CWT–04
Booklet No.:
Booklet Series:
03092014
Hydrology & Irrigation Engg.
(Civil Engineering)
A
Student Name:
Roll Number:
Duration: 90 Minutes
PAPER
MAXIMUM MARKS: 60
INSTRUCTIONS
1.
IMMEDIATELY AFTER THE COMMENCEMENT OF THE EXAMINATION, YOU SHOULD CHECK THAT THIS TEST BOOKLET
DOES NOT HAVE ANY UNPRINTED OR TORN OR MISSING PAGES OR ITEMS ETC. IF SO, GET IT REPLACED BY A
COMPLETE TEST BOOKLET.
2.
This Test Booklet contains 30 questions. Each question comprises four responses (answers). You will select the
response which you want to mark on the Answer Sheet. In case you feel that there is more than one correct
response, mark the response which you consider the best. In any case, choose ONLY ONE response for each item.
3.
You have to mark all your response ONLY on the separate Answer Sheet provided.
4.
All items carry equal marks.
5.
Before you proceed to mark in the Answer Sheet the response to various items in the Test Booklet, you have to fill
in some particulars in the Answer Sheet as per instructions.
6.
Each questions 2 marks and 2/3 negative mark is assigned for the wrong answer.
QH ENGINEERS ZONE EDUCATION PVT. LTD.
65/C, Prateek Mar ket, Ne ar Canara Bank, Munirka, New Delhi -110067,
Ph(011) -26194869, Cell: 9873000903, 9873664427 , 8860182273:
E-mail: qhengineer zone@gmail.com ,website: www.qhengineerszone.com
1
QH Engineers Zone, 65/C, Near Prateek Market, Canara Bank, Munirka, New Delhi-110067,
P h ( 0 1 1 ) - 2 6 1 9 4 8 6 9 , C e l l : 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7 : E - ma i l : q h e n g i n e e r z o n e @ g ma i l . c o m
Hydrology & Irrigation Engg.
Hydrology Engineering
(1.)
Rainfall of intensity of 15 mm/h occurred over a area of 100 ha for 5 hours.
Measured direct runoff volume was found to be 15000 m3. The precipitation not
available for runoff in this case is:
(a.) 6 cm
(b.) 8 cm
(c.) 27.5 cm
(d.) 50 cm
Ans: a
Exp: Total volume of rainfall
 15  103  5  100  10000 m3
 75000 m3
 pp  n not available for runoff is
 75000  15000  60000 m3
60000
 0.06 m
10000  100

(2.)
6 cm
The double mass curve technique is adopted to:
(a.) Check the consistency of rain gauge records
(b.) To find the average rainfall over a number of years
(c.) To find the number of rain gauges required
(d.) To estimate the missing rainfall data
Ans: a
(3.)
The coefficient of variation of the rainfall for six rain gauge stations in
catchments was found to be 29.54%. The optimum number of stations in the
catchments for an admissible 10% error in the estimation of the mean rainfall
will be:
(a.) 3
(b.) 6
(c.) 9
(d.) 12
Ans: c
Exp: No. of optimum station
2
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Hydrology & Irrigation Engg.
 CV 
N 

 E 
2
4
 29.54 

  8.726
 10 
(4.)
9 stations
The shape of the recession limb of a hydrograph depends on:
(a.) Basin as well as storm characteristics
(b.) Storm characteristics only
(c.) Basin characteristics only
(d.) Base flow only
Ans: c
(5.)
Evaporation aspiration is confined
(a.) To day light hours
(b.) To night time only
(c.) Land surfaces only
(d.) None of these
Ans: d
(6.)
The rainfall on five successive days on a catchment was 2, 8, 9, 2 and 3 cm. If 
-index for the storm can be assumed to be 1.5 cm/day, the total direct runoff
from the catchment is
(a.) 26 cm
(b.) 12 cm
(c.) 13 cm
(d.) 14 cm
Ans: d
Exp: Direct runoff = (0.8 – 1.5) + (9 – 1.5) = 14 cm
(7.)
Interception losses includes:
(a.) Only evaporation loss
(b.) Evaporation, through flow & stem flow
(c.) Evaporation & transpiration
(d.) Only steam flow
Ans: c
(8.)
If the wind velocity at a height of 3 m above ground is 6 kmph, its value at a
height of 10 m above ground is:
3
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Hydrology & Irrigation Engg.
(a.) 8.5 Km/hr
(b.) 9.5 Km/hr
(c.) 7.1 Km/hr
(d.) 12 Km/hr
Ans: c
Exp: un  Ch 1/7
(9.)

6  C 3

C  5.129

u10  5.129  10 
1/7
1/7
 7.127 km/h
A colvert is designed for a flood magnitude of return period 120 years & has
expected life of 25 years. The risk in this hydrologic design is
(a.) 1  0.0125
(b.) 0.1  0.9925
(c.) 1  0.09120
(d.) 1  0.1025
Ans: b
Exp: We know that risk
n
1

R  1  1  

T
1 

 1  1 

 120 
25
 1  0.9925
(10.)
The DRH due to isolated storm is as shown in the figure, if the catchment area
is 1500 km2, the effective rainfall of the storm is
(a.) 2.5  104 cm
4
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Hydrology & Irrigation Engg.
(b.) 1.6  104 cm
(c.) 2.6  104 cm
(d.) 1.5  104 cm
Ans: d
Exp: Total runoff = Area of figure
 20  20 
1
1
 80  50   10  20
2
2
 2300 m3  R
Area of catchment
A  1500 km2
 Eff. Rainfall 

R
A
2300
1500  106
 1.533  104 cm
(11.)
In a linear reservoir, the
(a.) Volume varies linearly with elevation
(b.) Outflow rate varies linearly with storage
(c.) Storage varies linearly with time
(d.) Storage varies linearly with inflow rate
Ans: b
Exp: For linear reservoir the equation is S = KQ. Where S = Storage, Q = out
flow rate.
(12.)
If the base period of a 6 hour hydrograph of a basin is 84 hours, then a 12
hours unit hydrograph derived from this 6 hour with hydrograph will have a
base period of
(a.) 75 hours
(b.) 78 hours
(c.) 84 hours
(d.) 90 hours
Ans: d
Exp: 84 + 6 = 90 hours.
5
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Hydrology & Irrigation Engg.
(13.)
A DRH due to a storm over a basin has a time base of 80 hours with straight
line portions of the hydrograph with flow rates of 0, 20, 70, 90, 40 & 0 m3/s at
elapsed durations of 0, 10, 30, 40, 50 & 80 hours as indicated on the diagram
given below, respectively. The catchment area is 250 km2. What is the rainfall
excess in this storm?
(a.) 2.91 cm
(b.) 3.91 cm
(c.) 3.51 cm
(d.) 3.67 cm
Ans: (d)
Exp: Area under DRH

1
 20  70 
 70  90 
 10  20  
  20  

2
2
2




1
 90  40 
10  
  10   30  40
2
2


 2550
m3
 hr  9.18  106 m3
s
 Rainfall excess 
9.18  106 m3
250  106
= 0.037m = 3.672 cm.
(14.)
An instantaneous unit hydrograph is a hydrograph of(a.) Unit duration & infinitely small rainfall excess
(b.) Infinitely small duration & of unit rainfall excess
(c.) Infinitely small duration & of unit rainfall excess of an infinitely small area
(d.) Unit rainfall excess on infinitely small area
Ans: b
(15.)
The Muskingum channel routing equation is written for the outflow from the
reach Q in terms of the inflow I & coefficients C0, C1 & C2 as:
(a.) Q2  C0 I 0  C1Q1  C2 I 2
(b.) Q2  C0 I 2  C1I1  C2Q2
6
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Hydrology & Irrigation Engg.
(c.) Q2  C0 I 0  C1I1  C2 I 2
(d.) Q2  C0Q0  C1Q1  C2 I 2
Ans: b
Irrigation Engineering
(16.)
Given that the base period is 100 days & the duty of canal is 1000 hectares per
cumecs, the depth of water will be
(a.) 0.864 cm
(b.) 8.64 cm
(c.) 86.4 cm
(d.) 864 cm
Ans: c
Exp:   D  864  B
D  100 ha/cumec
B  100 days

(17.)

864  100
 86.4 cm
1000
For a culturable command area of 1000 hectare with intensity if irrigation of
50%, the duty on field for a certain crop is 2000 hectare/cumec. What is the
discharge required at head of water course with 25% losses of water?
(a.) 3/16 cumec
(b.) 1/4 cumec
(c.) 1/3 cumec
(d.) 1/2 cumec
Ans: c
Exp: The proposed area to be irrigated
= CCA × Intensity of irrigation
= 1000 × 0.5 = 500 hectare
The discharge needed 
500
= 0.25 cumec
2000
If 25% water is lost the discharge required

7
0.25
 100
75
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Hydrology & Irrigation Engg.

(18.)
1
cumec
3
Assertion (A): In the boarder strip method of irrigation, the size of the strip
depends on soil characteristics, slope of the land & discharge.
Reason (R): Border strip method is a controlled type of subsurface irrigation
method.
(a.) Both A and R are true & R is the correct explanation of A
(b.) Both A and R are true & R is not correct explanation of A
(c.) A is true but R is false
(d.) A is false but R is true
Ans: c
(19.)
What is the SAR value for irrigation water with following characteristics:
NA – 22 meq/e, Ca – 3 meq/e, Mg – 1.5 meg/l
(a.) 14.67
(b.) 19.67
(c.) 20.67
(d.) 18.67
Ans: a
(20.)
The delta for a crop when its duty is 864 hectares/cumec. On the field, the base
period of this crop is 100 days is:
(a.) 120 cm
(b.) 102 cm
(c.) 100 cm
(d.) 104.8 cm
Ans: c
Exp:   864
(21.)
B 864  100

 100 cm
D
864
Duty:
(a.) Increases as one moves downstream from the head of the main canal
towards the head of the water courses
(b.) Decreases as one moves downstream from the head of the main canal
towards the head of the water courses
(c.) Remains constant as one moves downstream from the head of the main
canal towards the head of the water courses
8
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Hydrology & Irrigation Engg.
(d.) Will be zero at downstream at the head of water courses
Ans: d
(22.)
Assertion (A): Lining a canal is always beneficial & economical.
Reason (R): The seepage losses are greatly reduced & extra water is available
for irrigation.
(a.) Both A and R are true & R is the correct explanation of A
(b.) Both A and R are true & R is not correct explanation of A
(c.) A is true but R is false
(d.) A is false but R is true
Ans: d
(23.)
The permanent wilting point is the water content at which
(a.) The growth of plant is optimum
(b.) The plant dies
(c.) The plants growth is maximum
(d.) The plants growth is minimum
Ans: b
(24.)
The cost of construction of cross drainage works decreases if:
(a.) Canal is aligned along the slope
(b.) Canal is aligned along the watershed or ridge line
(c.) Canal is aligned along the contours
(d.) None of the above
Ans: b
(25.)
Statement-1: Kennedy stated all the channels to be in state of regime provided
they did not silt or scour.
Statement-2: Lacey differentiated between two regime conditions i.e. initial
regime & final regime.
Of the two statements above:
(a.) Both (1) and (2) are correct
(b.) Both (1) and (2) are wrong
(c.) (1) is wrong and (2) is correct
(d.) (2) is wrong and (1) is correct
Ans: a
(26.)
9
Leaching requirement is given as:
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Hydrology & Irrigation Engg.
(Where Dd = Depth of water drained out per unit area and Di = Depth of
irrigation water applied per unit of area).
(a.)
Dd
Di
(b.)
Di
Dd
(c.) Dd  Di
(d.) Dd  Di
Ans: a
(27.)
For canal head regular following statements are made:
1. It regulates the supply of water entering the canal
2. It controls the entry of silt in the canal
3. It prevents the river floods from entering the canal
Which of the above statements are correct:
(a.) 1, 2 and 3
(b.) Only 1
(c.) Only 1 and 2
(d.) Only 2 and 3
Ans: a
(28.)
Canal Fall is a:
(a.) Diversion structure
(b.) Cross-drainage structure
(c.) Regulating structure
(d.) River training structure
Ans: c
(29.)
The uplift pressure acting on the base of the dam;
(Where H’ = Hydrostatic pressure at toe, H = Hydrostatic pressure at heel)
(a.) Is constant
(b.) Is equal to hydrostatic pressure at heel & toe & joined by a straight line in
between
(c.) Varies linearly with maximum at heel & zero at toe
1


(d.) Is given by the equation   w H      w H   w  H   
3


10
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Hydrology & Irrigation Engg.
Ans: b
(30.)
In a concrete gravity dam, with a sloping upstream face, the resisting force is
provided by the:
(a.) Weight of the dam
(b.) Weight of the water supported on the upstream slope
(c.) Both (a) and (b)
(d.) None of these
Ans: c
(31.)
(a.)
(b.)
(c.)
(d.)
Ans:
11
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m