Previous Year Paper

BPSC AE Paper 6 (Civil Engineering) 2019 Official Paper (Previous Year Paper)

Held in 2019

50 questions60 minutes
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Technical (50 questions)

Question 1

According to Indian standards, the number of rain gauge stations for an area of 5200 km2 in plain should be

  1. AOption (a)

    10

  2. BOption (b)

    15

  3. COption (c)

    20

  4. DOption (d)

    40

View answer & explanation

Correct answer

Answer: (a)A.

10

Why this answer is correct

Concept:

Rain gauge is an instrument used to measure the rainfall at the location where it is installed. It is known by various other names such as – Pluviometer, ombrometer, hyetometer, and odometer.

In order to have a correct estimate about the amount of rainfall over a particular area, proper distribution of rain gauge stations are required. For this, a term called rain gauge density is defined which means that no. rain gauge stations required per square km2.

As per IS a recommendation, the rain gauge density depending on the terrain of the area and it is given below:

LocationRain Gauge Density
Plain region1 station per 520 km2
Average elevation of 1000m.1 station per 260-390 km2
Hilly and heavy rainfall areas1 station per 130 km2

 

∴ Number of required rain gauge stations = Area × Rain gauge density

Calculation:

As it can be seen from the above table, for plain 1 no of rain gauge required for an area of 520 km2.

Hence, number of rain gauge stations required for an area of 5200 km2 in planes

∴ N = 5200/520 = 10 stations.

Question 2

The maximum superelevation on hill roads should not exceed 

  1. AOption (a)

    7%

  2. BOption (b)

    8%

  3. COption (c)

    9%

  4. DOption (d)

    10%

View answer & explanation

Correct answer

Answer: (d)D.

10%

Why this answer is correct

Concept:

The centrifugal force exerted on the vehicles while traversing through the curves is counteracted by providing superelevation, which is given by:​​

;e=V2127R ;e = \frac{{{V^2}}}{{127R}}

However, it is assumed that the centrifugal force is completely nullified if the vehicle is travelling at its 75% of the vehicle design speed.

∴ ;e=(0.75V)2127R=V2225R ;e = \frac{{{(0.75V)^2}}}{{127R}} = \frac{{{V^2}}}{{225R}}

According to IRC,

Maximum Superelevation
Plain & Rolled Terrain7 %
Hilly Terrain10%
Urban Roads4%

Question 3

End of speed limit is

  1. AOption (a)

    regulatory sign

  2. BOption (b)

    warning sign

  3. COption (c)

    informatory sign

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

regulatory sign

Why this answer is correct

As per IRC: 67-2012, Mandatory Sign/ Regulatory sign

All mandatory sign or regulatory sign is circular in shape. These signs are with a red colour circular ring and a diagonal bar with black symbol or arrow or letter on white background.

The red ring indicates prohibitory regulation while the red diagonal bar prohibits the action and movement indicates by the black symbol.

End of speed limit, Stop, U-turn prohibited are examples of regulatory sign.

Note:

Cautionary/ warning signs are triangular in shape with the red border and black symbols on white background. For example, hairpin bend, narrow bridge, school ahead, etc.

Informative signs also called guide signs, are provided to assist the drivers to reach their desired destinations.

Question 4

The stopping sight distances(S) of a vehicle for Indian highways is given by: 

(Here V is the speed of the vehicle, t is the brake reaction time and η is the efficiency of brakes)

  1. AOption (a)

    S=0.28V.t+0.01V2ηS = 0.28V.t + \frac{{0.01{V^2}}}{\eta }

  2. BOption (b)

    S=0.28V.t+0.01ηV2S = 0.28V.t + \frac{{0.01\eta }}{{{V^2}}}

  3. COption (c)

    S=0.01V.t+0.28V2ηS = 0.01V.t + \frac{{0.28{V^2}}}{\eta }

  4. DOption (d)

    S=0.01V.t+0.28ηV2S = 0.01V.t + \frac{{0.28\eta }}{{{V^2}}}

View answer & explanation

Correct answer

Answer: (a)A.

S=0.28V.t+0.01V2ηS = 0.28V.t + \frac{{0.01{V^2}}}{\eta }

Why this answer is correct

Stopping Sight Distance:

The stopping sight distances (S) of a vehicle is the minimum distance open to the vision of the driver to intended to stop the vehicle to avoid collision with an obstruction (moving or stationary).

∴ S = Lag distance + Braking distance

Lag distance is the distance traveled by vehicle during the total reaction time of the driver.

Let us assume the vehicle is traveling “V kmph” & “t” in seconds is the total reaction time of the driver.

Lag distance (in meter), L = 0.28 V × t  

Braking distance (in meter), B= V22gfηB=~\frac{{{V}^{2}}}{2gf\eta }  where V is in m/s

Assuming g = 9.81 m/s2 and V is in kmph

We get, \(B=\frac{0.01{{V}^{2}}}{f\eta }\)

Here, f is the coefficient of longitudinal friction and η is breaking efficiency.

 \(S=0.28Vt+\frac{0.01~{{V}^{2}}}{f\eta }\)

None of the given option matches, however, the most closet is option ‘1’.

Note: Co-efficient of friction incorporates in braking efficiency.

Question 5

The perimeter discharge (P-Q) relation is given by the equation

  1. AOption (a)

    P = 2.25Q1/2

  2. BOption (b)

    P = 2.25Q3/2

  3. COption (c)

    p = 4.75Q1/2

  4. DOption (d)

    P = 4.75Q3/2

View answer & explanation

Correct answer

Answer: (c)C.

p = 4.75Q1/2

Why this answer is correct

Concept:

Lacey Regime's theory is applicable to final regime channels. The final regime is achieved when slope, width, depth is also adjusted as per requirements.

According to Lacey, the design formulas for canal design is as follows:

1) silt factor ⇒ f=1.76dmm\rm{f = 1.76\sqrt {{d_{mm}}}}

  1. velocity of flow ⇒ V=[Qf2140]16V = {\left[ {\frac{{Q{f^2}}}{{140}}} \right]^{\frac{1}{6}}}

3) hydraulic mean depth ⇒ R=5V22fR = \frac{{5{V^2}}}{{2f}}

  1. wetted perimeter ⇒ P=4.75QP = 4.75\sqrt Q

5) bed slope ⇒ S=f533340×Q16S = \frac{{{f^{\frac{5}{3}}}}}{{3340 \times {Q^{\frac{1}{6}}}}}

  1. Lacey's Normal Regimed depth ⇒ Rr′=0.477[Qf]13R_r' = 0.477{\left[ {\frac{{Q}}{{f}}} \right]^{\frac{1}{3}}}

  2. Lacey's Normal depth ⇒ Rr′=1.35[Q2f]13R_r' = 1.35{\left[ {\frac{{Q^2}}{{f}}} \right]^{\frac{1}{3}}}

Question 6

Δψ between two streamlines represents___ (where ψ is a stream function)

  1. AOption (a)

    velocity 

  2. BOption (b)

    discharge 

  3. COption (c)

    head

  4. DOption (d)

    pressure 

View answer & explanation

Correct answer

Answer: (b)B.

discharge 

Why this answer is correct

Concept:

Stream Function: It is the scalar function of space and time.

The partial derivative of stream function with respect to any direction gives the velocity component perpendicular to that direction.

& Stream function defines only for two-dimensional flow.

Properties of stream function:

If ψ exists, it follows the continuity equation and the flow may be rotational or irrotational.

If ψ satisfies the Laplace equation, then the flow is irrotational.

Signification:

  1. Stream function represents streamline and it is constant along streamline.
  2. The difference between any two-stream function give discharge per unit depth.

Question 7

A block of wood 2 m long, 2 m wide and 1 m deep is floating horizontally in water. If density of wood is 800 kg/m3, then the volume of water displaced will be

  1. AOption (a)

    3.2 m3

  2. BOption (b)

    2.6 m3

  3. COption (c)

    2 m3

  4. DOption (d)

    6 m3

View answer & explanation

Correct answer

Answer: (a)A.

3.2 m3

Why this answer is correct

Concept:

When a body is either wholly or partially immersed in a fluid, a lift is generated due to the net vertical component of hydrostatic pressure forces experienced by the body. This lift is called the buoyant force and the phenomenon is called buoyancy.

The Archimedes principle states that the buoyant force on a submerged body is equal to the weight of the liquid displaced by the body and acts vertically upward through the centroid of the displaced volume.

Thus,  the net weight of the submerged body, (the net vertical downward force experienced by it) is reduced from its actual weight by an amount that equals the buoyant force.

FB = ρghA = ρgV

Calculation:

Let ‘h’ is the height of wood block submerged in water. Block is floating implies that weight of block is balanced by the Buoyant force (FB).

(ρwood gVwood) = weight of water displaced by wood

(ρwood × Vwood) g = ρwater × Vdisp × g

800 × (2 × 2 × 1) = 1000 × (h × 2 × 2) × 1

h = 0.8 m

Volume of water displaced by wood is

∴ Vdisp = 0.8 × 2 × 2 = 3.2 m2

Question 8

The rate of filtration of a slow sand filter ranges from

  1. AOption (a)

    10 to 100 L/h/m2

  2. BOption (b)

    100 to 200 L/h/m2

  3. COption (c)

    200 to 400 L/h/m2

  4. DOption (d)

    400 to 1000 L/h/m2

View answer & explanation

Correct answer

Answer: (b)B.

100 to 200 L/h/m2

Why this answer is correct

Slow sand filter: They consist of fine sand, supported by gravel. They capture particles near the surface of the bed and are usually cleaned by scraping away the top layer of sand that contains the particles.

Cleaning:

  • For cleaning, the top layer of sand is scrapped and 1.5 to 3 cm of sand is removed in each cleaning.
  • The frequency of cleaning is 1 to 3 months.
  • The rate of filtration is 100-200 l/hr/m2 of filter area.
  • It is used for smaller plants in villages.
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Rapid sand Filter: They consist of larger sand grains supported by gravel and capture particles throughout the bed. They are cleaned by backwashing water through the bed to 'lift out' the particles.

Cleaning:

  • Cleaning of the filter can also be done using surface wash but it should be accompanied by backwash.
  • Washing period = 24 to 48 hrs.
Slow sand filterRapid sand filter
Pre-treatmentEffluents from plain sedimentation coagulation, not req.Coagulation, flocculation, and sedimentation is the must.
Filter SandD10 b/w 0.2 and 0.4 mm Cu ⇒ 1.8 to 3. Gravel -30-75 cm thickD10 = 0.35 mm - 0.55 mm Cu = 1.2 to 1.8 Gravel-60-90 cm thick
Size of each unitLarge such as 30m × 60 m. An area from 100-2000 m2. Less flexible on shock loadingSmall 5m × 8m. Area < 100m2 More flexible on shock loading
Rate of filtration100 to 200 litres/hr/m23000 to 6000 L/hr/m2
EfficiencyVery efficient in removing bacteria but less efficient in removing colour.Less efficient in bacteria (80-90%), very efficient in colour removal.
Post-treatmentDisinfection is not must. Almost pure water obtained. Turbidity removal up to 50 ppmDisinfection is the must. Turbidity up to 35 ppm.
Method of cleaningScrapping and removing the top 1.5 to 3 cm thick layer and washing down by hoses.Backwashing with or without compressed air.
Qty of wash water req.0.2 to 0.6% of total filtered water is req. in every 1 to 3 monthsA large amount of total filtered water is req. in every 1 to 3 days. (1-5%)

Question 9

In designing Imhoff tanks, the usual retention period is

  1. AOption (a)

    2 hours

  2. BOption (b)

    8 hours

  3. COption (c)

    14 hours

  4. DOption (d)

    20 hours

View answer & explanation

Correct answer

Answer: (a)A.

2 hours

Why this answer is correct

Imhoff tank: It is a robust and effective settler that causes a suspended solids reduction of 50 to 70%, COD reduction of 25 to 50%, and leads to potentially good sludge stabilization. The retention period of the Imhoff tank is 2 to 4 hours.

It consists of two compartments:

Upper Compartment: This compartment has a circular or rectangular shape with V-shaped walls and a slot at the bottom, allowing solids to settle into the digestion compartment, while preventing foul gas from rising up and disturbing the settling process. It is exposed to atmosphere, and thus working condition is aerobic.

Lower Compartment: In this compartment, sludge accumulates and is compacted and partially stabilized through anaerobic digestion. Thus, working condition is anaerobic

Question 10

The maximum spacing of laterals in a rapid sand filter can be 

  1. AOption (a)

    10 cm

  2. BOption (b)

    30 cm 

  3. COption (c)

    50 cm 

  4. DOption (d)

    100 cm

View answer & explanation

Correct answer

Answer: (b)B.

30 cm 

Why this answer is correct

Explanation:

The Standard design parameters for Rapid Sand filter are:

  • Maximum length of lateral is not less than 60 times its diameter.
  • Spacing of holes: For 6 mm holes at 7.5 cm c/c and for 13 mm holes at 15 cm c/c.
  • Cross sectional area of lateral shall be not less than 2 times area of perforations.
  • Cross sectional area of manifold should be 2 times total area of laterals.
  • Maximum loss of head = 2 to 5 m.
  • Spacing of laterals = 15 to 30 cm c/c. ⇒ Maximum spacing is 30 cm.
  • Pressure of wash water at perforations should not be greater than 1.05 kg/cm2.
  • Velocity of flow in lateral = 2 m/s.
  • Velocity of flow in manifold = 2.25 m/s.
  • Velocity of flow in manifold for wash-water= 1.8 to 2.5 m/s.
  • Velocity of rising wash-water= 0.5 to 1.0 m/min.
  • Amount of wash-water = 2 to 4% of total filtered water.
  • Time of back washing = 10 to 15 min.
  • Head of water over the filter = 1.5 to 2.5 m.
  • Free board = 60 cm.

Question 11

If W is total BOD, V is filter volume and F is recirculation factor in a trickling filter, then unit organic loading is obtained by 

  1. AOption (a)

    u=WFVu = \frac{{WF}}{V}

  2. BOption (b)

    u=VFWu = \frac{{VF}}{W}

  3. COption (c)

    u=WVFu = \frac{{WV}}{F}

  4. DOption (d)

    u=WVFu = \frac{{W}}{VF}

View answer & explanation

Correct answer

Answer: (d)D.

u=WVFu = \frac{{W}}{VF}

Why this answer is correct

Concept:

Organic loading rate (OLR) is defined as the how much organic matter is entering in to trickling in one day per unit volume. It is expressed in kg/m3/day. It is given as:

u=WVFu = \frac{W}{{VF}}

Where W is BOD in Kg/day; V is the volume of a trickling filter in m3 and F is the recirculation factor.

This recirculation factor, F is given as

F=1+R(1+0.1R)2F = \frac{{1 + R}}{{{{\left( {1 + 0.1R} \right)}^2}}}

Where R is the recirculation ratio.

Other points about OLR:

  1. It decides the efficiency of the trickling filter. Higher OLR means lower the efficiency of trickling filters.
  2. For Standard Rate Trickling Filter (SRTF) the OLR is 0.08 – 0.32 kg/m3/day and for High rate trickling filter (HRTF), the OLR is 0.32-1 kg/m3/day.

Question 12

In water supply pipes, wrought iron and cast iron pipes have relationship as

  1. AOption (a)

    Life of wrought iron pipes > life of cast iron pipes 

  2. BOption (b)

    Life of cast iron pipes > life of wrought iron pipes 

  3. COption (c)

    Both life spans are equal 

  4. DOption (d)

    Life of wrought iron pipes = 2 (life of cast iron pipes)

View answer & explanation

Correct answer

Answer: (b)B.

Life of cast iron pipes > life of wrought iron pipes 

Why this answer is correct

The difference between cast iron pipes and wrought-iron pipes is given below:

Cast Iron pipesWrought Iron pipes
Carbon content : 2.0 – 4.5%Carbon content – 0.25 %.
It is strong in compression, but weak in tension and bendingWrought iron High in elasticity and in tensile strength.
It can be hardened by heating & sudden cooling.It Cannot be hardened or tempered.
Cast iron pipes are normally made from 5 cm to 120 cm in diameter.Wrought iron pipes are made from 12 mm to 15 cm in diameter.
Cast iron pipes are durable and can last up to 100 years.Wrought iron pipes are less durable and their design life is about less than 50 years.

 

∴ Life of cast iron pipes > life of wrought iron pipes

Question 13

When the bed level of canal is higher than the highest flood level (HFL) of discharge, then the cross discharge work is said to be

  1. AOption (a)

    aqueduct

  2. BOption (b)

    super-passage 

  3. COption (c)

    canal syphon 

  4. DOption (d)

    under tunnel

View answer & explanation

Correct answer

Answer: (a)A.

aqueduct

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 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.

 

  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.

​

  1. 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 14

The bed of canal is lowered in case of 

  1. AOption (a)

    syphon aqueduct

  2. BOption (b)

    canal syphon

  3. COption (c)

    level crossing 

  4. DOption (d)

    All of the above 

View answer & explanation

Correct answer

Answer: (b)B.

canal syphon

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 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.

 

  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.

​

  1. 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 15

Which of the following hydraulic units is used for transmitting increased or decreased torque to the driven shaft?

  1. AOption (a)

    Hydraulic ram 

  2. BOption (b)

    Hydraulic intensifier 

  3. COption (c)

    Hydraulic torque converter 

  4. DOption (d)

    Hydraulic accumulator 

View answer & explanation

Correct answer

Answer: (c)C.

Hydraulic torque converter 

Why this answer is correct

Hydraulic Units:

Hydraulic accumulators are energy storage devices. They are analogous to rechargeable batteries in electrical systems, they store and discharge energy in the form of pressurized fluid and are used to improve hydraulic-system efficiency.

A hydraulic intensifier is a device used to increase the intensity of pressure of water by means of energy available from a large quantity of water at a low pressure.

Hydraulic ram is a device used to lift small quantity of water to a greater height when a large quantity of water is available at a smaller height.

Hydraulic torque converter is a device used for transmitting fluid power from machine to shaft bye changing (increasing or decreasing) the torque on system.

Question 16

When the length of body wall of a fall is less than the normal width of canal, it is called 

  1. AOption (a)

    Notch fall 

  2. BOption (b)

    sarda fall 

  3. COption (c)

    flumed fall 

  4. DOption (d)

    ogee fall

View answer & explanation

Correct answer

Answer: (c)C.

flumed fall 

Why this answer is correct

Canal Falls:

Whenever the available natural ground slope is steep than the designed bed slope of the channel, the difference is adjusted by constructing vertical ‘falls’ or ‘drops’ in the canal bed at suitable intervals, but such a drop in a natural canal bed will not be stable and, therefore, in order to retain this drop, a masonry structure is constructed. Such a structure is called a Canal Fall or a Canal drop.

Depending on the ground level conditions and shape of the fall the various types of fall are:

1. Ogee Fall: This type of fall has gradual convex and concave surfaces i.e. in the ogee form.

2. Stepped Fall: It consists of a series of vertical drops in the form of steps. This is suitable in places where sloping ground is very long and requires a long glacis to connect the higher bed level u/s with lower bed level d/s.

3. Vertical Fall (Sarda Fall): It is suitable when canal u/s bed is on the level of upstream curtain wall, canal u/s bed level is below the crest of curtain wall.

4. Rapid Fall: When the natural ground level is even and rapid, this rapid fall is suitable. It consists of long sloping glacis.

  1. Notch Fall: In this type a body or foundation wall across the channel consisting of several trapezoidal or rectangular notches between side pier and intermediate pier is constructed.
  2. Flumed Fall: In the flumed type length of the body wall of a fall is less than the normal canal width. The section is restricted at the site of the fall. The narrowing of the section is done gradually.

Question 17

Cross regulators are provided 

  1. AOption (a)

    to rise the water level to its upstream during the periods of low discharge in parent channel 

  2. BOption (b)

    to help in closing the supply to downstream of the parent channel 

  3. COption (c)

    to absorb fluctuations in various sections of the channel systems 

  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:

Cross regulator and distributary head regulator are provided to control the supplies passing down the parent channel and the offtaking channel respectively. A cross regulator is provided on the parent channel at the d/s of the offtake to head up the parent channel at the channel to draw the required supply. A distributary head regulator is provided at the head of the offtaking channel (or distributary) to control the supplies entering the offtaking channel.

Cross Regulator is a regulator constructed in the main or parent canal at downstream side of an off take canal is called cross-regulator as shown in figure:

Functions of Cross- regulator are:

  1. To Control the flow of water or fluctuations in canal system.
  2. To feed the off taking Canals. During the low Discharge, the cross regulator raises water level on the u/s so that the off taking channel can takes its full supply.
  3. It stops the water supply in downstream section of parent canal for the purpose of repair.
  4. To provide roadway for vehicular traffic.

Question 18

When the reservoir is full, the maximum compressive force in a gravity dam is produced 

  1. AOption (a)

    at the toe

  2. BOption (b)

    at the heel

  3. COption (c)

    within the middle third of the base

  4. DOption (d)

    at the center of the base

View answer & explanation

Correct answer

Answer: (a)A.

at the toe

Why this answer is correct

The maximum compressive stresses occur at heel (mostly during reservoir empty condition) or at the toe (at reservoir full condition) and on planes normal to the face of the dam.

∴ For reservoir empty condition maximum compressive force will be at the heel.

∴ For reservoir full condition maximum compressive force will be at the toe of the dam.

Figure below is the pressure distribution diagram for reservoir full condition:

Question 19

Sewage treatment units are normally designed for:

  1. AOption (a)

    5 – 10 years 

  2. BOption (b)

    15 – 20 years

  3. COption (c)

    30 – 40 years 

  4. DOption (d)

    40 – 50 years

View answer & explanation

Correct answer

Answer: (b)B.

15 – 20 years

Why this answer is correct

Concept:

Design period:

  • It may be defined as the time period for which the component remains functional.
  • As per Manual on Sewerage and Sewage Treatment Systems - 2013, the following table shows the design period of sewerage components:
ComponentDesign period (years) from base year
Land Acquisition≥ 30
Conventional Sewers30
Non – conventional Sewers15
Pumping mains30
Pumping stations30
Pumping machinery15
Sewage treatment plants15
Effluent Disposal30
Effluent Utilisation15

Hence, the design period treatment unit is taken as 15 - 20 years.

Question 20

In water supply for public, threshold odour should be

  1. AOption (a)

    1

  2. BOption (b)

    between 1 and 3

  3. COption (c)

    3

  4. DOption (d)

    more than 3

View answer & explanation

Correct answer

Answer: (b)B.

between 1 and 3

Why this answer is correct

Odour and taste:

It is one of the parameters which define the physical property of water

The odour in water is measured by a device known as osmoscope and expressed in terms of Threshold Odour Number (TON)

TON is a dilution ratio at which odour is just detectable

TON=A+BA{\rm{TON}} = \frac{{{\rm{A}} + {\rm{B}}}}{{\rm{A}}}

Where,

A = Volume of water sample tested in 'ml'

B = Volume of distilled water used in dilution in 'ml'

Capacity of osmoscope = A + B = 200 ml

There is no test for taste but senses of odour and taste are closely assosciated with each other. When water smell bad it also tastes bad.

The taste is expressed in terms of Flavoured Threshold Number (FTN) 

FTN=A+BA{\rm{FTN}} = \frac{{{\rm{A}} + {\rm{B}}}}{{\rm{A}}}

According to Drinking water standard, TON should not exceed 3 and preferably 1.

Question 21

Septic tank is a

  1. AOption (a)

    settling tank

  2. BOption (b)

    digestion tank 

  3. COption (c)

    Both (A) and (B)

  4. DOption (d)

    None of the above 

View answer & explanation

Correct answer

Answer: (c)C.

Both (A) and (B)

Why this answer is correct

Septic Tank:

A septic tank can be defined as a primary sedimentation tank with large detention time (12 to 36hrs).

A septic tank combines the function of sedimentation tank, a sludge digestion tank and a sludge storage tank.

A septic tank is usually provided with a brick wall in which cement mortar [not less than 20cm (9 inches)] thick and the foundation floor is of cement concrete 1 : 2 : 4.

Dimensions of Septic Tank:

  1. Width = 650 mm
  2. Length = 2 to 4 times the width
  3. Depth = 1 to 1.3 m
  4. Capacity = 1 cubic meter

Detention Period:

The detention period is taken to be 24 hrs.

Question 22

The pipe which is used to carry the discharge from sanitary fitting like bathrooms, kitchens, etc. is called _______

  1. AOption (a)

    Waste pipe

  2. BOption (b)

    Soil pipe

  3. COption (c)

    Vent pipe

  4. DOption (d)

    Anti-siphonage pipe

View answer & explanation

Correct answer

Answer: (a)A.

Waste pipe

Why this answer is correct

Pipes transporting wastes from households to the sewer are given as:

a) Waste Pipe: It carries waste from kitchens, basins, bathrooms, etc.

b) Soil Pipe: It carries night soil from toilets

Waste pipes are provided in a different range of sizes, from 32 mm to 50 mm.

Question 23

The pipes for water supply are tested for

  1. AOption (a)

    pressure

  2. BOption (b)

    leakage 

  3. COption (c)

    dimensions

  4. DOption (d)

    All of the above

View answer & explanation

Correct answer

Answer: (d)D.

All of the above

Why this answer is correct

Water supply pipes are tested for:

  1. Pressure: The stress-induced in the wall of a pipe when pressure is applied using water as a medium. The hydrostatic stress is related to the applied pressure, P, the wall thickness at any point, e, and the mean outside diameter, d , of a pipe and calculated using the following approximation equation

σ=P(d−e)2eσ =\frac{P\left( d-e \right)}{2e} where P and σ are in same unit.

  1. Dimensions: To check the conformity of the wall thickness of the pipe throughout its entire length, it is necessary to measure the wall thickness of the pipe at any point along its length. This shall be done by cutting the pipe at any point along its length and measuring the wall thickness as above. Alternatively, to avoid the destruction of the pipe, non-destructive testing methods such as the use of ultrasonic wall thickness measurement gauges shall be used at any four points along the length of the pipe. 
  2. Leakage: The water supply pipe should be tested for leakage as it will lead to loss of water.

Question 24

In rapid sand filters, the depth of tank varies in between

  1. AOption (a)

    1 m to 2 m

  2. BOption (b)

    2.5 m to 3.5 m

  3. COption (c)

    3 m to 3.5 m

  4. DOption (d)

    0.6 m to 0.9 m

View answer & explanation

Correct answer

Answer: (b)B.

2.5 m to 3.5 m

Why this answer is correct

They consist of larger sand grains supported by gravel and capture particles throughout the bed. They are cleaned by backwashing water through the bed to 'lift out' the particles.

The Standard design parameters for Rapid Sand filter are:

  1. Maximum length of lateral is not less than 60 times its diameter
  2. Spacing of holes: For 6 mm holes at 7.5 cm c/c and for 13 mm holes at 15 cm c/c
  3. Cross-sectional area of lateral shall be not less than 2 times area of perforations
  4. Cross-sectional area of manifold should be 2 times total area of laterals
  5. Depth of tank = 2.5 m to 3.5 m 6. Spacing of laterals = 15 to 30 cm c/c
  6. Pressure of wash water at perforations should not be greater than 1.05 kg/cm2
  7. Velocity of flow in lateral = 2 m/s
  8. Velocity of flow in manifold = 2.25 m/s
  9. Velocity of flow in manifold for wash-water= 1.8 to 2.5 m/s
  10. Velocity of rising wash-water= 0.5 to 1.0 m/min
  11. Amount of wash-water = 1 - 5% of total filtered water
  12. Time of backwashing = 10 to 15 min
  13. Head of water over the filter = 1.5 to 2.5 m
  14. Free board = 60 cm

Question 25

Seepage through embankment in an earth dam is controlled by 

  1. AOption (a)

    drainage filters

  2. BOption (b)

    drain trench 

  3. COption (c)

    relief wells 

  4. DOption (d)

    Provision of downstream berms

View answer & explanation

Correct answer

Answer: (a)A.

drainage filters

Why this answer is correct

Explanation:

The seepage control measures are necessitated to prevent the adverse effect of water percolating through the foundation to the embankment. The following devices are used to control seepage through earthen dam:

1) Embankment seepage control:

  • Toe filter
  • Horizontal drainage filter
  • Protective filter d/s of the toe
  • D/S coarse section
  • Chimney drain extending upward into the embankment

2) Foundation seepage control:

  • Impervious cutoff
  • U/S impervious blanket
  • D/S seepage berms
  • Drainage trench
  • Relief wall

Note: Apart from the horizontal drainage filter, all the given options are examples of foundation seepage control not embank seepage control.

Question 26

A chamber made of concrete, fibre glass, PVC or plastic, through which domestic waste water, sewage flows for secondary treatment is called

  1. AOption (a)

    drainage tank

  2. BOption (b)

    septic tank

  3. COption (c)

    pit latrine tank

  4. DOption (d)

    water harvesting tank

View answer & explanation

Correct answer

Answer: (b)B.

septic tank

Why this answer is correct

Concept:

A septic tank is an underground chamber made of concrete, fiberglass or plastic through which domestic wastewater (sewage) flows for basic treatment. Action within the septic tank is anaerobic in nature.

The term "septic" refers to the anaerobic bacterial environment that develops in the tank which decomposes the waste discharged into the tank.

Followings are the classification of secondary treatment units

MethodContact MechanismDecomposition
Trickling filterAttached growthAerobic
Rotating biological contactorAttached growthAerobic
Activated sludge processSuspended growthAerobic
Oxidation pondSuspended growthAerobic
Septic tankSuspended growthAnaerobic
Imhoff tankSuspended growthAnaerobic

Question 27

Septic tank is usually consists of brick wall in cement not less than

  1. AOption (a)

    20 cm

  2. BOption (b)

    100 cm

  3. COption (c)

    80 cm

  4. DOption (d)

    200 cm

View answer & explanation

Correct answer

Answer: (a)A.

20 cm

Why this answer is correct

Septic tank:

It works on the principle of anaerobic digestion. It is a primary sedimentation tank having a longer detention period of 12 to 36 hours, in which raw sewage is directly fed into it.

Construction Detail: It usually consists of a brick wall in cement not less than 20 cm thick and the foundation floor is of cement concrete having proportion in 1 : 3 : 6 or 1 : 2 : 4. Both inside and outside faces of wall and floor are plastered with a minimum thickness of 12 mm thick cement mortar 1 : 3 and all inside corners are rounded.

The following should be remembered about the septic tank.

  1. The minimum capacity of the septic tank is 8 to 10 persons or 2250 liters and the maximum capacity of the septic tank is usually 300 persons.
  2. BOD removal efficient – 25-65 %.
  3. Suspended solids removal efficiency – 60-70%.
  4. Rate of accumulation of sludge – 30 litres/person/year.
  5. Sewage flow without sullage- 40-70 lcd.
  6. Sewage flow with sullage: 90-150 lcd
  7. Cleaning period or removal of sludge: 6 to 12 months, mostly once in a year i.e. removal of sludge is less frequent.

Question 28

The BOD after the filtration of sewage from the low-rate trickling filter is 

  1. AOption (a)

    40-50%

  2. BOption (b)

    80-90%

  3. COption (c)

    70-80%

  4. DOption (d)

    90-99%

View answer & explanation

Correct answer

Answer: (b)B.

80-90%

Why this answer is correct

Trickling Filter:

Trickling filter is an aerobic attached growth system. This filter consists of 2 m to 3 m deep circular bed of packed granular stone or gravel about 50 to 70 mm in size. Over this trickling filter, the settled sewage is allowed to sprinkle or trickle with the help of spray nozzles or rotary distributor.

The Slime layer is the layer of micro-organisms develops on the granular bed of the trickling filter.

There are 2 types of trickling filter:

  1. Standard rate trickling filter (SRTF)
  2. High rate trickling filter (HRTF)

 

The standard rate trickling filters (SRTFs) are also called conventional or low rate trickling filters.

Following are the design parameters and efficiency of SRTFs:

  1. The effluent obtained from SRTF is highly nitrified and stabilized.
  2. The BOD is reduced to 70 to 90 % of the original value.
  3. The BOD left is less than 20 ppm.
  4. The sludge obtained is heavy and easily digestible and has moisture content of 92 %.
  5. Hydraulic Loading rate: 1 - 4 m3/m2/day.
  6. Organic Loading rate: 0.08 - 0.32 kg/m3/day.
  7. Depth: 1.8 to 3 m.

Question 29

The organic loading in a trickling filter is measured in

  1. AOption (a)

    m3/day

  2. BOption (b)

    gm/m2/day

  3. COption (c)

    kg/hectare-meter/day

  4. DOption (d)

    kg/hectare/day

View answer & explanation

Correct answer

Answer: (c)C.

kg/hectare-meter/day

Why this answer is correct

Concept:

Trickling Filter:

Trickling filter is an aerobic attached growth system. This filter consists of 2 m to 3 m deep circular bed of packed granular stone or gravel about 50 to 70 mm in size. Over this trickling filter, the settled sewage is allowed to sprinkle or trickle with the help of spray nozzles or rotary distributor.

The Slime layer is the layer of micro-organisms develops on the granular bed of the trickling filter.

There are 2 types of trickling filter:

  1. Standard rate trickling filter (SRTF)
  2. High rate trickling filter (HRTF)

Organic loading rate (OLR) is defined as how much organic matter is entering into trickling in one day per unit volume. It is expressed in kg/m3/day. It is given as:

u=WVFu = \frac{W}{{VF}}

Where W is BOD in Kg/day; V is the volume of a trickling filter in m3 and F is the recirculation factor.

∴ kg/hectare-metre/day is a measuring unit of trickling filter.

Other points about OLR:

  1. It decides the efficiency of the trickling filter. Higher OLR means lower the efficiency of trickling filters.
  2. For Standard Rate Trickling Filter (SRTF) the OLR is 0.08 – 0.32 kg/m3/day and for High rate trickling filter (HRTF), the OLR is 0.32-1 kg/m3/day.

Question 30

The recirculation factor in a low-rate trickling filter is 

  1. AOption (a)

    0

  2. BOption (b)

    1

  3. COption (c)

    10

  4. DOption (d)

    100

View answer & explanation

Correct answer

Answer: (b)B.

1

Why this answer is correct

Trickling Filter:

Trickling filter is an aerobic attached growth system. This filter consists of 2 m to 3 m deep circular bed of packed granular stone or gravel about 50 to 70 mm in size. Over this trickling filter, the settled sewage is allowed to sprinkle or trickle with the help of spray nozzles or rotary distributor.

The Slime layer is the layer of micro-organisms develops on the granular bed of the trickling filter.

There are 2 types of trickling filter:

  1. Standard rate trickling filter (SRTF)
  2. High rate trickling filter (HRTF)

 

The recirculation means returning a portion of treated sewage to the treatment unit. By doing so, efficiency of trickling filter improves and it also reduces the load on filter.

The recirculation ratio is connected to another term called recirculation factor (F) to obtain the efficiency of filter and it is given as,

F=1+R(1+0.1R)2F = \frac{{1 + R}}{{{{\left( {1 + 0.1R} \right)}^2}}}

The ratio of the Volume of sewage re-circulated (O) to the volume of Raw sewage (I) is called recirculation ratio, R.

R=OIR = \frac{O}{I}

The low rate trickling filter does not have any recirculation system.

⇒ O = 0 ⇒ R = 0, substitute this in above equation ⇒ F = 1

Question 31

The diameter of a domestic sewer pipe laid at gradient 1 in 100 is recommended

  1. AOption (a)

    100 mm

  2. BOption (b)

    150 mm

  3. COption (c)

    200 mm

  4. DOption (d)

    175 mm

View answer & explanation

Correct answer

Answer: (b)B.

150 mm

Why this answer is correct

As, per IS 1972 (1983), Clause 4.6.1.2 The approximate gradients which give this velocity for the sizes of pipes likely to be used in building drainage and the corresponding discharges when flowing half-full are as follows:

Dia. of Sewer (mm)Minimum Gradient
1001 in 57
1501 in 100
2001 in 145
2301 in 175
2501 in 195
3001 in 250

Question 32

Which of the following is not a requirement for site selection of hydroelectric power plant?

  1. AOption (a)

    Availability of water

  2. BOption (b)

    Large catchment area

  3. COption (c)

    Rocky land 

  4. DOption (d)

    Sedimentation 

View answer & explanation

Correct answer

Answer: (d)D.

Sedimentation 

Why this answer is correct

Site selection of hydroelectric power plant:

Site selection and factors affecting the location of dam of hydroelectric power plants are storage of water, cost and type of land, transportation facilities, constituents of a hydro-electric plant, hydraulic structure and a schematic arrangement of hydro-electric power station and its operation.

The principle of the hydropower plants is that the potential energy is converted into kinetic energy. Because the water stored at great heights in the dam so it’s allowing the water to flow at high speed. Then the kinetic energy of flowing water used to generate electricity.

Sedimentation is not a requirement for site selection of hydroelectric power plants because sedimentation may reduce the water storage capacity of reservoir and may also cause damage to the turbine blades.

Availability of water, large catchment area and rocky land are the primary requirements for site selection of hydroelectric power plants.

Question 33

The amount of electrical energy that can be generated by a hydroelectric power plant depends upon

  1. AOption (a)

    head of water

  2. BOption (b)

    quantity of water 

  3. COption (c)

    specific weight of water 

  4. DOption (d)

    efficiency of alternator 

View answer & explanation

Correct answer

Answer: (a)A.

head of water

Why this answer is correct

Explanation:

The following factors should be considered while selecting the site for a hydroelectric power plant.

Availability of water:

  • To know the available energy from a given stream or river, the discharge flowing and its variation with time over a number of years must be known. The design and capacity of the hydroelectric power plant greatly depend on the amount of water available at the site. The estimates of the average quantity, maximum and minimum quality of water available at the proposed site should be made available to decide the need and capacity for storage, the plant capacity, the size of the headwords, and the dam to be built with adequate spillway and to set up the interconnected peak load thermal plant.

Water storage:

  • There is a wide variation in rainfall throughout the year; hence, to have a uniform power output, water storage is needed so that excess flow at certain times may be stored to make it available at times of low flow. The storage capacity can be estimated with the help of a mass curve. To select the site of the dam, a careful study should be made of the geology and topography of the catchment area to see if the natural foundations could be found.

Available water head:

  • A large quantity of water at a sufficient head should be available in order to generate the desired quantity of power. The level of water in the reservoir for a proposed plant should always be within limits throughout the year. For a given output, the quantity of water required to be supplied to the turbines reduces with an increase in the effective head.

Access to the site:

  • Good access and transport facilities to the site of the plant is always desirable factor. The inaccessible site or terrain will impede the movement of men and material.

Distance from load center:

  • Mostly the electric power generated in a hydroelectric power plant has to be transmitted to some considerable distance from the site of the plant. Hence, the routes and the distances for transmission of electric power should be carefully considered to reduce the cost of the erection of transmission lines and their maintenance. The cost of transmission lines and the transmission losses will be reduced if the site is close to the load center.

Geological investigation:

  • The dam constructed at the site should have a large catchment area to store water at a high head. The foundation rocks of the masonry dam and the other structures must be firm, stable, impervious, and strong enough to withstand the stresses in the structure and the thrust of water when the reservoir is full. The area must be free from earthquakes. The land of the site should be cheap and rocky:

Water pollution:

  • The water should be of good quality; otherwise, it may corrode and damage the plant structure and machinery.

Sedimentation:

  • Silting must be small as it reduces the capacity of the reservoir due to the deposition of sediments. Silting from the forest-covered area is small.

Question 34

Hydroelectric power plant is

  1. AOption (a)

    non-renewable source of energy 

  2. BOption (b)

    conventional source of energy 

  3. COption (c)

    non-conventional source of energy 

  4. DOption (d)

    continuous source of energy 

View answer & explanation

Correct answer

Answer: (b)B.

conventional source of energy 

Why this answer is correct

Explanation:

A hydroelectric power plant:

  • Utilizes the potential energy of stored water (usually in dams) to generate electricity.
  • It is:
  • Renewable (since water cycle is naturally replenished),
  • Conventional (widely used for many decades),
  • Not non-conventional (non-conventional refers to newer sources like solar, wind, tidal),
  • Can be continuous, but depends on seasonal water availability.

Question 35

What type of flow can be taken for granted in a pipe of uniform cross-section?

  1. AOption (a)

    Steady 

  2. BOption (b)

    Unsteady 

  3. COption (c)

    Uniform 

  4. DOption (d)

    Non-uniform 

View answer & explanation

Correct answer

Answer: (c)C.

Uniform 

Why this answer is correct

Explanation:

Steady and Unsteady flow:

If the flow parameters such as velocity, acceleration, the discharge do not change with time then flow is called steady flow, otherwise, flow is un-steady.

Uniform and Non-uniform flow:

If flow parameters such as velocity, acceleration, discharge changes with space, then flow are called non-uniform flow or varied flow, otherwise the flow is uniform.

According to the continuity equation, ρaV =constant

where ρ = density, a = cross-sectional area of flow, V = velocity of flow.

Justification for answer:

For a pipe of a uniform cross-section, no matter what the rate of flow is, the velocity of flow inside the pipe will always remain constant. Hence, it’ll always be a uniform flow.

It’ll be a steady flow if and only if the water level is maintained at a constant level by supplying water at the same rate as it gets discharged, else the water level will keep decreasing with time leading to an unsteady flow.

Question 36

What is the most common medium for sediment transport?

  1. AOption (a)

    ice 

  2. BOption (b)

    Human 

  3. COption (c)

    Wind

  4. DOption (d)

    Water 

View answer & explanation

Correct answer

Answer: (d)D.

Water 

Why this answer is correct

Water is the most common medium for sediment transport. It plays a significant role in the erosion, transportation, and deposition of sediment in various environments, including rivers, streams, oceans, and lakes. Water can move sediment particles through its flow, whether it is in the form of rivers, waves, or currents.

Option 1) Ice: While ice can transport sediment in the form of glaciers and ice sheets, it is not as widespread or common as water as a medium for sediment transport.

Option 2) Human: Human activities can certainly influence sediment transport through activities such as dredging or construction, but it is not the most common natural medium for sediment transport.

Option 3) Wind: Wind can transport sediment, particularly smaller particles like sand and dust, in arid or coastal environments. However, in terms of overall sediment transport on a global scale, water is more significant and widespread.

Question 37

How many types of weirs are there based on the shape of the crest?

  1. AOption (a)

    6

  2. BOption (b)

    4

  3. COption (c)

    5

  4. DOption (d)

    3

View answer & explanation

Correct answer

Answer: (b)B.

4

Why this answer is correct

Classification of weir based on shape of crest:

  1. Sharp crested weir: The crest shape of the weir is very sharp. It is most widely used weir.

  2. Broad-crested weir: These are constructed only in a rectangular shape and are suitable for the larger flows.

  3. Narrow-crested weir: It is similar to rectangular weir with a narrow shaped crest at the top. The discharge over the narrow crested weir is similar to discharge over the rectangular weir.

  4. Ogee-shaped weir: Generally ogee shaped weirs are provided for the spillway of a storage dam. The crest of the ogee weir is slightly risen and falls into parabolic form.

Question 38

Triangular weir is also called

  1. AOption (a)

    trigonometric 

  2. BOption (b)

    ogee

  3. COption (c)

    v-notch

  4. DOption (d)

    isolated

View answer & explanation

Correct answer

Answer: (c)C.

v-notch

Why this answer is correct

Concept:

A V-notch weir is also called the triangular notch or weir. The discharge over a triangular weir or notch is given by the:

Q=815Cdtanθ2;2g;H5/2Q = \frac{8}{{15}}{C_d}tan\frac{\theta }{2};\sqrt {2g} ;{H^{5/2}}

Where, Q = discharge of fluid

Cd = Coefficient of discharge

H = height of water above the notch

Note:

The discharge through a rectangular notch weir is,

Q=23CdL;2g;H3/2Q = \frac{2}{3}{C_d}L;\sqrt {2g} ;{H^{3/2}}

Question 39

Weirs are normally used to calculate

  1. AOption (a)

    volume

  2. BOption (b)

    headloss

  3. COption (c)

    discharge

  4. DOption (d)

    velocity 

View answer & explanation

Correct answer

Answer: (c)C.

discharge

Why this answer is correct

Concept:

Weir is a physical structure of masonry constructed across the channel width to calculate the discharge of the channel section.

Rectangular weir:

The discharge through a rectangular notch weir is,

Q=23CdL;2g;H3/2Q = \frac{2}{3}{C_d}L;\sqrt {2g} ;{H^{3/2}}

Where, Q = discharge of fluid, L = Length of the notch weir, Cd = Coefficient of discharge and H = height of water above the notch

Triangular weir:

A V-notch weir is also called the triangular notch or weir. The discharge over a triangular weir or notch is given by the:

Q=815Cdtanθ2;2g;H5/2Q = \frac{8}{{15}}{C_d}tan\frac{\theta }{2};\sqrt {2g} ;{H^{5/2}}

Where, Q = discharge of fluid, Cd = Coefficient of discharge, θ = Notch angle and H = height of water above the notch

Trapezoidal weir:

The Cippoletti weir is a trapezoidal weir, having side slopes 1 horizontal to 4 vertical. The purpose of the slope, on the sides, is to obtain an increased discharge through the triangular portions of the weir, which otherwise would have been decreased due to end contractions in the case of rectangular weirs.

A type of contracted weir that is related to the rectangular sharp-crested weir is Cippoletti weir, which has a trapezoidal cross-section with side slopes 1: 4 (H: V).

Question 40

Bearings are provided in the bridges to 

  1. AOption (a)

    allow translation and rotation in bridges

  2. BOption (b)

    transfer forces from superstructure to substructure

  3. COption (c)

    isolate superstructure from substructure

  4. DOption (d)

    All of the above

View answer & explanation

Correct answer

Answer: (d)D.

All of the above

Why this answer is correct

Bridge bearings function:

Bridge bearings are used to transfer forces from the superstructure to the substructure.

This transfer of load allows the following types of movements of the superstructure:

(a) Translation movements

Displacements in vertical and horizontal directions.

(b) Rotational movements cause due to moments.

Allow free movement of girders due to temperature variations (contractions and expansion).

Prevent excessive damage to the bridge if any pier sinks slightly.

It is located between the substructure and superstructure of bridge

The types of bridge bearings used are as follows:

  1. Pin
  2. Roller
  3. Rocker
  4. Metal sliding bearings

Question 41

Which of the following loads contributes to longitudinal forces in bridges?

  1. AOption (a)

    Dead load

  2. BOption (b)

    Wind load 

  3. COption (c)

    Temperature load 

  4. DOption (d)

    Both 1 and 3

View answer & explanation

Correct answer

Answer: (c)C.

Temperature load 

Why this answer is correct

Longitudinal forces in bridge:

Longitudinal forces mean force acting on a bridge along its length or along with the motion of traffic.

The longitudinal forces are caused by braking or accelerating of the vehicle on the bridge. When the vehicle stops suddenly or accelerates suddenly it induces longitudinal forces on the bridge structure, especially on the substructure. IRC recommends 20% of the live load should be considered as a longitudinal force on the bridges.

Dead load is due to the weight of the bridge and it always acts in the direction of gravity i.e. perpendicular to the roadway of the bridge.

Thermal stresses are caused due to temperature. When the temperature is very high or very low they induce stresses in the bridge elements, especially at bearings and deck joints. They induce both vertical and horizontal forces.

Wind load acts in every direction of the bridge. However, their impact as lateral wind force is more as compared to the longitudinal direction.

Other types of forces and load on bridges:

1. Centrifugal forces: If bridge is to be built on horizontal curves, then the movement of the vehicle along curves will cause centrifugal force on to the superstructure.

2. Buoyancy Effect: Buoyancy effect is considered for substructures of large bridges submerged under deep water bodies.

3. Seismic Loads: They induce both vertical and horizontal forces during the earthquake.

4. Erection Stresses: Erection stress are induced by the construction equipment during the bridge construction.

Question 42

Development of surges in open channel is 

  1. AOption (a)

    Gradually varied flow 

  2. BOption (b)

    rapidly varied flow

  3. COption (c)

    steady flow

  4. DOption (d)

    normal flow

View answer & explanation

Correct answer

Answer: (b)B.

rapidly varied flow

Why this answer is correct

Steady and Unsteady flow:

If the flow parameters such as velocity, acceleration, discharge do not change with time then flow is called steady flow, otherwise, flow is un-steady.

Gradually varied and Rapidly varied flow:

If flow parameters such as velocity, acceleration, discharge changes with space, then flow is called non-uniform flow or varied flow. If the change in depth of flow is gradually along the length, then is it called gradually varied flow and if the change in depth of flow is sudden then it is called rapidly varied flow.

The examples for above kind of flows are given below:

Type of flowExample
Steady and uniform flowFlow in prismatic channel
Steady but gradually varied flowBack water curve due to obstruction
Steady but rapidly varied flowHydraulic jump
Unsteady and Gradually varied flowFlood flow in river
Unsteady and rapidly varied flowSurge

 

The surge means a sudden powerful forward or upward movement of water or powerful disturbance in water which could not be possible in gradually varied flow and steady flow.

∴ Development of surges in open channel is rapidly varied flow.

Question 43

Irrigation canals are generally aligned along: 

  1. AOption (a)

    Ridge line

  2. BOption (b)

    Contour line

  3. COption (c)

    Valley line

  4. DOption (d)

    Straight line

View answer & explanation

Correct answer

Answer: (a)A.

Ridge line

Why this answer is correct

Watershed canal or ridge canal:

The dividing ridgeline between the catchment areas of two streams (drains) is called the watershed or ridge canal. Thus between two major streams, there is the main watershed (ridge line), which divides the drainage area of the two streams. Similarly, between a mainstream and any of its tributary, there are subsidiary watersheds (ridge lines), dividing the drainage between the two streams on either side. The canal which is aligned along any natural watershed (ridge line) is called a watershed canal, or a ridge canal. Aligning a canal (main canal or branch canal or distributary) on the ridge ensures gravity irrigation on both sides of the canal. Since the drainage flows away from the ridge, no drainage can cross a canal aligned on the ridge. Thus, a canal aligned on the watershed saves the cost of construction of cross-drainage works.

Question 44

For small discharge and high heads which pump is preferred

  1. AOption (a)

    Centrifugal pump

  2. BOption (b)

    Reciprocating pump

  3. COption (c)

    Axial flow pump

  4. DOption (d)

    Radial flow type

View answer & explanation

Correct answer

Answer: (b)B.

Reciprocating pump

Why this answer is correct

Explanation:

Pumps

  • The hydraulic machines which convert the mechanical energy into hydraulic energy are called pumps.
  • If the mechanical energy is converted into pressure energy by means of the centrifugal force acting on the fluid, the hydraulic machine is called a centrifugal pump.
Centrifugal pumpsReciprocating pumps
The discharge is continuous and smoothThe discharge is fluctuating and pulsating
It can handle a large quantity of liquid with low pressureIt handles a small quantity of liquid only with high pressure
It can be used for lifting highly viscous liquidsIt is used only for lifting pure or less viscous liquids
It is used for large discharge through smaller headsIt is meant for a small discharge and high heads
The cost of a centrifugal pump is less as compared to a reciprocating pumpThe cost of a reciprocating pump is approximately four times the cost of a centrifugal pump
The Centrifugal pump runs at high speed. They can be coupled to an electric motorIt runs at a low speed. Speed is limited due to consideration of separation and cavitation
The operation of a centrifugal pump is smooth and without much noise. The maintenance cost is lowThe operation of a reciprocating pump is complicated and with much noise. The maintenance cost is high.
It requires a smaller floor area and installation cost is lowIt requires a larger floor area and installation cost is high
Efficiency is highEfficiency is low

Question 45

A hyetograph is a graphical representation of

  1. AOption (a)

    rainfall intensity and time 

  2. BOption (b)

    rainfall depth and time

  3. COption (c)

    discharge and time

  4. DOption (d)

    cumulative rainfall and time 

View answer & explanation

Correct answer

Answer: (a)A.

rainfall intensity and time 

Why this answer is correct

Rainfall hyetograph shows Intensity of rainfall v/s time. For example,

Note:

Type of CurveMeaning
Mass CurveCumulative rainfall vs Time
Double Mass CurveCumulative rainfall of nearby stations vs Cumulative rainfall of concerned station
HydrographDischarge vs time
HyetographRainfall Intensity vs time
DAD - Depth Area duration CurvesRainfall depth vs Area &Time

Question 46

Kirpich equation is used to determine which one of the following?

  1. AOption (a)

    Runoff from a given rainfall

  2. BOption (b)

    Base time for unit hydrograph 

  3. COption (c)

    Time of concentration in runoff hydrograph 

  4. DOption (d)

    None of the above 

View answer & explanation

Correct answer

Answer: (c)C.

Time of concentration in runoff hydrograph 

Why this answer is correct

Concept:

Kirpich equation is an empirical equation used to determine the time of concentration in runoff hydrograph and it is given as:

t=0.0194L0.77S−0.385t = 0.0194{L^{0.77}}{S^{ - 0.385}}

Where,

t = time of concentration in minutes

L = the maximum length of water travel in m.

S = the slope of the catchment and it is given as

S=ΔHLS = \frac{{{\rm{\Delta }}H}}{L}

ΔH is the difference in elevation between most remote points on the catchment outlet.

Note:

Time of Concentration is the time required by the entire drainage area to contribute to the runoff is called time of concentration or time required by the most extreme point in the drainage to reach the point of interest.

Question 47

The top of the ground on which foundation of road rests is called

  1. AOption (a)

    Soling 

  2. BOption (b)

    base

  3. COption (c)

    Either 1 or 2

  4. DOption (d)

    None of the above 

View answer & explanation

Correct answer

Answer: (d)D.

None of the above 

Why this answer is correct

Foundation of the road:

A pavement may be constructed in a number of layers and the top layer has to be of the best quality to sustain maximum compressive stress, in addition, to wear and tear. The lower layers will experience the lesser magnitude of stress and low-quality material can be used.

∴ The top of the ground on which foundation of road rests is subgrade.

Question 48

According to IRC recommendations, absolute minimum radius of curve for safe operation for a design speed of 100 kmph is

  1. AOption (a)

    100 m

  2. BOption (b)

    200 m

  3. COption (c)

    300 m

  4. DOption (d)

    400 m

View answer & explanation

Correct answer

Answer: (d)D.

400 m

Why this answer is correct

Concept:

Absolute Minimum Radius =(Minimum;Design;Speed)2127×(0.07+0.15)= \frac{{{{\left( {Minimum;Design;Speed} \right)}^2}}}{{127 \times \left( {0.07 + 0.15} \right)}}

Ruling Minimum Radius =(Ruling;Design;Speed)2127×(0.07+0.15)= \frac{{{{\left( {Ruling;Design;Speed} \right)}^2}}}{{127 \times \left( {0.07 + 0.15} \right)}}

Minimum Radius beyond which there is no need to provide super-elevation

=Design;Speed225×Camber= \frac{{Design;Speed}}{{225 \times Camber}}

Information about Design speed:

Class of RoadTerrain
PlainRolling
RulingMinimumRulingMinimum
NH/SH100808065
MDR80656550
ODR65505040
VR50404035

 

where, NH = National Highway, SH = State Highway, MDR = Main District Road, ODR = Ordinary District Road, VR = Village Road

Calculations:

Absolute Minimum Radius =(Minimum;Design;Speed)2127×(0.07+0.15)= \frac{{{{\left( {Minimum;Design;Speed} \right)}^2}}}{{127 \times \left( {0.07 + 0.15} \right)}}

Absolute Minimum Radius =(100)2127×(0.07+0.15)=357.9;m= \frac{{{{\left( {100} \right)}^2}}}{{127 \times \left( {0.07 + 0.15} \right)}} = 357.9;m

∴ Option 4 is more appropriate

Question 49

Critical load position in critical load combination for rigid pavement design is taken as ___

  1. AOption (a)

    interior loading

  2. BOption (b)

    edge loading

  3. COption (c)

    corner loading

  4. DOption (d)

    interior, edge and corner loading

View answer & explanation

Correct answer

Answer: (d)D.

interior, edge and corner loading

Why this answer is correct

Explanation:

(i) Rigid pavement slab has finite length and width, either the character or the intensity of maximum stress induced by the application of a given traffic load is dependent on the location of the load on the pavement surface.

(ii) The three typical locations are the interior, edge, and corner of the slab, where differing conditions of slab continuity exist. These locations are termed as critical load positions.

(iii) Maximum stress at the interior (SRi) and edge (SRe) of the rigid slab is given by

Case 1: Winter + Midday

SRi = Si + Swi + Sfi and SRe = Se + Swe + Sfe

Case 2: Summer + Midday

SRi = Si + Swi - Sfi and SRe = Se + Swe - Sfe

Maximum stress at the corner (SRc) of the rigid slab is given by

Case 3: Winter + MidNight

SRc = Sc + Swc

where,

Si = Direct loading stress due to wheel load at the interior of the slab

Swi = Warping stress at the interior of the slab

Sfi = Frictional stress at the interior of the slab

Se = Direct loading stress due to wheel load at the edge of the slab

Swe = Warping stress at the edge of the slab

Sfe = Frictional stress at the edge of the slab

Sc = Direct loading stress due to wheel load at the corner of the slab

Swc = Warping stress at the corner of the slab

Confusion Points

Question is asked about general condition, so given answer is correct if In question it is mentioned that the combination of temperature and wheel load then in that condition design for edge and check for a corner.

Question 50

The most suitable material for highway embankments is 

  1. AOption (a)

    Granular soil

  2. BOption (b)

    Organic soil

  3. COption (c)

    Silt

  4. DOption (d)

    Clay

View answer & explanation

Correct answer

Answer: (a)A.

Granular soil

Why this answer is correct

Concept:

The most suitable material for highway embankments is granular soil like sand because it has the following characteristics:

  1. Higher permeability as compared to clay, so that water gets easily drained off.
  2. The amount of primary consolidation is less.
  3. If well consolidated and confined, they form a foundation that is almost as stable as a rock.
  4. It posses higher shear strength due to mechanical interlocking of particles and having a higher angle of internal friction as compared to cohesive soils like clay.

Note:

Organic soils are highly compressible and hence, cannot be used as a construction material. They are used for agriculture purposes only.

Silty soil is usually more fertile than other types of soil i.e. it is good for growing crops.

50 questions in the full paper