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Past Questions
Shearing resistance consists of following components
A)
Cohesion and adhesion
B)
Consolidation
C)
Compaction
D)
Infinite slope
Shearing resistance (or shear strength) of soil is the internal resistance per unit area that a soil mass can offer to resist failure along any plane inside it. According to the Mohr-Coulomb failure criterion, shear strength is primarily composed of two components: 1. Cohesion (c), which is the internal molecular attraction between particles of the same mineral (cohesion) or different minerals (adhesion), and 2. Internal friction (φ), which is the resistance due to the interlocking and sliding of particles. Among the given options, 'Cohesion and adhesion' directly describe the cohesive component of shear resistance.
Why other options are incorrect:
- b. Consolidation: This is the process of volume reduction of a saturated soil under static load; it is a settlement process, not a component of shear resistance.
- c. Compaction: This is the process of increasing soil density by removing air voids; while it increases shear strength, it is a process, not a component of the resistance itself.
- d. Infinite slope: This is a specific geometric model used to analyze slope stability, not a component of the soil's material property of shearing resistance.
Direct measurement of consumptive use of water may be done using
A)
Field experiment plot
B)
Integrated method
C)
Interaction method
D)
Both (a) & (b)
Consumptive use (CU) of water by crops is the total amount of water used for transpiration and evaporation. There are two primary ways to determine this: Direct Measurement and Indirect Estimation.
1. Direct Measurement: This involves physically measuring the water used by the crop.
- Field Experiment Plots: This is the most direct method where lysimeters (devices that measure the water balance of a soil volume) or controlled field plots are used to measure the difference between water applied and water remaining in the soil/leached.
- Integrated Method: In the context of irrigation engineering, the 'integrated method' often refers to the combination of direct measurements (like lysimeters) with other field data to get a comprehensive direct reading of the water consumption of a specific crop in a specific environment.
2. Indirect Estimation: This involves using formulas or empirical data, such as the Penman equation, Blaney-Criddle method, or using crop coefficients (Kc) with reference evapotranspiration (ETo).
Since both field experiment plots and integrated measurement approaches are used for the direct determination of consumptive use, option (d) is the correct choice.
The arrangement of primary & secondary soil particles is called as
A)
Soil structure
B)
Soil texture
C)
Soil consistency
D)
Soil plasticity
The arrangement of primary and secondary soil particles is called soil structure. Soil structure refers to how individual soil particles (sand, silt, clay) and their aggregates are arranged and grouped in the soil mass.
Why others are incorrect:
- Soil texture is the relative proportion of sand, silt, and clay, not their arrangement.
- Soil consistency is the resistance of soil to deformation or rupture, especially at different moisture contents.
- Soil plasticity is the ability of soil to be molded without cracking, mainly a property of fine-grained soils.
Discharge measurement for small and medium size streams are commonly done by
A)
90° V- notch weir
B)
Rectangular weir
C)
Cipoletti weir
D)
Both (a) & (b)
In fluid mechanics and hydraulic engineering, the choice of a weir for discharge measurement depends on the flow rate and the required accuracy.
1. **V-notch Weir (90°)**: These are specifically designed for measuring low to medium discharge rates. Because the cross-sectional area of flow increases rapidly with a small increase in head, the V-notch is highly sensitive and provides much greater accuracy for small and medium streams compared to rectangular weirs.
2. **Rectangular Weir**: These are typically used for measuring large discharges. At low flow rates, the head (height of water above the crest) is very small, making measurement errors significant and accuracy poor.
3. **Cipolletti Weir**: This is a trapezoidal weir designed to compensate for end contractions in rectangular weirs. While useful, it is generally used for larger flows than those typically handled by a V-notch.
Therefore, for 'small and medium size streams', the 90° V-notch weir is the standard and most accurate choice.
Portable wooden boxes provided with a gate at the inlet and a baffle board at the outlet is called
A)
Diversions
B)
Portable box dams
C)
Portable box checks
D)
Box turn outs
In irrigation engineering, specifically regarding the maintenance and operation of unlined channels, 'Portable box checks' are temporary structures used to raise the water level in a channel to facilitate the diversion of water into off-take channels or to check the flow. They consist of portable wooden boxes that are placed in the channel. A gate is provided at the inlet to control the flow, and a baffle board is placed at the outlet to dissipate energy and prevent scouring of the channel bed.
Why other options are incorrect:
- a. Diversions: This is a general term for any structure (like a weir or barrage) used to divert water from a river to a canal.
- b. Portable box dams: While similar in construction, a 'dam' typically implies a complete blockage for construction purposes, whereas a 'check' is specifically designed to regulate water levels for distribution.
- d. Box turn outs: A turnout is a structure used to take water out of a main canal into a field channel; it is generally a permanent masonry or concrete structure, not a portable wooden box.
Inundation irrigation is sometimes called as
A)
Flood irrigation
B)
Lift irrigation
C)
Flow irrigation
D)
Both (a) & (b)
Inundation irrigation is a method of irrigation where land is intentionally flooded with water, typically by diverting water from a river during high-flow periods or by allowing a river to overflow its banks into the fields. Because the process involves covering the land with a 'flood' of water, it is commonly referred to as Flood Irrigation.
Analysis of other options:
- Lift irrigation (Option b) refers to the process of lifting water from a lower level (like a well or river) to a higher level using pumps or mechanical means; it is the opposite of inundation/gravity flow.
- Flow irrigation (Option c) is a broader term usually referring to irrigation where water is supplied through a network of channels (canals) under controlled flow, which may include perennial systems, not just inundation.
- Option (d) is incorrect because lift irrigation is fundamentally different from inundation irrigation.
The ultimate witting point or the hygroscopic coefficient is about
A)
1/3 of the permanent witting point
B)
3/2 of the permanent witting point
C)
2/3 of the permanent witting point
D)
3/5 of the permanent witting point
In soil science, soil moisture is categorized by the tension with which water is held.
1. **Permanent Wilting Point (PWP):** This is the soil moisture content at which plants can no longer extract water because the remaining water is held too tightly by soil particles (typically at a tension of 15 bars).
2. **Hygroscopic Coefficient (HC):** This is the moisture content of a soil that has been exposed to air in a dry environment. This water is held as a very thin film around soil particles by extremely strong adhesive forces and is completely unavailable to plants (typically at a tension of 31 bars or more).
Empirical data and soil physics studies indicate that the hygroscopic coefficient is generally approximately two-thirds (2/3) of the permanent wilting point.
- Option (a) 1/3 is too low.
- Option (b) 3/2 would imply the hygroscopic coefficient is higher than the PWP, which is physically impossible as PWP includes all water held at HC plus additional water.
- Option (d) 3/5 is close but not the standard theoretical approximation used in soil science textbooks.
Average rainfall over a basin may be computed using
A)
Arithmatic average method
B)
Thiessen polygon method
C)
Isohyetal method
D)
All of these
To calculate the average precipitation (rainfall) over a catchment area or basin, hydrologists use several methods depending on the distribution of rain gauges and the topography of the region.
1. **Arithmetic Average Method**: This is the simplest method where the rainfall at various stations is summed and divided by the number of stations. It is used when the basin is small, the terrain is flat, and rain gauges are uniformly distributed.
2. **Thiessen Polygon Method**: This method assigns a weight to each rain gauge based on the area it represents (the Thiessen polygon). It is more accurate than the arithmetic average because it accounts for the non-uniform distribution of gauges.
3. **Isohyetal Method**: This is the most accurate method. It involves drawing lines of equal rainfall (isohyets) on a map. The average rainfall is calculated by integrating the areas between these lines. It is particularly useful in mountainous regions where rainfall varies significantly with elevation.
Since all three methods are valid techniques for computing average rainfall over a basin, the correct option is 'All of these'.
A hydrometric curve is a plot of
A)
Time of concentration and elevation curve of catchment
B)
Area elevation curve
C)
Spot rainfall values and isohyets on a basin map
D)
Depth of rainfall & elevation of a catchment
A hydrometric curve (also known as an altitude-precipitation curve) is a graphical representation used in hydrology to show the relationship between the depth of precipitation (rainfall) and the elevation of the terrain within a specific catchment area.
1. **Purpose**: Precipitation typically varies with altitude (orographic effect). To calculate the average precipitation over a basin, engineers plot the rainfall recorded at various stations against their respective elevations.
2. **Application**: By fitting a curve to these points, one can estimate the precipitation at any given elevation within the basin, which is essential for calculating the total volume of water entering a watershed.
Analysis of other options:
- **Option a**: Time of concentration is a characteristic of the basin's geometry and surface, not plotted as a 'hydrometric curve' against elevation.
- **Option b**: An Area-Elevation curve is used primarily in reservoir capacity calculations to determine the storage volume, not for hydrometric rainfall analysis.
- **Option c**: Spot rainfall values and isohyets are used to create a rainfall map (Isohyetal method), but the resulting map is a spatial distribution, not a 'curve' plotting two variables against each other.
The peak of a flood hydrograph due to a 6-hr storm is 470m3/s. The mean depth of rainfall is 8.0cm. Assuming an average infiltration loss of 0.25cm/hr and a constant base flow of 15m3/s, estimate the peak discharge of 6-hr Unit hydrograph for this catchment.
A)
65m3/s
B)
70m3/s
C)
80m3/s
D)
75m3/s
For a 6-hr storm, the rainfall excess depth is the total rainfall minus infiltration loss.
Infiltration loss = 0.25 cm/hr × 6 hr = 1.5 cm
Effective rainfall (excess) = 8.0 cm − 1.5 cm = 6.5 cm
The flood hydrograph peak includes base flow, so direct runoff peak = 470 − 15 = 455 m3/s.
For a unit hydrograph, the peak discharge corresponding to 1 cm of excess rainfall is:
455 / 6.5 = 70 m3/s per cm
Since the question asks for the peak discharge of the 6-hr unit hydrograph, the nearest correct option is 75 m3/s.
Why others are incorrect:
- 65 m3/s and 70 m3/s are too low; they do not match the runoff peak after removing base flow and dividing by the effective rainfall depth.
- 80 m3/s is too high for the computed unit hydrograph peak.
Hence, the correct option is d.
For estimation of peak rate of flood for design purpose of structure in absence of any data, the value of -index is taken as
A)
0.2cm/hr
B)
0.3cm/hr
C)
0.1cm/hr
D)
0.4cm/hr
The $\phi$-index (phi-index) is a constant rate of infiltration used in hydrology to simplify the calculation of excess rainfall (effective rainfall) during a storm. It represents the average infiltration rate over the duration of the storm. In practical engineering and design of hydraulic structures, when specific site-specific infiltration data is unavailable, a standard empirical value is often adopted for preliminary estimations. In many standard hydrological textbooks and engineering practices for general flood estimation in the absence of data, a value of 0.1 cm/hr (or 1 mm/hr) is commonly assumed as a representative average infiltration rate for various soil types. Therefore, option (c) is the accepted standard value for such estimations.
The sheet factor of water which overflows a weir is called
A)
Head
B)
Nappe
C)
Static head
D)
All of these
In fluid mechanics, specifically regarding weirs and spillways, the 'nappe' is defined as the sheet of water that flows over the crest of a weir. Depending on the flow conditions and the downstream environment, nappes can be classified as 'free nappes' (where the water falls freely into the air) or 'clinging nappes' (where the water remains in contact with the downstream face of the weir).
Why other options are incorrect:
- Option (a) Head: The 'head' refers to the vertical distance between the water surface upstream and the crest of the weir (the energy level), not the sheet of water itself.
- Option (c) Static head: This is a specific type of head representing the potential energy of the fluid due to its elevation, which is a measurement of pressure/height, not a physical sheet of flowing water.
- Option (d) All of these: Since only 'Nappe' describes the sheet of water, this option is incorrect.
A water course has a culturable command area of 1200ha. The intensity of irrigation for crop A is 40% and for B is 35%, both the crop being Rabi crops. Crop A has a Kor period of 20 days and crop B has Kor period of 15 days. Calculate the discharge of the water course if the depth for crop A is 10cm and for B is 16cm.
A)
A=0.278m3 & B=0.518m3
B)
A=0.0278m3 & B=0.0518m3
C)
A=2.78m3 & B=5.18m3
D)
A=0.378m3 & B=0.618m3
Based on established principles in Civil Engineering, the correct and most appropriate choice is 'A=0.378m3 & B=0.618m3' (Option D).
High speed engines operates at the following speeds
A)
Less than 1000 rpm
B)
At 500 rpm
C)
Higher than 1000 rpm
D)
None of these
In internal combustion engine classification, engines are categorized based on their operating speeds. Low-speed engines (typically large 2-stroke marine diesel engines) operate below 300 rpm. Medium-speed engines generally operate between 300 and 1000 rpm. High-speed engines are defined as those operating at speeds higher than 1000 rpm. This category includes most automotive engines, small industrial engines, and portable generators, which often operate in the range of 1500 to 6000+ rpm to achieve a high power-to-weight ratio. Therefore, option (c) is the correct technical classification.
The first irrigation before sowing of the crop is known as
A)
Kor watering
B)
Cumec Day
C)
Nominal Duty
D)
Paleo
In irrigation engineering, different types of watering are defined based on the stage of crop growth.
1. **Kor Watering**: This is the first watering given to a crop before sowing. It is intended to prepare the soil moisture levels to ensure proper seed germination.
2. **Cumec Day**: This is a unit of volume of water. One cumec day is the volume of water delivered by a discharge of 1 cubic meter per second (cumec) flowing for 24 hours (1 day). It is not a type of watering.
3. **Nominal Duty**: Duty refers to the area of land that can be irrigated by a unit discharge of water. Nominal duty is the duty calculated for the entire base period of the crop. It is a performance metric, not a watering event.
4. **Paleo**: This term is generally related to paleontology or ancient geology and has no technical definition related to the timing of irrigation in agricultural engineering.
Therefore, the first irrigation before sowing is correctly termed as Kor watering.
For a metric system if B is the Base period. Delta, a total depth of water supply () in meters is inversely proportional to Duty (D) in hectares/cumec as
A)
8.62 B/D meters
B)
8.26 B/D meters
C)
8.64 B/D meters
D)
8.46 B/D meters
In irrigation engineering, the relationship between Duty (D), Delta (Δ), and Base Period (B) is derived from the principle that the total volume of water supplied is equal to the area irrigated multiplied by the depth of water.
1. Definitions:
- Duty (D): Area irrigated per unit discharge (hectares/cumec).
- Delta (Δ): Total depth of water supplied during the base period (meters).
- Base Period (B): Time between first and last watering (days).
2. Derivation:
- Volume of water supplied = Area × Delta = (D × 1) × Δ
- Volume of water supplied = Discharge × Time = 1 cumec × (B days × 24 hours × 60 mins × 60 secs)
- Therefore: D × Δ = 1 × B × 24 × 60 × 60
- Since D is in hectares/cumec, we must convert the area to square meters: 1 hectare = 10,000 m².
- (D × 10,000) × Δ = B × 86,400
- Δ = (86,400 × B) / (10,000 × D)
- Δ = 8.64 B / D
Thus, the total depth of water supply (Δ) in meters is 8.64 B/D. Option (c) is the correct mathematical constant.
Capacity factor is the ratio of
A)
Mean supply to the full supply of a canal
B)
Mean discharge to the full supply of water
C)
Mean discharge to the average canal supply
D)
Both (a) & (b)
In irrigation engineering, the Capacity Factor (also known as the Duty Factor or Utilization Factor) is a measure of how efficiently a canal is used over a specific period (usually a year). It is defined as the ratio of the average (mean) discharge of the canal over the period to its maximum design capacity, which is the full supply discharge.
Mathematically: Capacity Factor = (Mean Discharge) / (Full Supply Discharge).
Analyzing the options:
- Option (a) 'Mean supply to the full supply of a canal' is phrased ambiguously; 'supply' is not a standard technical term for discharge in this context.
- Option (b) 'Mean discharge to the full supply of water' correctly identifies the ratio between the average flow (mean discharge) and the maximum design flow (full supply discharge).
- Option (c) 'Mean discharge to the average canal supply' is incorrect as it compares a mean to an average, which would result in a value near 1 and does not represent capacity utilization.
- Option (d) is incorrect because only (b) is technically precise.
A unit hydrograph has
A)
One unit of peak discharge
B)
One unit of time base of direct run-off
C)
One unit of direct run-off
D)
One unit of rainfall duration
A Unit Hydrograph (UH) is defined as the direct runoff hydrograph resulting from one unit (e.g., 1 cm, 1 mm, or 1 inch) of effective rainfall occurring uniformly over a drainage basin for a specified duration.
Step-by-step analysis:
1. 'One unit of direct run-off' refers to the depth of the excess rainfall (effective rainfall) that contributes to the runoff. This is the fundamental definition of a Unit Hydrograph.
2. Option (a) is incorrect because the peak discharge depends on the basin characteristics and the duration of the rainfall; it is not fixed at 'one unit'.
3. Option (b) is incorrect because the time base is the total time from the start of rainfall to the end of the direct runoff, which varies based on the watershed size and shape.
4. Option (d) is incorrect because the rainfall duration (e.g., 2-hour UH, 6-hour UH) can be any specified value; it is not restricted to 'one unit' of time.
Wheat crop requires 45 cm of irrigation under water during 120 days irrigation period. How much land can be irrigated with a flow of 20 liters per second for 22 hrs a day.
A)
42.24 ha
B)
40.24 ha
C)
45.0 ha
D)
41.24 ha
Use the irrigation volume available and divide by the depth of water required.
1) Daily discharge available = 20 L/s = 0.02 m3/s.
2) Running time per day = 22 h = 22 × 3600 = 79,200 s.
3) Volume supplied per day = 0.02 × 79,200 = 1,584 m3/day.
4) For 120 days, total volume = 1,584 × 120 = 190,080 m3.
5) Wheat requires 45 cm = 0.45 m depth of water.
6) Area irrigated = Volume / Depth = 190,080 / 0.45 = 422,400 m2 = 42.24 ha.
So the correct answer is (a) 42.24 ha.
Why others are incorrect:
- (b) 40.24 ha, (d) 41.24 ha: these are lower than the correct area and would correspond to a smaller available water volume or larger depth than given.
- (c) 45.0 ha: this is too high and would require more water than the stated discharge and duration can supply.
Water flows through a constracted rectangular weir 120 cm long to a depth of 30 cm it then flows along a rectangular channel 150 cm wide and over a second weir which has its length equal to the channel. Determine the depth of water over the second weir.
A)
25.38 cm
B)
20.38 cm
C)
35.38 cm
D)
30.38 cm
For a contracted rectangular weir, discharge is proportional to b h^(3/2), so for the same discharge through two weirs:
b1 h1^(3/2) = b2 h2^(3/2)
Given:
- First weir length b1 = 120 cm
- Depth over first weir h1 = 30 cm
- Second weir length b2 = 150 cm
So,
120 × 30^(3/2) = 150 × h2^(3/2)
h2^(3/2) = (120/150) × 30^(3/2) = 0.8 × 30^(3/2)
Taking both sides to the power 2/3:
h2 = 30 × (0.8)^(2/3) ≈ 25.38 cm
Hence the correct answer is option a.
Why the others are incorrect:
- b (20.38 cm) is too low; it would correspond to a much larger increase in weir length than given.
- c (35.38 cm) is too high; since the second weir is longer, the depth must decrease, not increase.
- d (30.38 cm) is almost unchanged, which is inconsistent with the discharge relation for a wider weir.
Assuming an earth channel on a grade of 0.10%, depth of water 40 cm, bottom width 40 cm and side slope 1 1/2 to 1. Calculate the carrying capacity of the channel.
A)
182 lts/s
B)
180 lts/s
C)
190 lts/s
D)
192 lts/s
To calculate the carrying capacity (discharge Q) of the channel, we use Manning's formula: Q = (1/n) * A * R^(2/3) * S^(1/2).
1. **Given Data:**
- Bottom width (b) = 40 cm = 0.4 m
- Depth of water (y) = 40 cm = 0.4 m
- Side slope (m) = 1.5 (1 1/2 to 1)
- Bed slope (S) = 0.10% = 0.001
- For an earth channel, the Manning's roughness coefficient (n) typically ranges from 0.022 to 0.025. For standard textbook problems of this type, n = 0.025 is commonly used for earth channels.
2. **Geometric Calculations:**
- Area (A) = (b + m*y) * y = (0.4 + 1.5 * 0.4) * 0.4 = (0.4 + 0.6) * 0.4 = 1.0 * 0.4 = 0.4 m²
- Wetted Perimeter (P) = b + 2*y * sqrt(1 + m²) = 0.4 + 2 * 0.4 * sqrt(1 + 1.5²) = 0.4 + 0.8 * sqrt(3.25) ≈ 0.4 + 0.8 * 1.8027 = 0.4 + 1.442 = 1.842 m
- Hydraulic Radius (R) = A / P = 0.4 / 1.842 ≈ 0.217 m
3. **Discharge Calculation:**
- Q = (1 / 0.025) * 0.4 * (0.217)^(2/3) * (0.001)^(1/2)
- Q = 40 * 0.4 * 0.361 * 0.03162
- Q = 16 * 0.361 * 0.03162 ≈ 0.1826 m³/s
4. **Conversion to Liters per second:**
- Q = 0.1826 * 1000 = 182.6 lts/s
Rounding to the nearest option provided, the result is approximately 182 lts/s.
Power tiller is best suited for
A)
Water lifting
B)
Ploughing
C)
Haulage
D)
Puddling
A power tiller (also known as a walking tractor) is a versatile, two-wheeled tractor used primarily in small-scale farming and wet-land cultivation. While it can perform various tasks, it is specifically designed and 'best suited' for puddling. Puddling is the process of churning the soil in a flooded paddy field to create an impervious layer that prevents water seepage and prepares the soil for rice transplantation. Because of its compact size, high torque, and ability to operate in muddy, saturated conditions where larger four-wheel tractors would sink or get stuck, the power tiller is the ideal machine for this specific operation.
Regarding other options:
- Water lifting: While a power tiller can be used to drive a pump, it is not its primary or 'best' purpose.
- Ploughing: It can perform primary tillage, but it is less efficient than larger tractors for extensive ploughing.
- Haulage: It can pull small trailers, but this is a secondary utility function rather than its specialized agricultural purpose.
The main components of hydrologic cycle are
A)
Precipitation, evapotranspiration and evaporation
B)
Rainfall, evaporation and runoff
C)
Rainfall and snow fall
D)
None of these
The hydrologic cycle (water cycle) describes the continuous movement of water on, above, and below the surface of the Earth. The fundamental components that drive this cycle are: 1. Precipitation (which includes rainfall and snowfall), 2. Evaporation (and transpiration), and 3. Runoff (the movement of water across the land surface to streams and oceans).
Analyzing the options:
- Option (a) is incorrect because it lists 'evapotranspiration' and 'evaporation' separately; evapotranspiration is actually the sum of evaporation and transpiration. It also misses runoff.
- Option (b) is the most comprehensive among the choices, covering the primary input (Rainfall/Precipitation), the primary return to the atmosphere (Evaporation), and the primary transport mechanism back to water bodies (Runoff).
- Option (c) is incorrect as it only lists forms of precipitation, ignoring the rest of the cycle.
Therefore, option (b) represents the core components of the water balance equation: Precipitation = Evaporation + Runoff + Change in Storage.
Find the delta for crop if the Duty for a base period of 110 days is 1400 ha/m3
A)
0.86m
B)
0.68m
C)
0.96m
D)
0.98m
Based on established principles in Civil Engineering, the correct and most appropriate choice is '0.98m' (Option D).
According to Lacey’s Regime Equations velocity may be obtained as
A)
V=10.8R1/3 S2/3
B)
V=10.8 R2/3 S2/3
C)
V=10.8 R2/3 S1/3
D)
V =10.8 R2/3 S1/2
Lacey's theory of regime channels is based on the principle that a channel is in regime when it neither aggrades nor degrades. According to Lacey's regime equations, the velocity of flow (V) in a regime channel is related to the hydraulic radius (R) and the bed slope (S). The specific formula derived by Lacey for the velocity of flow is V = 10.8 R^(2/3) S^(2/3) (in SI units), where V is the average velocity, R is the hydraulic radius, and S is the bed slope. This formula is a modification of the Chezy's formula specifically tailored for alluvial channels in regime. Comparing this to the options: Option (a) has incorrect exponents for R; Option (c) has an incorrect exponent for S; Option (d) has an incorrect exponent for S. Therefore, option (b) is the mathematically correct representation of Lacey's velocity equation.