lateral shift in photographic position of an image due to its elevation above or below datum.

⇒d=rh2H−h1\Rightarrow {\rm{d}} = \frac{{{\rm{r}}{{\rm{h}}_2}}}{{{\rm{H}} - {{\rm{h}}_1}}}

Where,

d = Relief displacement, r = distance of image of top of tower, h2 = height of tower station, h1 = Bottom height above datum, and H = Flying altitude

Given: r = 4 cm, h2 = 500 m, h1 = 0 m, H = 2000 m

d=4×10−2×5002000=10−2m{\rm{d}} = \frac{{4 \times {{10}^{ - 2}} \times 500}}{{2000}} = {10^{ - 2}}{\rm{m}}

∴ d = 10 mm

Question 27

As per Indian Standard, the length of one link in the 30 m chain should be

  1. AOption (a)

    20 cm

  2. BOption (b)

    30 cm

  3. COption (c)

    40 cm

  4. DOption (d)

    10 cm

View answer & explanation

Correct answer

Answer: (a)A.

20 cm

Why this answer is correct

The 30 m chain is divided into 150 links. So, each link is of 0.2 m.

The tallies are provided after every 25 links (5 metres). A round brass ring is fixed after every metre as can be observed in the figure above.

Question 28

In plane tabling failure of fix occurs when 

  1. AOption (a)

    The plane table is inside the great triangle

  2. BOption (b)

    The plane table is inside the great circle

  3. COption (c)

    The plane table is outside the great circle

  4. DOption (d)

    The plane table is on the great circle

View answer & explanation

Correct answer

Answer: (d)D.

The plane table is on the great circle

Why this answer is correct

Explanation:

In plane tabling failure of fix occurs when The plane table is on the great circle.

The estimation of the location of the plane table depends on its position relative to the well-defined points considered for this purpose. Depending on their relative positions, three cases may arise :

(i) The position of the plane table is inside the great triangle;

(ii) The position of the plane table is outside the great triangle;

(iii) The position of the plane table lies on or near the circumference of the great circle.

In the case of (iii), the solution to the three-point problem becomes indeterminate or unstable.

But for cases (i) and (ii), Lehmann's rules are used to estimate the location of the plane table.

Question 29

In leveling, the correction due to refraction may be taken as.

Where Cc = Correction due to curvature

  1. AOption (a)

    12Cc\frac{1}{2}{C_c}

  2. BOption (b)

    13CC\frac{1}{3}{C_C}

  3. COption (c)

    15CC\frac{1}{5}{C_C}

  4. DOption (d)

    17Cc\frac{1}{7}{C_c}

View answer & explanation

Correct answer

Answer: (d)D.

17Cc\frac{1}{7}{C_c}

Why this answer is correct

Curvature Correction:

The curvature correction is given by:

CC=d22R{C_C} = \frac{{{d^2}}}{{2R}}

where

d = distance between any two given points (in km)

R = Radius of Earth in the same unit as that of d

The curvature correction is always negative.

Refraction Correction:

The refraction correction is given by:

CR=17×d22R{C_R} = \frac{1}{7} \times \frac{{{d^2}}}{{2R}}

The refraction correction is always positive.

Combined Correction due to curvature and refraction:

The combined correction due to curvature and refraction is given by:

CB=d22R−17×d22R=67×d22R{C_B} = \frac{{{d^2}}}{{2R}} - \frac{1}{7} \times \frac{{{d^2}}}{{2R}} = \frac{6}{7} \times \frac{{{d^2}}}{{2R}}

Hence, the correction in elevations due to the curvature and refraction is proportional to d2R\frac{{{d^2}}}{R}.

Question 30

The length of log chord in a circular curve is equal to:

Where ‘R’ is the radius of the curve, ‘φ’ is the defection angle.

  1. AOption (a)

    R sin φ

  2. BOption (b)

    R cos φ

  3. COption (c)

    2Rsinφ22{\rm{Rsin}}\frac{{\rm{\varphi }}}{2}

  4. DOption (d)

    2Rcos⁡φ22{\rm{R}}\cos \frac{{\rm{\varphi }}}{2}

View answer & explanation

Correct answer

Answer: (c)C.

2Rsinφ22{\rm{Rsin}}\frac{{\rm{\varphi }}}{2}

Why this answer is correct

Concept:

For the given curve:

Tangent length (T)=Rtan⁡ !!Δ!! 2\left( \text{T} \right)=\text{R}\tan \frac{\text{ }!!\Delta!!\text{ }}{2}

Length of curve (l)= !!π;!! R ;!!Δ!! 180\left( \text{l} \right)=\frac{\text{ }!!\pi ;!!\text{ R };!!\Delta!!\text{ }}{180}

Long chord (L)=2Rsin⁡ !!Δ!! 2\left( \text{L} \right)=2\text{R}\sin \frac{\text{ }!!\Delta!!\text{ }}{2}

External distance (E)=R(sec⁡ !!Δ!! 2 !! !! −1)\left( \text{E} \right)=\text{R}\left( \sec \frac{\text{ }!!\Delta!!\text{ }}{2}\text{ }!!~!!\text{ }-1 \right)

Mid-ordinate (M)=R(1−cos⁡ !!Δ!! 2 !! !! )\left( \text{M} \right)=\text{R}\left( 1-\cos \frac{\text{ }!!\Delta!!\text{ }}{2}\text{ }!!~!!\text{ } \right)

Chainage of T1 = Chainage of P.I. – T, and Chainage of T2 = Chainage of T1 + l

Question 31

The maximum rate of change of radial acceleration allowed on transition curves is

  1. AOption (a)

    100 mm/sec3

  2. BOption (b)

    300 mm/sec3

  3. COption (c)

    400 mm/sec3

  4. DOption (d)

    500 mm/sec3

View answer & explanation

Correct answer

Answer: (d)D.

500 mm/sec3

Why this answer is correct

Concept:

The rate of change of radial acceleration is given by,

\({\bf{C}} = \frac{{80}}{{75 + {\bf{V}}}};{\bf{m}}/{\bf{se}}{{\bf{c}}^3} = {\rm{;}}\left{ {\begin{array}{{20}{c}} {\not\le0.5;{\bf{m}}/{\bf{se}}{{\bf{c}}^3}}\ {\not\ge0.8;{\bf{m}}/{\bf{se}}{{\bf{c}}^3}} \end{array}} \right} = {\rm{;}}\left{ {\begin{array}{{20}{c}} {\not\le500;{\bf{mm}}/{\bf{se}}{{\bf{c}}^3}}\ {\not\ge 800;{\bf{mm}}/{\bf{se}}{{\bf{c}}^3}} \end{array}} \right}\)

Here, V = velocity of the vehicle (kmph)

∴ According to the given option, The maximum rate of change of radial acceleration allowed on transition curves is 500 mm/sec3.

Question 32

In a centered triangle the equations of condition are 

  1. AOption (a)

    four angle conditions

  2. BOption (b)

    three angle conditions and one side condition

  3. COption (c)

    four angle conditions and one side condition

  4. DOption (d)

    three angle conditions and two side conditions only

View answer & explanation

Correct answer

Answer: (b)B.

three angle conditions and one side condition

Why this answer is correct

Triangulation: It is the method of surveying based on the trigonometric proposition that if one side and two angles of a triangle are known, the remaining sides can be computed.

A triangulation system consists of a series of joined or overlapping triangles in which an occasional side is measured and remaining sides are calculated from angles measured at the vertices of the triangles.

Interpretation:

As triangle consists of maximum three angles, the requirement of four angle measurement is just unnecessary and hence option 1 and 3 is eliminated.

For plotting a triangle accurately, even if all three angles are given, the triangle cannot be plotted without at least data of any one side of the triangle.

∴ Along with three angles, for centered triangle we requires at least one side length and hence three angle conditions and one side condition is necessary as the equation of condition for centered triangle.

Question 33

In case of gravity dam subjected to earthquake; the hydrodynamic pressure variation curve is taken to be ( as per Von Korman)

  1. AOption (a)

    Parabolic

  2. BOption (b)

    Elliptical

  3. COption (c)

    Triangular

  4. DOption (d)

    Elliptical cum parabolic

View answer & explanation

Correct answer

Answer: (a)A.

Parabolic

Why this answer is correct

Concept:

The horizontal acceleration increases the water pressure as the dam accelerates towards the reservoir and water resists the movement due to its inertia.

The pressure distribution is parabolic in nature.

The magnitude and its location from the base of the dam is given in the figure below:

Where,

Kh = Horizontal seismic coefficient,

γw = Unit weight of water (kN/m2),

H = Height of the dam (m), and

Pe = Hydrodynamic pressure (kN/m2).

As per Von Karman it is parabolic 

 As per the Zanger method, it is Elliptical cum parabolic

Question 34

A hydraulic jump is formed when 

  1. AOption (a)

    a subcritical flow meets a supercritical flow.

  2. BOption (b)

    a subcritical flow meets a subcritical flow.

  3. COption (c)

    a supercritical flow meets a supercritical flow.

  4. DOption (d)

    a supercritical flow meets a subcritical flow.

View answer & explanation

Correct answer

Answer: (d)D.

a supercritical flow meets a subcritical flow.

Why this answer is correct

Answer:

The hydraulic jump occurs when there is a transition from supercritical flow to subcritical flow and its an example of rapidly varied flow where the flow properties change in a short reach.

Explanation:

Steady and Unsteady flow:

When the flow properties (such as velocity, density, etc.) do not change with respect to the time it is called steady flow and vice versa

Uniform and Non-uniform flow:

When the flow properties (such as velocity, density, etc.) do not change with respect to space i.e. along with the flow, it is called uniform flow and vice versa.

Varied flow:

When the flow properties (such as velocity, density, etc.) change with respect to space i.e. along with the flow, it is called varied flow.

Depending upon the changes it can be either gradually or rapidly varied flow.

Important Note:

Question 35

If the flood discharge flowing in a river is 3600 m3/s, its perimeter as per Lacey’s theory is likely to be

  1. AOption (a)

    360 m

  2. BOption (b)

    300 m

  3. COption (c)

    285 m

  4. DOption (d)

    270 m

View answer & explanation

Correct answer

Answer: (c)C.

285 m

Why this answer is correct

Concept:

Wetted perimeter as per Lacey’s Theory-

P=4.75Q;m{\bf{P}} = 4.75\sqrt {\bf{Q}} ;{\rm{m}}

Where,

Q = Discharge through river in m3/s

Solution:-

P=4.753600=4.75×60=285;m{\rm{P}} = 4.75\sqrt {3600} = 4.75 \times 60 = 285{\rm{;m}}

Other Related formulae given by Lacey:-

a. Avg Velocity V=(Qf2140)1/6;m/sec{\rm{V}} = {\left( {\frac{{{\rm{Q}}{{\rm{f}}^2}}}{{140}}} \right)^{1/6}}{\rm{;m}}/{\rm{sec}}

b. Silt factor f=1.76dia;of;particles;in;mm{\rm{f}} = 1.76\sqrt {{\rm{dia;of;particles;in;mm}}}

c. Hydraulic mean Radius R=52V2f;m;;{\rm{R}} = \frac{5}{2}\frac{{{{\rm{V}}^2}}}{{\rm{f}}}{\rm{;m;;}}

d. Bed slope ;S=f533340Q16{\rm{;S}} = \frac{{{{\rm{f}}^{\frac{5}{3}}}}}{{3340{{\rm{Q}}^{\frac{1}{6}}}}}

Question 36

The hydraulic failure of an ‘earth dam’ includes the following:

  1. AOption (a)

    over topping and wave erosion

  2. BOption (b)

    toe erosion and gullying

  3. COption (c)

    piping and sloughing

  4. DOption (d)

    both (a) and (b)

View answer & explanation

Correct answer

Answer: (d)D.

both (a) and (b)

Why this answer is correct

Concept:

Failures of earthen embankment dams are grouped into three categories

a) Hydraulic Failure: They occur from the uncontrolled flow of water over and adjacent to the embankment and due to the erosive action of water on the embankment slopes.Various hydraulic failures are:-

i) Overtopping:

 

(ii) Wave erosion: Notching of the upstream face by wave action that reduces the embankment cross section thickness and weakens embankment material.

(iii) Toe erosion: Erosion of downstream toe of the earth slope caused by misdirected spillway outlet discharge.

(iv) Gullying: Rainfall erosion of embankment slopes. Also caused by traffic from people and vehicles.

b) Seepage Failure: Seepage, if uncontrolled, can erode fine soil material from the downstream slope or foundation and continue moving towards the upstream slope to form a pipe or cavity to the pond or lake often leading to a complete failure of the embankment. This action is known as “piping.”

c) Structural Failure: Sloughs, bulges, cracks or other irregularities in the embankment are examples of structural failure.

Hence over-toppling, wave erosion, toe erosion and gullying all are the failures of an earthen dam.

Question 37

Crest level is kept low with large gates in the following:

  1. AOption (a)

    Sluice

  2. BOption (b)

    Escape

  3. COption (c)

    Regulator

  4. DOption (d)

    Barrage

View answer & explanation

Correct answer

Answer: (d)D.

Barrage

Why this answer is correct

Concept:

Barrage is a structure similar to weir with the only difference that the crest is kept at a low level and ponding of water is accomplished mainly by means of gates. During floods these gates can be raised above HFL and thus enable the high flood to pass with minimum afflux.

Regulator are located just u/s of a barrage in order to regulate the discharge flowing into off-taking channel and to control the entry of sediment in the channel.

Escape is a structure constructed on an irrigation channel for the disposal of surplus water from the channel. It is also known as surplus water escape or canal surplus escape. These are also called the safety valves of an irrigation channel system.

Question 38

In an aqueduct, natural drainage is kept at the following of a canal

  1. AOption (a)

    at the same level

  2. BOption (b)

    below

  3. COption (c)

    above

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (b)B.

below

Why this answer is correct

Concept:

The drainage water intercepting the canal can be classified in the following ways:

i) By passing the canal over the drainage

a) Aqueduct

b) Syphon aqueduct

ii) By passing the canal below the drainage

a) Superpassage

b) Syphon

 

iii) By the drain through canal ⇒ a) level crossing b) inlets and outlets

Question 39

The uplift pressure on upstream floor of a hydraulic structure determined by Bligh’s theory as compared to Khosla’s theory is

  1. AOption (a)

    same

  2. BOption (b)

    more

  3. COption (c)

    less

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

same

Why this answer is correct

Concept:

Bligh’s theory: It states that water creeps along the bottom contour of the structure.

The length of creep refers to the length of the path of water and the loss of head is proportional to the length. This theory does not discriminate between horizontal and vertical creeps.

Khosla’s Theory: It states that seepage of water does not creep along the bottom contour. It moves along a set of stream lines. In order to calculate the uplift pressure and exit gradient theory takes into account the flow pattern impermeable base of hydraulic structure.

Interpretation:

Since in both the cases, the seepage loss occurs as the water seeps. Also uplift pressure is proportional to the residual head which is maximum at the upstream of the hydraulic structure.

Uplift pressure, 

U=;γwH;kN/m2{\rm{U}} = {\rm{;}}{{\rm{\gamma }}_{\rm{w}}}{\rm{H;kN}}/{{\rm{m}}^2}

Note 1: For negligible thickness of the bed surface, uplift pressure will remain the same for both the theories.

Note 2: For considerable thickness, correction for the thickness of the slab should be considered. For such case the u/s uplift pressure will be more when calculated using Bligh’s creep theory always.

Confusion:

Option 1 and 2 both are correct. As in the question it is known given whether the thickness of the floor is to be considered. We have to select the most appropriate option out of given option. Hence it will be option 1 as in general the thickness of the slab is not considered.

Question 40

The exit gradient determined by Khosla’s theory in case of design of weir working under head ‘H’ with impervious floor length ‘b’ and depth of downstream sheet pile as d is _____

Where λ is a function of b and d?

  1. AOption (a)

    Hd1πλ\frac{{\rm{H}}}{{\rm{d}}}\frac{1}{{{\rm{\pi }}\sqrt {\rm{\lambda }} }}

  2. BOption (b)

    dH1πλ\frac{{\rm{d}}}{{\rm{H}}}{\rm{}}\frac{1}{{{\rm{\pi }}\sqrt {\rm{\lambda }} }}

  3. COption (c)

    Hd1λ\frac{{\rm{H}}}{{\rm{d}}}\frac{1}{{\sqrt {\rm{\lambda }} }}

  4. DOption (d)

    Hdcos⁡−1(λ−1λ)\frac{{\rm{H}}}{{\rm{d}}}{\cos ^{ - 1}}\left( {\frac{{{\rm{\lambda }} - 1}}{{\rm{\lambda }}}} \right)

View answer & explanation

Correct answer

Answer: (a)A.

Hd1πλ\frac{{\rm{H}}}{{\rm{d}}}\frac{1}{{{\rm{\pi }}\sqrt {\rm{\lambda }} }}

Why this answer is correct

In the Khosla’s theory complex profile of weir or barrage is broken down into a number of simple profiles and mathematical solution is obtained for such profiles.

 

Exit gradient (E) according to this theory based on mathematical solutions adopted by Khosla is given as:-

Exit;Gradient;(Ge)=Hd1πλ{\bf{Exit}};{\bf{Gradient}};\left( {{{\bf{G}}_{\bf{e}}}} \right) = \frac{{\bf{H}}}{{\bf{d}}}\frac{1}{{{\bf{\pi }}\sqrt {\bf{\lambda }} }}

Where,

λ=(1+1+α2)2{\rm{\lambda }} = \frac{{\left( {1 + \sqrt {1 + {{\rm{\alpha }}^2}} } \right)}}{2}

α = b/d

Important Point:

According to khosla water doesn’t creep along the bottom of the floor as stated as per Bligh’s creep theory but it moves along a fixed set of stream lines. The seepage in the vertical plane for a homogenous soil can be expressed by the Laplace equation.

∂2ϕ∂x2+∂2ϕ∂y2=0\frac{{{\partial ^2}\phi }}{{\partial {{\rm{x}}^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {{\rm{y}}^2}}} = 0

Corrections adopted in Khosla’s theory are:-

a) Correction for modular interference of Piles

b) Correction for floor thickness

c) Correction for slope

Question 41

Which of the following is a type of semi-modular outlet?

  1. AOption (a)

    Submerged pipe outlet

  2. BOption (b)

    Open flume outlet

  3. COption (c)

    Both (2) and (4)

  4. DOption (d)

    Kennedy’s Gauge outlet

View answer & explanation

Correct answer

Answer: (c)C.

Both (2) and (4)

Why this answer is correct

Explanation:

Outlets may be classified under the following three heads:

  1. Non-modular Outlet
  2. Semi-module or Flexible Module
  3. Rigid Module.

Non-modular Outlet:

  • A non-modular outlet is the one in which the discharge depends upon the difference in level between the water levels in the distributing channel and the water course. The discharge through such an outlet varies in wide limits with the fluctuations of the water levels in the distributing and the field channels.
  • The common examples under this category are submerged pipe outlet, masonry sluice and orifices, and wooden shoots.

Semi-module or Flexible outlet:

  • A flexible outlet or semi- module is the one in which the discharge is affected by the fluctuations in the water level of the distributing channel while the fluctuations in the water levels of the field channel do not have any effect on its discharge.
  • The various outlets in common use that fall under this category are pipe outlet: Kennedy's gauge outlet, Crump's open flume outlet and pipe-cum-open flume outlet.

Rigid Module:

  • A rigid module is the one which maintains constant discharge, within limits, irrespective of the fluctuations in the water levels in the distributing channel and/or field channel.
  • The most common outlet that falls under this category is the Gibb's rigid module.

Question 42

The ratio of the head recovered to the head put in, is known as-

  1. AOption (a)

    Modular limit

  2. BOption (b)

    Flexibility

  3. COption (c)

    Efficiency

  4. DOption (d)

    Sensitivity

View answer & explanation

Correct answer

Answer: (c)C.

Efficiency

Why this answer is correct

Modular limit: It is related to the energy dissipation required for obtaining modular flow in weir and flumes. It is defined as the value of the submergence ratio at which the real discharge deviates by 1% from the discharge calculated by the head-discharge equation.

The modular limit of weirs and flumes depends basically on the degree of streamline curvature at the control section and on the reduction of losses of kinetic energy if any, in the downstream expansion.

Efficiency: In irrigation engineering, it is defined as the ratio of the head recovered to the head put in. Lesser is the head required for the functioning of the outlet; more efficient the outlet will be. Efficiency is the measure of the conversation of the head by the outlet.

Flexibility is the ratio of the rate of change of discharge at an outlet to the rate of change of discharge of the distributary channel.

Flexibility, F =(dqq)(dQQ) = \frac{{\left( {\frac{{dq}}{q}} \right)}}{{\left( {\frac{{dQ}}{Q}} \right)}}

where,

q = discharge passing through the outlet = C1Hm

Q = flow rates in the distributary channels = C2yn

y = depth of flow in the channel, H = Head on the outlet

m = outlet index,  n = channel index

C1 and C2 = constant 

F =(dqq)(dQQ) = \frac{{\left( {\frac{{dq}}{q}} \right)}}{{\left( {\frac{{dQ}}{Q}} \right)}}= =mn×yH×dHdy = \frac{m}{n} \times \frac{y}{H} \times \frac{{dH}}{{dy}}

F = 0 for modular outlet

F = =mn×yH = \frac{m}{n} \times \frac{y}{H} for semi-modular channel, dy = dH

Sensitivity is the ratio of the rate of change of discharge through an outlet to the rate of change in the water surface level of distributary channel with respect to the normal depth of the flow in the channel.

Sensitivity =(dqq)(dSy) = \frac{{\left( {\frac{{dq}}{q}} \right)}}{{\left( {\frac{{dS}}{y}} \right)}}        where, S = Gauge reading 

Proportionality is the property of the channel and an outlet is said to be proportional when the rate of change of outlet discharge equals to the rate of change of channel discharge.

Proportionality =(dqq)(dQQ) = \frac{{\left( {\frac{{dq}}{q}} \right)}}{{\left( {\frac{{dQ}}{Q}} \right)}} = 1 =mn=Hy = \frac{m}{n} = \frac{H}{y}= =Outlet;IndexChannel;Index = \frac{{Outlet;Index}}{{Channel;Index}}

Question 43

If the discharge in canal equals to 70 m3/s with it silt factor √2, the velocity of flow in canal as per Lacey’s theory is

  1. AOption (a)

    0.5 m/s

  2. BOption (b)

    0.75 m/s

  3. COption (c)

    1.0 m/s

  4. DOption (d)

    1.25 m/s

View answer & explanation

Correct answer

Answer: (c)C.

1.0 m/s

Why this answer is correct

Concept:

According to Lacey's theory, the following formulae has been given below:

Velocity of flow, VV=(Qf2140)1/6V = {\left( {\frac{{Q{f^2}}}{{140}}} \right)^{1/6}}
Hydraulic mean depth, RR=52V2fR = \frac{5}{2}\frac{{{V^2}}}{f}
Area of the channel section, AA=QVA = \frac{Q}{V}
Wetted perimeter, PP=4.75QP = 4.75\sqrt Q
Bed Slope, SS=f5/33340Q1/6S = \frac{{{f^{5/3}}}}{{3340{Q^{1/6}}}}

where,

f = Silt factor and f=1.76dmmf = 1.76\sqrt {{d_{mm}}}

dmm = average size of particle in mm

Q = Discharge (cumecs) 

Calculation:

Given:

Q = 70 m3/s, f = √2

v=[Qf2140]16v = {\left[ {\frac{{Q{f^2}}}{{140}}} \right]^{\frac{1}{6}}}

⇒v6=(70×(2)2)140\Rightarrow {{\rm{v}}^6} = \frac{{\left( {70 \times {{\left( {\sqrt 2 } \right)}^2}} \right)}}{{140}}

 

⇒ v = 1 m/s

Difference between Lacey and Kennedy theory are:-

Lacey TheoryKennedy’s Theory
Lacey considered a semi-elliptical sectionKennedy considered a trapezoidal section
Lacey considered boundary material and defined the silt factorKennedy did not give any importance to boundary material.
Lacey gave the equation for the bed slopeKennedy didn’t give any such equation
This theory applies to both the alluvial canal and riverThis theory applies to alluvial canal only

Question 44

In a Sarda type fall, the width B of the trapezoidal crest is given by

Where ‘H’ = depth of water and ‘d’ = drop-in bed level

  1. AOption (a)

    B = 0.44 H+d\sqrt {{\rm{H}} + {\rm{d}}}

  2. BOption (b)

    B = 0.44 H−d\sqrt {{\rm{H}} - {\rm{d}}}

  3. COption (c)

    B = 0.55 H+d\sqrt {{\rm{H}} + {\rm{d}}}

  4. DOption (d)

    B = 0.55 H−d\sqrt {{\rm{H}} - {\rm{d}}}

View answer & explanation

Correct answer

Answer: (c)C.

B = 0.55 H+d\sqrt {{\rm{H}} + {\rm{d}}}

Why this answer is correct

Concept:

This type of fall is constructed on Sarda canal in Uttar Pradesh. It is a fall with raised crest and with vertical impact.

Body wall is kept rectangular when discharge of canal is less than 14 m3/sec otherwise it is kept trapezoidal.

For rectangular body wall,

Top width b=0.552d;{\rm{b}} = 0.552\sqrt {\rm{d}} {\rm{;}} and Bottom width B=;H+dp{\rm{B}} = {\rm{;H}} + \frac{{\rm{d}}}{{\sqrt {\rm{p}} }}

For trapezoidal body wall,

Top width b=0.552(H+d){\bf{b}} = 0.552\sqrt {\left( {{\bf{H}} + {\bf{d}}} \right)}  and Bottom width B=H+d/p{\rm{B}} = {\rm{H}} + {\rm{d}}/\sqrt {\rm{p}}

The edges are rounded with radius of 0.3m

Here, H – depth of water above the crest of the fall in m and d – height of the crest above downstream bed level in m.

Discharge formula for this type of fall under free fall condition is given by –

Q=CLH(Hb)1/6m3/s{\rm{Q}} = {\rm{CLH}}{\left( {\frac{{\rm{H}}}{{\rm{b}}}} \right)^{1/6}}{{\rm{m}}^3}/{\rm{s}}

Where,

L = Length of crest in m and \({\rm{C}} = {\rm{;}}\left{ {\begin{array}{*{20}{c}} {2;for;trapezoidal;crest}\ {1.85;for;rectangular;crest} \end{array}} \right}\)

Question 45

When the drain is over the canal, and FSL of canal is below the bed level of drain the structure provided is known as

  1. AOption (a)

    aqueduct

  2. BOption (b)

    canal syphon

  3. COption (c)

    super-passage

  4. DOption (d)

    syphon aqueduct

View answer & explanation

Correct answer

Answer: (c)C.

super-passage

Why this answer is correct

Concept:

A cross drainage work is a structure constructed for carrying a canal across a natural drain or river intercepting the canal to dispose drainage water without mixing with the continuous canal supplies.

Types of Cross Drainage Works:

Based on the relative bed levels, water levels of the canal and the drain and their relative discharge the Cross Drainage works are of the following types:-

1) Cross drainage works carrying the natural drain over the canal-

(i) Super passage

A super passage is also similar to a bridge in which the natural drain is carried over the canal.

  • A super passage is reverse of an aqueduct.

  • A super passage is constructed where the bed of the drain is well above the canal F.S.L

  • The canal flows at atmospheric pressure under the work.

  • In this case it is not possible to provide an inspection road along the canal.

(ii) Syphon

  • A syphon is similar to a syphon aqueduct with the difference that in the case of a syphon the canal water is carried through the barrels under the drain.
  • A syphon is constructed where the full supply level of the canal is higher than the bed of the drain.
  • The barrels in this case also act as inverted syphons through which the canal water flows under pressure.

​

(2) Cross drainage works carrying the Canal over the Natural Drain:

(i) Aqueduct

  • An aqueduct is a hydraulic structure which carries a canal (through a trough) across and above the drainage similar to a bridge in which instead of the road or a railway, a canal is carried over a natural drain.

  • In the case of an aqueduct, HFL (highest flood level) of the drainage should remain lower than the level of the underside of the canal trough.

  • The drain flows at atmospheric pressure under the work.

  • Generally, an inspection road is provided along with the trough

(ii) Syphon aqueduct

  • A syphon aqueduct is a cross drainage structure similar to an aqueduct except that the streambed is depressed locally where it passes under the trough of the canal and the barrels discharges the streamflow under pressure.
  • A syphon aqueduct is constructed where the water surface level of the drain at high flood is higher than the canal bed.
  • The horizontal floor of the barrels is provided with slopes at its ends to join the drain bed on either side.
  • The drain water flows under pressure through the barrels which act as inverted syphons and hence this cross drainage work is known as syphon aqueduct.

3) Cross drainage works admitting the drain water into the canal-

In this type of cross drainage works, the canal water and the drain water are allowed to intermingle with each other. This may be achieved by the following two types of cross drainage works:-

(i) Level crossing

(ii) Inlet and outlet

Question 46

For an annual flood series arranged in decreasing order of magnitude, the return period for a magnitude listed at position m’ in a total of N entries is

  1. AOption (a)

    m/N

  2. BOption (b)

    m/(N + 1)

  3. COption (c)

    (N + 1)/m

  4. DOption (d)

    N/(m +1)

View answer & explanation

Correct answer

Answer: (c)C.

(N + 1)/m

Why this answer is correct

Concept:

Return period (T):- It is defined as the average time interval after which a flood of given discharge is equalled or exceeded.

T=1p{\bf{T}} = \frac{1}{{\bf{p}}}

Where,

p = probability of occurrence or exceedance of an event with rank ‘m’.

The probability of occurrence or exceedance is given by three methods:-

MethodProbability(p)
Weibull MethodmN+1\frac{{\rm{m}}}{{{\rm{N}} + 1}}
California MethodmN\frac{{\rm{m}}}{{\rm{N}}}
Hazen Method(m−0.5)N\frac{{\left( {{\rm{m}} - 0.5} \right)}}{{\rm{N}}}

 

N = Total number of events

Thus as per Weibul method,

p=mN+1{\rm{p}} = \frac{{\rm{m}}}{{{\rm{N}} + 1}}

∴T=1p=(N+1)m\therefore {\rm{T}} = \frac{1}{{\rm{p}}} = \frac{{\left( {{\rm{N}} + 1} \right)}}{{\rm{m}}}

Hence the required return period is (N+1)/m

Important Point:

Hydraulic structures design are based on risk assessment and this is done by analyzing the return period of the design discharge.

Risk is defined as the probability of exceedance at least once in the designed life of the structure and Reliability is the probability of non-occurrence of the event in the design life.

Question 47

The total rainfall in a catchment area 1200 km2 during a 6 hours storm is 16 cm. While the surface run-off due to storm is 1.2 × 108 m3. The φ index is

  1. AOption (a)

    0.1 cm/hr

  2. BOption (b)

    1.0 cm/hr

  3. COption (c)

    0.2 cm/hr

  4. DOption (d)

    cannot be estimated 

View answer & explanation

Correct answer

Answer: (b)B.

1.0 cm/hr

Why this answer is correct

Concept:

∅−index=P−Rt\emptyset - {\rm{index}} = \frac{{{\rm{P}} - {\rm{R}}}}{{\rm{t}}}

Where,

R (Runoff) (m): It is defined as the part of water cycle that flows over land as surface water instead of being absorbed into the ground water or evaporated. Runoff is that part of precipitation that appears in uncontrolled surface streams, rivers drains or sewers.

P (Precipitation) (m): It is any liquid or frozen water that forms in the atmosphere and falls back to the earth. Precipitation is any product of the condensation of atmospheric water vapour that falls under gravity from clouds. The main forms of precipitation include drizzle, rain, sleet, snow, ice pellets, graupel and hail.

Infiltration Capacity: It is the rate at which water infiltrates into ground is called infiltration capacity. For consistency, in hydrological calculations, a constant value of infiltration rate for the entire storm duration is adopted.

The average infiltration rate is called the infiltration index and two types of infiltration indices commonly used:

  1. 𝛗 – Index: It is defined as the rate of infiltration above which volume equals the runoff volume.
  2. w – index: It is the average infiltration rate during the time when the rainfall intensity exceeds the infiltration rate.

Calculation:

Given, Rainfall (P) = 16cm and t = 6 hours

Runoff R=;1.2;×1081200×106=0.1m=10cm{\rm{R}} = {\rm{;}}\frac{{1.2{\rm{;}} \times {{10}^8}}}{{1200 \times {{10}^6}}} = 0.1{\rm{m}} = 10{\rm{cm}}

∅−index=P−Rt=16−106=66=1cm/hr;\emptyset - {\rm{index}} = \frac{{{\rm{P}} - {\rm{R}}}}{{\rm{t}}} = \frac{{16 - 10}}{6} = \frac{6}{6} = 1{\rm{cm}}/{\rm{hr;}}

Question 48

Thickness of concrete lining is governed by

  1. AOption (a)

    requirement of imperviousness.

  2. BOption (b)

    requirement of imperviousness & structural strength. 

  3. COption (c)

    Thumb rule for providing nominal thickness. 

  4. DOption (d)

    slop of bank.

View answer & explanation

Correct answer

Answer: (b)B.

requirement of imperviousness & structural strength. 

Why this answer is correct

Concept:

Lining of canal: It is the process of reducing seepage loss of irrigation water by adding an impermeable layer to the edges of the canal trenches.

Concrete canal lining:

  1. It is often used due to its high structural strength.
  2. They are tough, durable, relatively impermeable and hydraulically efficient.
  3. It is suitable for both small and large channels and both high and low flow velocities.

The lining of canal is performed for the following purposes:

  1. To minimize seepage loss in canal
  2. To increase the discharge in the canal section by increasing the velocity.
  3. To reduce maintenance of canals and weed growth
  4. To prevent the erosion of bed and sides of canals.

Question 49

Silt ejectors are provided as a silt control device on

  1. AOption (a)

    upstream of a spillway

  2. BOption (b)

    downstream of spillway

  3. COption (c)

    upstream of a canal head regulator

  4. DOption (d)

    downstream of a canal head regulator.

View answer & explanation

Correct answer

Answer: (d)D.

downstream of a canal head regulator.

Why this answer is correct

Concept:

Silt regulation works are required to preformed to control the erosion along the banks.

Two popular silt regulation works are as silt excluders and ejectors.

Silt excluders are constructed on the bed of the river, upstream of the head regulator.

While silt ejectors are constructed on the bed of the canal and a little distance downstream from the head regulator.

Silt excluders remove silt from the water before it enters the canal and silt ejectors extract the silt from the canal water after the silted water has travelled a certain distance in the off-take canal.

Silt ejectors also called silt extractors, and it extracts the silt from the canal water after the silted water has travelled a certain distance in the off-take canal.

These works are, therefore, constructed on the bed of the canal, and little distance downstream from the head regulator.

Question 50

The uplift pressure on the roof of a siphon aqueduct is maximum when

  1. AOption (a)

    canal is running at full supply level and drain is dry.

  2. BOption (b)

    drain is running at high flood level and canal is dry.

  3. COption (c)

    canal is running dry

  4. DOption (d)

    drain is running dry. 

View answer & explanation

Correct answer

Answer: (b)B.

drain is running at high flood level and canal is dry.

Why this answer is correct

Syphon Aqueduct:

 In a syphon aqueduct, canal water is carrier above the drainage but the high flood level (HFL) of drainage is above the canal trough. The drainage water flows under syphonic action and there is no presence of atmospheric pressure in the natural drain.

As the syphon aqueduct passes from below the super structure, if the drain is running at high flood level then it will generate maximum pressure at the roof. Hence, to counter this pressure, maximum uplift pressure is generated in the form of reactions.

Important points:

HFL of drain is much higher above canal bed in case of syphon aqueduct.

Water surface level of flood is depressed when it passes under the canal trough.

Bed of the drain is also lowered.

Question 51

Out of total liquid water on earth, the brackish water is

  1. AOption (a)

    above 95 percent 

  2. BOption (b)

    around 75 percent 

  3. COption (c)

    around 50 percent

  4. DOption (d)

    estimates are not available.

View answer & explanation

Correct answer

Answer: (d)D.

estimates are not available.

Why this answer is correct

Concept:

Brackish water:

It is the water having more salinity than freshwater, but not as much as seawater. It may result from mixing seawater with fresh water together, as in estuaries, or it may occur in brackish fossil aquifers.

The total volume of water on Earth is estimated at 1.386 billion km³, out of which 97.5% being salt water and 2.5% being fresh water.

Salinity level for different types of water are as follows:

  1. Fresh water (< 0.05%)
  2. Brackish water (0.05–3%)
  3. Saline water (3–5%)
  4. Brine (> 5% up to 26%-28% max)

Brackish water is just a type of salty water. Thus, there is no appropriate estimate available for computing brackish water on Earth.

Question 52

The suitable method for irrigating highly undulating land is

  1. AOption (a)

    drip irrigation

  2. BOption (b)

    furrow irrigation

  3. COption (c)

    sprinkler irrigation

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

sprinkler irrigation

Why this answer is correct

Concept:

The different types of irrigation systems are as follows:

1. Drip irrigation: In this method irrigation, the water is conveyed on the surface in 12 to 16 mm diameter tubes, fed from the large feeder pipes.

The water is allowed to drip or trickle slowly through the nozzle or orifices at practically zero pressure and thus the soil in the root-zone of crops is constantly kept wet.

Drip irrigation allows optimum utilisation of irrigation water by reducing percolation and evaporation losses and thus maintains the appropriate water content in the root-zone of plants.

The main drawback of this method is its high cost. This method is suitable for all types of land profile.

2. Furrow irrigation: This method is best suited when crops are grown in rows.

Most of the crops are grown and planted in rows and thus this is the most commonly used method.

In this method, unlike flooding, wetting is done of only a part of the field. The spacing of furrows varies according to the nature of the crop.

3. Sprinkler irrigation: In this method irrigation, the water is applied to the land in the form of a spray and it can be used for all type of soil and for any topography. It is highly recommended for soils having large undulations and easily erodible soils and it is mostly adopted in tea and coffee gardens.

There is no erosion of soil and fertilizers are used economically since they can be injected through the water.

Question 53

Water logging occurs when the water table is

  1. AOption (a)

    up to root zone of crops

  2. BOption (b)

    3.0 m below G.L

  3. COption (c)

    4.0 m below G.L

  4. DOption (d)

    4.5 m below G.L

View answer & explanation

Correct answer

Answer: (a)A.

up to root zone of crops

Why this answer is correct

Concept:

Water logging: It is the condition of land in which the soil profile is saturated with water either temporarily or permanently. In waterlogged lands, the water table rises to an extent that the soil pores in the crop root zone are saturated resulting in restriction of the normal circulation of air.

This restriction causes a decline in the level of oxygen and increases the level of the carbon dioxide. 

Below are the different values of water table to specify water logging:

NomenclatureDepth of water table
Water logged< 2 m
Potential area for water logging2-3 m
Safe< 3 m

 

∴ For a land, if the water table is 2 meters below the surface, it is considered as waterlogged. 

It is necessary to keep the highest position of water-table at least about 3 m below the ground surface.

Important points:

Causes of waterlogging:

Inadequate surface drainage, seepage from canal system, Over irrigation of fields, Obstruction of a natural drainage etc.

Effects of water logging:

Inhibiting activity of soil bacteria, decrease in available capillary water, rise in level of salt in the surface soil, growth of wild flora, fall in soil temperature etc.

Question 54

The Muskingham’s method of flood routing through a river reach is primarily a

  1. AOption (a)

    one-parameter model

  2. BOption (b)

    two-parameter model

  3. COption (c)

    three-parameter model

  4. DOption (d)

    model having no parameters

View answer & explanation

Correct answer

Answer: (b)B.

two-parameter model

Why this answer is correct

Concept:

Muskingham’s method is the most common hydrologic method of channel routing in which the channel storage in a reach is expressed as a function of both inflow and outflow discharge.

S=k(xIm+(1−x)Qm){\bf{S}} = {\bf{k}}\left( {{\bf{x}}{{\bf{I}}^{\bf{m}}} + \left( {1 - {\bf{x}}} \right){{\bf{Q}}^{\bf{m}}}} \right)

Where,

S = Storage in a channel reach, I = Inflow discharge, O = Outflow discharge, and k = Storage time coefficient

It has a unit of time and it is approximately equal to the time of travel of flood wave through the channel reach.

x = Weightage factor. Its range varies from 0 to 0.5.

m= 1, for naturally occurring channels.

Thus the two parameters are ‘k’ and ‘x’

Important Point:

For x = 0, storage is a function of outflow discharge only. Such reservoirs are known as linear storage or linear reservoirs.

Question 55

The ground water movement in alluvial soils is analysed with the help of

  1. AOption (a)

    Kennedy’s equations and relationship

  2. BOption (b)

    Lacey’s equations

  3. COption (c)

    Darcy’s equations

  4. DOption (d)

    Froude’s criterion

View answer & explanation

Correct answer

Answer: (c)C.

Darcy’s equations

Why this answer is correct

Concept:

The ground water movement in soils is called as percolation, which is analysed by Darcy’s equation.

Darcy Law:

For laminar flow through saturated soil Darcy established experimentally that the rate of flow ‘q’ across a sectional area ‘A’ of soil is proportional to the hydraulic gradient”.

q = k × i × A

q/A = k × i and V = k × i

Where,

V = Flow velocity, k = Coefficient of permeability, and i = Hydraulic gradient

Darcy’s law is valid as long as the flow is laminar and the soil is saturated. It is applied to soil fraction finer than fine gravels.

Important points:

Kennedy’s and Lacey’s equations are used for the designing of canals.

Question 56

The water utilizable by plants is available in the form of:

  1. AOption (a)

    Gravity water

  2. BOption (b)

    Hydroscopic water

  3. COption (c)

    Capillary water

  4. DOption (d)

    Chemical water

View answer & explanation

Correct answer

Answer: (c)C.

Capillary water

Why this answer is correct

Explanation:

Capillary water: 

Capillary water is held in the capillary pores (micropores). Capillary water is retained on the soil particles by surface forces. It is held so strongly that gravity cannot remove it from the soil particles. The molecules of capillary water are free and mobile and are present in a liquid state. Due to this reason, it evaporates easily at ordinary temperature though it is held firmly by the soil particle; plant roots are able to absorb it.

Gravitational water:

Gravitational water occupies the larger soil pores (macropores) and moves down readily under the force of gravity. Water in excess of the field capacity is termed gravitational water. Gravitational water is of no use to plants because it occupies the larger pores.

Hygroscopic water:

The water that is held tightly on the surface of the soil colloidal particle is known as hygroscopic water. It is essentially non-liquid and moves primarily in the vapour form. Hygroscopic water held so tenaciously by soil particles that plants cannot absorb it.

Question 57

The downstream curve of the ogee profile for an ogee shaped spillway crest is given by Where terms have usual meaning.

  1. AOption (a)

    (xH)0.85=2(Yx){\left( {\frac{{\rm{x}}}{{\rm{H}}}} \right)^{0.85}} = 2\left( {\frac{{\rm{Y}}}{{\rm{x}}}} \right)

  2. BOption (b)

    x0.85 = 2H1.85y

  3. COption (c)

    x1.85 = 2Hy0.85

  4. DOption (d)

    x1.85 = 2H1.85y

View answer & explanation

Correct answer

Answer: (a)A.

(xH)0.85=2(Yx){\left( {\frac{{\rm{x}}}{{\rm{H}}}} \right)^{0.85}} = 2\left( {\frac{{\rm{Y}}}{{\rm{x}}}} \right)

Why this answer is correct

Concept:

Ogee spillway: The shape of Ogee spillway is similar to the downstream face of the weir. It is an improved form of a straight drop spillway.

In this case, the downstream face of the weir is constructed corresponding to the shape of freely falling water jet which is in ogee shape.

In a high-overflow section, the crest profile merge with the straight downstream section of slope, as shown in the figure below.

Equation, x1.85 = 2H0.85y, represents the ogee shape of the spillway

Also can be written as (xH)0.85=2(Yx){\left( {\frac{{\rm{x}}}{{\rm{H}}}} \right)^{0.85}} = 2\left( {\frac{{\rm{Y}}}{{\rm{x}}}} \right)

where,

X = Horizontal distance from apex to the downstream tangent point and H = design head at upstream of spillway. Y = vertical distance from apex to the downstream tangent point.

 

Important points:

The downstream end of the spillway is in the form of a reverse curve which turn the flow into the apron of a stilling basin or into the spillway discharge channel.

When ogee spillway is subjected to greater flow, excessive turbulence and hence separation occurs on the surface.

Question 58

The indicator used in C.O.D. test is

  1. AOption (a)

    starch

  2. BOption (b)

    phenolphthalein

  3. COption (c)

    ferrous ammonium sulphate

  4. DOption (d)

    ferroin

View answer & explanation

Correct answer

Answer: (d)D.

ferroin

Why this answer is correct

Concept:

Chemical Oxygen Demand (COD) test is used to measure the content of organic matter of waste water, both biodegradable and non-biodegradable.

The oxygen equivalent of organic matter that can be oxidised is measured by using a strong chemical (K2Cr2O7) oxidising agent in an acidic medium.

The excess potassium dichromate is titrated with ferrous ammonium sulphate (FAS) until all of the excess oxidizing agent has been reduced to Cr3+.

The oxidation-reduction indicator ferroin is also added during this titration as it changes its color from red to pale blue when oxidized.

Important point:

COD test is also called dichromate-oxygen demand test.

Chemical Oxygen Demand is the total measurement of all chemicals (organics & in-organics) in the water/waste water.

Question 59

The best sewer material to resist hydrogen sulphide corrosion is 

  1. AOption (a)

    R.C.C

  2. BOption (b)

    brick masonry

  3. COption (c)

    asbestos cement 

  4. DOption (d)

    glazed stoneware 

View answer & explanation

Correct answer

Answer: (d)D.

glazed stoneware 

Why this answer is correct

Concept:

RCC:

  • For larger diameter pipes, plain cement concrete pipes are manufactured with steel reinforcement called as RCC pipes.
  • These pipes are quite resistant to erosion and abrasion. These pipes are strong in tension as well as compression (for withstanding external loads).

Brick masonry:

  • Brick sewers are lined by cement mortar plaster to make them smooth and hydraulically smooth. The stoneware and ceramic coating also help in resisting sulphide corrosion which is not possible with ordinary cement plaster as this can be attacked by sewer gases like hydrogen sulphide very easily.

Asbestos cement:

  • Asbestos cement pipes are manufactured from a mixture of asbestos fibre, silica and cement, converted under pressure to a dense homogeneous material, possessing considerable strength, called asbestos cement.
  • These are light in weight and hence easy to transport. They can be easily cut and assembled without skilled labour. Their interior is exceptionally smooth; thus, a hydraulically efficient sewer is provided. But they are not structurally strong.

Glazed Stoneware or Vitrified clay sewers:

  • These are manufactured from clays and shales of special qualities, which are pulverised then mixed thoroughly with water. These pipes are highly resistant to sulphide corrosion, and therefore, preferred for carrying polluted sewage and industrial wastes.
  • These are quite cheap, durable, easily available. But they are heavy and brittle and cannot be used under pressure environment.

Question 60

The bacteria’s which may survive with or without free oxygen, are called

  1. AOption (a)

    aerobic bacteria

  2. BOption (b)

    anaerobic bacteria

  3. COption (c)

    facultative bacteria

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

facultative bacteria

Why this answer is correct

Bacteria: These are the microscopic unicellular plants or organisms. For the study of sanitary engineering, they are divided into three groups are as follows:

Aerobic bacteria: The aerobic bacteria require light and free oxygen for their existence and development.

Anaerobic bacteria: The aerobic bacteria do not require light and free oxygen for their existence and development.

Facultative bacteria: The facultative bacteria can exist in the presence or absence of oxygen, but they grow in plenty in the absence of air.

Question 61

The maximum depth of sedimentation tanks is limited to

  1. AOption (a)

    2 m

  2. BOption (b)

    4 m

  3. COption (c)

    5 m

  4. DOption (d)

    6 m

View answer & explanation

Correct answer

Answer: (d)D.

6 m

Why this answer is correct

Concept:

Sedimentation tanks are designed to remove a part of organic matter from the sewage effluent coming out from the grit removal chambers or detritus chambers.

A plain sedimentation tank under normal conditions may remove about 60-65% of the suspended solids, and satisfy 30-35% of the BOD from the sewage.

There are two types of sedimentation tanks:

Intermittent settling tank: It also called as quiescent type tanks and these are simple settling tanks which store sewage for a certain period and keep it in complete test. The depth of intermittent tank is 2.5 to 6 m.

Continuous flow type tank: The supply in such tanks is consistent. In these tanks, water enters from one end and comes out from the other end. The depth of continuous tank is 3 to 4.5 m.

Hence, the maximum depth of sedimentation tank is 6 meters.

Question 62

COD/BOD ratio of fresh water is

  1. AOption (a)

    less than 1

  2. BOption (b)

    more than 1

  3. COption (c)

    equal to 1

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (c)C.

equal to 1

Why this answer is correct

Concept:

BOD: It refers to Biochemical oxygen demand i.e. oxygen required to oxidise the Biodegradable organics.

COD: It refers to the Chemical oxygen demand i.e. oxygen required to oxidise the biodegradable as well as non-biodegradable organic matter.

The ratio COD/BOD is defined as the portion of biodegradable organics compared to total (bio + non bio) degradable organics.

This ratio is generally greater than 1, but can be equal to 1 if sample has only biodegradable organics.

For fresher water, as both BOD and COD should be negligible, hence the ratio can be equal to 1.

Important points:

The COD/BOD ratio for the different forms of waste water are as follows:

COD/BOD <2 ⇒ Readily biodegradable effluent

2 < COD/BOD < 4 ⇒ Moderately biodegradable effluent

COD/BO > 4: hardly biodegradable effluent

Question 63

The proportion of solids in sewage is about

  1. AOption (a)

    2.5% or more

  2. BOption (b)

    1% or more

  3. COption (c)

    0.1% or less

  4. DOption (d)

    zero

View answer & explanation

Correct answer

Answer: (c)C.

0.1% or less

Why this answer is correct

Concept:

Sewage is a waste water from a community, containing solid and liquid excreta, derived from houses, streets, factories and industries.

It contains 99.9 percent water and 0.1 percent solids (partly organic and partly inorganic).

The strength of the sewage may be expressed in terms of Biochemical oxygen demand (BOD), Chemical oxygen demand (COD) and Suspended solids.

Important points:

Sewage with a BOD value of 300 ppm or above and suspended solids content 500 mg/l or more is called as strong sewage.

Sewage with a BOD of 100 ppm or less and suspended solids content 100 mg/l or less is called as weak sewage.

Question 64

Disinfection of water results in:

  1. AOption (a)

    Removal of turbidity

  2. BOption (b)

    Removal of hardness

  3. COption (c)

    Killing of disease bacteria

  4. DOption (d)

    Complete sterilization

View answer & explanation

Correct answer

Answer: (c)C.

Killing of disease bacteria

Why this answer is correct

Disinfection of water is carried out to remove as kill disease causing microorganisms like Bacteria.

Sterilization is an absolute term which denotes the complete killing or removal of micro-organisms of all kinds, while disinfection is a relative term indicating a mere removal of burden of pathogenic micro-organisms of the article.

Sterilization is the process of freeing of an article form all living organisms, including viruses, bacteria and their spores etc. both pathogenic and non-pathogenic.

Turbidity is removed by Sedimentation process etc.

Question 65

On standard silica scale, the turbidity should be limited to

  1. AOption (a)

    30 ppm

  2. BOption (b)

    20 ppm

  3. COption (c)

    10 ppm

  4. DOption (d)

    50 ppm

View answer & explanation

Correct answer

Answer: (c)C.

10 ppm

Why this answer is correct

Concept:

Turbidity: It is defined as the measurement of degree to which water losses its transparency due to presence of suspended solids.

It is an optical property of water which measures the amount of light scattered or absorbed depending upon the working principle.

The standard unit of Turbidity is Silica Turbidity Unit (STU) or Jackson’s turbidity unit (JTU).

STU is measured on standard silica scale.

On a standard silica scale, the permissible value of turbidity for drinking water is 10 ppm.

In terms of another turbidity unit i.e. Nephelometric turbidity unit (NTU), the permissible value of turbidity is 1 ppm and cause of rejection is 4 ppm.

Question 66

If the total hardness of water is greater than its total alkalinity, the carbonate hardness will be equal to

  1. AOption (a)

    total alkalinity

  2. BOption (b)

    total hardness

  3. COption (c)

    total hardness - total alkalinity

  4. DOption (d)

    noncarbonate hardness

View answer & explanation

Correct answer

Answer: (a)A.

total alkalinity

Why this answer is correct

Total hardness is the sum of carbonate hardness and non-carbonate hardness.

Total;hardness=[Ca+2]×5020+[Mg+2]×5012{\rm{Total;hardness}} = \left[ {{\rm{C}}{{\rm{a}}^{ + 2}}} \right] \times \frac{{50}}{{20}} + \left[ {{\rm{M}}{{\rm{g}}^{ + 2}}} \right] \times \frac{{50}}{{12}}

Total alkalinity is not the sum of all alkalinity, it means the equivalent alkalinity in mg/l as of CaCO3.

Total;alkalinity;(mg/l;as;CaCO3)=[CO32−]×5030+[HCO3−]×5061{\rm{Total;alkalinity;}}\left( {{\rm{mg}}/{\rm{l;as;CaC}}{{\rm{O}}_3}} \right) = \left[ {{\rm{CO}}_3^{2 - }} \right] \times \frac{{50}}{{30}} + \left[ {{\rm{HCO}}_3^ - } \right] \times \frac{{50}}{{61}}

Case 1: If Total hardness is greater than Total alkalinity

Carbonate hardness = Total alkalinity and Non-carbonate hardness = Total hardness - Carbonate Hardness

Case 2: If Total hardness is less than Total alkalinity

Carbonate hardness = Total hardness and Non-carbonate hardness = 0 (zero)

Question 67

The D.O level in natural unpolluted water at normal temperature is found to be

  1. AOption (a)

    1 mg/litre

  2. BOption (b)

    10 mg/litre

  3. COption (c)

    100 mg/litre

  4. DOption (d)

    1000 mg/litre

View answer & explanation

Correct answer

Answer: (b)B.

10 mg/litre

Why this answer is correct

Concept:

Dissolve oxygen (D.O.) is the amount of oxygen present in water naturally.

DO sustains the aquatic life in water bodies and its content in water depends on the temperature.

Dissolved oxygen present in sewage is very important for respiration of aerobic microorganism as well as for all other aerobic life forms.

If the temperature of sewage is more, then DO content is less and vice versa.

Saturated dissolved oxygen: It is the maximum quantity of DO that can remain in water at a particular temperature.

Typical values of dissolved oxygen of fresh water are:

Dissolved Oxygen (mg/L) at Saturation in freshwater
Temperature (C°)        (F°)Dissolved Oxygen at Saturation (mg/L)
0°32°14.6
5°41°12.8
10°50°11.3
15°59°10.1
20°68°9.1
25°77°8.3
30°86°7.6
<br>

Hence, at normal temperature, DO content is nearly 10 mg/l.

Important points:

The DO content of sewage is generally determined by the Winkler’s Test.

It is necessary to ensure at least 4 ppm of dissolved oxygen in the water body after disposal of sewage into it, otherwise fish and other aquatic life are likely to be killed.

Question 68

The length of rectangular sedimentation tank, of breadth B, should not be more than

  1. AOption (a)

    B

  2. BOption (b)

    2B

  3. COption (c)

    4B

  4. DOption (d)

    6B

View answer & explanation

Correct answer

Answer: (c)C.

4B

Why this answer is correct

A typical long rectangular tank has a length ranging from 4 times their width. The bottom is slightly sloped to facilitate sludge scraping.

The general dimension of a sedimentation tank are as follows:

Length = 4 × Width

Height = 3.5 m to 5 m

Slide depth = 0.8 m to 1.2 m

Freeboard = 0.5 m

Question 69

The colour in water is generally due to

  1. AOption (a)

    suspended impurities

  2. BOption (b)

    dissolved impurities

  3. COption (c)

    colloidal impurities

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

dissolved impurities

Why this answer is correct

Concept:

Colour caused by suspended and dissolved matter in water referred as apparent colour or fake colour.

After suspended matter causing colour is removed by filtration, centrifugation etc., the colour obtained is called true colour which is caused by dissolved matter only.

Measurement of colour is done in units of TCU (True colour unit) or Hazen unit where, 1 TCU is equal to colour produced by 1 mg per litre of platinum in the form of chloroplatinate ions.

Instruments used to measure colour: Nessler’s tube or Tintometer.

Important points:

Acceptable limit is 5 TCU

Cause for rejection is 25 TCU.

Phenolic compounds with chlorine produces taste and odour.

Question 70

A rectangular tank 15 × 6m × 2 m has to treat 2 million litres of water per day. The detention time of the tank should be

  1. AOption (a)

    2.16 hours

  2. BOption (b)

    5.63 hours

  3. COption (c)

    12.0 hours

  4. DOption (d)

    24 hours

View answer & explanation

Correct answer

Answer: (a)A.

2.16 hours

Why this answer is correct

Concept:

Detention time: It is the theoretical time required for a small amount of water to pass through a tank at a specified flow rate.

Volume of the tank (V) = Discharge (Q) × detention time (td)

Given:

Volume = 15 × 6 × 2 = 180 m3

Discharge = 2 × 106 L/D = 2000 m3/day

Thus,

Detention;time=VolumeDischarge=1802000×24;hours=2.16;hours{\rm{Detention;time}} = \frac{{{\rm{Volume}}}}{{{\rm{Discharge}}}} = \frac{{180}}{{2000}} \times 24{\rm{;hours}} = 2.16{\rm{;hours}}

Important Point:

The efficiency of the tank depends upon the detention time. If detention time is very low proper settling cannot occur and if very high may lead to biological reactions in the secondary sedimentation tank.

Question 71

Zero hardness of water is achieved by

  1. AOption (a)

    using lime soda process

  2. BOption (b)

    excess lime treatment

  3. COption (c)

    ion exchange method

  4. DOption (d)

    using excess alum dosage

View answer & explanation

Correct answer

Answer: (c)C.

ion exchange method

Why this answer is correct

Concept:

  • Water softening: It is the process of hardness removal from the water. It is caused by multivalent cation and affects water quality.
  • Lime soda method: It is a water softening method in which lime and soda ash are added to the water, which causes the precipitation of multivalent cation as CaCO3.
  • Precipitation of CaCO3 occurs only when the pH of water is greater than 9, so in case of less pH alkalinity is added to the water. In this process small amount of Ca2+ and Mg2+ precipitates very late, which will create incrustation in the pipe, so to avoid this recarbonation is done to dissolve back this small amount of cation.
  • Due to this, the method does not give zero hardness.
  • Ion Exchange Process: Ion-exchange resin, (zeolite) exchanges one ion from the water being treated for another ion that is in the resin (sodium is one component of softening salt, with chlorine being the other). Zeolite resin exchanges sodium for calcium and magnesium. It can produce water with zero hardness.

Question 72

In chlorination, due to rise in temperature of water, death rate of bacteria

  1. AOption (a)

    Increases

  2. BOption (b)

    Decreases

  3. COption (c)

    Remains unaffected

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

Increases

Why this answer is correct

The disinfection efficiency varies w.r.t the change in the temperature and contact time. 

At lower temperatures, bacterial kill tends to be slower and higher doses are needed. The effect of low temperatures is greater with combined chlorine than with free available chlorine.

For higher temperatures, lower chlorine would be required.

Contact time is as important as the chlorine dose in determining the efficiency of water treatment chlorination. Contact time is the amount of time which the chlorine has to react with the microorganisms in the water, which will equal the time between the moment when chlorine is added to the water and the moment when that water is used by the consumer. The longer the contact time, the more efficient the disinfection process is. When using chlorine for disinfection a minimum contact time of 30 minutes is required for adequate disinfection.

Question 73

The biochemical oxygen demand of treated water should be

  1. AOption (a)

    nil

  2. BOption (b)

    less than 30 mg/l

  3. COption (c)

    less than 150 mg/l

  4. DOption (d)

    always more than 9 mg/l

View answer & explanation

Correct answer

Answer: (a)A.

nil

Why this answer is correct

Explanation:

The biochemical oxygen demand for treated water should be zero.

Since it represents the amount of biodegradable organic matter present in water which is food for bacteria or microorganism, if not treated can cause several water-borne diseases.

Note:

A water supply with a BOD level of 3-5 ppm is considered moderately clean.

In water with a BOD level of 6-9 ppm, the water is considered somewhat polluted.

Question 74

Which of the following retards the self-purification of stream?

  1. AOption (a)

    Higher temperature 

  2. BOption (b)

    sunlight 

  3. COption (c)

    Satisfying oxygen demand

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (d)D.

None of the above

Why this answer is correct

Concept:

Self-purification of streams: The ability of streams to purify itself naturally and to bring back the polluted water into its original purified condition is called self-purification of streams.

This is achieved in presence of dissolved oxygen (DO) by oxidizing the micro-organisms (or bacteria) under the presence of light. So, more is the DO content, more will be the self-purification.

At higher temperature, the activity of micro-organisms are more as they are being oxidized more. Thus at a higher temperature, oxidation rate is more and so is the self-purification rate.

Sunlight has a bleaching action and it helps in photosynthesis of some water borne micro-organisms. They consumes CO2 and release O2, therefore increasing the DO content in the stream. So, the self-purification increases with the presence of sunlight.

Oxygen demand: Satisfying the oxygen demand means supplying the required DO for oxidation of micro-organisms. So, as the time gradually passes excess DO will present in the stream and it will increase self-purification capacity.

∴ Higher temperature, sunlight and satisfying oxygen demand all three will increase self-purification of streams.

Question 75

Which of the following is not coagulant?

  1. AOption (a)

    Alum

  2. BOption (b)

    Ferric chloride

  3. COption (c)

    Gelatin

  4. DOption (d)

    SO2

View answer & explanation

Correct answer

Answer: (d)D.

SO2

Why this answer is correct

The most effective chemical to be used as coagulants are alum, ferrous sulphate, ferric chloride, ferric chloride, and ferric sulphate.

a) Use of Alum (Al2 (SO4)3.18H2O) as coagulant:

Alum or filter alum has proved to be a very effective coagulant, and it is now extensively used throughout the world. It is quite cheap, forms an excellent stable floc, and does not require any skilled supervision for handling.

Al2(SO4)3.18H2O + 3Ca(HCO3)2 → 3CaSO4 + 2Al(OH)3 + 6 CO2

The only drawback in the use of alum is that the effective pH range for its use is small, i.e. 6.5 to 8.3, and in many cases it requires the addition of external alkali salts, thereby rendering it costlier.

b) Use of Copperas (FeSO4.7H2O) as coagulant:

Copperas (FeSO4.7H2O) is extensively used as a coagulant for raw waters that are not coloured. It is generally cheaper than alum and functions effectively in the pH range of 8.5 and above.

FeSO4.7H2O+Ca(OH)2→CaSO4+Fe(OH)2+7H2O{\bf{FeS}}{{\bf{O}}_4}.7{{\bf{H}}_2}{\bf{O}} + {\bf{Ca}}{\left( {{\bf{OH}}} \right)_2} \to {\bf{CaS}}{{\bf{O}}_4} + {\bf{Fe}}{\left( {{\bf{OH}}} \right)_2} + 7{{\bf{H}}_2}{\bf{O}}

it is not used for coloured raw waters. The quantity of copperas required is almost the same as that of alum.

c) Use of Chlorinated copperas as coagulant:

The resultant combination of ferric sulphate and ferric chloride is known as chlorinated copperas, and is a valuable coagulant of removing colours, especially where the raw water has a low pH value.

6(FeSO4.7H2O) + 3Cl2 → 2Fe2(SO4)3 + 2FeCl3 + 42H2O

Ferric sulphate is quite effective in the pH range of 4 to 7 and above 9, whereas Ferric chloride is quite effective in the pH range of 3.5 to 6.5 and above 8.5. The combination has, therefore, proved to be a very effective coagulant for treating low pH waters.

d) Use of sodium aluminate (Na2 Al2 O4) as coagulant:

Na2Al2O4+Ca(HCO3)2→CaAl2O4+Na2CO3+CO2↑+H2O{\rm{N}}{{\rm{a}}_2}{\rm{A}}{{\rm{l}}_2}{{\rm{O}}_4} + {\rm{Ca}}{\left( {{\rm{HC}}{{\rm{O}}_3}} \right)_2} \to {\rm{CaA}}{{\rm{l}}_2}{{\rm{O}}_4} + {\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3} + {\rm{C}}{{\rm{O}}_2} \uparrow + {{\rm{H}}_2}{\rm{O}}

This coagulant is about 1½ times costlier than alum, and is, therefore, generally avoided for treating ordinary public supplies, but, however, it is very useful for treating waters which do not have the naturally desired alkalinity, and thus cannot be treated with pure alum.

NOTE:

Gelatin is a translucent, colorless, flavorless substance derived from collagen, which is a protein found in the connective tissues, bones, and skin of animals, primarily cows and pigs. Gelatin can be used in the coagulation process of wastewater treatment, although it is not as common as traditional chemical coagulants like aluminum sulfate or ferric chloride.

∴ SO2 is not a coagulant.

Question 76

The pH value of fresh sewage is usually

  1. AOption (a)

    7

  2. BOption (b)

    more than 7

  3. COption (c)

    less than 7

  4. DOption (d)

    9

View answer & explanation

Correct answer

Answer: (b)B.

more than 7

Why this answer is correct

Concept:

  • The negative logarithm of hydrogen ion concentration is known as pH.
  • pH = - log10[H+]
<br>

If any solution is acidic then pH is less than 7 and if the solution is alkaline pH is more than 7.

  • The fresh sewage is generally alkaline in nature which means pH of fresh sewage is generally more than 7.
  • The exact value of pH for fresh sewage entirely depends upon the composition and percentage of sewage constituents and normally ranges from 7.5 to 9.5.
  • However as the time passes, its pH tends to fall due to production of acids by bacterial action and fresh sewage turns into septic condition which is acidic in nature.
<br>

Confusion Point:

pH of fresh sewage is more than 7 (average range is 7.5-9.5) and fresh sewage is alkaline in nature.

pH of septic sewage is less than 7 (average range is 6.5-4.5) and septic sewage is acidic in nature.

Question 77

For the COD test of sewage, organic matter is oxidized by Potassium dichromate (K2Cr2C7) in the presence

  1. AOption (a)

    Sulphuric Acid 

  2. BOption (b)

    Nitric Acid

  3. COption (c)

    Hydrochloric Acid

  4. DOption (d)

    Acetic Acid

View answer & explanation

Correct answer

Answer: (a)A.

Sulphuric Acid 

Why this answer is correct

Concept:

Chemical oxygen demand (COD): It is defined as the amount of oxygen required to oxidise the organic matters by a strong acidic agent at an elevated temperature into carbon dioxide, water and ammonia.

The COD test is widely used as an alternative to BOD test to estimate the strength of domestic and industrial wastes as COD test can give results in few hours as compared to BOD test which takes normally 5 days.

The COD test uses potassium dichromate (K2Cr2O7) in presence of concentrated sulfuric acid (H2SO4) solution that oxidizes both organic (predominate) and inorganic substances in a waste water sample.

The regents for COD test are followings:

a) Standard potassium dichromate solution (0.25N)

b) Concentrated sulfuric acid

c) Standard ferrous ammonium sulphate titrant (0.1N)

d) Ferroin indicator solution.

Question 78

After cleaning a slow sand filter, filtered water is not used for.

  1. AOption (a)

    6 to 12 h

  2. BOption (b)

    12 to 18 h

  3. COption (c)

    18 to 24 h

  4. DOption (d)

    24 to 36 h

View answer & explanation

Correct answer

Answer: (d)D.

24 to 36 h

Why this answer is correct

Concept:

  • In case of slow sand filter, after continuously using for some time impurities are started to accumulate within the filter and filtering quality deceases. So, they must be cleaned periodically to maintain the quality of filtration.
  • The micro-organisms present in slow sand filters utilise organic impurities as food and convert them into harmless compounds known as schmutzdecke by the process of biological metabolism.
  • This schmutzdecke layer helps in absorbing and straining out the impurities and increases the quality of filtration.
  • While cleaning the filter the schmutzdecke layer is also scraped of and it takes 24 to 36 hours (or 1day to 1.5 days) to form again. Thus the filtered water obtained before this period will not be properly filtered and will contain impurities.
<br>

∴ That’s why after cleaning a slow sand filter, raw water admitted into filter but filtered water is not used for 24 hr to 36 hr.

Question 79

Most important source of water for public water supply is from

  1. AOption (a)

    Lakes

  2. BOption (b)

    Ponds

  3. COption (c)

    Streams

  4. DOption (d)

    Rivers

View answer & explanation

Correct answer

Answer: (d)D.

Rivers

Why this answer is correct

Concept:

The whole process of supplying water to public is necessarily comprised of three main parts:

a) Collection of raw water from a water source.

b) Treatment of the raw water to make it usable, consumable and drinkable.

c) Distribution of treated water to public.

The good criteria for a water source is that there should be continuous and large supply of the raw water and the self-cleansing capacity of the source should be as more as possible.

In case of lakes and ponds, there is limited quantity of water available and there is no continuous supply of water and also due to stagnation of water, impurities and organic bacteria is more in these water source, so the treatment cost is more. That’s why these kind of water sources are not preferred.

In case of streams and narrow channels, water supply may be continuous but the self-cleansing capacity is not very good. So the treatment cost is more. That’s why these kind of water sources are not preferred.

In case of rivers, main advantage is that the water supply is continuous. Also due to current in the rivers proper dilution and mixing of impurities with the river water is ensured. So, the treatment cost is low compared to others

∴ Rivers are the most preferred and most important source for public water supply.

Important Point:

In case of sea water, it contains a large amount of salinity and chloride. So, treating this sea water to make it consumable and drinkable is highly expensive. That’s why sea water as source of supply for public water is not also preferred.

Question 80

Water may not contain much impurities if its source is

  1. AOption (a)

    reservoirs

  2. BOption (b)

    stream flowing in plains.

  3. COption (c)

    lake in lower regions.

  4. DOption (d)

    spring along hill slopes. 

View answer & explanation

Correct answer

Answer: (d)D.

spring along hill slopes. 

Why this answer is correct

Concept:

In water reservoirs and other water bodies like lakes in lower regions, due to stagnation of water nutrients, impurities and organic micro-organisms present are more. And often utilizing these micro-organisms and nutrients excessive algal growth takes place in water reservoirs which is a major source of impurity and pollution in water.

Rivers and streams flowing in plains are significantly polluted due to industrial and agricultural waste products. Chemical waste products from industrial processes are discharged into rivers. In agriculture fertilizers and pesticides are used which are washed through the soil by rain, to end up in rivers. So, in general streams and rivers flowing in plains contains considerable impurities and waste products.

But in case of springs in hilly slopes the water source is ground water and not so much agricultural and industrial pollutant is available due to less civilization and less agriculture in hilly areas.

∴ The impurities in springs in hilly regions are very less.

Confusion Point:

The springs which discharges hot water generally contains sulphur in hot water, hence it has impurity as sulphur and generally not used for water supply.

Question 81

The most important water quality parameter for domestic use of water is

  1. AOption (a)

    Carbonate hardness

  2. BOption (b)

    Non-Carbonate hardness

  3. COption (c)

    Coliform group of organisms

  4. DOption (d)

    Chlorides

View answer & explanation

Correct answer

Answer: (c)C.

Coliform group of organisms

Why this answer is correct

Concept:

Carbonate Hardness: Presence of carbonates and bicarbonates of Ca and Mg in the water leads to carbonate or temporary hardness. When such water are boiled out, CO2 gas escapes out and insoluble CaCO3 get precipitated and thus carbonate hardness leads to scaling in boilers. Therefore carbonate hardness is an important parameter for industrial water use.

Non-Carbonate Hardness: Presence of sulphates, chlorides and nitrates of Ca and Mg leads to non-carbonate hardness or permanent hardness. This non-carbonate hardness is removed by either Lime-Soda process or Base-Exchange process which generally leads to incrustation of pipes. Therefore non-carbonate hardness is also an important parameter for industrial water use.

Coliform group of organisms: These are the rod shaped non-pathogenic bacteria, whose presence or absence in water indicates presence or absence of fecal pollution. These coliform group of organisms are widely used to detect fecal pollution in water and used as the indicator for drinking water purposes. So, it an important parameter for drinking and domestic purposes.

Chlorides: Chlorides are generally present in water in the form of sodium chloride (or common salt). High concentrations of chlorides produce a noticeable salty taste in drinking water thus objectionable, but it can be removed by chlorination process.

∴ Even though the carbonate and non- carbonate hardness increases detergent expenses and chlorides provides salty taste, coliforms are the most important water quality parameter for domestic use of water because they indicates pollution in water.

Question 82

Hardness of water is caused by 

  1. AOption (a)

    presence of soap lather 

  2. BOption (b)

    Presence of chlorides and sulphates of sodium and potassium

  3. COption (c)

    Presence of CO3, SO4 or chlorides of Calcium and Magnesium

  4. DOption (d)

    Turbidity 

View answer & explanation

Correct answer

Answer: (c)C.

Presence of CO3, SO4 or chlorides of Calcium and Magnesium

Why this answer is correct

Concept:

Hardness of water:

  • Hardness of water is that characteristic which prevents the formation of sufficient leather or foam. It occurs due to presence of carbonates, bicarbonates, sulphates, chlorides of calcium and magnesium.
  • The carbonates and bicarbonates of calcium and magnesium are present in the water, then it leads to carbonate or temporary hardness.
  • The sulphates, chlorides and nitrates of calcium and magnesium are present in the water, then it leads to permanent hardness.

 

Turbidity: It is a measure of the degree to which the water loses its transparency due to the presence of suspended particulates. The more total suspended solids in the water, the murkier it appears and the higher the turbidity.

∴ Hardness of water caused by presence carbonates, bicarbonates, sulphates, chlorides of Ca and Mg and not by soap lather or turbidity.

Confusion Point:

Hardness of water occurs due to presence of carbonates, bicarbonates, sulphates, chlorides of divalent metallic ions (predominantly Ca2+ & Mg2+ & very occasionally Fe2+ & Mn2+) but hardness never occurs due to those salts of monovalent metallic ions (Na+ & K+).

Question 83

A heavy stone is suitable for

  1. AOption (a)

    Arches

  2. BOption (b)

    Rubble masonry

  3. COption (c)

    Roads

  4. DOption (d)

    Retaining walls

View answer & explanation

Correct answer

Answer: (d)D.

Retaining walls

Why this answer is correct

Concept

The stone masonry can be classified into two types:

Rubble Masonry & Ashlar Masonry

Rubble Masonry:

The type of stone masonry in which either undressed or roughly dressed stone is laid in a suitable mortar is called rubble masonry. In this masonry, the joints are not of uniform thickness.

Ashlar Masonry:

It is the type of stone masonry in which finely dressed stones are laid in cement or lime mortar is known as ashlars masonry. In this masonry are the courses are of uniform height, all the joints are regular, thin and have a uniform thickness. This type of masonry is much costly as it requires dressing of stones.

Three types of Rubble Masonry are as follows:

a) Un-coursed Random Rubble Masonry:

The random rubble masonry in which stones are laid without forming courses is known as un-coursed random rubble masonry. This is the roughest and cheapest type of masonry and is of varying appearance.

It is used for the construction of walls of low height in the case of ordinary buildings.

b) Coursed Random Rubble Masonry:

The random rubble masonry in which stones are laid in layers of equal height is called random rubble masonry. In this masonry, the stones are laid in somewhat level courses. Headers of one coursed height are placed at certain intervals. The stones are hammer dressed. CRRM is used for the construction of residential buildings, boundary walls, etc.

c) Squared Rubble Masonry:

The rubble masonry in which the face stones are squared on all joints and beds by hammer dressing or chisel dressing before their actual laying is called squared rubble masonry.

There are two types of squared rubble masonry:

  1. Coursed Square Rubble Masonry:

The square rubble masonry in which chisel dressed stones laid in courses is called coarse square rubble masonry. It is used for the construction of public buildings, hospitals, schools, markets, modern residential buildings, etc. and in hilly areas where good quality of the stone is easily available.

  1. Un-coursed square rubble masonry:

The squared rubble in masonry which hammer dressed stones are laid without making courses are called un-coursed square rubble masonry. It consists of stones that are squared on all joints and beds by hammer dressing. All the stones to be laid are of different sizes. It is used for the construction of ordinary buildings in hilly areas where a good variety of stones are cheaply available.

Dry rubble masonry:

The rubble masonry in which stones are laid without using any mortar is called dry rubble masonry or sometimes shortly as "dry stones". It is an ordinary masonry and is recommended for constructing walls of height not more than 6 m.

A heavy stone is useful for the construction of retaining walls.

Question 84

Quantity of Gypsum in ordinary Portland cement is

  1. AOption (a)

    3%

  2. BOption (b)

    4%

  3. COption (c)

    5%

  4. DOption (d)

    6%

View answer & explanation

Correct answer

Answer: (a)A.

3%

Why this answer is correct

Gypsum is basically a Hydrated calcium sulphate (CaSo4.2H2O). It acts as a retarder in the cement and increase the setting time of cement. In places where setting time of cement required to be more, Gypsum is added in to that cement.

Note:

Gypsum is added 3% to 4% in ordinary portland cement and reduced by 2.5% in Quick setting cement.

Question 85

The minimum depth of foundation below ground level for a residential building according to IS: 1904 should be

  1. AOption (a)

    100 cm

  2. BOption (b)

    80 cm

  3. COption (c)

    60 cm

  4. DOption (d)

    50 cm

View answer & explanation

Correct answer

Answer: (d)D.

50 cm

Why this answer is correct

As per Cl. 7.2, IS:1904,

All foundations shall extend to a depth of at least 50 cm below the natural ground level.

Important Point:

The design of building foundations is performed based on prescribed serviceability and strength criteria laid down by various standard codes adopted by different countries.

Generally, the total settlement of a foundation is described under the serviceability criterion; whereas a strength criterion is described by bearing capacity of the soil or rock underlying the foundation.

As per Rankine formula, minimum depth of foundation is given as

Df=qγ×(1−sinϕ1+sinϕ)2{{\rm{D}}_{\rm{f}}} = \frac{{\rm{q}}}{{\rm{\gamma }}} \times {\left( {\frac{{1 - {\rm{sin}}\phi }}{{1 + {\rm{sin}}\phi }}} \right)^2}

Where,

q = Gross bearing capacity, γ = Unit weight of soil, and ϕ = Angle of internal friction of soil

Question 86

The minimum height of plinth of a building with respect to the level of the centre of the road passing at its front or back should be 

  1. AOption (a)

    10 cm

  2. BOption (b)

    15 cm

  3. COption (c)

    30 cm

  4. DOption (d)

    60 cm

View answer & explanation

Correct answer

Answer: (c)C.

30 cm

Why this answer is correct

As per Building Bye-Laws, the following points are recommended:

  1. In order to be free from weathering effects and for better stability, the minimum depth of foundation is kept 1 m below the ground level. The width of the foundation should be such that the stress on the soil is within its safe bearing capacity.

  2. Plinth height may be 300 mm to 600 mm but 450 mm is more common. For water logging prone and important buildings, the minimum plinth height is 600 mm.

  3. The plinth area should be a minimum of 1.2 times the floor area of the building. Minimum width of the footing is 100 m more than the one above and a minimum depth of footing is the depth of the standard brick 100 mm. 

  4. The thickness of the damp proof courses 20 mm to 25 mm.

  5. The thickness of the cement concrete floor is a minimum of100 mm plus floor finish over it. Floor finishes are generally with 10 mm to 20 mm thick cement mortar or tiles.

6) The carpet area should be 50 to 65% of the floor area, etc.

Question 87

The function of cleats in a roof truss is:

  1. To support purlins

  2. To support common rafter

  3. To prevent the purlins from tilting.

The correct answer is

  1. AOption (a)

    Only 1

  2. BOption (b)

    Only 3

  3. COption (c)

    Both 2 and 3

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

Only 3

Why this answer is correct

Concept:

The truss arrangement is depicted below:

Cleats: These are a very small bracket kinds of member which do not require any welding in site, and itself contains holes for bolting connection in site. The main function is to prevent the tilting of the purlins.

Purlins: These are the members which spans on the roof frames to support the roof coverings and runs parallel to ridge to connect different trusses situated in the longitudinal direction. Purlins are supported on rafters.

Rafters: These are the series of sloped structural members that extend from the ridge to the bottom chords, and that are designed to support the roof deck and its associated loads. Rafters are supported on bottom chords members.

When angle sections or channel sections are used as purlins, cleats are used to connect the purlins to the rafters. Because angle sections and channel sections has very less resistance against torsion, and when heavy wind comes it has a high tendency to tilt. To increase torsional resistance and to reduce the tilting tendency of purlins the cleat angles are given in roof truss.

Question 88

Which of the following type of door can allow air circulation and privacy, even when it is closed?

  1. AOption (a)

    Wire gauge

  2. BOption (b)

    Louvered

  3. COption (c)

    Part panelled, part glazed

  4. DOption (d)

    flush door 

View answer & explanation

Correct answer

Answer: (b)B.

Louvered

Why this answer is correct

Concept:

The different types of doors are as follows:

For wire gauge doors:

In these doors wires are gauged into the frame of the doors to make a mesh of wires which provide free air circulation through the doors but prevents flies, mosquitos and insects from entering inside, there is no privacy.

For Louvered doors:

These are used when privacy with natural ventilation and quietness for rest is desired, as they has a fixed or movable wooden louvers which permit open ventilation while preserving the privacy. These are mostly used in bathroom, water closets, in residential and public buildings where good ventilation is desired.

Partly panelled and partly glazed doors:

These doors are divided into several parts and some of the parts are glazed by laminated glass or any other glazing materials. These kind of doors cannot provide air circulation but sunlight can enter and privacy in dependent upon the glazing materials.

Flush doors:

These are simple door designs that have plain facings on both sides.  Flush doors may have a solid, hollow or stave core. They preserve full privacy but no air circulation and sunlight can be allowed.

∴ Louvered doors can allow air circulation and privacy, even when it is closed.

Question 89

Efflorescence in building is due to sulphates of

  1. AOption (a)

    Calcium

  2. BOption (b)

    Sodium

  3. COption (c)

    Iron

  4. DOption (d)

    both (a) and (b)

View answer & explanation

Correct answer

Answer: (d)D.

both (a) and (b)

Why this answer is correct

Concept:

Efflorescence:

  • It is a whitish coloured powdered deposition of salts on the concrete surface that is formed due to evaporation of water from the concrete.
  • It is caused when water soluble salts are present in the concrete material, which comes on to the surface while evaporation of water from the concrete.
  • These salts are sulphate and carbonate salts of calcium and sodium and can come from bricks, cement, aggregates, water, or admixtures.
  • The following water soluble salts are generally leads to efflorescence:
<br>

Calcium Sulphate (CaSO4), Sodium Sulphate (Na2SO4), Calcium Carbonate (CaCO3), Sodium Carbonate (Na2CO3), etc.

∴   Sulphates and carbonates of sodium, and calcium leads to efflorescence, but not due to those of iron.

Question 90

Quick setting cement is produced by adding ________.

  1. AOption (a)

    less amount of gypsum is very fine powdered form

  2. BOption (b)

    more amount of gypsum is very fine powdered from

  3. COption (c)

    aluminium sulphate in very fine powdered from

  4. DOption (d)

    pozzolana in very fine powdered form

View answer & explanation

Correct answer

Answer: (c)C.

aluminium sulphate in very fine powdered from

Why this answer is correct

Quick setting cement means the cement which sets early.

Quick setting cement is prepared by adding small amount of alumina in finely grinded cement clinker and reducing the proportion of calcium sulphate (Gypsum).

Applications:

  1. Grouting
  2. Under water concreting.

Important point:

Quick setting cement should not be confused with rapid hardening cement. Rapid hardening cement develops higher rate of gain of strength while quick setting cement sets quickly only and its rate of gain of strength is similar to ordinary Portland cement. 

Note:

This is SSC JE Official question and answer. Option 1 and 3 both can be correct as per concept but, Option 3 is more appropriate.

Question 91

Finer the cement, more is the

  1. AOption (a)

    strength of the cement 

  2. BOption (b)

    workability 

  3. COption (c)

    Shrinkage cracking

  4. DOption (d)

    all of the above

View answer & explanation

Correct answer

Answer: (d)D.

all of the above

Why this answer is correct

Concept:

  • Finer cement has more specific surface area which increases the reactivity and thus the cement will participate more in hydration reaction and initial strength of the cement will increase but does not changes the final strength.

Note: In option, it is not mentioned, that it initial or final strength, so we have to go for inial strength, as the other 2 options are correct

  • As finer cement has more reactivity the water demand for given water cement ratio will increase. So, to maintain the water cement ratio extra water is to be added that will increase the workability of concrete. Also, as the cement becomes finer the cement particles starts to work as ball bearing to provide more workability to concrete by minimising inter particle friction.
  • As in the finer cement more water is used, so it is more likely to found water available in the void spaces. Also as the finer cement has more reactivity it will produce more heat of hydration which will evaporate the pore water more rapidly and consequently shrinkage cracking will increase.

∴ Use of finer cement will increase strength of cement, workability and shrinkage cracking of concrete.

Question 92

As per IS specification, the minimum compressive strength of 1st class bricks should be

  1. AOption (a)

    75 kg/cm2

  2. BOption (b)

    90 kg/cm2

  3. COption (c)

    100 kg/cm2

  4. DOption (d)

    125 kg/cm2

View answer & explanation

Correct answer

Answer: (c)C.

100 kg/cm2

Why this answer is correct

Explanation:

Clay bricks are classified as first class, second class, third class and fourth class based on their physical and mechanical properties.

  1. First Class Bricks:
  1. The crushing strength of the brick should not be less than 10 N/mm2. This limit varies with different Government organizations around the country.
  2. The surface should be smooth and rectangular, with parallel, sharp and straight edges and square comers.
  3. These should be free from flaws, cracks, and stones.
  4. These should have a uniform texture.
  5. No impression should be left on the brick when a scratch is made by a fingernail.
  6. The fractured surface of the brick should not show lumps of lime.
  7. A metallic or ringing sound should come when two bricks are struck against each other.
  8. Water absorption should be 20 % of its dry weight when immersed in cold water for 24 hours. 9. These are thoroughly burnt and are of deep red, cherry or copper color.

Uses: First class bricks are recommended for pointing, exposed face work in masonry structures, flooring and reinforced brickwork.

  1. Second Class Bricks are supposed to have the same requirements as the first-class ones except that
  1. Small cracks and distortions are permitted.
  2. A little higher water absorption of about 22 % of its dry weight is allowed.
  3. The crushing strength should not be less than 7.0 N/mm2.

Uses: Second class bricks are recommended for all important or unimportant hidden masonry works and centering of reinforced brick and reinforced cement concrete (RCC) structures.

  1. Third Class Bricks are under burnt. They are soft and light-colored producing a dull sound when struck against each other. Water absorption is about 25 percent of dry weight.

Uses: It is used for building temporary structures.

Fourth Class Bricks are over burnt and badly distorted in shape and size and are brittle in nature.

Uses: The ballast of such bricks is used for foundation and floors in lime concrete and road metal.

Question 93

If ‘D’ is the duration ‘ES’ and ‘EF’ are then earliest start and finish times, ‘LS’ and ‘LF’ are latest start and latest finish times, the following relationship holds good.

  1. AOption (a)

    F = LF - EF

  2. BOption (b)

    IS = LF + D

  3. COption (c)

    LF = LS - ES

  4. DOption (d)

    EF = ES + D

View answer & explanation

Correct answer

Answer: (d)D.

EF = ES + D

Why this answer is correct

Concept:

In case of project management the followings are definition of various terms:

Duration (D): Duration is the estimated or actual time required to complete a task or an activity.

Earliest start time (EST): It is defined as the earliest possible time at which an activity can start. It is calculated by moving from first to last event in a network diagram.

Earliest finish time (EFT): It is the earliest possible time at which an activity can finish.

Relationship between earliest start time, earliest finish time and duration is given by,

Earliest finish time = Earliest start time + duration of that activity

∴ EFT = EST + D

Latest start time (LST): It is defined as the latest time an activity can start and still allow the project to be completed on estimated project completion time.

Latest finish time (LFT): It is the latest possible time at which an activity can finish and still allow the project to be completed on estimated project completion time.

Relationship between latest start time, latest finish time and duration is given by,

Latest finish time = Latest start time + duration of that activity

∴ LFT = LST + D

∴ Correct relationship is EF = ES + D.

Question 94

Cant deficiency becomes an inevitable consideration on a main line and Branch line moving in

  1. AOption (a)

    same direction

  2. BOption (b)

    opposite direction

  3. COption (c)

    cross direction

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

same direction

Why this answer is correct

Standard terms in Railway Engineering are as follows:

1. Cant or superelevation: It is the amount by which one rail is raised above the other rail. It is positive when the outer rail on a curved track is raised above inner rail and is negative when the inner rail on a curved track is raised above the outer rail.

2. Equilibrium speed: It is the speed at which the centrifugal force developed during the movement of the vehicle on a curved track is exactly balanced by the cant provided. 3. Cant deficiency: It occurs when a train travels around a curve at a speed higher than the equilibrium speed. It is the difference between the theoretical cant required for such higher speed and actual cant provided. 4. Cant excess: It occurs when a train travels around a curve at a speed lower than the equilibrium speed. It is the difference between the actual cant and the theoretical cant required for such a lower speed. 5. Maximum permissible speed of the curve: It is the highest speed which may be permitted on a curve taking into consideration the radius of the curvature, actual cant, cant deficiency, cant excess and the length of transition. When the maximum permissible speed on a curve is less than the maximum sectional speed of the section of a line, permanent speed restriction becomes necessary. 6. Cant gradient and cant deficiency gradient: This indicate the amount by which cant or deficiency of cant is increased or reduced in a given length of transition e.g., 1 in 1000 means that cant or deficiency of cant of 1 mm. is gained or lost in every 1000 mm. of transition length. 7. Rate of change of cant or rate of change of cant deficiency: It is the rate at which cant or cant deficiency is increased or reduced per second, at the maximum permissible speed of the vehicle passing over the transition curve, e.g., 35 mm. per second means that a vehicle when traveling at a maximum speed permitted will experience a change in cant or deficiency of cant of 35 mm. in each second of travel over the transition.

Question 95

The value of super-elevation will be more when

  1. AOption (a)

    speed is more

  2. BOption (b)

    radius of curvature is less

  3. COption (c)

    both (A) & (b) of the above

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

both (A) & (b) of the above

Why this answer is correct

Concept:

Super-elevation: It is the inward transverse slope provided throughout the length of the horizontal curves to counteract the centrifugal force which tends to pull out the vehicle in outward direction and therefore to check the tendency of the vehicle to overturn or skid.

The value of super-elevation e is given by,

;e=V2gR;{\rm{;e}} = \frac{{{{\rm{V}}^2}}}{{{\rm{gR}}}};

Where,

V is speed of the vehicle, R is the radius of the horizontal curve and g is acceleration due to gravity.

∴ e ∝ V2

∴ Value of super elevation is more when speed is more.

∴ e ∝ 1/R

∴ Value of super elevation is more when radius is less.

Question 96

A pavement is classified as flexible or rigid based on its -

  1. AOption (a)

    sub-grade

  2. BOption (b)

    sub-base

  3. COption (c)

    base

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (d)D.

None of the above

Why this answer is correct

Flexible Pavement:

For rigid pavement:

For rigid pavement: Since there is only one layer of material between the concrete and the sub-grade, this layer can be called as a base or sub-base course.

These two classifications are done on the basis of the structural performance of the pavement.

Hence none of these is the best-suited option.

Question 97

In the absence of super-elevation, the formation of pot holes is generally found

  1. AOption (a)

    on the outer edge of road

  2. BOption (b)

    in the inner edge of road

  3. COption (c)

    in the middle of the road

  4. DOption (d)

    anywhere along the width of the road.

View answer & explanation

Correct answer

Answer: (a)A.

on the outer edge of road

Why this answer is correct

Concept:

Super-elevation:

  • It is the inward transverse slope provided throughout the length of the horizontal curves to counteract the centrifugal force which tends to pull out the vehicle in outward direction and therefore to check the tendency of the vehicle to overturn or skid.

  • To provide super elevation outer edge of the pavement is raised with respect to the inner edge of the pavement.

  • In the following diagram method of providing super elevation is depicted:

  • Now, if the super elevation is not provided, then all the vehicles will tend to pull out in outward direction and will negotiate the curve by travelling through the proximity of outer edge are and also  the tyre pressure will be in outer edge area.

  • That’s why near the outer edge area of the horizontal curve more wear and tear will take place and pot holes will form.

<br>

∴ In the absence of super-elevation, the formation of pot holes is generally found on the outer edge of the horizontal curve on the roads.

Question 98

Vehicle damage factor (VDF) as given by I.R.C., is used in

  1. AOption (a)

    Westergaard’s analysis

  2. BOption (b)

    CBR method of pavement design

  3. COption (c)

    design of drainage system

  4. DOption (d)

    design of dowel bars

View answer & explanation

Correct answer

Answer: (b)B.

CBR method of pavement design

Why this answer is correct

Concept:

Vehicle Damage Factor (VDF): Different commercial vehicles have different weights and the conversion of the design number of commercial vehicles during the design life of the road into a number of standard axle repetitions. This is done using VDF.

The vehicle damage factor (VDF) is a multiplier for converting the number of commercial vehicles of different axle loads and axle configurations to the number of standard axle-load repetitions.

Mathematically,

VDF=Total;number;of;standard;axlesTotal;number;of;vehicles;surveyed{\rm{VDF}} = \frac{{{\rm{Total;number;of;standard;axles}}}}{{{\rm{Total;number;of;vehicles;surveyed}}}}

This is used in the estimation of cumulative standard axles during the design life of the road which in turn is used in CBR method of design of flexible pavements.

Important Point:

  • CBR method of design of flexible pavements is an empirical method.
  • CBR testing of soil subgrade is done to determine the material property for design.
  • Penetration resistance corresponding to 2.5 mm and 5 mm penetration is determined and hence CBR is calculated. If CBR corresponding to 2.5 mm penetration is greater it is considered as CBR value and if CBR corresponding to 5mm penetration is greater then the test is repeated and then again if this value comes to be greater, it is adopted as CBR value.
  • As per IRC 37:2001, the thickness of different layers of pavements corresponding to the CBR value of soil subgrade and cumulative standard axles in msa is determined.

Question 99

Ballast is used in Railway section to serve as

  1. AOption (a)

    an elastic bed

  2. BOption (b)

    foundation of rail track

  3. COption (c)

    both (a) and (b)

  4. DOption (d)

    protection form animals

View answer & explanation

Correct answer

Answer: (c)C.

both (a) and (b)

Why this answer is correct

Concept:

Ballast: Ballast is layer of broken stone, gravel or any other suitable material placed under and around the sleepers to distribute the wheel loads from sleepers to the underlying subgrade foundation. Location of ballast along with other formation layers of rail cross section are shown in following diagram: 

Some of the main functions of ballast are followings:

a) To provide firm and level bedded foundation for the sleepers and rails to rest on

b) To protect the surface of subgrade and to form an elastic bed

c) To transmit and distribute the loads from the sleepers to the subgrade

d) To allow for maintaining correct track level without disturbing the rail road bed

e) To hold the sleepers in position during the passage of trains

f) To provide lateral stability to the track as a whole.

∴ Ballast is used in Railway section to serve both as an elastic bed and foundation of rail track.

Question 100

In railways, the disc signals are provided for the purpose of

  1. AOption (a)

    dead slow movement 

  2. BOption (b)

    indicating busy platform

  3. COption (c)

    possible danger ahead

  4. DOption (d)

    shunting

View answer & explanation

Correct answer

Answer: (d)D.

shunting

Why this answer is correct

Concept:

Disc Signals:

  • The disc signals are in form a circular disc with a red band on a white background and these discs can rotate in a vertical plane**.** By rotating the disc different operations are instructed.
  • In railways, disc signals are used to instruct various operations like changing the locomotive of a train, attaching/detaching of wagons, changing the order of wagons, etc. These kinds of operations are collectively known as shunting or shunt moves.
  • The indications of a disc signal is depicted in following diagram:

a) When the red band of the disc is in a horizontal position, it indicates “STOP”.

b) When the red band of the disc is in an inclined position, it indicates “PROCEED” slowly for shunting.

∴ In railways, the disc signals are provided for the purpose of shunting.

Important Information:

Generally, semaphore signals are used to indicate dead slow movement or to indicate busy platform or a possible danger ahead.

100 questions in the full paper