MES Jr. Grade • Paper-III
Rankine’s theory of active earth pressure assumes
A)
Soil mass is homogenous, dry and cohesionless
B)
Ground surface is plane which may be horizontal or inclined
C)
Back of the wall is vertical and smooth
D)
All of these
Rankine’s active earth pressure theory assumes: (1) the backfill soil is homogeneous, dry, and cohesionless; (2) the ground surface of the backfill is plane, either horizontal or uniformly inclined; and (3) the retaining wall back is vertical and smooth, so wall friction is neglected. Since options a, b, and c are all true assumptions of Rankine’s theory, the correct answer is d (All of these).
MES Jr. Grade • Paper-III
Surveys which are carried out to depict mountains, rivers, water bodies, wooded areas and other cultural details are known as
A)
Cadastral surveying
B)
Geodetic surveying
C)
Topographical surveying
D)
Plane survey
Topographical surveying is carried out to represent the natural and man-made features of an area, such as mountains, rivers, water bodies, forests/wooded areas, roads, buildings, and other cultural details. Hence, it is the correct answer.
Why the others are incorrect:
- Cadastral surveying: used to determine property boundaries, land parcels, and ownership details, not general terrain and feature depiction.
- Plane survey: a broad type of surveying that assumes the earth’s surface as plane for small areas; it is not specifically defined by the purpose of showing terrain and cultural features.
- Geodetic surveying: used for large areas with high precision, accounting for the earth’s curvature; it is mainly for control and reference, not for depicting local topographic features.
MES Jr. Grade • Paper-III
For testing a saturated clay for shear strength, the test recommended is
A)
Direct shear test
B)
Triaxial compression test
C)
Unconfined compression test
D)
All the above
For a saturated clay, the recommended test for determining shear strength is the **triaxial compression test**. It is the most reliable and widely used test because it allows control of drainage and confining pressure, which are important for saturated cohesive soils. In particular, it can simulate field stress conditions and measure shear strength under different drainage conditions.
Why the others are not correct:
- **Unconfined compression test**: Suitable mainly for saturated clays in undrained conditions, but it gives only the unconfined compressive strength and is less general/reliable than the triaxial test for shear strength determination.
- **Direct shear test**: More commonly used for cohesionless soils; it is less suitable for saturated clays because the failure plane is predetermined and drainage/stress conditions are not well controlled.
- **All the above**: Incorrect because not all listed tests are equally recommended for saturated clay shear strength testing.
MES Jr. Grade • Paper-III
The safe stopping distance may be calculated from the equation
A)
D = 0.278Vt + V^2 / 254f
B)
D = 0.254Vt + V^2 / 278f
C)
D = 0.254Vt + V^2 / 225f
D)
D = 0.225Vt + V^2 / 254f
The safe stopping distance (SSD) on a level road is given by the standard highway engineering formula: D = 0.278 V t + V^2 / (254 f), where V is speed in km/h, t is reaction time in seconds, and f is the coefficient of friction. Hence option (a) is correct.
Why the others are incorrect:
- (b) uses 0.254 instead of 0.278 in the reaction-distance term and 278 instead of 254 in the braking-distance term, so both constants are wrong.
- (c) has both constants incorrect: 0.254 should be 0.278, and 225 should be 254.
- (d) has 0.225 instead of 0.278 and 0.225 is not the standard coefficient used in the SSD formula; the braking term is correct only in the denominator 254f, but the first term is wrong.
Therefore, the correct answer is (a).
MES Jr. Grade • Paper-III
Remote sensing techniques are being usefully employed for the purpose of
A)
Improving natural resource management
B)
Land use
C)
Protection of environment
D)
All of these
Remote sensing is widely used for land use/land cover mapping, improving natural resource management (such as water, forests, soils, and agriculture), and environmental protection/monitoring (such as pollution, deforestation, and disaster assessment). Therefore, all the listed purposes are valid, making option d correct. Options a, b, and c are not incorrect in themselves; each describes one real application of remote sensing, but they are only partial answers. Since the question asks for the overall purpose, the best choice is d, All of these.
MES Jr. Grade • Paper-III
The smaller horizontal angle between the true meridian and a survey line is known as
A)
Declination
B)
Azimuth
C)
Dip
D)
Bearing
The smaller horizontal angle between the true meridian and a survey line is called the **bearing** of the line. Bearing is measured with respect to a meridian and is usually taken as the smaller angle (0° to 90°) from north or south toward east or west.
Why the others are incorrect:
- **Azimuth**: the horizontal angle measured clockwise from the north meridian, usually from 0° to 360°, not specifically the smaller angle.
- **Dip**: the vertical angle a line makes with the horizontal; it is not a horizontal angle in plan.
- **Declination**: the angle between the true meridian and the magnetic meridian, not between the meridian and a survey line.
MES Jr. Grade • Paper-III
The soils most susceptible to liquefaction are
A)
Saturated dense sands
B)
Saturated fine and medium sands of uniform particle
C)
Saturated clays of uniform size
D)
Saturated gravels and cobbles
Liquefaction is most likely in loose to medium-dense, saturated, cohesionless soils with uniform particle size, especially fine to medium sands. Such soils can build up excess pore water pressure under cyclic loading (e.g., earthquakes), causing a sudden loss of shear strength. Option (b) matches this condition.
Why the others are incorrect:
- (a) Saturated dense sands are less susceptible because dense packing provides higher resistance to volume change and pore-pressure buildup.
- (d) Saturated gravels and cobbles are generally too coarse and permeable to develop the high excess pore pressures needed for liquefaction.
- (c) Saturated clays do not typically liquefy in the classic sense; liquefaction is mainly a problem of cohesionless granular soils like sands and silts.
MES Jr. Grade • Paper-III
The drains which is provided parallel to roadway to intercept and divert the water from hill slopes is known as
A)
Sloping drain
B)
Catch-water drain
C)
Side drain
D)
Cross drain
Catch-water drain is the drain provided parallel to a roadway on the hill side to intercept runoff coming from the uphill slopes and divert it safely away from the road. This prevents water from reaching and damaging the roadway. Cross drain is used to carry water across the road alignment, side drain runs along the side of the road to collect road surface runoff, and sloping drain is not the standard term for this function.
MES Jr. Grade • Paper-III
Contour interval is
A)
The vertical distance between two consecutive contours
B)
The horizontal distance between two consecutive contours
C)
The vertical distance between two points on same contours
D)
The horizontal distance between two points on same contours
Contour interval is the vertical difference in elevation between two successive contour lines on a map. Therefore, option (a) is correct. Option (b) is incorrect because the horizontal distance between contours is called contour spacing, not contour interval. Option (d) is incorrect because it refers to the horizontal distance between two points on the same contour, which is not the definition of contour interval. Option (c) is incorrect because points on the same contour have the same elevation, so the vertical distance between them is zero, not the contour interval.
MES Jr. Grade • Paper-III
Planimeter is used for measuring
A)
Volume
B)
Area
C)
Contour gradient
D)
Slope angle
A planimeter is a surveying instrument used to measure the area of a plane figure on a map, drawing, or plan. Hence, option **b (Area)** is correct. The other options are incorrect because a planimeter does not measure **volume** (a), **contour gradient** (c), or **slope angle** (d); those require different instruments or calculations.
MES Jr. Grade • Paper-III
Site order book is used for recording
A)
Instruction of the executive engineer
B)
Construction measurements
C)
Requisition of plants and equipment
D)
Indents for materials to be ordered.
The site order book is maintained at the work site to record instructions, directions, and orders issued by the executive engineer or other supervising officers. It serves as an official communication record for site execution.
Why others are incorrect:
- b) Construction measurements are recorded in the measurement book, not the site order book.
- c) Requisition of plants and equipment is handled through requisition/procurement records, not the site order book.
- d) Indents for materials to be ordered are entered in indent or material requisition registers, not in the site order book.
MES Jr. Grade • Paper-III
The Terzaghi’s general bearing capacity equation for a continuous footing is given by
A)
qf = cNc + γDNq + 0.5γBNγ
B)
qf = cNc - γDNq + 0.5γBNγ
C)
qf = cNc + γDNq - 0.5γBNγ
D)
None of these
For a continuous (strip) footing, Terzaghi’s general bearing capacity equation is qf = cNc + γDNq + 0.5γBNγ, where c is cohesion, γ is unit weight of soil, D is depth of footing, B is footing width, and Nc, Nq, Nγ are bearing capacity factors. Hence option (a) is correct.
Option (b) is incorrect because the surcharge term γDNq must be positive, not negative, and the Nγ term is also positive.
Option (c) is incorrect because the Nγ term should be +0.5γBNγ, not negative.
Option (d) is incorrect because option (a) matches the standard Terzaghi equation.
MES Jr. Grade • Paper-III
In construction management PERT technique of network analysis is mainly useful for
A)
Small project
B)
Large and complex project
C)
Deterministic activity
D)
Research and Development project
PERT (Program Evaluation and Review Technique) is mainly used for projects with uncertain activity durations, especially research and development projects, where time estimates are probabilistic rather than fixed. It helps plan, schedule, and control such non-repetitive, uncertain work.
Why others are incorrect:
- a) Small project: PERT is generally not needed for small projects; simpler scheduling methods are usually sufficient.
- b) Large and complex project: This is more commonly associated with CPM/PERT network analysis in general, but PERT is specifically most useful when activity times are uncertain, as in R&D.
- c) Deterministic activity: PERT is not meant for deterministic activities; CPM is better suited when activity durations are known with certainty.
MES Jr. Grade • Paper-III
Slack time refers to
A)
An activity
B)
An event
C)
Both activity and event
D)
None of these
Slack time refers to the allowable delay in an event (or node) without affecting the overall project completion time. In CPM/PERT terminology, slack is associated with events, while float is generally used for activities. Therefore, option (b) is correct. Option (a) is incorrect because an activity has float, not slack. Option (c) is incorrect because slack is not used for both activity and event in the standard sense. Option (d) is incorrect because a correct option is present.
MES Jr. Grade • Paper-III
Which one of the following is not excavating and moving type equipment?
A)
Bulldozer
B)
Clam shell
C)
Scraper
D)
Dump truck
Dump truck is not an excavating and moving machine; it is a hauling/transport equipment used to carry already excavated material from one place to another. Clam shell is an excavating tool used for digging and handling loose material, scraper both excavates and transports soil, and bulldozer is earthmoving equipment used for pushing and moving soil over short distances.
MES Jr. Grade • Paper-III
Which of the following soil has more plasticity index
A)
Gravel
B)
Sand
C)
Silt
D)
Clay
Clay has the highest plasticity index because plasticity index (PI) measures the range of water content over which a soil remains plastic, and clay particles are very fine with a large specific surface area and strong water adsorption. This gives clay significant plastic behavior. Silt has some plasticity but much less than clay. Sand and gravel are coarse-grained soils and are essentially non-plastic, so their plasticity index is zero or negligible.
MES Jr. Grade • Paper-III
Weep holes are provided in retaining and breast walls
A)
To drain off the water from the filling
B)
To ventilate the stone masonry
C)
To add architectural beauty
D)
All of these
Weep holes are small openings provided in retaining and breast walls to allow accumulated water behind the wall to escape. This relieves hydrostatic pressure and prevents damage or instability. Hence, option (a) is correct. Option (b) is incorrect because weep holes are not meant for ventilating masonry. Option (c) is incorrect because they have no architectural purpose. Therefore, option (d) is also incorrect since only (a) is valid.
MES Jr. Grade • Paper-III
When the bituminous surfacing is done on already existing black top road or over existing cement concrete road, the type of treatment given is
A)
Seal coat
B)
Tack coat
C)
Prime coat
D)
Spray of emulsion
The correct treatment is **tack coat**. A tack coat is a thin bituminous bond coat applied **between two existing pavement layers**—for example, when laying bituminous surfacing over an already existing blacktop road or over a cement concrete surface—to ensure proper adhesion between the old surface and the new bituminous layer.
Why the others are incorrect:
- **Seal coat**: Used as a final thin layer over a bituminous surface to seal voids and improve waterproofing, not primarily for bonding old and new layers.
- **Prime coat**: Applied on a **granular base** before bituminous surfacing, not on an existing blacktop or concrete road.
- **Spray of emulsion**: This is a method/material of application, not the specific standard treatment name for bonding over an existing paved surface; the correct term remains tack coat.
MES Jr. Grade • Paper-II
An institution required to treat 4.2 m3/min of raw water for daily domestic supply. Flocculating particles are to be produced by chemical coagualation. A column analysis indicated that an overflow rate of 0.2mm/s will produce satisfactory particle removal in a settling basin at a depth of 3.5 m. The required surface area for settling is
A)
230 m2
B)
350 m2
C)
1280 m2
D)
23000 m2
For a settling basin, the required surface area is based on the overflow rate (surface loading rate):
A = Q / v_o
Given:
- Flow rate, Q = 4.2 m3/min = 4.2/60 = 0.07 m3/s
- Overflow rate, v_o = 0.2 mm/s = 0.0002 m/s
So,
A = 0.07 / 0.0002 = 350 m2
Hence, the correct answer is option b (350 m2).
Why the other options are incorrect:
- 230 m2: too small; would correspond to a higher overflow rate than given.
- 1280 m2: too large; not consistent with the stated flow and overflow rate.
- 23000 m2: far too large and clearly unrealistic for this problem.
Note: The basin depth of 3.5 m is not needed for calculating surface area from overflow rate.
MES Jr. Grade • Paper-II
If the straight sides of a triangular section of a lined canal with circular bottom of radius D, make an angle q with horizontal, the hydraulic mean depth is
A)
D
B)
D/2
C)
D/3
D)
D/4
For a lined canal section with a circular bottom of radius D and straight sides making an angle q with the horizontal, the hydraulic mean depth (hydraulic radius) is defined as R = A/P, where A is the flow area and P is the wetted perimeter. For this standard section, the geometry gives R = D/2, independent of the side slope angle q. Hence the correct option is b.
Why the others are incorrect:
- a (D): too large; hydraulic mean depth cannot equal the full radius for this section.
- c (D/3): not the standard result for this geometry.
- d (D/4): too small; it underestimates the area-to-wetted-perimeter ratio.
Therefore, the correct answer is b, D/2.
MES Jr. Grade • Paper-II
In laminar flow, the local instability occurs first at a point where
A)
u is maximum
B)
du/dy is maximum
C)
d^2u/dy^2 / (mu * du/dy) is maximum
D)
d^2u/dy^2 / (rho * du/dy) is maximum
Local instability in laminar flow is associated with the point where the ratio of the curvature of the velocity profile to the inertial effect is greatest. This is expressed by the quantity (d^2u/dy^2)/(rho * du/dy), which is the criterion used in stability analysis to identify the first point of instability. Hence option (d) is correct.
Why the others are incorrect:
- (a) u is maximum: The maximum velocity point is not where instability necessarily starts; instability depends on the shape of the velocity profile, not just the peak velocity.
- (b) du/dy is maximum: A large velocity gradient indicates strong shear, but local instability is not determined by shear alone.
- (c) d^2u/dy^2 / (mu * du/dy) is maximum: This uses viscosity mu instead of density rho. The relevant stability parameter involves inertial effects through rho, not mu in this form.
MES Jr. Grade • Paper-II
The ordinate of the instantaneous Unit Hydrograph (UH) of a catchment at any time t is
A)
Difference in slope of the S-curve with effective rainfall intensity of 1cm/hr
B)
The slope of the direct runoff unit hydrograph at that time
C)
The slope of the 1-hour unit hydrograph at that time
D)
The slope of the S-curve with effective rainfall intensity of 1cm/hr
The instantaneous unit hydrograph (IUH) is obtained from the S-curve method as the derivative (i.e., slope) of the S-hydrograph corresponding to a constant effective rainfall intensity of 1 cm/hr. Hence, the ordinate of the IUH at time t equals the slope of that S-curve at time t.
Why the others are incorrect:
- (a) Difference in slope of the S-curve: the IUH is not a difference of slopes; it is the slope itself.
- (b) Slope of the direct runoff unit hydrograph: a direct runoff hydrograph is the response to a finite-duration storm, not the instantaneous unit hydrograph definition.
- (c) Slope of the 1-hour unit hydrograph: a 1-hour UH is a finite-duration unit hydrograph, whereas the IUH is the limiting case as duration tends to zero.
MES Jr. Grade • Paper-II
Shear stress develops in a fluid element, if
A)
The fluid is at least
B)
The fluid container is subjected to uniform linear acceleration
C)
The fluid is inviscid
D)
The fluid is viscous and the flow is non-uniform
Shear stress in a fluid arises when adjacent fluid layers move with different velocities, i.e., when there is a velocity gradient. This requires a viscous fluid and a non-uniform flow. Hence option (d) is correct.
Why the others are incorrect:
- (c) Inviscid fluid: an inviscid fluid has zero viscosity, so it cannot sustain shear stress.
- (b) Uniform linear acceleration of the container: this may create a pressure gradient, but not necessarily shear stress; shear stress is associated with velocity gradients, not uniform acceleration alone.
- (a) The fluid is at least: this is incomplete/incorrect as stated and does not specify the necessary condition for shear stress.
MES Jr. Grade • Paper-II
Wheat crop requires 55 cm of water during 120 days of base period. The total rainfall during this period is 100 mm. Assume the irrigation efficiency to be 60%. The area (in ha) of the land which can be irrigated with a canal flow of 0.01 m3/s is
A)
13.82
B)
18.65
C)
23.42
D)
234.20
Water requirement of wheat = 55 cm = 550 mm. Effective rainfall during the period = 100 mm. So net irrigation depth needed = 550 - 100 = 450 mm = 0.45 m.
Canal discharge = 0.01 m3/s. In 120 days, total water supplied = 0.01 × 120 × 24 × 3600 = 103680 m3.
With irrigation efficiency = 60%, only 60% of canal water is effectively available to the field, so effective water = 0.60 × 103680 = 62208 m3.
Area irrigated = effective water / net depth = 62208 / 0.45 = 138240 m2 = 13.824 ha ≈ 13.82 ha.
Hence, the correct option is b.
Why others are incorrect:
- a (13.82) is close but not the exact option formatting? Actually this is the computed value; however in the given options, 13.82 corresponds to a, not b. Since the calculation gives 13.82 ha, the correct key is a.
- b (18.65), c (23.42), and d (234.20) do not match the computed irrigable area and arise from incorrect use of efficiency, rainfall, or unit conversion.
Correct option: a.
MES Jr. Grade • Paper-II
The difference in level between the top of a bank and supply level in a canal, is called
A)
berm
B)
free board
C)
height of bank
D)
none of these
The difference in level between the top of a canal bank and the full supply level (water level) in the canal is called the free board. It provides safety margin against overtopping due to waves, settlement, or fluctuations in water level.
Why others are incorrect:
- Height of bank: This is the total vertical height of the bank, not the margin above the supply level.
- Berm: A berm is a horizontal strip or shelf provided on the side slope or between sections, not the vertical difference asked here.
- None of these: Incorrect because the correct term is free board.