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Past Questions
The useful soil moisture for plant growth is
A)
chemical water
B)
hygroscopic water
C)
capillary water
D)
gravity water
In soil physics and hydrology, soil water is classified based on how it is held within the soil matrix.
1. **Capillary Water (Correct):** This is the water held in the small pores (capillaries) of the soil by surface tension and cohesive forces. It is held tightly enough to resist the pull of gravity but loosely enough that plant roots can extract it via osmotic pressure. Therefore, it is the primary source of moisture available for plant growth.
2. **Hygroscopic Water (Incorrect):** This is a thin film of water that adheres tightly to soil particles by adsorption. It is held with such high tension that plant roots cannot extract it.
3. **Gravity Water (Incorrect):** This is the water that fills the large pores (macropores) after heavy rain or irrigation. It moves downward through the soil profile due to gravity and is quickly drained away, making it unavailable for long-term plant use.
4. **Chemical Water (Incorrect):** This refers to water that is chemically combined in the structure of minerals; it is not available for biological processes.
If the degree of the curve (specified length 30m) is 153º, the radius of the curve is approximately
A)
573m
B)
382m
C)
1910m
D)
none of the above
1. Understanding the Definition: The 'degree of a curve' (D) is defined as the angle subtended at the center of the curve by an arc of a specified length (L). In this problem, the specified length L = 30m and the degree D = 153º.
2. Formula: The relationship between the radius (R), the arc length (L), and the central angle (D) in degrees is given by:
L = (π * R * D) / 180
Rearranging for R:
R = (L * 180) / (π * D)
3. Calculation:
L = 30 m
D = 153º
R = (30 * 180) / (3.14159 * 153)
R = 5400 / 480.66
R ≈ 11.23 m
4. Evaluation of Options:
Option (a) 573m, (b) 382m, and (c) 1910m are all significantly larger than the calculated value of 11.23m.
Note: If the degree was intended to be 153 minutes (153') instead of 153 degrees, the calculation would be:
D = 153 / 60 = 2.55º
R = (30 * 180) / (3.14159 * 2.55) ≈ 675m (Still not matching options).
If the degree was 15.3º:
R = (30 * 180) / (3.14159 * 15.3) ≈ 112.3m.
Based on the provided values (D = 153º, L = 30m), the radius is approximately 11.23m, which is not listed in options a, b, or c.
On the basis of the direction of flow of water through the runner, hydraulic turbines are classified as
A)
tangential flow turbine (Pelton wheel)
B)
radial flow turbine (Francis turbine)
C)
axial flow turbine (Kaplan turbine)
D)
mixed flow turbine (modern Francis turbine)
Hydraulic turbines are classified based on the direction of water flow relative to the axis of the runner as follows:
1. Tangential Flow: Water strikes the runner buckets tangentially to the periphery. The Pelton wheel is the primary example.
2. Radial Flow: Water flows radially inward toward the center of the runner. The early Francis turbines were purely radial.
3. Axial Flow: Water flows parallel to the axis of the runner. The Kaplan turbine is the primary example.
4. Mixed Flow: Water enters radially and leaves axially (or vice versa). Modern Francis turbines are typically mixed-flow turbines.
Note: In the provided options, each option correctly pairs a flow direction with its corresponding turbine. If this were a multiple-choice question from a competitive exam where only one option must be selected, it is likely a 'faulty' question or intended as a 'Select all that apply' or 'All of the above' type. However, since the prompt asks for the classification, all four statements are technically accurate definitions in Fluid Mechanics.
Synthetic unit hydrographs of Synder based on parameters
A)
Base time T
B)
Peak discharge Qp
C)
basin lag time tp
D)
all the above
Snyder's Synthetic Unit Hydrograph (SUH) method is used to derive a unit hydrograph for ungauged basins based on physical characteristics of the watershed. The method relies on several key parameters to define the shape and timing of the hydrograph:
1. **Basin Lag Time (tp):** This is the primary parameter in Snyder's method. It is the time from the centroid of the rainfall to the peak of the hydrograph. It is typically calculated using the formula tp = Ct(L)^m, where L is the length of the basin and C and m are regional constants.
2. **Peak Discharge (Qp):** Once the lag time is determined, the peak discharge of the unit hydrograph is calculated using the formula Qp = (2.78 * Ca * A) / tp, where Ca is another regional constant and A is the drainage area.
3. **Base Time (T):** The base time (the total duration of the hydrograph) is derived from the lag time. In Snyder's method, the base time is typically calculated as T = 5 * tp + t_r (where t_r is the duration of the unit rainfall).
Since the construction of the Snyder synthetic unit hydrograph requires the determination of the lag time, the peak discharge, and the base time to plot the curve, all three parameters are essential. Therefore, option 'd' is the correct choice.
For general engineering purposes, soils are classified as
A)
particle size system
B)
textural system
C)
High Way Research Board system
D)
unified soil classification system
In geotechnical and civil engineering, soils are classified based on their engineering properties to predict their behavior under load.
1. **Unified Soil Classification System (USCS):** This is the most widely used system for general engineering purposes. It classifies soils into categories such as gravels (G), sands (S), silts (M), and clays (C) based on grain size distribution and plasticity (Atterberg limits). It is designed specifically for engineers to communicate soil types consistently across projects.
2. **Textural System (Option b):** This is primarily used in agriculture and soil science to describe the proportion of sand, silt, and clay. It does not account for plasticity or engineering behavior.
3. **High Way Research Board (HRB) / AASHTO System (Option c):** While used in engineering, this system is specifically tailored for highway construction and pavement design, rather than 'general engineering purposes'.
4. **Particle Size System (Option a):** This is a component of classification (sieve analysis), but it is not a comprehensive classification 'system' on its own, as it ignores the chemical and plastic properties of fine-grained soils.
The process of keeping the plane table into a fixed direction so that a line representing a direction on the ground is
A)
centering
B)
orientation
C)
levelling
D)
resection
In plane table surveying, 'Orientation' is the process of setting the table such that the edge of the drawing sheet is parallel to a specific line (usually the meridian or a known baseline) or such that the table is kept in a fixed direction so that lines drawn on the paper are parallel to the corresponding lines on the ground.
Here is why the other options are incorrect:
- Centering: This is the process of setting the point on the paper exactly over the corresponding point on the ground using a plumbing fork.
- Levelling: This is the process of making the table top horizontal using a spirit level.
- Resection: This is a method of locating the position of the plane table station on the map by sighting known points, rather than by measurement from a known point.
The scour depth of a river during flood may be calculated from the Lacey’s equation with usual notations, is
A)
D = 0.47(Q/f)^1/2
B)
D = 0.47(Q/f)^1/3
C)
D = 0.47(Q/f)^2/3
D)
D = 0.47(Q/f)^1/4
According to Lacey's theory of regime channels, the depth of scour (R) for a river during a flood is calculated based on the discharge and the silt factor. The standard formula for the regime depth (R) is given by R = 0.47 (Q/f)^(1/3), where Q is the discharge in cubic meters per second (m³/s) and f is the silt factor. The silt factor depends on the average particle size of the bed material.
Analyzing the options:
- Option (a) suggests a power of 1/2, which is incorrect.
- Option (b) correctly identifies the power as 1/3, matching the established Lacey's equation for scour depth.
- Option (c) suggests a power of 2/3, which is incorrect.
- Option (d) suggests a power of 1/4, which is incorrect.
The canal fall which has gradual convex and concave surfaces, is called
A)
Inglis fall
B)
ogee fall
C)
straight glacis fall
D)
stepped fall
A canal fall is a structure used to lower the bed level of a canal when the natural ground slope is steeper than the designed bed slope of the canal.
1. **Ogee Fall**: This type of fall is characterized by a profile that consists of a gradual convex curve followed by a concave curve. This S-shaped profile is designed to ensure a smooth transition of water flow, reducing turbulence and minimizing the risk of erosion (scouring) at the toe of the fall.
2. **Inglis Fall**: This is a specific type of glacis fall designed by Sir Inglis, typically featuring a straight glacis with a specific slope and a cistern to dissipate energy.
3. **Straight Glacis Fall**: As the name suggests, the sloping part (glacis) is a straight inclined plane, not a curved surface.
4. **Stepped Fall**: This consists of a series of steps (like a staircase) to dissipate energy through turbulence at each step, rather than a smooth convex/concave surface.
Therefore, the description of 'gradual convex and concave surfaces' uniquely identifies the Ogee fall.
In hydroelectric project, contour map is used for
A)
reservoir capacities
B)
reservoir area
C)
power house
D)
dam design
In the context of a hydroelectric project, a contour map represents the topography of the land. When a dam is constructed, the valley behind it fills with water to form a reservoir. To determine the 'reservoir capacity' (the volume of water the reservoir can hold at various water levels), engineers use the contour map to calculate the area of the water surface at different elevations. By applying the prismoidal formula or the trapezoidal rule to these areas across different contour intervals, the total volume (capacity) of the reservoir is calculated. While contour maps are used to determine the reservoir area (option b), the primary engineering objective in hydroelectric planning is to determine the capacity (volume) to ensure power generation stability. Dam design (option d) and power house location (option c) require detailed site surveys and geological data beyond just a general contour map.
The concept of instantaneous unit hydrograph (IUH) was introduced first by
A)
C.O.Clark
B)
J.E.Nash
C)
L.K.Sherman
D)
V.T.Chow
The Instantaneous Unit Hydrograph (IUH) is a theoretical hydrograph that represents the runoff response of a watershed to a unit volume of effective rainfall occurring instantaneously (over an infinitesimally small duration). This concept was introduced by J.E. Nash in 1958 to simplify the analysis of rainfall-runoff processes by treating the watershed as a linear system.
Analysis of other options:
- C.O. Clark: Known for the Clark Unit Hydrograph model, which uses a routing approach (translation and attenuation).
- L.K. Sherman: Introduced the original concept of the Unit Hydrograph (UH) in 1932, but for a finite duration of rainfall (e.g., 1 hour, 6 hours), not an instantaneous one.
- V.T. Chow: A renowned author and expert in applied hydrology who refined and popularized many of these theories in textbooks, but did not introduce the IUH concept.
Determination of modification of flood wave parameters with time and distance is called
A)
flood forecasting
B)
flood routing
C)
flood frequency
D)
all the above
Flood routing is the process of determining the change in the shape, magnitude, and timing of a flood wave (hydrograph) as it moves downstream through a river reach or is stored in a reservoir. It specifically analyzes how the flood wave parameters are modified with respect to time and distance due to effects like attenuation (reduction in peak flow) and lag (delay in peak arrival).
Why other options are incorrect:
- a. Flood forecasting: This is the broader practice of predicting the occurrence and magnitude of a flood before it happens, often using routing as a tool, but it is not the specific term for the modification of wave parameters.
- c. Flood frequency: This refers to the statistical analysis of the probability of a flood of a certain magnitude occurring within a specific time period (e.g., a 100-year flood).
- d. All the above: Incorrect because only flood routing specifically describes the modification of wave parameters over time and distance.
Pitot tube is an instrument to measure
A)
pressure of flow
B)
discharge of liquid
C)
velocity of flow
D)
total energy of flowing liquid
A Pitot tube is a pressure-measurement instrument used to calculate the fluid flow velocity. It works based on Bernoulli's principle. The tube is placed in the flow stream such that the fluid enters the tube perpendicularly. At the tip of the tube, the fluid is brought to a complete stop, converting all the kinetic energy of the fluid into pressure energy. This point is known as the 'stagnation point'. By measuring the difference between the stagnation pressure (total pressure) and the static pressure of the fluid, the local velocity (v) can be calculated using the formula: v = sqrt(2 * (Ps - Pp) / ρ), where Ps is stagnation pressure, Pp is static pressure, and ρ is the fluid density.
Why other options are incorrect:
- (a) Pressure of flow: While it measures pressure to find velocity, it is specifically designed to find the velocity, not just the static pressure.
- (b) Discharge of liquid: Discharge (flow rate) is the product of velocity and cross-sectional area. While a Pitot tube can be used to find discharge if the area is known, the instrument itself directly measures velocity.
- (d) Total energy: Total energy includes potential energy (elevation), which a Pitot tube does not measure.
The survey instrument named optical square is used to measure
A)
distance
B)
angles
C)
setting right angles
D)
area
An optical square is a small, handheld surveying instrument used to set out perpendicular lines (right angles) from a given baseline. It works on the principle of double reflection, utilizing two mirrors set at an angle of 45 degrees to each other. When the surveyor looks through the eyepiece, the mirror reflects a line of sight that is exactly 90 degrees to the baseline, allowing the user to align a point (like a peg) to create a right angle.
Why other options are incorrect:
- a. Distance: Distance is measured using tapes, chains, or Electronic Distance Measurement (EDM) instruments.
- b. Angles: General angles are measured using a theodolite, transit, or sextant. While a right angle is a specific angle, the optical square is specialized only for 90-degree offsets, not general angle measurement.
- d. Area: Area is a calculated value derived from linear measurements or coordinates, not measured directly by a handheld optical instrument.
Bearing capacity of a soil depends on
A)
size of particles
B)
shape of particles
C)
cohesive properties of particles
D)
internal frictional resistance of particles
E)
all the above
The bearing capacity of soil is the capacity of the soil to support the loads applied to the ground. It is governed by the shear strength of the soil, which is influenced by several physical and mechanical properties:
1. **Internal Frictional Resistance (Option d):** According to the Mohr-Coulomb failure criterion, shear strength depends on the angle of internal friction (φ). This resistance is a primary component of bearing capacity for cohesionless soils (like sand).
2. **Cohesive Properties (Option c):** For fine-grained soils (like clay), the cohesion (c) provides significant shear strength, directly increasing the bearing capacity.
3. **Shape of Particles (Option b):** The angularity of particles affects the interlocking mechanism. Angular particles generally provide higher internal friction and thus higher bearing capacity compared to rounded particles.
4. **Size of Particles (Option a):** Particle size distribution (grading) affects the packing density and the void ratio. Well-graded soils typically have higher density and higher bearing capacity than poorly graded soils.
Since all these factors contribute to the overall shear strength and stability of the soil under a foundation, the bearing capacity depends on all of them.
Low head turbine is
A)
Kaplan turbine
B)
Pelton wheel
C)
modern Francis turbine
D)
all the above
Hydraulic turbines are classified based on the head (the vertical distance the water falls) they are designed to operate under.
1. **Kaplan Turbine (Option a):** This is an axial-flow reaction turbine. It is specifically designed for low-head applications (typically below 30-40 meters) and high flow rates. It features adjustable blades to maintain high efficiency over a range of flow conditions, making it the definitive 'low head turbine'.
2. **Pelton Wheel (Option b):** This is an impulse turbine. It requires a very high head (typically above 300 meters) to accelerate water to high velocities through a nozzle to drive the buckets. It is a 'high head turbine'.
3. **Francis Turbine (Option c):** This is a mixed-flow reaction turbine. It is designed for medium head applications (typically between 30 and 300 meters). While 'modern' Francis turbines have a wide operating range, they are fundamentally categorized as medium head turbines.
Since only the Kaplan turbine is designed for low head, option (a) is the correct answer.
The most popular and widely used theory of sediment in channel and stream is
A)
Brahms and Aiery’s theory
B)
Jaffrey’s lift concept theory
C)
White’s theory
D)
Shields’ tractive force theory
In fluvial hydraulics and sediment transport, the initiation of motion of particles on a bed is a critical concept. A.M. Shields developed the 'Tractive Force Theory' (also known as the Critical Shear Stress concept), which posits that sediment particles begin to move when the dimensionless shear stress (Shields parameter) exceeds a critical value. This theory is the most widely accepted and popular foundation for calculating the threshold of motion for bed load in streams and channels.
Comparing the options:
- Option (a) Brahms and Aiery: These are not recognized theories for sediment transport in hydrology.
- Option (b) Jaffrey’s lift concept: While lift forces are considered in sediment transport, Jaffrey's work is not the primary or most popular general theory for channel sediment.
- Option (c) White’s theory: While H.C. White contributed to fluid mechanics, there is no 'White's theory' that serves as the standard for sediment transport initiation.
- Option (d) Shields' tractive force theory: This is the industry and academic standard for determining the onset of sediment motion based on the balance between the tractive force of the flow and the resisting force of the particle's weight.
If the irrigation efficiency is 80%, conveyance losses is 20% and actual depth of watering is 16 cm, the depth of water to be supplied at the outlet is
A)
25 cm
B)
20 cm
C)
30 cm
D)
35 cm
Outlet depth = field depth / irrigation efficiency =16/0.8=20cm; with 20% conveyance loss source would need 25cm.
Choose the correct statement from the following
A)
Rivers, oceans, lakes and springs get water from rain
B)
Rain water is obtained by evaporation from lakes, oceans and lakes
C)
Water remains in atmosphere as vapours and clouds
D)
Hydrologic cycle is a continuous process of evaporation and precipitation of water in atmosphere
E)
all the above
The question asks to identify the correct statement regarding the hydrologic cycle. Let's analyze each option:
1. Option (a): Rivers, oceans, lakes, and springs are primary recipients of precipitation (rain/snow), which replenishes their water levels. This is a correct statement.
2. Option (b): Rain water originates from water vapor in the atmosphere, which is produced through the evaporation of water from oceans, lakes, and other water bodies. This is a correct statement.
3. Option (c): Water exists in the atmosphere in two primary forms: as an invisible gas (water vapor) and as liquid droplets or ice crystals (clouds). This is a correct statement.
4. Option (d): The hydrologic cycle is defined as the continuous movement of water on, above, and below the surface of the Earth, driven primarily by the processes of evaporation (and transpiration) and precipitation. This is a correct statement.
Since all statements (a, b, c, and d) are scientifically accurate descriptions of the hydrological process, the correct choice is 'all the above'.
The Reynolds number Re in pipe flow is less than 2000. The flow within the pipe is
A)
laminar
B)
viscous
C)
turbulent
D)
(a) and (b)
The Reynolds number (Re) is a dimensionless quantity used in fluid mechanics to predict flow patterns. It is defined as the ratio of inertial forces to viscous forces (Re = ρvd/μ). For internal flow in a pipe, the flow regime is generally categorized as follows:
1. Laminar Flow: Re < 2000. In this regime, the fluid flows in parallel layers (laminae) with no disruption between them.
2. Transition Flow: 2000 < Re < 4000. The flow exhibits characteristics of both laminar and turbulent flow.
3. Turbulent Flow: Re > 4000. The flow is characterized by chaotic property changes and eddies.
Since the question specifies that Re is less than 2000, the flow is strictly classified as laminar. While laminar flow is dominated by viscous forces, 'viscous' is a property of the fluid or a description of the force, not a classification of the flow regime itself. Therefore, 'laminar' is the precise technical term for the flow state.
According to Kennedy’s theory, the critical velocity in metres in an irrigation canal which keeps the channel free from silting and scouring, is
A)
0.55mD0.64
B)
0.84mD0.64
C)
0.84mD0.54
D)
0.55mD0.54
Kennedy's formula Vc = 0.55*m*D^0.64 where m is critical velocity ratio and D depth.
Low specific speed turbine is
A)
Francis turbine
B)
Kaplan turbine
C)
Pelton wheel
D)
all the above
Specific speed (Ns) of a turbine is a dimensionless index used to classify turbines based on their shape and performance characteristics. It is defined as the speed of a geometrically similar turbine which would produce the same power under a unit head.
1. Pelton Wheel: This is an impulse turbine designed for high head and low discharge. Because it operates at high heads, its specific speed is very low (typically ranging from 10 to 50). Therefore, it is classified as a low specific speed turbine.
2. Francis Turbine: This is a mixed-flow reaction turbine designed for medium head and medium discharge. Its specific speed is moderate (typically ranging from 60 to 300).
3. Kaplan Turbine: This is an axial-flow reaction turbine designed for low head and high discharge. It has a high specific speed (typically ranging from 300 to 1000).
Comparing the three, the Pelton wheel has the lowest specific speed.
Hydraulic machine in which the energy of large quantity of water falling through a small height is utilized to lift small quantity of water to a greater height without any external force is called
A)
hydraulic ram
B)
hydraulic press
C)
hydraulic lift
D)
hydraulic accumulator
A hydraulic ram is a cyclic pump that uses the energy of a large volume of water falling through a low head (small height) to lift a small volume of water to a much higher head. It operates on the principle of the 'water hammer' effect. When a valve closes suddenly, the kinetic energy of the flowing water is converted into a high-pressure surge, which forces a small portion of the water into a delivery pipe to a higher elevation without the need for an external power source (like electricity or fuel).
Why other options are incorrect:
- b. Hydraulic press: A machine used to create a compressive force using Pascal's law; it requires an external input force/power.
- c. Hydraulic lift: A device used to raise heavy loads using fluid pressure; it requires an external pump or force to operate.
- d. Hydraulic accumulator: A pressure storage reservoir in which a non-compressible hydraulic fluid is held under pressure; it stores energy rather than lifting water using a fall.
Symon’s rain gauge is
A)
non-recording gauge
B)
tipping bucket gauge
C)
flat recording gauge
D)
weighing type gauge
Symons' rain gauge is the standard non-recording type rain gauge used in India. It consists of a cylindrical vessel with a funnel that collects rainwater and directs it into a bottle. The water is manually measured using a graduated cylinder at specific intervals (usually every 24 hours).
Why other options are incorrect:
- b. Tipping bucket gauge: This is a recording gauge that uses a mechanism to tip a bucket once a specific amount of rain is collected, sending an electrical pulse to a recorder.
- c. Flat recording gauge: This refers to recording gauges (like the floating type) that produce a chart (mass curve) of rainfall over time.
- d. Weighing type gauge: This is a recording gauge that weighs the collected precipitation (including snow) and records the weight change over time.
Performance of turbine under unit head is expressed in
A)
unit speed (Nu)
B)
unit power (Pu)
C)
unit discharge (Qu)
D)
all the above
In hydraulic turbine analysis, 'unit quantities' are non-dimensional or normalized parameters used to describe the performance of a turbine regardless of the actual head (H) available. These are essential for scaling and comparing turbines of different sizes.
1. Unit Speed (Nu): Defined as the speed of the turbine divided by the square root of the head (Nu = N / √H). It represents the performance of the turbine under a unit head.
2. Unit Discharge (Qu): Defined as the discharge divided by the product of the square root of the head and the diameter squared (Qu = Q / (D²√H)). It represents the flow rate under a unit head.
3. Unit Power (Pu): Defined as the power developed divided by the product of the head to the power of 3/2 and the diameter squared (Pu = P / (D²H^(3/2))). It represents the power output under a unit head.
Since all three parameters (speed, discharge, and power) are expressed as 'unit' quantities to describe turbine performance under a unit head, option 'd' is the correct choice.
Yield of well is determined by the method
A)
recuperation test
B)
constant level test
C)
(a) and (b)
D)
none of the above
The yield of a well refers to the quantity of water that can be discharged from the well per unit of time without causing an excessive drop in the water level. To determine this, several hydraulic tests are performed:
1. **Recuperation Test (Recovery Test):** This test involves pumping the well to a certain drawdown and then stopping the pump to observe the rate at which the water level recovers (recuperates) to its original static level. The rate of recovery is used to calculate the permeability of the aquifer and the potential yield of the well.
2. **Constant Level Test (Constant Rate Test):** In this method, the well is pumped at a constant discharge rate for a prolonged period. The drawdown is measured over time until a steady-state condition is reached. This allows engineers to determine the specific capacity and the sustainable yield of the well.
Since both the recuperation test and the constant level test are standard engineering methods used to determine the yield and hydraulic properties of a well, option (c) is the correct choice.