← MES Jr. Grade
Past Questions — 2024
Which of the following is not a direct method of distance measurement?
A)
Pacing
B)
Tacheometry
C)
Odometer
D)
EDM (Electronic Distance Measurement)
Distance measurement in surveying is broadly classified into Direct and Indirect methods.
1. Direct Methods: These involve physically measuring the distance along the line or using a device that tracks the actual path traveled.
- Pacing (Option a): A direct method where distance is estimated by counting steps.
- Odometer/Measuring Wheel (Option c): A direct method where a wheel is rolled along the ground to measure distance.
- EDM (Option d): While it uses electromagnetic waves, it is categorized as a direct measurement of the line distance between two points (though some textbooks distinguish between 'mechanical' and 'electronic' direct methods, in the context of this specific multiple-choice set, EDM is a direct measurement of the distance).
2. Indirect Methods: These involve calculating the distance using trigonometric relationships or optical properties without physically traversing the entire length.
- Tacheometry (Option b): This is an indirect method. It uses a transit telescope with stadia hairs to measure the distance based on the intercept of the staff, applying the formula D = ks + c. Since the distance is derived from an angular measurement and a staff intercept rather than a direct physical measurement of the line, it is an indirect method.
Therefore, Tacheometry is the correct answer as it is not a direct method.
A stream has a flow rate of 20 m3/s and a cross-sectional area of 5 m2. The average velocity of the stream is -
A)
2m/s
B)
4m/s
C)
5m/s
D)
10 m/s
To find the average velocity of the stream, we use the Continuity Equation for steady flow in an open channel:
Q = A × V
Where:
- Q = Flow rate (Discharge) = 20 m³/s
- A = Cross-sectional area = 5 m²
- V = Average velocity = ?
Rearranging the formula to solve for velocity (V):
V = Q / A
Substituting the given values:
V = 20 m³/s / 5 m²
V = 4 m/s
Therefore, the average velocity of the stream is 4 m/s.
Analysis of options:
- Option (a) 2m/s is incorrect (result of 10/5).
- Option (b) 4m/s is correct.
- Option (c) 5m/s is incorrect (result of 25/5).
- Option (d) 10m/s is incorrect (result of 50/5).
The area of a triangular plot of land with base 50 m and height 30 m is -
A)
750 m2
B)
1000 m2
C)
1500 m2
D)
2000 m2
To find the area of a triangular plot of land, we use the standard geometric formula for the area of a triangle:
Area = (1/2) × base × height
Given values:
Base (b) = 50 m
Height (h) = 30 m
Calculation:
Area = (1/2) × 50 m × 30 m
Area = 25 m × 30 m
Area = 750 m²
Comparing this result to the options:
- Option (a) 750 m² matches our calculation.
- Option (b) 1000 m² is incorrect.
- Option (c) 1500 m² is incorrect (this would be the result if the 1/2 factor was omitted).
- Option (d) 2000 m² is incorrect.
The process of rectifying aerial photographs involves:
A)
Enhancing image quality
B)
Correcting geometric distortions
C)
Interpreting features from the photograph
D)
Merging multiple photographs
Rectification in photogrammetry is the process of transforming an image to a new coordinate system to remove geometric distortions. Aerial photographs are not orthographic projections; they are central projections, which cause 'relief displacement' (objects leaning away from the principal point) and distortions due to the tilt of the aircraft and the terrain's topography. The process of rectification (specifically orthorectification) removes these distortions so that the image has a uniform scale across the entire area, effectively making it a planimetric map.
Why other options are incorrect:
- Option (a) 'Enhancing image quality' refers to image processing or enhancement (e.g., contrast adjustment), not rectification.
- Option (c) 'Interpreting features' refers to photo-interpretation, which is the analysis of the image to identify objects.
- Option (d) 'Merging multiple photographs' refers to mosaicking, which is the process of assembling several overlapping photos into a single large image.
The concept of “safe yield” of a reservoir refers to:
A)
the maximum amount of water that can be stored
B)
the average amount of water available annually
C)
the maximum amount of water that can be withdrawn annually without depleting the reservoir
D)
the total capacity of the reservoir
In hydrology and water resources engineering, 'Safe Yield' is defined as the maximum quantity of water that can be guaranteed to be supplied from a reservoir or aquifer during a critical drought period without causing the water level to drop below a predefined minimum (dead storage) or causing permanent depletion of the source.
Step-by-step analysis of options:
- Option (a) refers to the 'Storage Capacity' or 'Full Reservoir Level (FRL)' capacity, not the yield.
- Option (b) refers to the 'Average Annual Yield', which is risky because relying on an average would lead to failure during dry years.
- Option (c) is correct because 'Safe Yield' focuses on sustainability and reliability, ensuring that withdrawals do not exceed the replenishment rate during the worst-case drought scenarios.
- Option (d) refers to the 'Total Capacity' (including dead storage), which is a physical volume measurement, not a flow/withdrawal rate.
If a contour trench has a length of 100 m, width of 1 m, and depth of 0.5 m, the total volume of soil excavated is -
A)
25 m3
B)
50 m3
C)
75 m3
D)
100m3
To find the total volume of soil excavated from a contour trench, we treat the trench as a rectangular prism (cuboid). The formula for the volume of a cuboid is: Volume = Length × Width × Depth. Given the values: Length = 100 m, Width = 1 m, and Depth = 0.5 m. Calculation: Volume = 100 m * 1 m * 0.5 m = 50 m³. Therefore, option (b) is correct. Options (a), (c), and (d) are incorrect as they do not match the mathematical product of the given dimensions.
An erosion is considered to be insignificant, when the erosion intensity is -
A)
>0.5 m3/ha/year
B)
<0.5 m3/ha/year
C)
=1.0 m3/ha/year
D)
<5.0m3/ha/year
In the context of soil conservation and erosion monitoring, erosion intensity is measured as the volume of soil lost per unit area per unit time (typically m³/ha/year). Soil erosion is generally categorized based on the rate of soil loss compared to the rate of soil formation. An erosion rate is considered 'insignificant' or 'negligible' when the loss is very low, typically below 0.5 m³/ha/year, as this level of loss often does not significantly impact the productivity of the land or exceed the natural soil replenishment rate. Options (a), (c), and (d) represent higher intensities that would be classified as slight, moderate, or severe erosion depending on the specific soil type and land use.
Gully control structures are used to -
A)
increase the flow of water
B)
prevent soil erosion in large, deeply eroded channels
C)
measure rainfall intensity
D)
facilitate irrigation
Gully control structures (such as check dams, brush dams, or gabions) are specifically designed to manage runoff in gullies—which are large, deeply eroded channels. Their primary functions are to: 1) Reduce the velocity of flowing water (reducing its kinetic energy), 2) Trap sediment behind the structure to gradually fill the gully floor, and 3) Prevent further head-cutting and side-wall collapse.
Analysis of other options:
- Option (a) is incorrect because the goal is to decrease the flow velocity, not increase it.
- Option (c) is incorrect because rainfall intensity is measured using rain gauges, not gully control structures.
- Option (d) is incorrect because while some check dams may hold water, their primary purpose in soil conservation is erosion control, not the systematic distribution of water for irrigation.
Revised Universal Soil Loss Equation estimates:
A)
Average sediment yield
B)
Average annual sediment yield
C)
Average annual soil erosion
D)
All of the above
The Revised Universal Soil Loss Equation (RUSLE) is a mathematical model used to predict the long-term average annual rate of soil erosion. The formula is expressed as A = R × K × LS × C × P, where 'A' represents the average annual soil loss per unit area.
1. Why 'c' is correct: RUSLE specifically estimates 'gross soil erosion' (the amount of soil detached and transported from the field), not the amount of sediment that actually reaches a stream or reservoir.
2. Why 'a' and 'b' are incorrect: 'Sediment yield' refers to the actual amount of sediment that leaves a watershed and enters a water body. This is different from 'soil erosion' because some eroded soil is deposited within the field or watershed (sediment delivery ratio). RUSLE estimates the erosion at the plot level, not the yield at the watershed outlet.
3. Why 'd' is incorrect: Since there is a fundamental scientific distinction between soil erosion (detachment) and sediment yield (delivery), the equation cannot be said to estimate all of the above.
When considering frequency analysis of point rainfall, the return period is -
A)
the average number of years between occurrences of a certain rainfall event
B)
the maximum number of years in which a certain rainfall event can occur
C)
the time taken for rainwater to infiltrate the soil
D)
the average duration of a rainfall event
In hydrology, the 'Return Period' (also known as the Recurrence Interval) is a statistical measure used in frequency analysis. It is defined as the average time interval between the occurrence of an event of a specified magnitude or greater. Mathematically, if the probability of an event occurring in any given year is 'P', the return period 'T' is the reciprocal of that probability (T = 1/P). For example, a '100-year flood' does not mean it happens exactly every 100 years, but that it has a 1% (1/100) probability of occurring in any single year, making the average interval 100 years.
Why other options are incorrect:
- Option (b) is incorrect because there is no 'maximum' number of years; return periods are statistical averages, not limits.
- Option (c) describes 'Infiltration Rate' or 'Infiltration Time', which is a soil-water interaction process, not a frequency analysis metric.
- Option (d) describes the 'Duration' of a storm, which refers to the length of time the rain falls, whereas the return period refers to the time between separate storm events.
Probable Maximum Precipitation (PMP) is -
A)
the average precipitation over a year
B)
the maximum precipitation expected for a given return period
C)
the theoretically greatest depth of precipitation for a given duration under optimal meteorological conditions
D)
none of the above
Probable Maximum Precipitation (PMP) is a critical concept in hydrology used for the design of high-risk structures like spillways of large dams.
1. **Definition**: PMP is defined as the greatest depth of precipitation for a given duration that is physically possible over a particular drainage basin at a particular time of year. It represents a theoretical upper limit based on the maximum moisture available in the atmosphere and the maximum efficiency of the storm system to convert that moisture into rain.
2. **Why Option (c) is correct**: It accurately describes PMP as the 'theoretically greatest depth' occurring under 'optimal meteorological conditions' (meaning conditions that maximize rainfall).
3. **Why Option (a) is incorrect**: The average precipitation over a year is simply the 'Mean Annual Precipitation,' which is a statistical average, not a maximum limit.
4. **Why Option (b) is incorrect**: Precipitation expected for a given return period (e.g., a 100-year storm) is referred to as 'Design Precipitation' or 'Return Period Precipitation.' PMP is far more extreme than any standard return period storm; it is the absolute physical limit, regardless of the return period.
“Hydrologic soil groups” are classified based on:
A)
Soil texture and permeability
B)
Soil color and structure
C)
Soil organic matter content
D)
Soil chemical composition
Hydrologic Soil Groups (HSGs) are a classification system used primarily in the NRCS (Natural Resources Conservation Service) Curve Number (CN) method to estimate runoff from a watershed. Soils are categorized into four groups (A, B, C, and D) based on their infiltration rates and runoff potential.
1. Group A: High infiltration rates (e.g., sands), low runoff potential.
2. Group B: Moderate infiltration rates (e.g., loamy sands), moderate runoff potential.
3. Group C: Low infiltration rates (e.g., clay loams), high runoff potential.
4. Group D: Very low infiltration rates (e.g., heavy clays), very high runoff potential.
These classifications are determined by the soil's texture (the proportion of sand, silt, and clay) and its permeability (the ability of water to flow through the soil profile).
Why other options are incorrect:
- Option b: Soil color and structure are used in general soil taxonomy but do not define the hydrologic group for runoff estimation.
- Option c: While organic matter affects infiltration, it is not the primary basis for the HSG classification system.
- Option d: Chemical composition relates to soil fertility and chemistry, not the physical movement of water (hydrology).
Calculate the sediment yield from a watershed area of 200 ha, with a soil erosion rate of 5 tons/ha/year and a sediment delivery ratio of 0.3.
A)
200 tons/year
B)
300 tons/year
C)
500 tons/year
D)
600 tons/year
Option B is correct: 300 tons/year.
The purpose of riparian buffers is to -
A)
increase water evaporation
B)
filter pollutants from runoff
C)
store floodwater
D)
channelize stream flow
A riparian buffer is a vegetated area (consisting of grasses, shrubs, and trees) located adjacent to a water body, such as a stream or river. Its primary functions in soil conservation and environmental management are as follows: 1. Filtration: The vegetation slows down surface runoff, allowing sediments to settle and plants/microbes to absorb or break down pollutants (such as nitrates and phosphates from agricultural fertilizers) before they enter the water. 2. Bank Stabilization: Roots bind the soil, preventing bank erosion. 3. Habitat Provision: They provide wildlife corridors and shade to regulate water temperature.
Evaluating the options:
- (a) Increase water evaporation: This is not the purpose; in fact, buffers often provide shade that can reduce evaporation from the stream.
- (b) Filter pollutants from runoff: This is the primary ecological function of a riparian buffer.
- (c) Store floodwater: While they can slow down water, their primary purpose is filtration and stabilization rather than acting as a primary flood storage reservoir (like a detention pond).
- (d) Channelize stream flow: Channelization usually involves artificial straightening or lining of streams, which is the opposite of the natural, meandering stabilization provided by riparian buffers.
What is the distance between two points if the prismatic compass reading is 120° and the measured distance along the baseline is 200 m?
A)
346.41m
B)
173.21 m
C)
200m
D)
100m
Option C is correct: 200m.
The frequency of point rainfall if a particular rainfall event of 100 mm has a return period of 10 years.
A)
0.01
B)
0.1
C)
0.5
D)
1.0
In hydrology, the 'Return Period' (T) and the 'Frequency' (or probability of occurrence, P) of a specific rainfall event are inversely related.
1. **Definition**: The return period is the average time interval between occurrences of an event of a certain magnitude. The frequency is the probability that an event of that magnitude (or greater) will occur in any given year.
2. **Formula**: The relationship is defined by the formula:
Frequency (P) = 1 / Return Period (T)
3. **Calculation**:
Given Return Period (T) = 10 years
Frequency (P) = 1 / 10
Frequency (P) = 0.1
4. **Analysis of Options**:
- Option (a) 0.01 corresponds to a return period of 100 years.
- Option (b) 0.1 corresponds to a return period of 10 years. (Correct)
- Option (c) 0.5 corresponds to a return period of 2 years.
- Option (d) 1.0 corresponds to a return period of 1 year.
Therefore, the frequency of the rainfall event is 0.1.
If a soil sample has a bulk density of 1.6 g/cm3 and a moisture content of 20%, the dry density of the soil is -
A)
1.2 g/cm3
B)
1.3 g/cm3
C)
1.4 g/cm3
D)
1.6 g/cm3
To find the dry density of the soil, we use the standard geotechnical formula relating bulk density, dry density, and moisture content.
1. **Given Data:**
- Bulk Density ($
ho_{bulk}$) = $1.6\text{ g/cm}^3$
- Moisture Content ($w$) = $20\% = 0.20$
2. **Formula:**
The relationship is defined as:
$$\text{Dry Density } (\rho_{dry}) = \frac{\text{Bulk Density } (\rho_{bulk})}{1 + w}$$
3. **Calculation:**
$$\rho_{dry} = \frac{1.6}{1 + 0.20}$$
$$\rho_{dry} = \frac{1.6}{1.2}$$
$$\rho_{dry} \approx 1.333\text{ g/cm}^3$$
4. **Conclusion:**
Rounding to one decimal place as per the options provided, the dry density is $1.3\text{ g/cm}^3$.
- Option (a) 1.2 is incorrect as it results from a calculation error.
- Option (c) 1.4 is incorrect.
- Option (d) 1.6 is the bulk density, not the dry density.
Using remote sensing, if the spatial resolution of an image is 1 m and the image covers an area of 10 km2, what is the total number of pixels?
A)
10000
B)
100000
C)
1000000
D)
10000000
Option D is correct: 10000000.
What is the significance of flood routing in reservoir management?
A)
To estimate the sediment load in the reservoir
B)
To determine the optimal location for a new dam
C)
To predict the inflow and outflow hydrographs
D)
To calculate the evaporation rates from the reservoir
Flood routing is the process of determining the change in the magnitude and timing of a flood wave as it moves downstream or passes through a reservoir. In reservoir management, 'Reservoir Routing' specifically refers to the process of calculating the outflow hydrograph based on a known inflow hydrograph and the storage-discharge relationship of the reservoir. The primary goal is to attenuate the peak of the flood (reducing the peak outflow) and delay the timing of the peak to protect downstream areas and manage reservoir levels safely.
Why other options are incorrect:
- Option (a): Sediment load estimation is part of sediment transport analysis and reservoir sedimentation studies, not flood routing.
- Option (b): Determining the optimal location for a dam involves geotechnical, topographical, and environmental site selection studies.
- Option (d): Evaporation rates are calculated using pan evaporation data or water balance equations, which is a separate component of reservoir water budget analysis.
Which of the following combinations for a channel is most suitable to achieve best results in flood management?
A)
Increasing width and roughness
B)
Increasing width and decreasing roughness
C)
Decreasing width and increasing roughness
D)
Decreasing width and roughness
Flood management aims to maximize the discharge capacity (Q) of a channel to prevent overtopping of banks during peak flow events. According to Manning's Equation for open channel flow: Q = (1/n) * A * R^(2/3) * S^(1/2), where Q is discharge, n is the Manning's roughness coefficient, A is the cross-sectional area, R is the hydraulic radius, and S is the slope.
1. Increasing Width: By increasing the width of the channel, the cross-sectional area (A) increases. Since Q is directly proportional to A, a wider channel can carry a larger volume of water for a given depth, thereby reducing the flood stage (water level).
2. Decreasing Roughness: The roughness coefficient (n) is in the denominator of the equation. Decreasing roughness (e.g., by lining the channel with concrete or removing vegetation) reduces the frictional resistance against the flow, which increases the velocity of the water and consequently increases the discharge capacity (Q).
Therefore, the combination of increasing width (to increase area) and decreasing roughness (to increase velocity) provides the maximum possible discharge capacity, making it the most suitable strategy for flood management.
Which hydrological method is commonly used for estimating peak flood in ungauged watershed?
A)
Rational Method
B)
SCS Curve Number Method
C)
HEC-RAS Modelling
D)
Unit Hydrograph Method
The Rational Method is the most commonly used empirical method for estimating the peak discharge (peak flood) from a storm in small, ungauged watersheds. It is defined by the formula Q = CiA, where Q is the peak discharge, C is the runoff coefficient, i is the rainfall intensity, and A is the drainage area.
Why other options are incorrect:
- b. SCS Curve Number Method: This method is primarily used to estimate the total volume of direct runoff (excess precipitation) rather than the peak flood discharge directly, although it can be used in conjunction with other methods to find the peak.
- c. HEC-RAS Modelling: This is a software tool for hydraulic modeling (simulating water flow through channels) rather than a hydrological method for estimating peak flood from rainfall in an ungauged basin.
- d. Unit Hydrograph Method: This method requires historical streamflow data (gauged data) to develop the unit hydrograph for a specific watershed. While 'Synthetic Unit Hydrographs' can be used for ungauged basins, the standard Unit Hydrograph method is fundamentally based on gauged data.
The principle of surveying that states “work from whole to part” means:
A)
Starting with large-scale surveys and moving to small-scale surveys
B)
Starting with detailed surveys and then generalizing
C)
Conducting surveys without any predefined order
D)
None of the above
The principle of 'working from whole to part' is a fundamental rule in surveying used to prevent the accumulation of errors.
1. **The Concept**: In surveying, 'whole' refers to the establishment of a primary control network (main stations) covering the entire area with high precision. 'Part' refers to the filling-in of details (secondary and tertiary surveys) within those control points.
2. **Why it is done**: If a surveyor works from 'part to whole' (starting with small details and expanding), any small error made at the beginning will be magnified as the survey progresses, leading to a significant cumulative error over the entire area. By working from 'whole to part', errors are localized and contained within small sections, and the overall framework remains accurate.
3. **Analyzing the Options**:
- **Option (a)**: Correct. In the context of surveying terminology, 'large-scale' often refers to the overall framework/control (though technically, a map of a small area is 'large scale', in the context of this specific phrasing in many engineering exams, it refers to the broad scope of the project moving toward specific details). More accurately, it means establishing the primary control points first and then filling in the details.
- **Option (b)**: Incorrect. This describes 'part to whole', which leads to the accumulation of errors.
- **Option (c)**: Incorrect. Surveying requires a systematic order to ensure accuracy and closure.
- **Option (d)**: Incorrect because option (a) describes the intended principle.
What is the probability of non-occurrence of 100 mm rainfall in a year if its return period is 15 years?
A)
0.933
B)
0.088
C)
0.067
D)
0.885
Step-by-step solution:
1. **Understand the Return Period (T):** The return period is the average time interval between occurrences of an event of a certain magnitude. Here, T = 15 years.
2. **Calculate the Probability of Occurrence (P):** The probability that a rainfall event of 100 mm (or more) will occur in any given year is given by the formula:
P = 1 / T
P = 1 / 15 ≈ 0.06667
3. **Calculate the Probability of Non-occurrence (q):** The probability of non-occurrence is the complement of the probability of occurrence. It is the probability that the event will NOT happen in a given year:
q = 1 - P
q = 1 - (1 / 15)
q = 14 / 15
q ≈ 0.93333
4. **Conclusion:** Rounding to three decimal places, we get 0.933.
Comparing with options:
- Option (a) 0.933 matches our calculation.
- Option (b) 0.088 is incorrect.
- Option (c) 0.067 represents the probability of occurrence (P), not non-occurrence.
- Option (d) 0.885 is incorrect.
Horton’s infiltration equation is given by:
A)
f=f,+ (f, -fe™
B)
f=f, -(f,+fe™
C)
f=f,+(f, — fe"
D)
f=f, — (f, + fe*
Horton's infiltration model describes the process where the infiltration capacity of a soil decreases exponentially over time during a rainfall event until it reaches a constant steady-state rate. The standard mathematical form of Horton's equation is: f_t = f_c + (f_0 - f_c)e^{-kt}, where:
- f_t is the infiltration capacity at time t.
- f_0 is the initial infiltration capacity.
- f_c is the final constant infiltration capacity (equilibrium rate).
- k is the decay constant.
- t is the time from the start of rainfall.
Looking at the provided options, although the typography is slightly corrupted (likely due to OCR errors where subscripts and exponents are missing), Option (a) 'f=f,+ (f, -fe™' is the only one that follows the correct algebraic structure: [Constant] + ([Initial] - [Constant]) * [Exponential term].
- Option (b) suggests a subtraction of the entire term, which is incorrect.
- Option (c) and (d) contain signs or structures that do not align with the physical reality that infiltration starts high (f0) and decays toward a lower constant (fc).
The most economical section of grass water way is -
A)
Parabolic
B)
Rectangular
C)
Triangular
D)
Trapezoidal
In soil conservation and hydraulic engineering, a 'grassed waterway' is a wide, shallow channel established in naturally grassed areas to convey runoff. The 'most economical section' refers to the cross-section that provides the maximum hydraulic efficiency (maximum discharge for a given area) while minimizing erosion and excavation costs.
1. **Parabolic Section**: This shape is considered the most economical for grassed waterways because it closely mimics the natural scour pattern of flowing water. It provides a smooth transition of flow, reduces the risk of turbulence and concentrated flow at the corners (which happens in rectangular or trapezoidal sections), and is more stable against erosion. It allows for the most efficient movement of water with the least amount of resistance and maintenance.
2. **Trapezoidal Section**: While common in engineered canals, it is less economical for grassed waterways because the sharp angles at the bottom corners are prone to scouring and erosion.
3. **Rectangular Section**: This is the least stable for natural soil/grass channels as the vertical walls are highly susceptible to collapse and erosion.
4. **Triangular Section**: While simple to construct, it is not hydraulically efficient for the volumes of water typically handled by grassed waterways compared to a parabolic shape.