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Past Questions — 2026
The drilling fluid commonly used in hydraulic rotary drilling is:
A)
Cement slurry
B)
Bentonite mud
C)
Diesel
D)
Lime water
In hydraulic rotary drilling, a drilling fluid (often called 'drilling mud') is circulated through the drill pipe and returned up the annulus. Bentonite, a type of swelling clay, is the most commonly used material for this purpose.
1. Why Bentonite Mud is correct: Bentonite creates a viscous fluid that serves several critical functions: it carries drill cuttings (rock fragments) from the bottom of the hole to the surface; it forms a 'filter cake' on the borehole walls to prevent the fluid from leaking into the surrounding formation and to prevent the walls from collapsing; and it cools and lubricates the drill bit.
2. Why other options are incorrect:
- a. Cement slurry: This is used for 'grouting' or sealing the annulus after the drilling is complete to prevent surface contamination of the aquifer, not as a drilling fluid during the drilling process.
- c. Diesel: While used in some specialized oil drilling operations (oil-based muds), it is not the common fluid for standard groundwater hydrology/water well drilling due to environmental contamination risks.
- d. Lime water: This does not possess the necessary viscosity or thixotropic properties required to transport cuttings or stabilize the borehole walls.
The end contraction in a rectangular weir causes:
A)
increase in discharge
B)
decrease in effective length of crest
C)
increase in head
D)
turbulent flow
End contraction in a rectangular weir occurs when the water approaches the crest from the sides and the flow contracts at the ends, reducing the effective width of the crest. Hence, it causes a decrease in the effective length of crest. Option (a) is incorrect because contraction does not increase discharge; it actually reduces discharge for a given head. Option (c) is incorrect because the head is not increased by end contraction; rather, for the same discharge, a greater head may be required due to reduced effective length. Option (d) is incorrect because end contraction is a geometric/flow-width effect, not the direct cause of turbulent flow.
In the design mix method, proportions of concrete are determined based on:
A)
Volume batching
B)
Experience of mason
C)
Laboratory tests and calculations
D)
Computer Analysis
In concrete technology, there are two primary ways to determine proportions: Nominal Mix and Design Mix.
1. **Nominal Mix:** Proportions are fixed based on standard ratios (e.g., 1:2:4) and are generally used for small, non-critical works. This relies on general guidelines rather than specific material properties.
2. **Design Mix:** This is a scientific approach where proportions are determined to achieve a specific strength (characteristic strength) and durability. This process involves:
- **Laboratory Tests:** Testing the specific properties of the available materials (e.g., specific gravity of cement, grading of aggregates, moisture content of sand).
- **Calculations:** Using standard codes (like IS 10262 in India) to calculate the water-cement ratio and the quantity of aggregates required.
- **Trial Mixes:** Casting cubes and testing them in a lab to verify if the target strength is achieved, adjusting the mix if necessary.
Why other options are incorrect:
- **Option a (Volume batching):** This is a method of measuring materials on site, not a method for determining the proportions themselves.
- **Option b (Experience of mason):** This is an unscientific approach and is not used in professional engineering design mixes.
- **Option d (Computer Analysis):** While software can assist in calculations, the 'Design Mix Method' fundamentally relies on empirical laboratory data and standardized calculations; computer analysis alone cannot replace the physical testing of materials.
A cavity-type tube well is generally preferred when:
A)
The water-bearing stratum is very thick.
B)
Water is available only through the sides.
C)
A strong clay layer exists to support the pipe.
D)
The sand particles are very fine
A cavity-type tube well (also known as a cavern well) is a specialized type of well used in specific geological conditions. In these wells, a cavity is created in the water-bearing stratum, and the well pipe is placed within this cavity. This design is specifically preferred when the aquifer consists of very fine sand or silty sand. In such cases, standard screen slots would be too large (allowing sand to enter the well) or too small (causing excessive friction loss and clogging). By creating a cavity and using a gravel pack or a specific filter arrangement, the fine particles are kept out while allowing water to flow freely into the well.
Analysis of other options:
- Option (a): Thick strata do not necessitate a cavity-type well; standard tube wells are typically used for thick aquifers.
- Option (b): Water entering through the sides is a characteristic of almost all tube wells (radial flow), not a specific reason to choose a cavity-type design.
- Option (c): A strong clay layer is generally an aquiclude (impermeable) and is not the primary reason for implementing a cavity-type design, which focuses on the permeability and grain size of the aquifer itself.
In the casing of a centrifugal pump, the kinetic energy of the water is converted into:
A)
Potential energy
B)
Pressure energy
C)
Heat energy
D)
Centrifugal moment.
In a centrifugal pump, the impeller rotates at high speed, imparting kinetic energy (velocity) to the water. As the water leaves the impeller and enters the pump casing (specifically the volute casing), the cross-sectional area of the flow path gradually increases. According to Bernoulli's principle, as the velocity of the fluid decreases due to the increasing area, the kinetic energy is converted into pressure energy. This increase in pressure allows the pump to lift the water to the required height or deliver it to a specific destination.
- Option (a) Potential energy: While the pump eventually provides head (potential energy), the immediate conversion within the casing is from kinetic to pressure energy.
- Option (c) Heat energy: This is a result of friction and turbulence (losses), not the intended purpose of the casing.
- Option (d) Centrifugal moment: This is a force/torque concept related to the impeller's rotation, not a form of energy conversion in the casing.
The IS Code for concrete mix design is:
A)
IS 1893
B)
IS 4990
C)
IS 10262
D)
IS 13920
The correct IS Code for the mix design of concrete is IS 10262. This code provides the guidelines and procedures for determining the proportions of cement, water, fine aggregate, and coarse aggregate to achieve a desired strength and durability.
Analysis of other options:
- IS 1893: This code deals with 'Criteria for Earthquake Resistant Design of Structures'.
- IS 4990: This is not a primary standard for concrete mix design (often confused with other material specifications).
- IS 13920: This code specifies the 'Ductile Design and Detailing of Reinforced Concrete Structures Subjected to Seismic Forces'.
Which of the scale is the smallest respectively?
A)
1 cm = 10 m
B)
1 cm = 1000 km
C)
RF = 1/100000
D)
1 : 10000
To determine which scale is the smallest, we must convert all options into a common format, specifically the Representative Fraction (RF), which is a dimensionless ratio (1 unit on map = X units on ground). A smaller scale has a larger denominator, meaning a smaller area of the ground is represented by a larger unit on the map, or conversely, a larger area of the ground is compressed into a smaller map space.
Step-by-step conversion to RF:
Option (a): 1 cm = 10 m.
Since 1 m = 100 cm, 10 m = 1000 cm.
RF = 1 / 1000
Option (b): 1 cm = 1000 km.
Since 1 km = 100,000 cm, 1000 km = 100,000,000 cm.
RF = 1 / 100,000,000
Option (c): RF = 1 / 100,000
Option (d): 1 : 10,000.
RF = 1 / 10,000
Comparing the denominators:
- Option (a): 1,000
- Option (d): 10,000
- Option (c): 100,000
- Option (b): 100,000,000
The largest denominator corresponds to the smallest scale. Since 100,000,000 is the largest value, Option (b) is the smallest scale.
Hooke’s law states that stress is directly proportional to strain within:
A)
Plastic range
B)
Elastic limit
C)
Strain hardening zone
D)
Necking zone
Option B is correct: Elastic limit.
Water absorption of a First Class Brick should not exceed _______of its dry weight after immersion in water for 24 hours.
A)
15%
B)
10%
C)
5-10%
D)
20-25%
Option D is correct: 20-25%.
Why is an ogee spillway preferred in high dams?
A)
It increases the flow velocity significantly
B)
It allows better aeration of water
C)
It reduces water pressure and improves flow condition
D)
It requires less construction material compared to other spillways
An ogee spillway is designed with an S-shaped profile that closely follows the lower nappe of a ventilated sheet of water flowing over a sharp-crested weir. This specific geometry is preferred in high dams for the following reasons: 1. Hydraulic Efficiency: The shape minimizes flow separation and turbulence, ensuring a smooth transition of water from the reservoir to the downstream channel. 2. Pressure Distribution: By matching the natural profile of the falling water, it prevents the formation of negative pressures (suction) on the spillway surface. If the profile were not ogee-shaped, sub-atmospheric pressures could develop, leading to cavitation, which can severely damage the concrete structure of a high dam. 3. Discharge Capacity: It provides a higher coefficient of discharge compared to other weir types, allowing more water to pass safely during flood events.
Analysis of other options:
- Option (a) is incorrect because increasing flow velocity excessively is generally avoided to prevent erosion and cavitation.
- Option (b) is incorrect because while aeration is important (often handled by aeration grooves), it is not the primary reason the ogee shape is preferred over other spillway types.
- Option (d) is incorrect because the preference for an ogee spillway is based on hydraulic performance and structural safety (cavitation prevention), not primarily on the volume of construction material.
In the designation M15, M refers to the mix and the number refers to:
A)
The specified 28-day cube strength of mix in N/mm2
B)
The specified 28-hours cube strength of mix in N/mm2
C)
The specified days-hours cube strength of mix in N/mm2
D)
The specified 8-hours cube strength of mix in N/mm2
In concrete technology, the designation 'M' stands for 'Mix' and the number following it represents the characteristic compressive strength of the concrete. According to IS 456:2000 (the Indian Standard code for plain and reinforced concrete), the characteristic strength is defined as the strength of a 150mm x 150mm x 150mm cube tested after 28 days of curing. Therefore, in M15, '15' indicates that the specified compressive strength of the concrete cube is 15 N/mm² (or MPa) at 28 days. Options b, c, and d are incorrect because strength is measured in days (specifically 28 days), not hours, to allow for the hydration process of cement to reach a standardized level of maturity.
Find out the wrong number in the given series : 6, 9, 14, 20, 30, 41
A)
9
B)
14
C)
20
D)
41
The series should follow an increasing pattern of differences: 6, 9 (+3), 14 (+5), 20 (+6), 30 (+10), 41 (+11). A more natural intended pattern is adding consecutive prime numbers: +3, +5, +7, +9, +11, which would give 6, 9, 14, 21, 30, 41. So 20 is the wrong term; it should be 21. The other terms 9, 14, 30, and 41 fit the pattern of cumulative additions, while 20 breaks it.
A centrifugal pump is required to lift water to a total head of 40m at the rate of 50ltr/s. If its overall efficiency is 62%, the power required for the pump will be:
A)
41.6kW
B)
51.6kW
C)
21.6Kw
D)
31.6kW
Discharge, Q = 50 L/s = 0.05 m^3/s.
Hydraulic power required:
P_h = ρgQH = 1000 × 9.81 × 0.05 × 40 = 19,620 W = 19.62 kW.
Overall efficiency, η = 62% = 0.62.
So, input power required:
P = P_h / η = 19.62 / 0.62 = 31.65 kW ≈ 31.6 kW.
Therefore, the correct option is d) 31.6 kW.
Why the others are incorrect:
- a) 41.6 kW: too high; it would correspond to a much lower efficiency or larger head/discharge.
- b) 51.6 kW: far too high for the given flow and head.
- c) 21.6 kW: close to hydraulic power, but ignores the 62% overall efficiency.
A small watershed has: Length of the longest flow path, L=500m and average slope of the watershed, S=0.01. Estimate the time of concentration using the Kirpich formula:
A)
14 minutes
B)
10 minutes
C)
23 minutes
D)
30 minutes
Option A is correct: 14 minutes.
As per IS 383, fine aggregate is one which passes through:
A)
10 mm sieve
B)
4.75 mm sieve
C)
2.36 mm sieve
D)
1.18 mm sieve
According to IS 383 (the Indian Standard specification for coarse and fine aggregates from natural sources for concrete), aggregates are classified based on their particle size. Fine aggregate is defined as the material that passes through a 4.75 mm IS sieve. Conversely, coarse aggregate is defined as the material that is retained on a 4.75 mm IS sieve. Therefore, the 4.75 mm sieve serves as the dividing line between fine and coarse aggregates in Indian Standards. Option (a) 10 mm is typically used to categorize coarse aggregates, while options (c) and (d) refer to smaller sieves used for grading zones of fine aggregates but not for the primary definition of fine aggregate itself.
Grassed waterways are designed to suit the average flow velocity of:
A)
1.0 to 1.5 m/s
B)
1.5 to 2.0 m/s
C)
2.0 to 2.5 m/s
D)
2.5 to 3.0 m/s
Grassed waterways are designed to convey runoff from agricultural fields safely without causing soil erosion. The primary objective is to maintain a 'non-erosive velocity.' In hydraulic engineering for agriculture, the permissible velocity for a well-established grass cover typically ranges from 1.0 to 1.5 m/s. If the velocity exceeds this range (as seen in options b, c, and d), the shear stress exerted by the flowing water on the soil surface becomes greater than the resisting force provided by the grass roots and stems, leading to 'scouring' or the stripping away of the vegetation and underlying soil. Therefore, the design velocity is kept low to ensure the stability of the waterway.
The value of Poisson’s ratio depends upon
A)
Nature of load, tensile or compressive
B)
Magnitude of load
C)
Material of the test specimen
D)
Dimensions of the test specimen
Poisson's ratio (ν) is defined as the ratio of lateral strain to longitudinal strain when a material is stretched or compressed. Mathematically, ν = -(lateral strain / longitudinal strain).
1. Why Option (c) is correct: Poisson's ratio is a fundamental material property. It depends entirely on the internal structure, bonding, and nature of the material (e.g., steel has a ratio of approx 0.3, rubber is approx 0.5, and cork is approx 0).
2. Why Option (a) is incorrect: For linear elastic materials, Poisson's ratio remains constant regardless of whether the load is tensile or compressive.
3. Why Option (b) is incorrect: Within the elastic limit, the ratio of strains remains constant regardless of the magnitude of the load applied.
4. Why Option (d) is incorrect: Poisson's ratio is an intensive property, meaning it does not depend on the size, shape, or dimensions of the specimen.
Rainfall recorded at 3 rain gauges in a catchment is given below. Raingauge A: 6 cm (Area 20 km2), B: 8 cm (Area 30 km2), C: 10 cm (Area 20 km2). Determine the average rainfall over the catchment using the Thiessen Polygon Method.
A)
10 cm
B)
9 cm
C)
12 cm
D)
8 cm
The Thiessen Polygon Method is used to calculate the average precipitation over a catchment area by assigning a weight to each rain gauge based on the area it represents. The formula for the average rainfall (P_avg) is: P_avg = (Σ (P_i * A_i)) / Σ A_i, where P_i is the rainfall at gauge i and A_i is the area associated with gauge i.
Step-by-step calculation:
1. Identify the data for each gauge:
- Gauge A: P_A = 6 cm, A_A = 20 km²
- Gauge B: P_B = 8 cm, A_B = 30 km²
- Gauge C: P_C = 10 cm, A_C = 20 km²
2. Calculate the total weighted rainfall (Σ P_i * A_i):
- (6 cm * 20 km²) + (8 cm * 30 km²) + (10 cm * 20 km²)
- 120 + 240 + 200 = 560 cm·km²
3. Calculate the total area of the catchment (Σ A_i):
- 20 km² + 30 km² + 20 km² = 70 km²
4. Calculate the average rainfall:
- P_avg = 560 / 70 = 8 cm.
Therefore, the average rainfall over the catchment is 8 cm. Option (d) is correct, while options (a), (b), and (c) are mathematically incorrect based on the weighted average formula.
What is the maximum percentage of silt allowed in sand to be used in concrete?
A)
5%
B)
10%
C)
8%
D)
2%
In concrete technology, silt refers to the fine particles (usually smaller than 75 microns) present in the sand. Excessive silt is detrimental to concrete because it increases the water demand for a given workability, weakens the bond between the cement paste and the aggregate, and increases the risk of shrinkage and cracking. According to standard specifications (such as IS 383 in the Indian context, which is typically followed in MPSC exams), the maximum permissible limit for silt/clay content in fine aggregates (sand) used for concrete is generally 8% by weight. If the silt content exceeds this limit, the sand must be washed before use to ensure the durability and strength of the concrete structure. Options a, b, and d are incorrect as they do not align with the standard engineering specifications for general concrete production.
A small head pump operates for 5hrs/day with an average discharge of 2.0 ltr/s. Find the area that can be irrigated in one day if the depth of application is 5cm and application efficiency is 75%.
A)
560m2
B)
540m2
C)
640m2
D)
660m2
Step-by-step calculation:
1. **Calculate the total volume of water pumped per day (V_total):**
- Discharge (Q) = 2.0 liters/second = 0.002 m³/s
- Time (t) = 5 hours/day = 5 × 3600 seconds = 18,000 seconds
- Total Volume = Q × t = 0.002 m³/s × 18,000 s = 36 m³
2. **Calculate the volume of water actually stored in the root zone (V_effective):**
- Application Efficiency (η) = 75% = 0.75
- V_effective = V_total × η = 36 m³ × 0.75 = 27 m³
3. **Calculate the area (A) that can be irrigated:**
- Depth of application (d) = 5 cm = 0.05 m
- The formula for volume is Volume = Area × Depth
- Area (A) = V_effective / d = 27 m³ / 0.05 m = 540 m²
Conclusion: The area that can be irrigated is 540 m², which corresponds to option (b).
Number of links in a 30 m metric chain is:
A)
180
B)
100
C)
200
D)
150
In surveying, a metric chain is standardized based on its total length and the length of its individual links.
1. Standard Metric Chains: The most common metric chains used in surveying are 20 meters and 30 meters in length.
2. Link Length: For both 20m and 30m metric chains, the standard length of one single link is 20 cm (or 0.2 meters).
3. Calculation for a 30m chain:
Number of links = Total length of chain / Length of one link
Number of links = 30 m / 0.2 m = 150 links.
Therefore, a 30m metric chain consists of 150 links.
Comparison with other options:
- Option (b) 100 links would correspond to a 20m chain if the link length were 20cm (incorrect, 20m chain has 100 links).
- Option (c) 200 links would correspond to a 40m chain (not a standard metric chain).
- Option (a) 180 links does not correspond to standard metric chain specifications.
The minimum curing period for concrete made with Ordinary Portland Cement (OPC) as per IS 456 under normal weather conditions is:
A)
3 days
B)
5 days
C)
7 days
D)
14 days
According to IS 456:2000 (Plain and Reinforced Concrete - Code of Practice), the curing period is specified based on the type of cement used and the environmental conditions. For concrete made with Ordinary Portland Cement (OPC), the minimum curing period under normal weather conditions is 7 days. If the concrete is made with mineral admixtures (like fly ash or GGBS) or if the exposure condition is 'severe', the curing period is typically extended to 10 or 14 days. Therefore, for standard OPC in normal conditions, 7 days is the mandated minimum to ensure the hydration process proceeds sufficiently to achieve the required strength and durability. Options a and b are insufficient for hydration, and option d is generally reserved for specific admixtures or extreme conditions.
Which chemical compostion has highest content in OPC?
A)
Alumina
B)
Silica
C)
Lime
D)
Iron Oxide
Ordinary Portland Cement (OPC) is manufactured by burning a mixture of calcareous (lime-bearing) and argillaceous (clay-bearing) materials. The chemical composition of OPC typically consists of the following approximate percentages by weight:
1. Lime (CaO): 60% to 67% - This is the primary constituent. It provides strength but too much lime causes unsoundness.
2. Silica (SiO2): 17% to 25% - It imparts strength to the cement.
3. Alumina (Al2O3): 3% to 8% - It acts as a flux, lowering the clinkering temperature.
4. Iron Oxide (Fe2O3): 0.5% to 6% - It gives the characteristic grey color to the cement.
5. Magnesia (MgO), Sulfur Trioxide (SO3), and Alkali oxides: Small percentages.
Comparing the options, Lime has the highest percentage (approx. 63%), followed by Silica, Alumina, and Iron Oxide. Therefore, Lime is the chemical composition with the highest content in OPC.
When the Froude number is less than 1, the flow is called:
A)
Supercritical flow
B)
Critical flow
C)
Subcritical flow
D)
Uniform flow
The Froude number (Fr) is a dimensionless parameter used in open channel hydraulics to describe the regime of flow. It is defined as the ratio of inertial forces to gravitational forces: Fr = V / √(gD), where V is the flow velocity, g is the acceleration due to gravity, and D is the hydraulic depth.
1. Subcritical Flow (Fr < 1): The flow velocity is less than the wave celerity. In this regime, the flow is deep and slow, and disturbances (waves) can travel upstream. This is also known as 'tranquil flow'.
2. Critical Flow (Fr = 1): This represents the transition state where the specific energy is at a minimum for a given discharge.
3. Supercritical Flow (Fr > 1): The flow velocity is greater than the wave celerity. The flow is shallow and fast, and disturbances cannot travel upstream. This is also known as 'rapid flow'.
Since the question specifies the Froude number is less than 1, the flow is classified as subcritical flow.
What is the bending moment of a cantilever beam subjected to Uniformly Distributed Load? Assume length of beam = L; load = w per unit length.
A)
wL
B)
wL / 2
C)
wL2 / 2
D)
wL2 / 4
Step-by-step derivation:
1. A cantilever beam is fixed at one end and free at the other. For a beam of length L subjected to a Uniformly Distributed Load (UDL) of 'w' per unit length, the total load acting on the beam is W = w * L.
2. The total load (wL) acts at the center of gravity of the distribution, which is at a distance of L/2 from the fixed support.
3. The maximum bending moment (M_max) occurs at the fixed support because that is where the internal resistance to rotation is greatest.
4. Bending Moment is calculated as Force × Perpendicular Distance. Therefore, M = (Total Load) × (Distance to centroid) = (w * L) * (L / 2) = wL²/2.
Analysis of options:
- Option (a) wL represents the total vertical load (shear force at support), not the moment.
- Option (b) wL/2 is incorrect as it lacks the squared length term required for a moment calculation.
- Option (c) wL²/2 is the correct formula for the maximum bending moment of a cantilever with UDL.
- Option (d) wL²/4 is incorrect; this value does not correspond to the maximum moment of this specific beam configuration.