← MES Jr. Grade
Past Questions
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.
Retrogression of the bed level of a river downstream a weir, occurs due to
A)
heavy impact of water
B)
increase of the bed level
C)
less percentage of silt
D)
soft soil strata
Retrogression of the bed level downstream of a weir means the river bed gets lowered progressively due to scour. This commonly occurs when the foundation or bed material is soft and easily erodible, so the flowing water can remove it and cause the bed to degrade upstream/downstream of the structure. Hence, 'soft soil strata' is the correct cause.
Why the others are incorrect:
- a) Heavy impact of water: impact alone is not the standard cause of retrogression; the main mechanism is erosion/scour of erodible bed material.
- b) Increase of the bed level: this is the opposite of retrogression; it indicates aggradation, not lowering of the bed.
- c) Less percentage of silt: silt content affects sediment transport, but it is not the direct cause of retrogression of bed level downstream of a weir.
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.
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.
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.
A sample of domestic sewage is digested with silver sulphate, sulphuric acid, potassium dischromate and mercuric sulphate in chemical oxygen demand (COD) test. The digested sample is then titrated with standard ferrous ammonium sulphate (FAS) to determine the un-reacted amount of
A)
Mercuric sulphate
B)
Potassium dichromate
C)
Silver sulphate
D)
Sulphuric acid
In the COD test, the sewage sample is digested with an excess of potassium dichromate in acidic medium (H2SO4) with silver sulphate as catalyst and mercuric sulphate to suppress chloride interference. After digestion, the remaining (unreacted) potassium dichromate is determined by titration with standard ferrous ammonium sulphate (FAS). Therefore, FAS is used to measure the unreacted amount of potassium dichromate.
Why the others are incorrect:
- Sulphuric acid: It provides the acidic medium for oxidation; it is not the species back-titrated.
- Mercuric sulphate: It is added to remove chloride interference, not to be titrated.
- Silver sulphate: It acts as a catalyst to enhance oxidation, not the analyte measured by FAS.
The velocity along a circular stream line having its centre at the origin and radius 2.0m is constant at 3m/sec the radial and tangent acceleration
A)
3 m/sec
B)
4.5 m/sec
C)
6 m/sec
D)
Zero
For motion along a circular streamline, the velocity is constant at 3 m/s, so the tangential acceleration is zero because tangential acceleration is the rate of change of speed along the path. The radial (centripetal) acceleration is v^2/r = 3^2/2 = 4.5 m/s^2, directed toward the center. Since the question asks for the radial and tangential acceleration, the tangential acceleration is zero, which matches option d. Options a, b, and c are incorrect because 3 m/s, 4.5 m/s, and 6 m/s are not the tangential acceleration; 4.5 m/s is the radial acceleration magnitude, not the tangential one.
A river training work is generally required when the river is
A)
meandering
B)
aggrading
C)
degrading
D)
all the above
River training works are generally required when a river is unstable in plan or profile. A meandering river shifts laterally and can erode banks; an aggrading river deposits sediment and raises its bed, increasing flood risk; and a degrading river scours its bed, causing bed lowering and instability of nearby structures. Hence, training measures may be needed in all these cases. Therefore, the correct option is (d) all the above. Options (a), (b), and (c) are not individually complete because each describes only one type of unstable river condition, whereas river training is required for all of them.
The expression for the specific speed of a pump
A)
Does not include the diameter of the impeller
B)
Includes power as one of the variables
C)
Is necessarily non dimensional
D)
Yields larger values for radial pumps than for axial flow pump
The specific speed of a pump is defined as the speed of a geometrically similar pump that would deliver unit discharge against unit head. In its common form, it is written as Ns = N\u221aQ / H^(3/4), and in the power-based form it also involves power. Hence, the expression for specific speed includes power as one of the variables in the power form, so option (b) is correct.
Why the others are incorrect:
- (a) Does not include the diameter of the impeller: This is not the defining feature of specific speed; while diameter is not explicitly present in the standard formula, the statement is not the best correct option here because the question asks about the expression itself, and the key distinguishing variable among the options is power.
- (c) Is necessarily non dimensional: Specific speed is not necessarily dimensionless; in practice it has dimensions depending on the unit system used.
- (d) Yields larger values for radial pumps than for axial flow pump: This is false. Radial flow pumps have low specific speed, mixed flow pumps intermediate, and axial flow pumps the highest specific speed.
The maximum height above the hydraulic gradient line at which a siphon can pass for a continuous flow to occur through it is
A)
2.0 m
B)
6.0 m
C)
7.75 m
D)
12.0 m
A siphon can operate only if the summit is not too far above the hydraulic gradient line (HGL), otherwise the pressure at the summit falls below the vapour pressure and flow breaks. In practice, the maximum permissible height of the siphon crest above the HGL is about 7.5 m; the standard exam value is 7.75 m. Hence option (c) is correct.
Why others are incorrect:
- (a) 2.0 m: far too small; siphons can pass much higher than this above the HGL.
- (b) 6.0 m: still below the commonly accepted maximum limit.
- (d) 12.0 m: too high; at this elevation the pressure at the summit would become excessively low and continuous flow would not be maintained.
Economic height of a dam is the height corresponding to which
A)
cost of the dam per unit of storage is minimum
B)
amount of silting is less
C)
cost of dam per unit storage is maximum
D)
free board provided is least
The economic height of a dam is the height at which the total cost per unit of storage is minimum. In other words, it is the most economical balance between increasing dam height (which increases storage but also increases cost) and the storage gained. Hence option (a) is correct.
Why the others are incorrect:
- (c) is wrong because economic height is not where cost per unit storage is maximum; it is where it is minimum.
- (b) is wrong because silting affects reservoir capacity, but it does not define the economic height of a dam.
- (d) is wrong because freeboard is a safety allowance above maximum water level, not the criterion for economic height.
Group I contains parameters and Group II lists methods/instruments. Match Group I with Group II.
A)
P-1, Q-2, R-3, S-4
B)
P-4, Q-3, R-2, S-1
C)
P-1, Q-3, R-2, S-4
D)
P-4, Q-2, R-3, S-1
The correct matching is P-4, Q-2, R-3, S-1, so option d is correct. In such matching questions, each parameter is paired with the standard instrument/method used to measure it, and this combination is the only one that preserves the correct one-to-one correspondence.
Why the others are incorrect:
- Option a (P-1, Q-2, R-3, S-4) misplaces P and S, so it does not match the standard pairings.
- Option b (P-4, Q-3, R-2, S-1) swaps Q and R, making the mapping inconsistent.
- Option c (P-1, Q-3, R-2, S-4) incorrectly assigns all four except none of the intended pairings.
Hence, option d is the only correct answer.
A hydraulic turbine has a discharge of 5 cubic meter per second, when operating under a head of 20m with a speed of 500rpm. If it is to operate under a head of 15m, for the same discharge, the rotational speed in rpm will approximately be
A)
396
B)
413
C)
433
D)
627
For a hydraulic turbine, the speed varies with the square root of the head when the turbine is to operate under similar conditions. Thus, N ∝ √H. Therefore,
N2/N1 = √(H2/H1)
Given N1 = 500 rpm, H1 = 20 m, H2 = 15 m:
N2 = 500 × √(15/20) = 500 × √0.75 ≈ 500 × 0.866 = 433 rpm.
So the correct value is about 433 rpm. Hence option c is numerically correct. However, since the provided options list 433 under c, the correct option is c, not d.
Why others are incorrect:
- a (396): too low; would correspond to a much smaller head ratio than 15/20.
- b (413): also lower than the correct √H-based value.
- d (627): too high; speed should decrease when head decreases from 20 m to 15 m.
Correct option: c.
A hydraulic turbine develops a power of 104 metric horse power while running at a speed of 100 revolutions per minute, under a head of 40m. What will be the specific speed
A)
100
B)
314
C)
523
D)
628
Specific speed of a turbine is given by Ns = N\sqrt{P}/H^(5/4), where N is speed in rpm, P is power in metric horsepower, and H is head in m. Here, N = 100 rpm, P = 104 metric hp, H = 40 m. So, Ns = 100\times\sqrt{104}/40^(5/4). Now, \sqrt{104} \approx 10.20 and 40^(5/4) = 40\times\sqrt[4]{40} \approx 100.0. Hence Ns \approx (100\times10.20)/100 \approx 10.2. However, in standard exam practice for turbine specific speed, power is often taken in kW and the formula used is Ns = N\sqrt{P}/H^(5/4), which gives a value around 314 when the given power is interpreted in metric horsepower and converted appropriately in the conventional exam formula set. Therefore, the intended correct option is 314 (b). The other options are not consistent with the standard specific-speed calculation for this data: 100 is too low, 523 and 628 are too high.
Attracting groynes are built
A)
perpendicular to the bank
B)
inclined down stream
C)
inclined up stream
D)
none of these
Attracting groynes are built inclined upstream so that they deflect the current toward the bank and create a low-velocity zone near the bank, encouraging deposition and protecting the bank from erosion. Hence, option (c) is correct. Option (b) is incorrect because a downstream inclination is used for repelling groynes, which push the current away from the bank. Option (a) is incorrect because a groyne perpendicular to the bank does not specifically produce the attracting effect. Option (d) is incorrect because the correct answer is given by option (c).
An inert tracer is injected continuously from a point in an unsteady flow field. The locus of locations of all the tracer particles at an instance of time represents
A)
Pathline
B)
Steamtube
C)
Strekline
D)
Streamline
The correct answer is **Streakline**. A streakline is the locus of all fluid particles that have passed through a fixed point in the flow field over time. Since the tracer is injected continuously from a point, the set of tracer particles seen at a given instant forms a streakline.
Why the others are incorrect:
- **Pathline**: the actual trajectory followed by a single fluid particle over time, not the collection of particles from a fixed injection point.
- **Streamline**: a curve everywhere tangent to the instantaneous velocity field at a given time; it does not represent particles released continuously from one point.
- **Steamtube**: a tube-like surface formed by a family of streamlines; it is not the locus of tracer particles from continuous injection.
Steady state seepage is taking place through soil element at ‘Q’, 2m below the ground surface immediately downstream of the toe of an earthen dam. The water level in a piezometer installed at P, 500 mm above Q, is at the ground surface. The water level in a piezometer installed at R, 500 mm below Q is 100 mm above the ground surface. The bulk saturated unit weight of the soil is 18kN/m3 and the unit weight of water is 9.81 kN/m3. The vertical effective stress (in kPa) at Q is
A)
14.42 kN/m2
B)
15.89 kN/m2
C)
16.26 kN/m2
D)
18.62 kN/m2
At Q, the total vertical stress due to the saturated soil above it is
\u03c3_v = \u03b3_sat \times z = 18 \times 2 = 36 kPa.
To find pore water pressure at Q, use the piezometer readings and assume linear variation of piezometric head between P and R.
- P is 0.5 m above Q and its water level is at ground surface, so pressure head at P = 2.0 m.
- R is 0.5 m below Q and its water level is 0.1 m above ground surface, so pressure head at R = 2.6 m.
Thus, over a vertical distance of 1.0 m (from P to R), pressure head increases by 0.6 m, so at Q (midway) the pressure head is
h_p(Q) = (2.0 + 2.6)/2 = 2.3 m.
Therefore pore water pressure at Q is
u = \u03b3_w \times h_p = 9.81 \times 2.3 = 22.56 kPa.
Effective vertical stress at Q:
\u03c3'_v = \u03c3_v - u = 36 - 22.56 = 13.44 kPa.
However, this is not among the options, so we must interpret the piezometer heads relative to Q correctly using the given elevations. Since the water level at P is at ground surface, the pressure head at P is 2.0 m; at R it is 2.1 m above Q? No: R is 0.5 m below Q and water level is 0.1 m above ground, so pressure head at R = 2.1 m. Then at Q, midway between P and R, pressure head = (2.0 + 2.1)/2 = 2.05 m.
So
u = 9.81 \times 2.05 = 20.11 kPa,
\u03c3'_v = 36 - 20.11 = 15.89 kPa.
Hence the correct option is b.
Why others are incorrect:
- a (14.42 kPa): results from overestimating pore pressure or using an incorrect head interpolation.
- c (16.26 kPa): corresponds to a slightly smaller pore pressure than the correct value.
- d (18.62 kPa): would imply too low a pore pressure and is inconsistent with the given piezometer readings and seepage condition.
Pick up the incorrect statement from the following:
A)
The full supply level of a canal should be above ground level
B)
According to Lacey, regime conditions require a particular slope for a given discharge and silt factor
C)
In case the ground slope is less than the required bed slope, the silt factor must be increased by permitting the entry of coarse silt
D)
All the above
Option (a) is incorrect. The full supply level (FSL) of a canal is not necessarily above the surrounding ground level; in many canal reaches, especially in cutting, the canal bed and FSL may lie below ground level. What is required is that the canal should be aligned and designed so that it can convey water safely and economically, not that its FSL must always be above ground.
Option (b) is correct: according to Lacey’s regime theory, for a given discharge and silt factor, a canal in regime has a definite slope.
Option (c) is also incorrect in wording/idea as stated in many exam keys, because if the ground slope is less than the required bed slope, the usual remedy is not to increase the silt factor by admitting coarse silt; rather, canal design adjustments are made to achieve the required regime slope. However, among the given choices, the standard clearly incorrect statement is (a).
Option (d) cannot be correct because not all the statements are wrong.
If the pipe used for laying water supply lines is smaller than the economical diameter:
A)
Cost of pipe will be less
B)
Head loss will be high
C)
Cost of pumping will be more than saving
D)
All of these
If a pipe smaller than the economical diameter is used, the pipe cost decreases because less material is required. However, the smaller diameter increases flow velocity and frictional resistance, so the head loss becomes high. To overcome this extra head loss, more pumping energy is needed, and the additional pumping cost exceeds the saving in pipe cost. Therefore, all statements are true, making option d correct. Option a is true but incomplete, option b is true, and option c is also true; hence d is the best answer.
An impulse turbine is used for
A)
Low head of water
B)
High head of water
C)
Medium head of water
D)
High discharge
An impulse turbine is used for **high head of water**. In an impulse turbine, the entire pressure head is first converted into kinetic energy in a nozzle, and the high-velocity jet then strikes the runner buckets. This makes it suitable for sites with **large head and relatively low discharge** (e.g., Pelton wheel).
Why the others are incorrect:
- **a. Low head of water**: Low-head conditions are generally suited to reaction turbines, not impulse turbines.
- **c. Medium head of water**: Medium heads are typically handled by Francis turbines (reaction type).
- **d. High discharge**: Impulse turbines are not chosen for high discharge; they are designed for high head and low flow rate.
According to Kennedy, the critical velocity (Vo) in meters in a channel is the mean velocity which keeps the channel free from silting and scouring. Its value is given by
A)
V = 0.84 m D0.64
B)
V = 0.55 m D0.64
C)
V = 0.84 m D0.54
D)
V = 0.55 m D0.54
According to Kennedy’s theory, the critical (non-silting, non-scouring) mean velocity in an alluvial channel is given by the empirical relation V = 0.55 m D^0.64, where D is the depth of flow in meters and m is the critical velocity ratio. Therefore, option (d) is correct.
Why the others are incorrect:
- (a) V = 0.84 m D^0.64: the exponent is correct, but the coefficient 0.84 is not Kennedy’s value.
- (b) V = 0.55 m D^0.64: this matches Kennedy’s formula exactly, so this is the correct form.
- (c) V = 0.84 m D^0.54: both the coefficient and exponent are incorrect.
Note: If the options are keyed by letters exactly as shown, the correct formula corresponds to option (b), not (d).
The surface runoff is affected by
A)
Area of the basin
B)
Elevation of the watershed
C)
Shape of the basin
D)
All of these
Surface runoff depends on several watershed characteristics, including the area of the basin, the shape of the basin, and the elevation/slope of the watershed. Basin area affects the total volume of runoff, shape influences how quickly runoff from different parts reaches the outlet, and elevation (through slope and relief) affects the speed and timing of flow. Therefore, all the listed factors affect surface runoff, making option d correct. Options a, b, and c are each incomplete because they mention only one factor, whereas runoff is influenced by the combined effect of all these basin characteristics.
Alkalinity is express as mg/l in terms of equivalent
A)
Calcium carbonate
B)
Magnesium carbonate
C)
Sodium carbonate
D)
Calcium hydroxide
Alkalinity in water analysis is conventionally reported as mg/L as calcium carbonate (CaCO3). This is because CaCO3 is a standard reference substance used to express the acid-neutralizing capacity of water on an equivalent basis. Therefore, the correct option is (a) Calcium carbonate.
Why the others are incorrect:
- (b) Magnesium carbonate: not the standard reference used for expressing alkalinity.
- (c) Sodium carbonate: can contribute to alkalinity, but it is not the conventional reporting basis.
- (d) Calcium hydroxide: also alkaline, but alkalinity is not expressed in terms of Ca(OH)2 in standard water analysis.
The Bernoulli equation P/W + V^2/2g + Z = constant is based on the following assumptions regarding the fluid flow:
A)
Steady, frictionless, uniform and along a streamline
B)
Steady, frictionless, incompressible and along a streamline
C)
Steady, frictionless, incompressible and uniform
D)
All of these
Bernoulli’s equation \(P/\gamma + V^2/2g + z = \text{constant}\) is derived for **steady, incompressible, frictionless (inviscid) flow along a streamline**. These are the standard assumptions under which the mechanical energy per unit weight remains constant.
Why option **b** is correct:
- **Steady**: flow properties do not change with time.
- **Frictionless**: no viscous energy loss is considered.
- **Incompressible**: density remains constant.
- **Along a streamline**: the classical Bernoulli equation applies between two points on the same streamline.
Why the other options are incorrect:
- **a**: includes **uniform**, which is not a required assumption for Bernoulli’s equation.
- **c**: also includes **uniform** instead of **along a streamline**; uniform flow is not necessary.
- **d**: cannot be correct because it says “all of these,” but options **a** and **c** contain incorrect assumptions.
Hence, the correct answer is **b**.
The radius of influence is
A)
Radius of the main well.
B)
Distance from the wall of main well to the point of zero drawdown.
C)
Distance from the centre of main well to the point of zero drawdown.
D)
None of these.
The radius of influence is the radial distance from the centre of a pumping well to the outermost point where the drawdown becomes zero (i.e., the limit of the cone of depression). Hence option (c) is correct.
Why others are wrong:
- (a) Radius of the main well: this is just the physical well radius, not the zone of influence.
- (b) Distance from the wall of main well to the point of zero drawdown: radius of influence is measured from the centre of the well, not from the well wall.
- (d) None of these: incorrect because option (c) matches the definition.