Category: Irrigation Engineering Objective Questions

250+ TOP MCQs on Method of Pricing Irrigation and Answers

Irrigation Engineering Multiple Choice Questions on “Method of Pricing Irrigation”.

1. On which factor does the pricing of irrigation water is done on a flat rate?
a) Type of Crops Sown
b) Amount of Irrigation Water Used
c) For the Whole Field Sown
d) Per Hectare

Answer: d
Clarification: When the pricing of irrigation water is done on the flat rate or crop rate basis, then these rates are specified per hectare of each sown crop, cultivated by the cultivators.

2. Groundnut is a cash crop but whereas wheat is a food crop.
a) True
b) False

Answer: a
Clarification: According to the pricing of irrigation water groundnut is priced four times the wheat crop even though groundnut uses only one-third of water used by wheat. But groundnut is priced more because the reason is that groundnut has a higher earning capacity compared to wheat. So, therefore from this we can say that groundnut is a cash crop and wheat is a food crop.

3. On how many factors does the specified rates in the crop rate system of pricing depends?
a) 4
b) 3
c) 2
d) 1

Answer: c
Clarification: The specified rates in the crop rate system of pricing depending on the factors like water requirement of each crop, and earning capacity of that crop.

4. Crop rate system of pricing does not help the cultivators for economizing the use of water.
a) True
b) False

Answer: a
Clarification: This system does help the cultivators for economizing in the use of water because the cultivators will be charged at a flat rate for the area of crops grown. So, therefore the cultivators will never bother to the wastage of water and will use the water recklessly.

5. What does the crop rate system may lead to?
a) Overutilization Water
b) Careful use of Water
c) Sufficient use of Water
d) Insufficient use of Water

Answer: a
Clarification: As this method does not encourage the cultivators to economically use the water, it leads to over utilization of water which in turn leads waterlogging, salinity of irrigated lands, and wastage of precious irrigation water, which ultimately leads to scarcity of irrigation water.

6. In crop rate revenue system field assessment is done by?
a) Canal Officer
b) IAS Officer
c) Tahsildar
d) Patwari

Answer: d
Clarification: The field assessment in the crop rate revenue system, the area required by each crop grown by the cultivators is carried out and recorded by patwari at the time of sowing, cultivating, and harvesting.

7. Who has the authority to grant remissions to the cultivators?
a) IAS Officer
b) Canal Officer
c) Patwari
d) Tahsildar

Answer: b
Clarification: We cannot expect a complete yield from the complete area of the crops sown. Due to natural calamities or due to human made mistakes the yield of the crop may decrease. So, when the failure of crop occurs due to the reasons which cannot be controlled by the cultivators, the canal officers have the authority to grant remissions to the cultivators in full or part.

8. In which method of pricing measuring devices are installed?
a) Direct Revenue Collection
b) Crop Rate System
c) Volumetric Method of Pricing
d) Crop Rate Revenue System

Answer: c
Clarification: In this type of system measuring devices like water meters, modular outlets, and venturi meters at the outlets of irrigation minors because in this pricing of irrigation water is done on basis of actual consumption of water by each cultivator or by a group of cultivators like society.

9. Which type of system is not in practice in India?
a) Crop Rate Revenue System
b) Direct Revenue Collection
c) Crop Rate System
d) Volumetric Method of Pricing

Answer: d
Clarification: This type of system is not in practice in India even though it provides inherent advantages towards the economic use of irrigation water due to the high installment and maintenance cost of measuring devices.

10. Which type of system can be easily used in tube well irrigation?
a) Direct Revenue Collection
b) Crop Rate Revenue System
c) Volumetric Method of Pricing
d) Crop Rate System

Answer: c
Clarification: Volumetric method of pricing can be efficiently used in tube well irrigation because the process billing can be done on the information of a number of units of electricity utilized in running a particular tube well supplying water to a particular society of cultivators.

250+ TOP MCQs on Sediment Transport – Mechanics and Answers

Irrigation Engineering Multiple Choice Questions on “Sediment Transport – Mechanics”.

1. The soil is assumed to be coherent in the study of the mechanism of sediment transport.
a) True
b) False
Answer: b
Clarification: Coherent means that there is a cohesive force between particles such as in the case of clays. The soil is assumed to be incoherent so that each soil grain is studied individually. Most of our river beds are made up of gravels and sands in which there is no cohesion.

2. The basic mechanism behind the phenomenon of sediment transport is ______________
a) drag force opposite to the direction of the flow
b) drag force in the direction of the flow
c) free motion of the sediment particles
d) force exerted by water vertically
Answer: b
Clarification: The force exerted by water in the direction of flow is the drag force or tractive force. If this force on the bed particles exceeds the force opposing their movement then, the bed particles will start moving to lead to the sediment transport.

3. Which of the following statement is wrong?
a) Threshold condition is the one in which a few particles on the bed will just start moving
b) Knowledge of critical velocity helps in designing stable non-scouring channels
c) The critical tractive force approach helps in designing unstable channels in alluviums
d) The knowledge of threshold condition is required for the computation of sediment load
Answer: c
Clarification: Shield was the first person to analyze the experimental data on incipient motion condition using the critical tractive force approach. The assumption of the entry of clean and clear water in the channel is used to develop non-scouring highest possible flow velocity at the peak flow.

4. For the bed of the canal, the average shear stress is equal to the tractive force per unit area.
a) True
b) False
Answer: a
Clarification: The unit tractive force in channels is uniformly distributed along the wetted perimeter. For the bed of the canal, the average shear stress is equal to the tractive force per unit area. For side slopes of the canal, average shear stress is equal to 0.75 times the tractive force per unit area.

5. A wide unlined channel carrying silt-free water has a depth of 2.0 m. The maximum slope that can be given to a channel is 1 in 10,000. Calculate the maximum tractive stress permissible on the bed to prevent scouring.
a) 0.2 kg/m2
b) 1.962 kg/m2
c) 2 kg/m2
d) 1 kg/cm2
Answer: a
Clarification: The maximum tractive stress = Yw. R. S
= 9.81 x 1000 x 2.0 x 10-4
= 1.962 kg/m2

6. The water flows at a depth of 0.6 m in a wide stream having a bed slope of 1 in 2500. The critical tractive stress is 0.53 N/m2. Determine the motion of soil grains and the average shear stress.
a) 2.35 N/m2 and soil grains will be stationary
b) 2.35 N/m2 and soil grains is in motion
c) 0.235 N/m2 and soil grains will be stationary
d) 0.235 N/m2 and soil grains is in motion
Answer: b
Clarification: The average shear stress = Yw. R. S
= 9.81 x 1000 x 0.6 x 1/2500 = 2.35 N/m2.
This value is more than 0.53 N/m2, the soil grains will not be stationary and sediment transport and scouring will occur.

250+ TOP MCQs on Water Requirements of Crops – Irrigation Efficiencies and Answers

Irrigation Engineering Multiple Choice Questions on “Water Requirements of Crops – Irrigation Efficiencies”.

1. Which efficiency is also called on-farm efficiency?
a) Water conveyance efficiency
b) Water application efficiency
c) Water storage efficiency
d) Water use efficiency
Answer: b
Clarification: The water application efficiency is the ratio of water stored in the root zone to the water actually delivered to the field. It takes into consideration the water lost in the farm hence, it is also known as on-farm efficiency.

2. What is water conveyance efficiency?
a) The ratio of the quantity of water delivered to the field and quantity of water pumped into the canal
b) The ratio of water stored in the root zone and the water delivered to the field
c) The ratio of water used beneficially and the water delivered to the field
d) The ratio of water stored in the root zone and the water needed before irrigation
Answer: a
Clarification: It is the ratio of water delivered into the fields to the water entering into the channel. This efficiency accounts for the conveyance or transit losses.

3. The quantity of water beneficially used is the sum of the water used by the crops and the water used for leaching.
a) True
b) False
Answer: a
Clarification: The beneficially used water accounts for the water required for leaching and by the crop as well. This is required in calculating water use efficiency.

4. The mean depth of water is 1.5 cm and the mean deviation from the mean is 0.1 cm. Determine its distribution efficiency.
a) 15%
b) 85%
c) 66.66%
d) 93%
Answer: d
Clarification: The formula for distribution efficiency is:
Nd = (1- d/D) x 100; where d = mean deviation and D = mean depth of water
Given, d=0.1 cm and D = 1.5 cm
Hence, Nd = 93%.

5. The presence of excess salts in the soil requires _____________
a) high water storage efficiency
b) low water storage efficiency
c) high water application efficiency
d) high water use efficiency
Answer: a
Clarification: Leaching is the supply of additional water to wash away the salts in a saline prone area. The presence of excess salts in water requires high water storage efficiency in order to keep the salts washed out of the soil.

6. Which of the following statement is wrong about consumptive use efficiency?
a) It is the ratio of normal consumptive use of water to the net amount of water depleted from the root zone
b) It accounts for the loss of water by deep percolation
c) Evaporation losses are considered
d) The losses due to percolation and evaporation are not considered
Answer: d
Clarification: It is the ratio of normal consumptive use of water to the net amount of water depleted from the root zone. It accounts for the loss of water by deep percolation and evaporation following irrigation.

7. The quantity of water that is pumped into a farm distribution system is 2 cumec. 1.6 cumec is delivered to a turn-out, 1km from the well. Compute the conveyance efficiency.
a) 80%
b) 20%
c) 50%
d) 60%
Answer: a
Clarification: By definition, Conveyance efficiency Nc = (output/input) x 100
Given, Output = 1.6 cumec and Input = 2 cumec
So, Nc = 80%.

8. If the water losses are more, the output and the efficiency is more.
a) True
b) False
Answer: b
Clarification: Efficiency is inversely proportional to the losses. If losses are more, the output is less and therefore, efficiency is less.

9. Determine the distribution efficiency, if the depths of water in the field are 1.1 cm and 1.8 cm.
a) 80%
b) 90%
c) 85%
d) 75%
Answer: d
Clarification: Mean depth (D) = (1.1 + 1.80) / 2 = 1.45
Deviation from mean depth (d) = 1.80 – 1.45 = 1.1 – 1.45 = 0.35 (neglecting negative sign)
Nd = (1 – d/D) x 100 = 75%.

10. Uniformity coefficient is also known as ____________
a) water conveyance efficiency
b) consumptive use efficiency
c) water distribution efficiency
d) water storage efficiency
Answer: c
Clarification: Water distribution efficiency is also known as the uniformity coefficient. It is the measure of how uniformly water is applied to the area being irrigated. The effectiveness of irrigation is measured by its uniformity coefficient.

11. The water distribution efficiency is 1.0 if _____________
a) the deviation from the mean depth is 1
b) the deviation from the mean depth is 0
c) the deviation from the mean depth is less than 1
d) the deviation from the mean depth is greater than 1
Answer: b
Clarification: It represents the extent to which the water has penetrated to a uniform depth throughout the area. If the water penetrates uniformly throughout the field, the deviation from the mean depth is zero and hence, water distribution efficiency is 1.0.

12. What is the correct formula for uniformity coefficient efficiency?
a) Nd = 1 – d/D
b) Nd = 1 + d/D
c) Nd = 1 – D/d
d) Nd = 1 + D/d
Answer: a
Clarification: The correct representation of the formula is Nd = 1-d/D.
Where Nd = water distribution efficiency, d = deviation from the mean depth and D = mean depth.

250+ TOP MCQs on Tank or Surplus Escape Weirs Design and Answers

Irrigation Engineering Multiple Choice Questions on “Tank or Surplus Escape Weirs Design”.

1. The usual difference between F.T.L and M.T.L is rarely allowed to exceed _____________
a) 0.3 m
b) 0.6 m
c) 0.9 m
d) 0.5 m
Answer: c
Clarification: This difference is fixed on a compromise basis in order to obtain the overall economy and efficiency. It is kept from 0.3 to 0.6 m for small and medium-sized tanks and is rarely allowed to exceed 0.9 m.

2. Which of the following type of tank weirs are provided with a number of vertical steps instead of horizontal or sloping downstream apron?
a) Type A: Masonry weir with a vertical drop
b) Type B: Rock-fill weirs with a sloping apron
c) Type C: Masonry weirs with a glacis
d) Type D: Combination of Type A and Type C
Answer: d
Clarification: Weirs of Type-D are called as weirs with stepped aprons as they are provided with a number of vertical steps as in case of a stepped fall instead of providing a horizontal or sloping downstream apron. Type-A and Type-D weirs are the most widely adopted.

3. The width of the horizontal floors of type A and D weirs from the foot of the drop wall to the d/s edge of the floor should never be less than ________________
a) 5 (D + H)
b) 3 (D + H)
c) 2.5 (D + H)
d) 2 (D + H)
Answer: d
Clarification: The width of the horizontal floor of Type-A and Type-D weirs should not be less than 2(D + H) and in important works, the width can be increased to 3(D + H). where D is the height of drop wall and H is the maximum head of water over the wall. The rough stone apron forming a talus below the last curtain wall generally vary from 2.5(D+ H) to 5(D + H) depending upon the nature of the soil and the velocity and annual probable quantity and intensity of run-off.

4. Which among the following is a correct representation of Ryve’s formula?
a) Qp = C1. A2/3
b) Qp = C1. A1/3
c) Qp = C1. L A2/3
d) Qp = C1. A
Answer: a
Clarification: All tank weirs generally in South India are designed on the basis of Ryve’s formula which is given by:
Qp = C1.A2/3 where Qp is the peak flood discharge, C1 is the Ryve’s coefficient, A is the catchment area
The formula is directly applicable for free catchments in all isolated tanks.

5. The capacity of an irrigation tank is sometimes increased by installing a temporary stone wall over the top of the tank weir. This fixture is known as____________________
a) saddle
b) dam stone
c) breast wall
d) divide wall
Answer: b
Clarification: A common practice has been resorted to in South India is to fix dam stones at the crest wall of the weir. The dam stones cause obstruction to the discharge and if no dam stones or shutters are provided above F.T.L, the flood water will start spilling over the surplus work immediately after the tank water level exceeds F.T.L.

6. The discharge through the sluice of a small irrigation tank is usually controlled by _______________
a) dam stone
b) plug
c) special balanced valves
d) shutters
Answer: b
Clarification: Flat shutters working in grooves and regulated by screw spears are used in the case of large tanks where the quantity of water to be released is great. Special balanced valves or shutters moving on rollers are generally installed for heads over 9 m or so. A Dam stone causes obstruction to flow and is used sometimes to increase the capacity.

7. What will the discharging capacity of a masonry waste weir of 55 m length and 1.5 m width provided in an earthen bund for passing a flood with 1.2 m water depth over its crest?
a) 120 cumecs
b) 254 cumecs
c) 59.4 cumecs
d) 65.8 cumecs
Answer: a
Clarification: The discharge over the weir without any velocity of approach is given by –
Q = C.L H3/2 where C = 1.66 when the width of rectangular crest > 0.9 m, L is the length of the weir and H is the head of water over the weir.
Q = 1.66 x 55 x 1.23/2 = 120 cumecs.

8. Pipe sluices are generally not adopted in tank bunds where the depth below F.T.L exceeds ____________
a) 1.5 m
b) 1 m
c) 2.5 m
d) 5 m
Answer: c
Clarification: Pipe sluices are earthenware or cement or cast iron pipes which may be used in place of masonry culverts in case of very small slices. The earthenware pipes may get fractured or leakage through their joints may take place resulting in a breach. The pipes can neither be examined nor repaired easily without cutting open the bund. They are not adopted when the depth below F.T.L exceeds 2.5 m or so.

9. Calculate peak discharge for a combined catchment of a tank forming a constituent unit of a group with the following given data –

Combined catchment 			     = 26 sq. km
Intercepted catchment 			     = 20 sq. km
Ryve’s coefficient for combined catchment    = 9.0
Ryve’s coefficient intercepted 		     = 1.8

a) 66 cumecs
b) 59 cumecs
c) 50 cumecs
d) 69 cumecs
Answer: a
Clarification: Using Ryve’s equation –
Peak flood discharge Qp = C1A2/3 – c1a2/3 where C1 = 9.0, c1 = 1.8, A = 26 sq.km and a = 20 sq.km
Qp = (9 x 262/3) – (1.8 x 202/3) = 65.9 cumecs.

10. What is the approximate value of Ryve’s coefficient for combined catchment having limited areas near hills?
a) 6.5
b) 8.5
c) 10.2
d) Up to 40
Answer: c
Clarification: For areas within 80 km from the east coast, the value of constant is taken as 6.5 and for areas within 80 to 160 km from the east coast; the value is taken as 8.5. In the case of limited areas near hills, the value of the coefficient is 10.2 and the actual observed values are always up to 40.

250+ TOP MCQs on Ogee Profile and Answers

Irrigation Engineering Multiple Choice Questions on “Ogee Profile”.

1. The discharge passing over an ogee spillway per unit length of its apex line is proportional ___________________
a) H
b) H2
c) H1/2
d) H3/2
Answer: d
Clarification: The discharge passing over the ogee spillway is given by the formula –
Q = C. Le. H3/2 where Le is the effective length of the spillway crest, C is the coefficient of discharge and H is the total head over the crest including the velocity head.
It is clear from the discharge equation that the discharge is proportional to H3/2.

2. When the crest of an ogee spillway is designed to be in accordance with the lower nappe of a free-falling water jet over a duly ventilated sharp-crested weir then theoretically
a) The pressure on the spillway crest will be zero at design head only
b) The pressure on the spillway crest will always be zero
c) The pressure on the spillway crest will always be negative
d) The pressure on the spillway crest will be always positive
Answer: a
Clarification: In actual practice development of negative pressure takes place beneath the nappe due to the removal of air by the falling jet. This causes the danger of cavitation and induces fluctuation and pulsation effects. To control this aeration pipes may be provided along spillway face.

3. When the gated sluices are provided through the body of the dam spillway then the ogee spillway structure shall normally be of Corbel type.
a) True
b) False
Answer: b
Clarification: The extra concrete is required for thickening of the section at the d/s end which can be saved by shifting the curve of the nappe in a backward direction until it becomes tangential to the d/s face. It can be affected by providing a corbel on u/s face.

4. The velocity of approach has found to be negligible effect upon discharge if ____________
a) the ratio H/Hd is more than 1.33
b) the ratio H/Hd is less than 1.33
c) the ratio H/Hd is equal to zero
d) the ratio H/Hd is more than 1
Answer: a
Clarification: The velocity of approach has been found to have a negligible effect upon discharge when the height of the weir is more than 1.33 times the design head. When this ratio is less than 1.33 as in low spillways, the velocity is having an appreciable effect.

5. If a flood enters a dam reservoir at F.R.L, the efficiency of its ogee spillway will ______________
a) increase with the increasing head
b) reduce with the increasing head
c) remain constant with the increasing head
d) vary with the increasing head
Answer: a
Clarification: The coefficient of discharge increases with the increasing head above the full reservoir level till it becomes about 2.2 at full design head i.e. MRL. Since the spillway starts working when the water level just crosses FRL, the coefficient of discharge is about 77% at full design head. As the water level increases from F.R.L to M.R.L, the efficiency increases from 77% to 100%.

6. The downstream apron of the ogee spillway is found to have a negligible effect on the coefficient of discharge when the value of ________________
a) (Hd + d) / He is less than 1.7
b) (Hd + d) / He exceeds 1.7
c) (Hd + d) / He is equal to 1
d) (Hd + d) / He is less than 1.33
Answer: b
Clarification: The d/s apron is found to have a negligible effect on the coefficient of discharge when the value of (Hd + d) / He exceeds 1.7. where d is the tailwater depth, Hd is the design head and He is the Design head including the velocity head. There can be a decrease in the coefficient due to tailwater submergence.

7. What is the value of the pier abutment coefficient for pointed nose piers?
a) 0.1
b) 0.02
c) 0.01
d) 0.0
Answer: d
Clarification: The value of pier contraction coefficient depends upon the shape of the piers.

S.NO. Pier Shape Contraction Coefficient
1 Square nosed piers without any rounding 0.1
2 Rounded nose piers and 900 cut water nosed piers 0.01
3 Pointed nose piers 0.0

8. What is the value of abutment contraction coefficient for square abutment with headwall perpendicular to the direction of flow?
a) 0.2
b) 0.1
c) 0.02
d) 0.01
Answer: a
Clarification: For square abutment with headwall at 90° to the direction of the flow, the abutment contraction coefficient is 0.2. For rounded abutment with headwall at 90° to the direction of flow, the value is taken as 0.1.

9. What is the assumption that is taken during the entire design of an overflow spillway?
a) The upper and lower nappe is subjected to negative pressures
b) The upper and lower nappe is subjected to full atmospheric pressure
c) The upper and lower nappe is subjected to zero atmospheric pressure
d) The upper and lower nappe is subjected to both positive and negative pressure
Answer: b
Clarification: In actual practice, the development of negative pressure due to insufficient aeration takes place beneath the nappe due to the removal of air by the falling jet. But the entire design has been done with the assumption that the upper and lower nappe is subjected to full atmospheric pressure.

10. The greater is the divergence from the streamline flow, greater is the contraction coefficient.
a) True
b) False
Answer: a
Clarification: The greater is the divergence, the greater is the contraction coefficient and lesser is the effective length of the crest. A 90° cut water nose pier is generally preferred as it has quite a low value of pier contraction coefficient and is most efficient.

11. Calculate the effective length of the spillway which consists of 6 spans having a clear width of 10 m each. The thickness of each pier may be taken as 2.5 m and the total design head on the crest including velocity head is 16.3 m. Assuming 90° cut water nose piers and rounded abutments.
a) 55 m
b) 60 m
c) 65 m
d) 75 m
Answer: a
Clarification: The effective length of the spillway is given by-
Le = L – 2[N.Kp + Ka].H
Where L = clear length of the spillway crest = 6 x 10 = 60, N = Number of piers = 5 and H = 16.3 m
For 90° cut water nose piers and rounded abutments; Ka = 0.1 and Kp = 0.01
Le = 60 – 2[5 x 0.01 + 0.1] x 16.3 = 55.1 m.

250+ TOP MCQs on Reservoir Capacity Determination Using Hydrograph and Mass-curve – 2 and Answers

Irrigation Engineering Questions and Answers for Experienced people on “Reservoir Capacity Determination Using Hydrograph and Mass-curve – 2”.

1. The design annual rainfall for the catchment of a proposed reservoir has been computed to be 99 cm. The catchment area contributing to the proposed reservoir is 1000 sq.km having a mean annual temperature of 20°C. Calculate the annual design catchment yield for the reservoir using Khosla’s formula.
a) 89.4 M.m3
b) 8.94 M.m3
c) 894 m3
d) 894 M.m3
Answer: d
Clarification: Using Khosla’s formula –
Q = P – 0.48 Tm where, Rainfall (P) = 99 cm and Mean annual temperature (Tm) = 20°C
Q = 99 – 0.48 x 20 = 89.4 cm = 0.894 m
The total yield produced from the given catchment = 0.894 x 1000 x 106 = 894 M.m3.

2. The reservoir capacity cannot exceed the catchment yield.
a) True
b) False
Answer: a
Clarification: The reservoir capacity is fixed at a value which is lesser of the value of –
i. The accessed gross storage required to meet the demand
ii. The accessed dependable yield for the reservoir site.

3. With the reduction in reservoir capacity over the passage of time, the trap efficiency ___________
a) increases
b) decreases
c) remains unaffected
d) may increase or decrease depending upon the reservoir characteristics
Answer: b
Clarification: If the reservoir capacity reduces with constant inflow value, the trap efficiency reduces. Hence, for small reservoirs having small capacity on large rivers having large inflow rates, the trap efficiency is extremely low.

4. Trap efficiency of a storage reservoir is defined as the ratio of ___________________
a) total annual sediment inflow to the reservoir capacity
b) total sediment deposited in a given period to the total sediment inflow in that period
c) total annual sediment deposited in the reservoir to the dead storage capacity of the reservoir
d) reservoir capacity to the total annual sediment
Answer: b
Clarification: Trap efficiency of a storage reservoir is defined as the ratio of the total sediment deposited in a given period to the total sediment inflow in that period. It can be defined as the percentage of sediment deposited in the reservoir in spite of taking precautionary control measures.

5. Capacity inflow ratio for a storage reservoir is defined as the ratio of _______________________
a) reservoir capacity to the average annual flood inflow
b) reservoir capacity to the average annual sediment inflow
c) the dead storage capacity of the reservoir to the average annual sediment deposited
d) total annual sediment inflow to the reservoir capacity
Answer: a
Clarification: Capacity-inflow ratio can be defined as the ratio of the reservoir capacity to the total inflow of water. The trap efficient is a function of the capacity-inflow ratio which is represented by a graph between them.

6. The capacity-inflow ratio for a reservoir _________________
a) is a constant factor overtime
b) increases with time
c) decreases with time
d) may increase or decrease with time
Answer: c
Clarification: The silting rate in the reservoir will be more in the beginning as its capacity reduces due to silting, the silting rate will also reduce. The trap efficiency is a function of capacity. The capacity reduces when trap efficiency reduces and lesser sediment is trapped.

7. A sequent peak algorithm is a plot between __________________
a) Accumulated flow v/s time
b) Discharge v/s time
c) (Cumulative Inflow – Cumulative Outflow) v/s time
d) Mass outflow v/s time
Answer: d
Clarification: It is a plot between time on X-axis and cumulative net flow on Y-axis. It is an excellent alternative to the mass-curve method of determining reservoir capacity. The positive value of cumulative net flow indicates a surplus of inflow and a negative value indicates a deficit of inflow.

8. A flow duration curve is a curve plotted between ________________
a) Accumulated flow v/s time
b) Discharge v/s time
c) (Cumulative Inflow – Cumulative Outflow) v/s time
d) Streamflow v/s Percent of time the flow is equaled or exceeded
Answer: d
Clarification: A flow-duration curve is a curve plotted between the streamflow and the percent of the time the flow is equaled or exceeded. It is also called a discharge-frequency curve and it represents the cumulative frequency distribution.

9. A steep slope of the flow duration curve indicates a stream with _________________
a) highly variable discharge
b) small variability of flow
c) considerable base flow
d) large flood plains
Answer: a
Clarification: A steep slope represents a stream with highly variable discharge and a flat slope represents a small variability of flow. The considerable base flow is indicated by a flat portion on the lower end of the curve and the upper end of the curve is of river basins having large flood plains.

10. The lowest portion of the capacity-elevation curve of a proposed irrigation reservoir draining 20 km2 of catchment is represented by the following data:
i. The rate of silting for the catchment = 300 m3 / km2 / year
ii. Life of the reservoir = 50 years
iii. Dead storage = 30
iv. The FSD of the canal at the head = 80 cm
v. The crop water requirement = 250 ha.m
vi. Dependable yield of the catchment = 0.29 m
Calculate the gross capacity of the reservoir.
a) 287.5 ha.m
b) 317.5 ha.m
c) 580 ha.m
d) 37.5 ha.m
Answer: b
Clarification: Net water demand = 250 ha.m and Reservoir losses = 15% x 250 = 37.5 ha.m
Live storage to meet the given demand = 250 + 37.5 = 287.5 ha.m
Gross storage required to meet the demand = live storage + dead storage = 287.5 + 30 = 317.5 ha.m
Dependable yield = 0.29 x 20 x 106 = 580 ha.m
The gross capacity is fixed at the lesser value of the gross storage and the dependable yield. Hence, the reservoir capacity = 317.5 ha.m.

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