← Legal Metrology
Past Questions — 2010
Which of the following mortars is most suitable for construction work in waterlogged areas?
A)
cement mortar
B)
lime mortar
C)
gauged mortar
D)
mud mortar
Step-by-step reasoning: 1. Analysis of the environment: Waterlogged areas subject construction materials to constant moisture, seepage, and potential chemical attacks from groundwater. 2. Evaluation of Cement Mortar: Cement mortar (a mixture of Portland cement and sand) is hydraulic, meaning it sets and hardens under water. It provides high compressive strength, low permeability, and superior durability in damp conditions. 3. Evaluation of Lime Mortar: Pure lime mortar is non-hydraulic (sets by absorbing CO2 from the air). In waterlogged areas, it may fail to set properly or may dissolve/leach over time, leading to structural failure. 4. Evaluation of Gauged Mortar: Gauged mortar is a mixture of cement and lime. While better than pure lime, it is generally less water-resistant and has lower strength than pure cement mortar in saturated conditions. 5. Evaluation of Mud Mortar: Mud mortar is completely soluble in water and would wash away or lose all structural integrity in a waterlogged area. Conclusion: Due to its hydraulic properties and strength, cement mortar is the most suitable choice for waterlogged environments.
Modular co-ordination of construction means
A)
planning
B)
designing
C)
execution
D)
all the above
Modular coordination is a system of standardization in the construction industry. It involves the use of a 'module' (a standard unit of measurement) to ensure that building components, materials, and dimensions are compatible.
1. Planning: Modular coordination starts at the planning stage to determine the overall dimensions of the structure based on standard module sizes to minimize waste.
2. Designing: During the design phase, architects and engineers use these modules to specify the sizes of walls, windows, doors, and other components so they fit together perfectly without the need for on-site cutting or modification.
3. Execution: During execution (construction), modular coordination allows for the use of prefabricated components that can be quickly and accurately assembled on-site, reducing labor time and material wastage.
Since modular coordination is integrated into every phase of the project lifecycle—from the initial concept to the final assembly—it encompasses planning, designing, and execution.
A simply supported beam of span L carries a uniformly distributed load of w/unit length. The maximum bending moment is
A)
wL2/4
B)
wL2/8
C)
wL2/12
D)
wL2/16
Step-by-step derivation for a simply supported beam with a uniformly distributed load (UDL):
1. **Reaction Calculation**: For a beam of length L with a UDL of w per unit length, the total load is W = w * L. Due to symmetry, the vertical reactions at the supports (R1 and R2) are equal: R1 = R2 = (wL) / 2.
2. **Bending Moment Equation**: The bending moment M at any distance 'x' from one support is given by the sum of moments of the forces to the left of that section:
M(x) = (R1 * x) - (w * x * x/2)
M(x) = (wL/2)x - (wx²/2)
3. **Finding Maximum Bending Moment**: The maximum bending moment occurs where the shear force is zero. For a symmetric simply supported beam with UDL, this occurs at the center (x = L/2).
4. **Substitution**: Substitute x = L/2 into the moment equation:
M_max = (wL/2)(L/2) - (w(L/2)²/2)
M_max = (wL²/4) - (wL²/8)
M_max = (2wL² - wL²) / 8
M_max = wL²/8
Therefore, option (b) is correct. Options (a), (c), and (d) represent incorrect coefficients resulting from incorrect integration or different loading/support conditions.
The bending moment is the maximum on a section where shear force is
A)
is equal
B)
is maximum
C)
is minimum
D)
changes sign
In the study of Strength of Materials and Structural Analysis, there is a fundamental mathematical relationship between the Shear Force (V) and the Bending Moment (M). The rate of change of the bending moment with respect to the distance (x) along the beam is equal to the shear force at that section. Mathematically, this is expressed as: dM/dx = V. According to calculus, a function (M) reaches its maximum or minimum value (stationary point) when its first derivative (dM/dx) is equal to zero. Therefore, the bending moment is maximum or minimum where the shear force (V) is zero. For the shear force to be zero, it must either be zero across a segment or, more commonly, cross the zero axis, which means it 'changes sign' (from positive to negative or vice versa). Thus, the maximum bending moment occurs at the section where the shear force changes sign.
The portion A of the reinforced concrete retaining wall shown below is called
A)
stem
B)
heel slab
C)
toe slab
D)
none of the
Without figure, Portion A typical denotes vertical stem of cantilever retaining wall; heel and toe are base slabs.
When plates of thickness t are exposed to weather, tacking rivets are provided at a pitch in line not exceeding
A)
8 t
B)
16 t
C)
24 t
D)
32 t
In steel structure design, specifically regarding riveted or bolted joints, 'tacking rivets' (or tacking bolts) are used to hold plates in position before the final riveting/bolting process is completed. When plates are exposed to weather, there is a risk of moisture entering the gap between the plates, which can lead to corrosion or 'rusting' of the internal surfaces. To prevent the plates from separating or shifting due to environmental factors and to maintain a tight fit, tacking rivets are provided. According to standard engineering practice and structural codes for steelwork, the pitch of these tacking rivets in a line should not exceed 32 times the thickness of the plate (32t). This ensures sufficient clamping force and stability across the length of the joint during the fabrication and exposure phase. Options a, b, and c provide pitches that are unnecessarily tight and not in accordance with standard structural specifications for this specific purpose.
A ferrous metal is
A)
steel
B)
cast iron
C)
wrought iron
D)
all the above
A ferrous metal is defined as a metal that contains iron (Fe) as its primary constituent.
1. Steel: An alloy of iron and carbon (usually with carbon content less than 2%), making it a ferrous metal.
2. Cast Iron: An alloy of iron and carbon (usually with carbon content greater than 2%), making it a ferrous metal.
3. Wrought Iron: A nearly pure form of iron with very low carbon content and small amounts of slag, making it a ferrous metal.
Since all three options (steel, cast iron, and wrought iron) are iron-based materials, they are all classified as ferrous metals.
The crushing strength of first class should be
A)
above 10 N/mm2
B)
below 10 N/mm2
C)
above 20 N/mm2
D)
none of the above
According to Indian Standard (IS) specifications for common burnt clay bricks, bricks are classified based on their compressive strength. First-class bricks are the highest quality bricks used for load-bearing walls and permanent structures. The minimum average compressive strength required for a first-class brick is 10.5 N/mm², which is generally rounded or referred to in competitive exams as 'above 10 N/mm²'.
Breakdown of options:
- Option (a) is correct because first-class bricks must exceed the 10 N/mm² threshold to ensure structural stability.
- Option (b) is incorrect because a strength below 10 N/mm² would categorize the brick as second, third, or fourth class.
- Option (c) is incorrect because while some specialized engineering bricks may reach 20 N/mm², the standard requirement for a 'first class' common brick is not that high.
- Option (d) is incorrect as option (a) accurately describes the standard requirement.
The test strength of the sample of concrete is taken as average of the strength of
A)
2 specimens
B)
3 specimens
C)
4 specimens
D)
5 specimens
According to standard codes of practice for concrete testing (such as IS 456:2000 and IS 516 in India, which govern RCC design and testing), the compressive strength of concrete is determined by testing cubes. To ensure statistical reliability and account for variations in casting and curing, a 'sample' is defined as a set of three specimens. The 'test strength' of that specific sample is calculated as the arithmetic average of the compressive strengths of these three specimens. If the difference between any one specimen and the average exceeds a certain percentage (usually 15%), the test results may be considered unreliable. Therefore, the standard procedure is to take the average of 3 specimens.
Vibrator used in concrete
A)
improves consolidation
B)
makes concrete stronger and durable
C)
removes voids in concrete
D)
all the above
Concrete vibration is a process used to compact freshly placed concrete. Here is the step-by-step reasoning for each option:
1. **Improves Consolidation (Option a):** Consolidation is the process of removing entrapped air from the concrete mix. Vibrators apply high-frequency vibrations that rearrange the concrete particles into a denser state, thereby improving consolidation.
2. **Removes Voids (Option c):** When concrete is poured, air pockets (voids) are naturally trapped. These voids act as weak points. Vibrators force these air bubbles to rise to the surface and escape, effectively removing voids and honeycombing.
3. **Makes Concrete Stronger and Durable (Option b):** Strength and durability are directly related to the density of the concrete. By removing voids and improving consolidation, the concrete becomes denser. A denser concrete matrix has lower permeability (reducing the ingress of harmful chemicals) and higher compressive strength.
Since all three statements are correct and interrelated (vibration leads to void removal, which leads to consolidation, which ultimately results in higher strength and durability), the correct answer is 'all the above'.
Trailers are used
A)
as hauling equipments
B)
to trail the excavation
C)
as compaction equipments
D)
all the above
In construction and civil engineering, trailers are non-powered vehicles designed to be towed by a tractor or a truck. Their primary function is the transportation (hauling) of heavy materials, equipment, or excavated soil from one location to another.
Analysis of options:
- Option (a) is correct because 'hauling' refers to the act of transporting loads, which is the sole purpose of a trailer.
- Option (b) 'to trail the excavation' is not a standard technical term or function in construction equipment operations.
- Option (c) 'as compaction equipments' is incorrect; compaction is performed by rollers, rammers, or vibratory plates, not by trailers.
- Option (d) is incorrect because options (b) and (c) are invalid.
The minimum pitch of rivet holes of diameter d should not be less than
A)
1.25 d
B)
1.5 d
C)
2.0 d
D)
2.5 d
In the design of steel structures (based on standard codes such as IS 800), the 'pitch' of rivets is defined as the distance from the center of one rivet to the center of the next rivet in the direction of the load. To ensure the structural integrity of the plate and prevent shear failure or tearing between the holes, a minimum spacing is required. According to standard specifications for riveted joints, the minimum pitch is stipulated as 2.5 times the diameter of the rivet hole (2.5d). This ensures there is sufficient net area of the plate remaining to carry the applied load. Options a, b, and c provide values that are too small and would lead to excessive stress concentrations and potential failure of the base metal between the rivets.
The basic material for manufacture of cast iron is
A)
pig iron
B)
wrought iron
C)
haematite
D)
none of the above
Cast iron is an alloy of iron and carbon (typically containing 2% to 4% carbon). The primary raw material used to manufacture cast iron is pig iron, which is the immediate product of the blast furnace. Pig iron has a very high carbon content (usually 3-4.5%) and is too brittle for most structural applications, so it is remelted in a cupola furnace or induction furnace, where impurities are removed and alloying elements are added to produce various grades of cast iron (such as grey cast iron, white cast iron, or ductile iron).
Why other options are incorrect:
- b. Wrought iron: This is the purest form of commercial iron with very low carbon content. It is not used as a base for cast iron; rather, it is a different category of iron altogether.
- c. Haematite: This is an iron ore (Fe2O3). While haematite is the raw material used to produce pig iron in a blast furnace, it cannot be used directly to manufacture cast iron without first being smelted into pig iron.
The kiln which may work regularly throughout the year is
A)
Bull’s kiln
B)
Hoffman’s kiln
C)
Clamp
D)
none of the above
The correct answer is Hoffman's kiln because it is designed for continuous operation. Here is the detailed breakdown:
1. **Hoffman's Kiln (Continuous Kiln):** This is a circular or oval kiln consisting of a series of interconnected chambers. The firing process is continuous; while some chambers are being loaded with raw bricks, others are being fired, and others are being cooled. The heat from the firing chambers is used to preheat the bricks in adjacent chambers, making it highly fuel-efficient and allowing it to work regularly throughout the year without needing to be cooled down completely.
2. **Bull's Trench Kiln (Intermittent/Semi-continuous):** While more efficient than a clamp, the Bull's kiln operates in cycles. It involves loading, firing, and cooling phases. Although it can produce large quantities, it is not as 'continuous' in its operational cycle as the Hoffman kiln.
3. **Clamp (Intermittent):** A clamp is a temporary structure where bricks are stacked and fired. Once the firing is complete, the clamp must be dismantled to retrieve the bricks. It is a seasonal, batch-process method and cannot work regularly throughout the year.
Therefore, due to its regenerative heating system and multi-chamber design, the Hoffman's kiln is the only option that allows for uninterrupted, year-round production.
The maximum twisting moment a shaft can resist, is a product of the permissible shear stress and
A)
moment of inertia
B)
polar moment of inertia
C)
polar modulus
D)
modulus of rigidity
The torsion equation for a circular shaft is given by: T/J = (τ/r) = (Gθ/L), where T is the twisting moment (torque), J is the polar moment of inertia, τ is the shear stress, r is the radius of the shaft, G is the modulus of rigidity, θ is the angle of twist, and L is the length. To find the maximum twisting moment (T) a shaft can resist based on a permissible shear stress (τ_max), we rearrange the formula: T = τ_max * (J/r_max). The term (J/r_max) is defined as the Polar Modulus (Zp). Therefore, T = τ_max * Zp.
Analysis of options:
- a. Moment of Inertia (I): Used for bending stress, not torsion.
- b. Polar Moment of Inertia (J): T is proportional to J, but the relationship with stress requires dividing by the radius.
- c. Polar Modulus (Zp): Correct, as it is the ratio of the polar moment of inertia to the maximum radius (J/r).
- d. Modulus of Rigidity (G): This relates to the stiffness/deformation (angle of twist), not the maximum stress capacity.
The lifting equipment used for heavy materials is
A)
pulley
B)
chain hoist
C)
winches
D)
crane
In construction equipment, lifting devices are categorized based on their capacity and range of motion.
1. **Crane (Option d):** This is the primary equipment designed specifically for lifting and moving heavy materials (such as steel beams, concrete slabs, and machinery) both vertically and horizontally across a construction site. Cranes are engineered for high load capacities and wide reach, making them the definitive answer for 'heavy materials'.
2. **Pulley (Option a):** A pulley is a simple machine used to change the direction of a force or provide a mechanical advantage. While it is a component of lifting systems, it is not a standalone piece of 'equipment' for heavy materials in a professional construction context.
3. **Chain Hoist (Option b):** A chain hoist is used for lifting heavy loads, but it is typically used for vertical lifting in a restricted area (like a workshop or a specific installation point) and lacks the mobility and massive capacity of a crane.
4. **Winches (Option c):** A winch is used primarily for pulling or hauling loads horizontally or lifting them vertically using a drum and cable. While powerful, they are generally used for tensioning or pulling rather than the comprehensive lifting and positioning of heavy construction materials.
A pug mill is used for
A)
softening brick earth
B)
moulding brick earth
C)
tempering brick earth
D)
all the above
A pug mill is a machine used in brick manufacturing to mix clay (brick earth) with water to achieve a uniform consistency and plasticity. This process is technically known as 'tempering'.
1. Why 'tempering' is correct: Tempering involves the kneading of the clay to remove air voids, break down lumps, and ensure the moisture is distributed evenly throughout the mass, making it suitable for moulding.
2. Why 'softening' is incorrect: While the clay becomes softer during the process, 'softening' is a general result, whereas 'tempering' is the specific engineering term for the process of preparing the clay for moulding.
3. Why 'moulding' is incorrect: Moulding is the subsequent step where the tempered clay is shaped into bricks using moulds (either manually or by a brick-making machine). The pug mill prepares the material for moulding but does not perform the moulding itself.
4. Since options (a) and (b) are either imprecise or incorrect, option (d) is also incorrect.
A column is regarded as long column if the ratio of its effected length and lateral dimension, exceeds
A)
25
B)
20
C)
15
D)
10
In Reinforced Concrete Design (RCC), columns are classified as 'short' or 'slender' (long) based on their slenderness ratio. The slenderness ratio is defined as the ratio of the effective length (le) to the least lateral dimension (D). According to IS 456:2000 (the Indian Standard code for plain and reinforced concrete), a column is considered a short column if the ratio of its effective length to its least lateral dimension is less than 12. Conversely, if the ratio exceeds 12, the column is regarded as a long or slender column. While the options provided in the question (10, 15, 20, 25) do not include the exact value of 12, in many competitive exam contexts based on older textbooks or specific simplified criteria, '12' is the standard threshold. However, looking at the provided options, '12' is closest to '10' or '15'. In the context of standard RCC codes (IS 456), the limit is 12. If forced to choose from the given options where 12 is missing, there is often a discrepancy in the question paper; however, technically, any value exceeding 12 makes it long. In many technical exams for this specific post, option (d) 12 is the intended answer, but since the option says 10, it is likely a typo in the original exam paper for 12. Based on the standard code IS 456, the threshold is 12.
Scraper is self sufficient machine used for
A)
digging
B)
loading
C)
hauling
D)
all the above
A scraper is a specialized piece of heavy construction equipment designed to perform multiple earthmoving functions in a single cycle, making it 'self-sufficient'.
1. Digging: The scraper has a cutting edge (bowl) that is lowered into the ground. As the machine moves forward, the blade slices into the soil, digging it up.
2. Loading: As the machine continues to move, the material sliced by the blade is forced upward and into the hopper (bowl), effectively loading the machine without the need for a separate loader.
3. Hauling: Once the bowl is full, the scraper transports (hauls) the material to the desired disposal or fill site.
Because it integrates the functions of a loader, a dump truck, and a grader (for spreading), it is used for digging, loading, and hauling. Therefore, option 'd' is the correct answer.
Henry Gantt developed Bar Charts for planning and scheduling of projects in
A)
1940
B)
1920
C)
1900
D)
1880
Henry Gantt, an American mechanical engineer and management consultant, developed the Gantt chart (a bar chart used for project scheduling) around 1910-1915, but the foundational work and the conceptual development of these charts began around 1900. In the context of standard Civil Engineering and Construction Management textbooks (such as those used for MPSC or UPSC exams), the development of the Gantt chart is typically associated with the period around 1900 to 1910. Among the given options, 1900 is the most accurate representation of the era in which he introduced these planning tools to improve industrial productivity. Options a (1940) and b (1920) are too late, as the charts were already widely used in World War I (1914-1918), and option d (1880) is too early for the formalization of this specific management tool.
Spacing of stirrups in a rectangular beam, is
A)
increased at the ends
B)
increased at the centre of the beam
C)
kept constant throughout the bar
D)
deceased towards the centre of the beam
In a simply supported rectangular beam, the shear force is maximum at the supports (ends) and minimum (often zero) at the center of the span. Stirrups are provided primarily to resist shear stress. Since the shear force is highest at the ends, stirrups must be placed closer together (decreased spacing) to provide sufficient shear strength. Conversely, as the shear force decreases towards the center of the beam, the requirement for shear reinforcement decreases, allowing the spacing of the stirrups to be increased. Therefore, stirrups are closely spaced near the supports and widely spaced (increased spacing) at the center.
P = 4*pi^2*EI / L^2 is the equation of Euler’s crippling load if
A)
both ends are hinged
B)
one end is fixed and other end is free
C)
both ends are fixed
D)
one end is fixed and other end is hinged
Euler load Pc=pi^2EI/Le^2, for both ends fixed Le=L/2 => Pc=4pi^2EI/L^2.
The critical path line moves along activities having total
A)
positive
B)
negative
C)
zero
D)
same
In project management (CPM - Critical Path Method), the 'Critical Path' is defined as the longest sequence of activities in a network diagram that determines the shortest possible duration of the project.
1. **Total Float (Slack):** This is the amount of time an activity can be delayed without delaying the project completion date. It is calculated as: Total Float = Latest Start (LS) - Earliest Start (ES) or Latest Finish (LF) - Earliest Finish (EF).
2. **Critical Activities:** Activities that lie on the critical path have no flexibility in their timing. Any delay in these activities will directly result in a delay of the entire project. Therefore, their Total Float is exactly zero.
3. **Non-Critical Activities:** These activities have a positive total float, meaning they can be delayed by a certain amount of time without affecting the project deadline.
Since the critical path consists of activities where there is no room for delay, the total float for these activities is zero.
The various activities of project are shown in bar charts by
A)
vertical lines
B)
horizontal lines
C)
crosses
D)
dots
In construction management, a bar chart (commonly known as a Gantt Chart) is a tool used for project scheduling. In this chart, the vertical axis (Y-axis) lists the various activities or tasks of the project, while the horizontal axis (X-axis) represents the time scale (days, weeks, or months). Each activity is represented by a horizontal bar (or line) whose length corresponds to the duration of that activity. The start and end points of the horizontal line indicate the scheduled start and finish dates of the task. Therefore, activities are shown by horizontal lines. Vertical lines are typically used as grid lines to mark specific dates or milestones, and crosses or dots are not standard representations for activity durations in bar charts.
The central deflection of a simply supported steel beam of length L with concentrated load W at the centre, is
A)
WL3 / 3EI
B)
WL3 / 12EI
C)
WL3 / 24EI
D)
WL3 / 48EI
For a simply supported beam of length L with a concentrated load W acting at the center, the maximum deflection occurs at the center of the span. This is a standard derivation in the Strength of Materials (Mechanics of Materials) using the double integration method or Moment-Area method.
1. The bending moment at a distance x from the support (for 0 ≤ x ≤ L/2) is M(x) = (W/2)x.
2. Using the differential equation of the elastic curve: EI(d²y/dx²) = M(x) = (W/2)x.
3. Integrating once: EI(dy/dx) = (W/4)x² + C1.
4. Integrating again: EIy = (W/12)x³ + C1x + C2.
5. Applying boundary conditions: At x=0, y=0 (so C2=0). At x=L/2, dy/dx=0 (due to symmetry), which gives C1 = -(W/4)(L/2)² = -WL²/16.
6. Substituting x=L/2 into the deflection equation: EIy_max = (W/12)(L/2)³ - (WL²/16)(L/2) = WL³/96 - WL³/32 = (WL³ - 3WL³)/96 = -2WL³/96 = -WL³/48.
Thus, the magnitude of the central deflection is WL³/48EI.
Comparing with options:
- Option (a) WL³/3EI is incorrect.
- Option (b) WL³/12EI is incorrect.
- Option (c) WL³/24EI is incorrect.
- Option (d) WL³/48EI is the correct standard formula.