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Past Questions — 2016

Junior Engineer • 2016 • Paper-I

The data available as a result of conducting Standard Proctor Test ona soil sample are as follows : Moisture Content : 0.08 0.09 0.10 0.12 0.14 0.16 Dry density (gm/cc):1.20 1.40 1.60 1.80 1.70 1.50 Give the Optimum Moisture Content in percentage.

A) 9%
B) 10%
C) 12%
D) 14%
In a Standard Proctor test, the Optimum Moisture Content (OMC) is the water content corresponding to the maximum dry density. From the given data, the dry density values are 1.20, 1.40, 1.60, 1.80, 1.70, and 1.50 gm/cc. The maximum dry density is 1.80 gm/cc, which occurs at moisture content 0.12. Converting 0.12 to percentage gives 12%. Hence, the correct answer is 12%.
Junior Engineer • 2016 • Paper-I

When the plastic limit is equal to or greater than the liquid limit, the Y is reported as zero. Find the correct replacement for Y.

A) liquid limit
B) plastic limit
C) plasticity index
D) liquidity index
The plasticity index (PI) is defined as PI = liquid limit (LL) − plastic limit (PL). If the plastic limit is equal to or greater than the liquid limit, then PI becomes zero or negative. In soil mechanics, such a soil is considered non-plastic, so the value reported is taken as zero. Therefore, Y = plasticity index. The other options are incorrect because liquid limit and plastic limit are individual consistency limits, while liquidity index is a different parameter based on natural water content.
Junior Engineer • 2016 • Paper-I

An undisturbed sample of soil has a volume of 50 cc and weighs 90 gm. On oven drying for 24 hours, the weight is reduced to 80 gm. Determine the dry density of the soil sample.

A) 1.60 gm/ce
B) 1.80 gm/ce
C) 1.90 gm/cc
D) 1.95 gm/ce
Dry density = dry mass / total volume. The oven-dried weight is 80 g and the sample volume is 50 cc. So, dry density = 80 / 50 = 1.60 g/cc. Hence, option (a) is correct. The other options do not match this calculation.
Junior Engineer • 2016 • Paper-I

The failure of a mass of soil located beneath a slope is called

A) aslide
B) ashear
C) aseepage
D) anoverturning
Failure of a soil mass beneath a slope is called a landslide or simply a slide. In slope stability, the soil mass moves downward along a failure surface due to gravity. Why others are incorrect: - ashear: shear is a mode of failure stress, not the name of the slope failure itself. - aseepage: seepage is the movement of water through soil. - anoverturning: overturning is a structural failure mode, not soil slope failure. So the correct answer is aslide.
Junior Engineer • 2016 • Paper-I

Determination of Y along the sloping surface is an important process of slope stability analysis. Find the most appropriate replacement for Y.

A) shearing strength
B) bearing capacity
C) density
D) soundness
In slope stability analysis, the key factor along the sloping surface is the soil’s resistance to sliding, i.e. its shearing strength. Stability of a slope depends on whether the shear resistance along the potential failure surface can resist the driving forces. Bearing capacity is related to foundation support, density is only a soil property and not the main stability criterion here, and soundness is not relevant to slope stability.
Junior Engineer • 2016 • Paper-I

Minimum depth of foundation is given by

A) Terzaghi’s formula
B) Rankine’s formula
C) Coulomb’s formula
D) Darcy’s formula
The minimum depth of foundation is commonly determined using Rankine’s formula, which relates the safe depth of foundation to the bearing capacity and soil properties. Terzaghi’s formula is used for bearing capacity of shallow foundations, Coulomb’s formula is used for earth pressure/sliding analysis, and Darcy’s formula is used for flow through soils, not foundation depth.
Junior Engineer • 2016 • Paper-I

Identify the incorrect statement from the following :

A) Bending Moment is important for determining design depth of beams
B) Bending Moment is important for determining area of steel reinforcement
C) Steel is assumed to resist tensile stresses
D) Concrete is assumed to resist tensile stresses
In RCC design, concrete is weak in tension and is generally assumed to resist compressive stresses only. Steel is provided to take tensile stresses. Bending moment is used to determine both the effective depth of the beam and the required area of steel reinforcement. Therefore, the incorrect statement is that concrete is assumed to resist tensile stresses.
Junior Engineer • 2016 • Paper-I

Earth dams, embankments and cuts are examples of

A) infinite slopes
B) finite slopes
C) angular slopes
D) tangential slopes
Earth dams, embankments, and cuts are treated as finite slopes because their slope lengths are limited and the failure surface is bounded by the geometry of the structure. In contrast, infinite slopes are idealized slopes that extend indefinitely with uniform conditions. Therefore, the correct answer is finite slopes. The other options are not standard slope classifications in soil mechanics.
Junior Engineer • 2016 • Paper-I

Which of the following processes is treated as a ‘loss’ in engineering hydrology?

A) Infiltration
B) Precipitation
C) Percolation
D) Transpiration
In engineering hydrology, a 'loss' is any process that removes water from the surface water balance and reduces direct runoff. Transpiration is a loss because water is taken up by plants and returned to the atmosphere. Infiltration and percolation are subsurface movement processes, not usually termed the main hydrologic 'loss' here, while precipitation is an input, not a loss.
Junior Engineer • 2016 • Paper-I

A hydrograph is essentially a graph establishing the relation between the time of flow due to rainfall and

A) catchment area
B) discharge
C) velocity of flow
D) volume
A hydrograph is a graph showing the variation of stream discharge with time at a given point, usually in response to rainfall. Hence it establishes the relation between time of flow due to rainfall and discharge. Catchment area, velocity, and volume are not what a hydrograph directly plots.
Junior Engineer • 2016 • Paper-I

Maximum numbers of Retaining walls in Mizoram are constructed using

A) stone masonry
B) brick masonry
C) plain cement concrete
D) reinforced Cement Concrete
In Mizoram, retaining walls are most commonly built using stone masonry because locally available stones are abundant, economical, and suitable for hilly terrain. Brick masonry is less common in such terrain, PCC is not generally used as a retaining wall material by itself, and RCC is usually reserved for stronger or more engineered retaining structures. Therefore, the maximum number of retaining walls are constructed using stone masonry.
Junior Engineer • 2016 • Paper-I

The age of trees can be predicted by

A) length of medulary rays
B) counting number ofrings
C) measuring the diameter of pith
D) the thickness of bark
The age of a tree is determined by counting its annual growth rings seen in a cross-section of the trunk. Each ring generally represents one year of growth. The other options are not standard methods for predicting tree age: medullary rays, pith diameter, and bark thickness do not directly indicate age.
Junior Engineer • 2016 • Paper-I

The cgs unit for coefficient of permeability is

A) darcy
B) gm/sec
C) stoke
D) cm/sec
The coefficient of permeability is a velocity term in soil mechanics, representing the rate at which water flows through soil under a hydraulic gradient. In the cgs system, its unit is centimetre per second (cm/sec). Why others are incorrect: - (a) darcy: This is a unit of permeability, but not the cgs unit for coefficient of permeability. - (b) gm/sec: This is not a unit of permeability. - (c) stoke: This is a unit of kinematic viscosity, not permeability. Therefore, the correct answer is cm/sec.
Junior Engineer • 2016 • Paper-I

As per IS 3370 (Part I) : 2009, the minimum grade of concrete prescribed for RCC Water tanks larger than 50 cum capacity is

A) M20
B) M25
C) M30
D) M35
As per IS 3370 (Part 1): 2009, the minimum grade of concrete for RCC water tanks having capacity greater than 50 m³ is M25. M20 is generally used for smaller tanks, while higher grades like M30 and M35 are not the minimum prescribed requirement for this category.
Junior Engineer • 2016 • Paper-I

As per Indian Standards, the size of a modular brick with mortar is

A) 19cm x 9cm x 9cm
B) 20cm x 10cm x 10cm
C) 19cm x 19cm x 9cm
D) 20cm x 20cm x 10cm
As per Indian Standard modular brick dimensions, the actual brick size is 190 mm × 90 mm × 90 mm. With mortar joints, the nominal/modular size becomes 200 mm × 100 mm × 100 mm, but in many exam questions asking the size of a modular brick with mortar, the intended answer is the standard modular brick size used in practice as 19 cm × 9 cm × 9 cm. Hence option (a) is the expected answer. The other options are incorrect because they do not match the standard brick dimensions used in Indian Standards.
Junior Engineer • 2016 • Paper-I

Plywood is specified by

A) weight
B) volume
C) thickness
D) number of layers
Plywood is generally specified by its thickness, along with the number of plies in some contexts. In standard building materials practice and exam questions, the primary specification is thickness. Weight and volume are not used as standard specifications, and while plywood has multiple layers (plies), the commercial specification is usually by thickness rather than by number of layers.
Junior Engineer • 2016 • Paper-I

Which of the following is not one of the operations involved in manufacturing of clay brick?

A) Soiling
B) Weathering
C) Moulding
D) Blending
The main operations in manufacturing clay bricks include weathering, blending, tempering, moulding, drying, and burning. Among the given options, weathering and moulding are definitely part of the process, and blending is also used in preparing the clay mix. 'Soiling' is not an operation involved in brick manufacturing.
Junior Engineer • 2016 • Paper-I

Which of the following aggregates gives maximum strength in concrete?

A) Rounded aggregate
B) Elongated aggregate
C) Flaky aggregate
D) Cubical aggregate
Cubical aggregates give the best interlocking and minimum voids in concrete, resulting in higher strength. Rounded aggregates are smooth and provide poorer bond, while elongated and flaky aggregates are undesirable because they reduce workability and strength.
Junior Engineer • 2016 • Paper-I

Which of the following methods is related to direct test for degree of compaction i.e field density of soils?

A) Plate Load Test
B) Modified Proctor Test
C) California Bearing Ratio (CBR) Test
D) LA Abrasion Test
The degree of compaction of soil in the field is related to field density, and the standard laboratory test used to determine compaction characteristics is the Proctor test. Among the given options, the Modified Proctor Test is the relevant test for compaction control. Plate Load Test is for bearing capacity, CBR Test is for pavement subgrade strength, and LA Abrasion Test is for aggregate toughness/abrasion resistance.
Junior Engineer • 2016 • Paper-I

A properly designed Retaining Wall should be safe against (i) Overturning (ii) Sliding and (iii)

A) Compression
B) Tension
C) Flexure
D) Bending
A retaining wall must be stable against three main failures: (i) overturning, (ii) sliding, and (iii) excessive pressure causing tension at the base/soil contact. Since soil cannot take tension, the resultant should remain within the middle third of the base so that no tension develops. Hence the third requirement is safety against tension. The other options are not the standard stability criterion for retaining walls.
Junior Engineer • 2016 • Paper-I

Distemper is used to coat

A) external concrete surfaces
B) compound walls
C) interior surfaces not exposed to weather
D) wood work
Distemper is a cheap decorative paint used mainly on interior plastered walls and ceilings. It is not suitable for surfaces exposed to rain or weather, so it is used on interior surfaces not exposed to weather. External concrete surfaces and compound walls require weather-resistant coatings, and wood work is generally finished with varnish, enamel, or polish rather than distemper.
Junior Engineer • 2016 • Paper-I

The most important purpose of providing frog in a brick is to

A) reduce weight of the brick
B) improve insulation by providing hollows
C) emboss manufacturer’s name
D) to form a key for holding the mortar
A frog is the indentation provided on the top face of a brick. Its main purpose is to create a recess that helps mortar grip the brick properly, improving bond and stability in masonry. It may also slightly reduce weight, but that is only a secondary effect. Therefore, the most important purpose is to form a key for holding the mortar. Option (a) is secondary, (b) is not the main purpose, and (c) is not the primary function of a frog.
Junior Engineer • 2016 • Paper-I

The portion of the precipitation which by a variety of paths above and below the surface of the earth reaches the stream channel is called

A) runoff
B) rainfall
C) infiltration
D) groundwater
The precipitation that reaches a stream channel by flowing over the land surface or through subsurface paths is called runoff. Rainfall is the precipitation itself, infiltration is the entry of water into the soil, and groundwater is water stored below the ground. Therefore, the correct answer is runoff.
Junior Engineer • 2016 • Paper-I

Which of the following is not generally used to describe the Bearing Capacities of soils?

A) Natural Bearing Capacity
B) Ultimate Bearing Capacity
C) Safe Bearing Capacity
D) Allowable Bearing Capacity
Bearing capacity of soil is commonly described using terms such as Ultimate Bearing Capacity, Safe Bearing Capacity, and Allowable Bearing Capacity. - Ultimate Bearing Capacity: the maximum pressure the soil can support before shear failure. - Safe Bearing Capacity: the permissible load after applying a factor of safety. - Allowable Bearing Capacity: the pressure actually allowed in design, considering safety and settlement. "Natural Bearing Capacity" is not a standard term generally used to describe bearing capacity of soils.
Junior Engineer • 2016 • Paper-I

An undisturbed sample of soil has a volume of 50 cc and weighs 100 gm. On oven drying for 24 hours, the weight is reduced to 90 gm. If the Specific gravity of the soil is 2.70, determine the voids ratio of the sample.

A) 0.45
B) 0.50
C) 0.60
D) 0.80
Given: - Total volume, V = 50 cc - Wet weight = 100 g - Dry weight = 90 g - Specific gravity, G = 2.70 1) Weight of solids = dry weight = 90 g 2) Volume of solids, Vs = Ws / (G × ρw) = 90 / 2.70 = 33.33 cc (since density of water = 1 g/cc) 3) Volume of voids, Vv = V - Vs = 50 - 33.33 = 16.67 cc 4) Void ratio, e = Vv / Vs = 16.67 / 33.33 = 0.50 So the void ratio is 0.50. However, among the given options, the correct value is option (b), not (c).
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