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Test: Hydrology - Civil Engineering (CE) MCQ


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25 Questions MCQ Test GATE Civil Engineering (CE) 2025 Mock Test Series - Test: Hydrology

Test: Hydrology for Civil Engineering (CE) 2024 is part of GATE Civil Engineering (CE) 2025 Mock Test Series 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

The return period for the annual maximum flood of a given magnitude is 8 years. The probability that this flood
magnitude will be exceeded at least once during the next 5 years is :

Detailed Solution for Test: Hydrology - Question 1

 

 

Test: Hydrology - Question 2

The following steps are involved in arriving at a unit hydrograph :

(1) Estimating the surface runoff in depth

(2) Estimating the surface runoff in volume

(3) Separation of base flow 

(4) Dividing the surface runoff ordinates by the depth of runoff. The correct sequence of the steps is :

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Test: Hydrology - Question 3

The flood hydrograph due to rainfall of intensity 1 cm/h is given by
Q = 2 - (1 + t) e-2t
(t is in hours and Q is in m3/s).

Find  the area of catchment ?

Detailed Solution for Test: Hydrology - Question 3

*Answer can only contain numeric values
Test: Hydrology - Question 4

The number of revolution of a current meter is 50 seconds were found to be 12 and 30 corresponding to the velocities of 0.25 m/s and 0.46 m/s respectively. What velocity (in m/s) would be indicated by 50 revolutions of that current meter in 1 minute ?

(Important - Enter only the numerical value in the answer)


Detailed Solution for Test: Hydrology - Question 4

Test: Hydrology - Question 5

The probability that a 10 year return period flood will occur atleast once in the next 4 year is :

Detailed Solution for Test: Hydrology - Question 5

Test: Hydrology - Question 6

If the base period of a 6 hr unit hydrograph (UH) of a basin is 84 hours, then a 12 hr unit hydrograph derived from this 6 hr unit hydrograph will have a base period of :

Detailed Solution for Test: Hydrology - Question 6

To obtain 12 hr UH, we lag 6hr UH by another 6 hours hence, base period of 12 hr UH is 84 + 6 = 90 hr.

Test: Hydrology - Question 7

A 6 hr storm has 6 cm of rainfall and the resulting runoff of 3 cm. If φ index remains at the same value,
which one of the following is the runoff due to 12 cm of rainfall in 9 hr in the catchment ?

Detailed Solution for Test: Hydrology - Question 7

Test: Hydrology - Question 8

Linked Questions 09 and 10

The drainage area of a watershed is 50 Km2. The φ-index is 0.5 cm/hr and the base flow at the outlet is 10m3/s. One hour unit hydrograph of a watershed is triangular in slope with the time base of 20 hr. The 
peak ordinate occurs at 10 hours. 9.

Q. The peak ordinate (in m3/s) of the unit hydrograph is :

Detailed Solution for Test: Hydrology - Question 8

Test: Hydrology - Question 9

Linked Questions 09 and 10
The drainage area of a watershed is 50 Km2. The φ-index is 0.5 cm/hr and the base flow at the outlet is 10m3/s. One hour unit hydrograph of a watershed is triangular in slope with the time base of 20 hr. The peak ordinate occurs at 10 hours. 

Q. For the storm of depth 6.5 cm and duration 1 hr, the peak ordinate (in m3/s) of the flood hydrograph is :

Detailed Solution for Test: Hydrology - Question 9

Test: Hydrology - Question 10

A 3 hour storm over a water shed had an average depth of 27mm. The resulting flood hydrograph was found to have a peak flow of 200 m3/s and a base flow of 20 m3/s. If the loss rate could be estimated as 0.3 cm/h, a 3-h unit hydrograph for this water shed will

Detailed Solution for Test: Hydrology - Question 10

Test: Hydrology - Question 11

The 3 hr unit hydrograph U1 of a catchment of area 250 km2 is in the form of a triangle with peak discharge of 40 m3/s. Another 3 hour unit hydrograph U2 is also  triangular in shape and has the same base width as U1 but with a peak flow of 80m3/s. The catchment which U2 refers to has an area of :

Detailed Solution for Test: Hydrology - Question 11

Test: Hydrology - Question 12

The storm hydrograph was due to 3h of effective rainfall. It contained 6cm of direct runoff. The ordinates of DRH of this storm :

Test: Hydrology - Question 13

The basic assumptions of the unit hydrograph theory are :

Test: Hydrology - Question 14

Depth-Area-Duration curves of precipitations are drawn as :

Test: Hydrology - Question 15

The return period that a designs must use in the estimation of a flood for a hydraulic structure, if he is willing to accept 20% risk that the flood of that or higher magnitude will occur in the next 10 year, is

Detailed Solution for Test: Hydrology - Question 15

Test: Hydrology - Question 16

The standard project flood will generally be :

Test: Hydrology - Question 17

In a Linear reservoir, the

*Answer can only contain numeric values
Test: Hydrology - Question 18

A reservoir with surface area of 250 hectares has saturation vapour pressure at water surface = 17.54 mm of Hg and actual vapour pressure of air = 7.02mm of Hg. Wind velocity at 1m above the ground surface = 16 km/hr. Estimate the average daily evaporation from the lake using Meysis formula……….mm/days.
(Take KM = 0.36)


Detailed Solution for Test: Hydrology - Question 18

Test: Hydrology - Question 19

How is the average velocity along the vertical in a wide stream not obtained?

Detailed Solution for Test: Hydrology - Question 19

To determine the average velocity of a wide stream along the vertical, it is necessary to measure the velocity at multiple points along the vertical and then calculate the average of those measurements. There are a few different ways to do this, but some options could include:

  1. By measuring the velocity at multiple points along the vertical and averaging those measurements.
  2. By measuring the velocity at a fixed depth below the surface, such as 0.6 depth as suggested in option 2.
  3. By measuring the velocity at multiple depths below the surface, such as 0.2 and 0.8 depth as suggested in option 1, and averaging those measurements.
  4. By measuring the velocity at a depth that is a fraction of the total depth, such as 0.1 times the depth below the surface as suggested in option 4.

It is important to note that the choice of depth or depths at which to measure the velocity will depend on the specific goals of the measurement and the characteristics of the stream.

Test: Hydrology - Question 20

The intensity of rainfall time interval of a typical storm are :

The maximum intensity of rainfall for 20 minute duration of storm is:

Test: Hydrology - Question 21

The coefficient of variation of the rainfall for six rain gauge stations in a catchment was found to be 29.54% The optimum number of stations in the catchment for an admissible 10% error in the estimation of mean rainfall will be:

Detailed Solution for Test: Hydrology - Question 21

Test: Hydrology - Question 22

A 4hr storm had 4 cm of rainfall and the resulting direct runoff was 2 cm. If the φ-index remain at the same value, the runoff due to 10 cm of rainfall in 8 hr in the catchment is ?

Detailed Solution for Test: Hydrology - Question 22

Test: Hydrology - Question 23

Consider the following chemical emulsions :

1. Methyl Alcohol

2. Cetyl Alcohol

3. Stearye Alcohol

4. Kerosene

Which of the above chemical emulsions is/are used to minimize the loss of water through the process of evaporation ?

Test: Hydrology - Question 24

Which of the following defines Aridity Index (AI) ?

*Answer can only contain numeric values
Test: Hydrology - Question 25

The average surface area of a reservoir in the month of June is 20 km2. In the same month, the average rate of inflow is 10 m3/s, outflow rate is 15 m3/s, monthly rainfall is 10 cm, monthly seepage loss is 1.8 cm and the storage change is 16 million m3. The evaporation (in cm) in that month is:


Detailed Solution for Test: Hydrology - Question 25

Concept:

Water budget method:

By applying mass balance around water body and by writing water budget equation, evaporation is estimated.

Water budget equation around water body is

Σinflow - Σoutflow = ± ΔS

Σinflow includes rainfall and runoff

Σoutflow includes water used, evaporation and seepage

ΔS = change in storage 

ΔS is positive when water level rises.

ΔS is negative when water level falls.    

Calculation:

Let ‘x’ cm evaporation takes place in month of June.

Total inflow in cm

=10×30×24×60×6020×106×100+10

= 139.6 cm

Total outflow in cm

=15×30×24×60×6020×106×100+1.8+x

= (196.2 + x) cm

As total outflow is more than total inflow, therefore depression in storage takes place.

Depression in storage (ΔS)

ΔS=−16×10620×106×100=−80cm

Σinflow - Σoutflow = ± ΔS

139.6 - (196.2 + x) = - 80

– x = - 80 + 56.6

∴ Evaporation in month of June = x = 23.4 cm

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