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Test: Hydrology - Civil Engineering (CE) MCQ


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15 Questions MCQ Test - Test: Hydrology

Test: Hydrology for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Test: Hydrology questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Hydrology MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Hydrology below.
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Test: Hydrology - Question 1

What is the process by which water enters the small pore spaces between particles in soil rock?

Detailed Solution for Test: Hydrology - Question 1

Infiltration:

  • Some of the water that falls on land soaks into the ground. Water penetrates the ground through joints or small pore spaces between particles via a process known as infiltration.
  • Infiltration is the process of water entry into the soil through the earth's surface. The water at the soil surface can originate from rain, snowmelt or anthropogenic activities (e.g. to regulate groundwater formation by artificial infiltration).

Transpiration:

  • Transpiration is the process of water movement through a plant and its evaporation from aerial parts, such as leaves, stems, and flowers. Water is necessary for plants but only a small amount of water taken up by the roots is used for growth and metabolism.

Precipitation:

  • In meteorology, precipitation is any product of the condensation of atmospheric water vapor that falls under gravitational pull from clouds. The main forms of precipitation include drizzle, rain, sleet, snow, ice pellets, graupel, and hail.

Sublimation:

  • Sublimation is the transition of a substance directly from the solid phase to the gas phase without passing through the intermediate liquid phase.
Test: Hydrology - Question 2

Penman’s evapo-transpiration equation is based on

Detailed Solution for Test: Hydrology - Question 2

Evapotranspiration Equation:
There are large number of evapotranspiration equations available, they are purely empirical and not theoretical. The most used equation is Penman's Equation.

Penman's Equation:
Penman's equation is based on sound theoretical reasoning and is obtained by combination of energy and mass transfer approach.

PET = Daily potential evapotranspiration (in mm per day)
A = Slope of the saturation vapour pressure vs. temperature curve at the mean air temperature (in mm of Hg per °C)
Hn = Net radiation
Ea = Parameter including wind velocity and saturation deficit.
γ = Psychromrtic constant = 0.49 mm of Hg/°C 

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Test: Hydrology - Question 3

The hydrologic risk of a 100 year flood occuring during the 2-year service life of a project is

Detailed Solution for Test: Hydrology - Question 3

Hydrologic risk

Or Risk (R) = 1 - (1 - P)n
Where T = Flood return period
P = Probability of occurring flood in a given year
n = Number of years of risk calculate
Calculation:
Given
Flood return period (T) = 100 years
Service life of the project (n) = 2 years

Risk (R) = 1 - (1 - 0.01)2
Risk (R) = 1 - (0.99)2
Risk (R) = 1 - 0.9801 = 0.0199
Risk (R) = 0.0199 × 100 = 1.99%
Risk (R) = 1.99%

Test: Hydrology - Question 4

The rainfall figures for successive 30 – minute intervals are 35, 40, 120, 85, 45, 45 and 30 mm/hr. If the ϕ index is 35 mm/hour, determine W-index.

Detailed Solution for Test: Hydrology - Question 4

Concept:

Where,
Pe = effective rainfall causing runoff, R = runoff, 
te = duration of effective rainfall, P = total rainfall and tr = total duration of rainfall
Calculation:
Given,
ϕindex = 35 mm/hr
∵  We know that,  runoff occurs when the intensity of rainfall (i) >  ϕindex 
∴ From rainfall intensity data, the intensity of 30 mm/hr < 35 mm/hr, hence ignored in the calculation of ϕindex
⇒ Pe = (35 + 40 + 120 + 85 + 45 + 45) × 0.5
⇒ Pe = 185 mm and te = 3 hr
∵ We know that, 

⇒ R = 80 mm
Now, P = (35 + 40 + 120 + 85 + 45 + 45 + 30) × (30/60)
⇒ P = 200 mm and tr = 3.5 hr
∵ We know that, 

⇒ Windex = 34.28 mm/hr

Test: Hydrology - Question 5

Which of the following statements is/are correct?
1. Isochrones are curves of equal pore water pressure
2. Isochrones depict the variation of the pore water pressure along with the depth of the soil sample
3. Isochrones vary with time

Detailed Solution for Test: Hydrology - Question 5

Isochrones:
Isochrones is a line on the map which connects points having an equal time of travel of the surface runoff to the catchment outlet. These are some properties of Isochrones:

  • Isochrones vary with time, It's most commonly used to depict travel times, such as drawing a 30-minute travel time perimeter around a start location. The isochrone below joins up all points within a 45-minute drive from the origin. 
  • Isochrones depict the variation of the pore water pressure along with the depth of the soil sample.
  • Isochrones are mainly used for transport planning, property search, sales territory planning, etc. ​
Test: Hydrology - Question 6

Select the correct option with regard to the following two statements (H1 and H2) pertaining to the hydrograph of a storm in a catchment.
H1: The rising limb of the hydrograph depends on the catchment characteristics only.
H2: The recession limb of the hydrograph depends on the storm characteristics and catchment characteristics.

Detailed Solution for Test: Hydrology - Question 6

Hydrograph:

  • A hydrograph is a plot between discharge and time at any given section of a river, channel, etc. 
  • It is a response of a given catchment to the rainfall input.
  • The shape of the rising limb of a hydrograph depends on both catchment characteristics and rainfall characteristics.
  • The shape of the falling or recession limb of a hydrograph depends only on catchment characteristics.
  • The inflation point on the falling limb is often assumed to be the point where direct runoff ends.
  • Time of Concentration is the time required by the entire drainage area to contribute to the runoff is called the time of concentration or time required by the most extreme point in the drainage to reach the point of interest.

Test: Hydrology - Question 7

During a storm, the rates of rainfall observed at a frequency of 15 min for one hour are 12.5, 17.5, 22.5 abd 7.5 cm/h. If the phi-index is 7.5 cm/h, then the total run-off will be

Detailed Solution for Test: Hydrology - Question 7

The ϕ – index is a rate of infiltration in which, the rate of infiltration exceeds the value at which volume of runoff become equals to the volume of rainfall.

where
PTotal = Total precipitation
Q = Runoff
Calculation:

ϕ - index = 7.5 cm/hr
ϕ - index is the average rate of rainfall such that the volume of rainfall in excess of that rate is equal to the volume of surface runoff
So runoff (R) in cm is,

Test: Hydrology - Question 8

Rainfall of intensity 20 mm/h occurred over a watershed of area 1 km2 for duration of six hours. It measures a direct runoff for a volume of 30,000 m3 in the stream. Find the precipitation NOT available for runoff in this case?

Detailed Solution for Test: Hydrology - Question 8

Concept:
Direct Runoff: It is the part of runoff that enters the stream immediately after the precipitation. It includes surface runoff, prompt interflow, and precipitation on the channel surface. It is sometimes termed as direct storm runoff or storm runoff.
The volume of runoff = Area of catchment × Depth of rainfall
Precipitation: 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.
Total infiltrated volume = Total rainfall – Runoff
Calculation:
Given that:
Rainfall intensity = 20 mm/hr = 0.02 m/hr
Watershed area = 1 km2
Total duration = 6 hour
Direct runoff = 30000 m3
Total rainfall = Rainfall intensity × Total duration area
Total rainfall = (0.02 × 6) × 1 × 106 = 1,20,000 m3
Precipitation not available for runoff means precipitation that is infiltrated.
Total infiltrated volume = Total rainfall – Runoff
Total infiltrated volume = 1,20,000 – 30,000 = 90,000 m3
For 1 km2 watershed area,

∴ Infiltrated depth is 9 cm.

Test: Hydrology - Question 9

The total quantity of water on the surface of earth is estimated as about

Detailed Solution for Test: Hydrology - Question 9
  • Three–fourths of the Earth's surface is occupied by water.
  • The total volume of water on Earth is estimated at 1.386 ×   109 km3= 1386 × 106 km3 ≈ 1400 × 106 km3 out of which 97.5% is salt water and 2.5% is fresh water.
  • The total mass of water both underground and on the surface of the Earth constitute the hydrosphere.
  • Around 61% of water is present in the Northern Hemisphere of Earth, while the Southern Hemisphere has around 81% water.
  • It includes rivers, lakes, seas and oceans which cover about 71% of the surface area of Earth.
  • The cryosphere is the frozen water part of the Earth water system.
  • Water resources available on earth are :
    • Freshwater: Glaciers, ice caps, and snow
    • Groundwater
    • Oceans, seas, streams, rivers, lakes, ponds
    • Wetlands- lagoons, swamps, and marshes
    • Precipitation- Rain, snow, and dew 
Test: Hydrology - Question 10

A culvert is designed for a flood frequency of 100 years and a useful life of 20 years. The risk involved in the design of the culvert (in percentage up to two decimal places)

Detailed Solution for Test: Hydrology - Question 10

Hydrologic risk

Or Risk (R) = 1 - (1 - P)n
Where T = Flood return period
P = Probability of occurring flood in a given year
n = Number of years of risk calculate or useful life of the structure.
Calculation:
Given
Flood return period (T) = 100 years
Service life of the project (n) = 20 years

Risk (R) = 1 - (1 - 0.01)20
Risk (R) = 1 - (0.99)20
Risk (R) = 1 - 0.8179 = 0.1820
Risk (R) = 0.1820 × 100 = 18.20%
Risk (R) = 18.20%
The risk involved in the design of the culvert is 18.20%.

Test: Hydrology - Question 11

Which one of the following points should be kept in mind while selecting the site for a rain gauge station?

Detailed Solution for Test: Hydrology - Question 11

While selecting the site for rain gauge stations the following points should be considered:

  • The site should be on level ground and on open space. It should never be on sloping ground.
  • The site should be such that the distance between the gauge station and the objects (like a tree, building, etc) should be at least twice the height of the objects.
  • In the hilly area, where level ground is not available, the site should be so selected that the station may be well shielded from high wind.
  • The site should be easily accessible to the observer.
  • The site should be well protected from cattle by wire fencing.
Test: Hydrology - Question 12

The observed annual runoff from a basin of area 500 km2 is 150 Mm3 and the corresponding annual rainfall over the basin during the same year is 750 mm. What is the runoff coefficient?

Detailed Solution for Test: Hydrology - Question 12

∵ We know that,
Runoff coefficient = Runoff/Rainfall
Given,
Runoff = 150 Mm3
Rainfall depth = 750 mm
Area = 500 km2
Rainfall depth in terms of volume = 0.75 × 500 = 375 Mm3 (1 million-m = 106 m)
Runoff coefficient = 150/375
= 0.4

Test: Hydrology - Question 13

A canal is 80 km long and has an average surface width of 15 m. If the evaporation measured in a class A pan is 0.5 cm/day, the volume of water evaporated in a month of 30 days is (in m3)

Detailed Solution for Test: Hydrology - Question 13

The volume of water evaporated (V) = E × L × B
Where, E = Evaporation measured per day, L = Stretch of evaporation  & B = Average surface width
Calculation:
Given, L = 80 km = 8 × 104
B = 15 m
Evaporation = 0.5 cm / day
Pan co-efficient = 0.7
Total volume of evaporation (V) = 8 × 104 × 15 × 0.5 x 10-2 × 0.7 × 30 m3
∴ V = 126000 m3

Test: Hydrology - Question 14

The instrument used for measuring evaporation is:

Detailed Solution for Test: Hydrology - Question 14
  • Atmometer is a device that is used to measure the rate of water evaporation.  It is also known as the evaporimeter.
  • Rotameter is used to measure discharge but the current meter is used to measure velocity in open channels. 

Important Points
Instruments used in the measurement

Test: Hydrology - Question 15

Rainfall with an intensity of 6 mm/h is classified as:

Detailed Solution for Test: Hydrology - Question 15

Precipitation- It is the fall of water in various forms on the earth from the clouds.
The usual forms of precipitation are as follows:

Note:
Rainfall intensity is measured by Rain Gauge, Pluviometer, Ombrometer or Hyetometer.

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