← MES Jr. Grade

Past Questions

MES Jr. Grade • 2014 • Paper-II

Φ index is defined as

A) The difference between maximum and minimum infiltration capacities
B) The difference between the total rainfall and the total runoff divided by the duration of the storm
C) The rainfall intensity above which the rainfall volume equals the observed runoff volume
D) The minimum infiltration rate during the storm
The Φ-index is a constant infiltration rate assumed for a storm such that the rainfall in excess of this rate equals the observed direct runoff. In other words, it is the rainfall intensity above which the rainfall volume becomes equal to the runoff volume. Option (d) describes the W-index more closely in some contexts, while (a) and (b) are not the definition of Φ-index.
MES Jr. Grade • 2014 • Paper-II

A flood is given a rank m in an observed record of n years. Then according to Weibul’s formula, its return period is given by

A) n/m
B) n/(m+1)
C) n/(m-1)
D) (n+1)/m
According to Weibull’s plotting position formula, the return period T of a flood ranked m in an observed record of n years is T = n/m. Here, m is the rank of the flood event in descending order of magnitude. The other options do not match Weibull’s standard formula.
MES Jr. Grade • 2014 • Paper-II

Mach’s number is defined as the square root of the ratio of

A) Inertia force to the pressure force
B) Inertia force to the surface tension force
C) Inertia force to the elastic force
D) None of these
Mach number is the ratio of flow velocity to the speed of sound. In fluid mechanics, it can be expressed as the square root of the ratio of inertia force to elastic force (compressibility effect). Therefore, the correct option is (c). Option (a) corresponds to Euler number, and option (b) corresponds to Weber number.
MES Jr. Grade • 2014 • Paper-II

The average domestic demand of water for an Indian city is

A) 135 lpcd
B) 270 lpcd
C) 500 lpcd
D) 750 lpcd
For an Indian city, the standard average domestic water demand is about 135 litres per capita per day (lpcd), as commonly used in water supply and environmental engineering norms. The other values are too high for average domestic demand: 270 lpcd and above are not typical Indian urban domestic norms.
MES Jr. Grade • 2014 • Paper-II

Two primary air pollutants are

A) sulphur oxide and ozone
B) nitrogen oxide and peroxyacetylnitrate
C) sulphur dioxide and hydrocarbon
D) ozone and peroxyacetylnitrate
Primary air pollutants are those emitted directly from a source into the atmosphere. Sulphur dioxide and hydrocarbons are directly released from combustion and industrial processes, so they are primary pollutants. Ozone and peroxyacetylnitrate are secondary pollutants because they are formed in the atmosphere by photochemical reactions. Nitrogen oxides are primary pollutants, but peroxyacetylnitrate is secondary, so option (b) is incorrect.
MES Jr. Grade • 2014 • Paper-II

Double mass curve technique is used

A) to prepare rainfall hyetograph
B) to derive the hydrograph
C) to derive the S-curve hydrograph
D) to check the consistency of rainfall records
The double mass curve technique is a hydrological method used to test the consistency of rainfall data. It compares cumulative rainfall at a station with cumulative rainfall of nearby stations or a reference group. If the slope changes, it indicates inconsistency due to changes in gauge location, exposure, instruments, or observing practices. It is not used to prepare a hyetograph, derive a hydrograph, or derive an S-curve hydrograph.
MES Jr. Grade • 2014 • Paper-II

The velocity of flow in a sewer should be between

A) 0.6 m/sec and 3 m/sec
B) 2 m/sec and 10 m/sec
C) 30 cm/sec and 90 cm/sec
D) 30 m/sec and 90 m/sec
In sewer design, the self-cleansing velocity should be high enough to prevent deposition of solids, but not so high that it causes abrasion or damage to the sewer. The commonly accepted range is about 0.6 m/s to 3 m/s. Hence, option (a) is correct. The other options are either too low to prevent silting or unrealistically high for sewer flow.
MES Jr. Grade • 2014 • Paper-II

The flow in open channel is laminar if the Reynold number is

A) 2000
B) less than 2000
C) less than 5000
D) none of these
For open channel flow, the Reynolds number criterion is commonly taken as: laminar flow for Re < 500, transitional flow for Re between 500 and 2000, and turbulent flow for Re > 2000. Therefore, the flow is laminar when the Reynolds number is less than 2000. Option (a) is not a condition but a value, and (c) is too high for laminar flow.
MES Jr. Grade • 2014 • Paper-I

The Young’s Modulus of elasticity of steel is

A) 150 kN/mm2
B) 200 kN/mm2
C) 250 kN/mm2
D) 300 kN/mm2
Young’s modulus of elasticity for steel is commonly taken as about 200 GPa, which is equal to 200 kN/mm². Hence option (b) is correct. The other values are not the standard modulus for steel.
MES Jr. Grade • 2014 • Paper-I

When HYSD bars are used the minimum quantity of reinforcement in a slab in each principal direction is

A) 0.12%
B) 0.13%
C) 0.16%
D) 0.60%
As per IS 456:2000, the minimum steel in a slab is: - 0.15% of the total cross-sectional area when mild steel bars are used - 0.12% of the total cross-sectional area when HYSD bars are used Since the question asks about HYSD bars, the correct minimum reinforcement in each principal direction is 0.12%. The other options do not match the codal provision.
MES Jr. Grade • 2014 • Paper-I

Numbers of bricks required in one cubic meter of brick masonry is

A) 400
B) 450
C) 500
D) 550
In standard estimation, the number of modular bricks required for 1 cubic meter of brick masonry is taken as about 500. This is based on the nominal size of bricks including mortar joints. Therefore, option (c) 500 is correct. The other options are not the standard approximate value used in civil engineering estimates.
MES Jr. Grade • 2014 • Paper-I

A steel rod 1 m long having square cross section is pulled under a tensile load of 8 tones. The extension in the rod was 1 mm only. If E =2x106 kg/cm2, the side of the rod is

A) 4 cm
B) 2 cm
C) 1 cm
D) 1.5 cm
Use the extension formula for a prismatic bar: \(\delta = \frac{PL}{AE}\). Given: - \(L = 1\,m = 100\,cm\) - \(P = 8\) tonnes \(= 8000\,kgf\) - \(\delta = 1\,mm = 0.1\,cm\) - \(E = 2\times 10^6\,kg/cm^2\) So, \[ A = \frac{PL}{E\delta} = \frac{8000 \times 100}{(2\times 10^6)\times 0.1} = \frac{800000}{200000} = 4\,cm^2 \] The rod has a square cross-section, so side \(= \sqrt{A} = \sqrt{4} = 2\,cm\). Hence, the correct option is (b) 2 cm. Other options do not match the area required by the given load, extension, and modulus of elasticity.
MES Jr. Grade • 2014 • Paper-I

An R.C.C column is treated as short column if its slenderness ratio is less than

A) 30
B) 35
C) 40
D) 50
In RCC design, a column is treated as a short column when its slenderness ratio is less than 12 for braced columns in general design practice, but in many exam contexts for RCC columns the commonly used limiting value for short columns is 30. Among the given options, 30 is the correct threshold. The other values (35, 40, 50) are higher and do not represent the standard short-column limit used in such MCQs.
MES Jr. Grade • 2014 • Paper-I

The strength of a riveted joint is equal to

A) shearing strength of the rivets
B) bearing strength of the rivets
C) tearing strength of the plate
D) least of (a), (b) and (c)
In a riveted joint, failure can occur by: (1) shearing of rivets, (2) bearing/crushing of rivets or plate, or (3) tearing of the plate. The actual strength of the joint is governed by the weakest mode, so it is equal to the least of these three strengths. Hence, option (d) is correct.
MES Jr. Grade • 2014 • Paper-I

A rectangular block of size 200mm x 100 mm x 50 mm is subjected to a shear stress of 100 N/mm2. If modulus of rigidity of material is 1x 105 N/mm2, strain energy stored will be

A) 10 N.m
B) 25 N.m
C) 50 N.m
D) 100 N.m
For a body under pure shear, strain energy stored is U = (τ^2 / 2G) × Volume. Given: τ = 100 N/mm² G = 1 × 10^5 N/mm² Volume = 200 × 100 × 50 = 1,000,000 mm³ So, U = (100² / (2 × 10^5)) × 1,000,000 = (10000 / 200000) × 1,000,000 = 0.05 × 1,000,000 = 50,000 N·mm = 50 N·m. Thus the correct answer is 50 N·m, which corresponds to option (c). Note: If the intended formula is used directly, the result is 50 N·m, not 25 N·m. So the correct option is (c).
MES Jr. Grade • 2014 • Paper-I

Steel beam theory is used for

A) design of simple steel beams
B) steel beams encased in concrete
C) doubly reinforced beams ignoring compressive stress in concrete
D) beams if shear exceeds 4 times of allowable shear stress
Steel beam theory is the basic elastic theory used for designing ordinary steel beams subjected to bending. It assumes the beam behaves elastically and is used for simple steel beam design. Option (b) refers to composite/encased beams, (c) is related to reinforced concrete beam theory, and (d) is not a standard use of steel beam theory.
MES Jr. Grade • 2014 • Paper-I

The type of window provided on the sloping side of a pitched roof is called

A) Gable window
B) Dormer window
C) Lantern
D) None of these
A window provided on the sloping side of a pitched roof is called a dormer window. It projects from the roof slope to admit light and ventilation. A gable window is located in the gable end wall, while a lantern is a raised structure on the roof used mainly for light and ventilation.
MES Jr. Grade • 2014 • Paper-I

Moment of Inertia of a circular section of diameter ‘d’ about an axis passing through its C.G. and lying in the plane of the section is given by

A) I = πd4 / 32
B) I = πd4 / 64
C) I = πd4 / 16
D) none of these
For a solid circular section, the moment of inertia about any centroidal axis lying in the plane of the section (diametral axis) is I = πd^4/64. This is the standard formula used in Strength of Materials. Option (a) πd^4/32 is the polar moment of inertia about the centroidal axis perpendicular to the plane, not the in-plane axis. Hence the correct answer is (b).
MES Jr. Grade • 2014 • Paper-I

In a well-seasoned timber, the moisture content will be in the range of

A) 20-25%
B) 15-20%
C) 10-12%
D) 5-7%
Well-seasoned timber is timber that has been dried sufficiently for use, and its moisture content is typically reduced to about 10–12%. This range gives good strength, durability, and dimensional stability. Higher moisture contents like 15–25% indicate less seasoning, while 5–7% is unusually low for ordinary well-seasoned timber.
MES Jr. Grade • 2014 • Paper-I

The depth of footing for an isolated column is governed by: 1) Maximum bending moment, 2) Shear Force, 3) Punching Shear; The Correct answer is :

A) Only (1)
B) Both (1) and (2)
C) Both (1) and (3)
D) All (1), (2) and (3)
For an isolated footing, the depth (effective depth/thickness) is selected to safely resist all critical structural actions: (1) bending moment, (2) one-way shear, and (3) punching shear. The footing must be thick enough so that flexural strength and both shear checks are satisfied. Therefore, all three govern the footing depth.
MES Jr. Grade • 2014 • Paper-I

Floor Area Ratio (FAR) is calculated as

A) Ratio of total covered area of all floors to the plot area
B) Ratio of plot area to the total covered area of all floors
C) Ratio of covered area of ground floor to the plot area
D) Ratio of plot area to the covered area of ground floor
Floor Area Ratio (FAR), also called Floor Space Index (FSI), is defined as the total built-up/covered floor area of all floors divided by the plot area. Therefore, option (a) is correct. Options (b) and (d) reverse the ratio, and option (c) considers only the ground floor, which is not the definition of FAR.
MES Jr. Grade • 2014 • Paper-I

Coefficient of friction depends on

A) Nature of surfaces only
B) Area of contact only
C) Both (a) and (b)
D) None of these
The coefficient of friction depends on the nature and condition of the two surfaces in contact, such as roughness and material. In basic engineering mechanics, it is considered independent of the apparent area of contact. Therefore, option (a) is correct. Option (b) is incorrect because area of contact does not affect the coefficient of friction in the ideal model, and option (c) is also incorrect for the same reason.
MES Jr. Grade • 2014 • Paper-I

The thickness at the edge of a reinforced concrete footing resting on soil shall not be less than

A) 25 cm
B) 20 cm
C) 15 cm
D) 10 cm
For reinforced concrete footings resting on soil, standard RCC design provisions specify that the thickness at the edge should not be less than 15 cm. This ensures adequate structural integrity, durability, and resistance to cracking at the footing edge. Hence, option (c) is correct. The other options are either too small for safe detailing or not the code-prescribed minimum.
MES Jr. Grade • 2014 • Paper-I

When an end of a continuous beam is fixed, in Kani’s method, the rotation contribution will be

A) zero
B) EI/l
C) 2EI/l
D) 4EI/l
In Kani’s method, a fixed end cannot rotate, so the rotation contribution at a fixed end is zero. Hence the correct option is (a) zero. The other values like EI/L, 2EI/L, or 4EI/L relate to member stiffness factors or distribution concepts, not the rotation contribution of a fixed support.
MES Jr. Grade • 2014 • Paper-I

The path traced by a projectile in the sky is

A) circular
B) parabolic
C) elliptical
D) catenary
For ideal projectile motion (neglecting air resistance), the horizontal motion is uniform and the vertical motion is uniformly accelerated due to gravity. Combining these two motions gives a parabolic trajectory. Hence, the path traced by a projectile is parabolic. Circular, elliptical, and catenary paths do not describe standard projectile motion.
Page 177
← Prev Next →