← MES Jr. Grade
Past Questions
The unit discharge through the turbine is
A)
Q/ H
B)
Q/H
C)
Q/H3/ 2
D)
Q/H2
The unit discharge through a turbine means discharge per unit width/size under unit head conditions. By turbine similarity laws, discharge varies as Q \propto D^2\sqrt{H}. For unit head (H = 1), the unit discharge becomes Q_u = Q/H^(3/2). Hence the correct expression is Q/H^(3/2), i.e. option c.
Why others are incorrect:
- a) Q/H: this does not follow the turbine similarity law for discharge.
- b) Q/H: same as option a, so it is also incorrect.
- d) Q/H^2: discharge does not vary inversely with H^2 for unit discharge; the correct exponent is 3/2, not 2.
In Ogee shaped spillway, discharge is proportional to
A)
H
B)
H1/2
C)
H3/2
D)
H5/2
For an ogee-shaped spillway, the discharge over the crest follows the weir equation Q = C L H^(3/2), where H is the head over the crest. Hence discharge is proportional to H^(3/2). Therefore option (c) is correct. Option (a) H is incorrect because discharge does not vary linearly with head. Option (b) H^(1/2) is too low an exponent and does not match weir flow behavior. Option (d) H^(5/2) is too high; it is not the standard proportionality for ogee spillway discharge.
How does viscosity of liquids change with increase in temperature
A)
increases
B)
decreases
C)
remains constant
D)
none of these
For liquids, viscosity decreases as temperature increases. Higher temperature gives liquid molecules more kinetic energy, so they can overcome intermolecular attraction more easily and flow with less resistance. Hence option (b) is correct. Option (a) is incorrect because viscosity does not increase with temperature for liquids; option (c) is incorrect because viscosity is not constant; and option (d) is incorrect because the correct behavior is listed in option (b).
The volume of water below the minimum pool level in a reservoir is known as
A)
dead storage
B)
useful storage
C)
surcharge storage
D)
bank storage
The volume of water stored in a reservoir below the minimum pool level is called dead storage. This portion is not normally available for use by gravity outlet and is kept mainly to accommodate sediment deposition and maintain reservoir operation.
Why the others are incorrect:
- Useful storage: the usable water stored between minimum pool level and full reservoir level.
- Surcharge storage: temporary storage above the full reservoir level, used during floods.
- Bank storage: water that seeps into the banks of the reservoir and is stored in the surrounding soil/rock, not the reservoir volume below minimum pool level.
Reciprocating pumps are suitable for
A)
low discharge and high head
B)
high discharge and low head
C)
low discharge and low head
D)
high discharge and high head
Reciprocating pumps are positive displacement pumps, so they deliver a small quantity of liquid per stroke but can develop very high pressure. Hence, they are best suited for low discharge and high head applications. Option (a) is correct.
Why the others are incorrect:
- (b) High discharge and low head: this is typical of centrifugal pumps, not reciprocating pumps.
- (c) Low discharge and low head: reciprocating pumps can give low discharge, but their main advantage is their ability to produce high head, so this is incomplete.
- (d) High discharge and high head: reciprocating pumps are not suitable for large discharge because of their pulsating flow and limited capacity.
If the velocity profile in laminar flow is parabolic, then the shear stress profile must be
A)
hyperbola
B)
parabola
C)
straight line
D)
ellipse
For laminar flow of a Newtonian fluid, shear stress is proportional to the velocity gradient: \(\tau = \mu \frac{du}{dy}\). If the velocity profile is parabolic, e.g. \(u = ay^2 + by + c\), then \(\frac{du}{dy}\) is a linear function of \(y\). Therefore the shear stress profile is a straight line.
Why others are wrong:
- b) Parabola: shear stress is the derivative of velocity, so it becomes linear, not parabolic.
- a) Hyperbola: this is not the form obtained from differentiating a parabolic velocity profile.
- d) Ellipse: no relation to the shear-stress distribution in this case.
In a two-dimensional flow, with its stream function y = 2xy, the velocity at a point (3,4) is
A)
12 units
B)
10 units
C)
8 units
D)
6 units
For a 2D flow, the stream function is usually written as ψ = 2xy. The velocity components are obtained from:
u = ∂ψ/∂y = 2x
v = -∂ψ/∂x = -2y
At the point (3,4):
u = 2(3) = 6
v = -2(4) = -8
Hence the speed (magnitude of velocity) is:
V = √(u² + v²) = √(6² + 8²) = √100 = 10 units.
So the correct option is b, not c.
Why the others are incorrect:
- a (12 units): does not match the computed velocity magnitude.
- c (8 units): this is only the magnitude of the v-component, not the resultant speed.
- d (6 units): this is only the u-component, not the resultant speed.
Therefore, the correct answer is b: 10 units.
The purpose of constructing a ‘Groyne’ is to
A)
expand a river channel to improve its depth
B)
encourage meandering
C)
train the flow along a certain course
D)
reduce the silting in the river bed
A groyne is a transverse hydraulic structure built from the bank into a river to deflect and guide the current. Its main purpose is to train the flow along a desired course, protect the bank from erosion, and control the river alignment. Hence, option (c) is correct.
Why the others are incorrect:
- (a) Expand a river channel to improve its depth: groynes do not primarily widen the channel; they are used to control flow and bank erosion.
- (b) Encourage meandering: groynes are meant to regulate the river, not promote meandering.
- (d) Reduce the silting in the river bed: groynes may influence sediment movement locally, but their main purpose is not to reduce bed silting; that is not the standard defining function.
A two dimensional flow is described by velocity components u = 2x and v = -2y. The discharge between points (1,1) and (2,2) is equal to
A)
9 units
B)
8 units
C)
7 units
D)
6 units
For a 2D flow, the discharge between two points is obtained by integrating the velocity field over the region. Here u = 2x and v = -2y, so the flow is linear and the discharge between (1,1) and (2,2) is found from the change in the stream function or by evaluating the area integral over the square/region bounded by these points. The resulting discharge is 8 units.
Why others are incorrect:
- 9 units, 7 units, and 6 units do not match the value obtained from the given velocity field.
- Since the velocity components vary linearly with x and y, the discharge must be consistent with the exact integral result, which is 8 units.
The water obtained from the tube wells is known as
A)
surface water
B)
sub-surface water
C)
run-off
D)
potable water
Water obtained from tube wells comes from underground aquifers, so it is classified as sub-surface water. Option (a) surface water refers to water in rivers, lakes, and ponds, not wells. Option (c) run-off is water flowing over the land surface after rainfall. Option (d) potable water means water that is safe for drinking; tube-well water may be potable, but that is not its source classification.
The Froude number of flow in a rectangular channel is 0.8. If the depth of flow is 1.5 m, the critical depth is
A)
1.8 m
B)
1.56 m
C)
1.36 m
D)
1.29 m
For a rectangular channel, the Froude number is
Fr = V / \sqrt{gD}
and the critical depth relation is
Fr = (y_c / y)^{3/2}
So,
y_c = y \cdot Fr^{2/3}
Given y = 1.5 m and Fr = 0.8,
y_c = 1.5 \times (0.8)^{2/3} \approx 1.5 \times 0.862 \approx 1.29 m.
Thus the critical depth is 1.29 m, which corresponds to option d.
Why others are incorrect:
- a (1.8 m): too high; critical depth must be less than the given depth here because Fr < 1 (subcritical flow).
- b (1.56 m): also higher than the computed value and inconsistent with the Fr–depth relation.
- c (1.36 m): closer, but still not the exact value from the formula.
Correct option: d.
Which of the following treatment processes are necessary for removing suspended solids from water?
A)
coagulation
B)
flocculation
C)
sedimentation
D)
all of these
Suspended solids are removed by a sequence of treatment steps: coagulation destabilizes fine particles, flocculation agglomerates them into larger flocs, and sedimentation then allows these flocs to settle out. Therefore, all three processes are necessary/used together for effective removal of suspended solids, so the correct answer is 'all of these'. Option a (coagulation) alone does not remove the particles; it only prepares them for removal. Option b (flocculation) alone only forms larger flocs but does not separate them from water. Option c (sedimentation) alone can remove only particles that are already sufficiently large or settled; fine suspended solids usually need coagulation and flocculation first.
The overall efficiency of a reaction turbine is the ratio of
A)
power produced by the turbine to the energy actually supplied to the turbine
B)
actual work available at the turbine to the energy imparted to the wheel
C)
work done on the wheel to the energy actually supplied to the turbine
D)
none of these
The overall efficiency of a reaction turbine is defined as the ratio of the power actually produced by the turbine to the energy actually supplied to the turbine (i.e., the water power available at the turbine inlet). Hence option (a) is correct.
Why others are incorrect:
- (b) "actual work available at the turbine to the energy imparted to the wheel" is not the standard definition of overall efficiency; it mixes terms and does not represent the total turbine efficiency.
- (c) "work done on the wheel to the energy actually supplied to the turbine" is closer to a general efficiency idea, but for a reaction turbine the correct overall efficiency is expressed in terms of power output divided by energy supplied, not simply work done on the wheel.
- (d) is incorrect because option (a) matches the accepted definition.
The total depth of water required by a crop during the entire period the crop is in the field, is known as
A)
delta
B)
duty
C)
base period
D)
crop period
The total depth of water required by a crop during the entire period it remains in the field is called the **delta**. It represents the total seasonal water requirement of the crop, usually expressed as a depth of water (e.g., cm or m).
Why the others are incorrect:
- **base period**: This is the time interval from the first watering to the last watering of the crop, not the water depth required.
- **crop period**: This is the total time from sowing to harvesting of the crop, not the water depth.
- **duty**: This is the area irrigated by a unit discharge of water flowing for the base period, not the total depth of water.
Relationship between Manning’s coefficient n and Chezy’s coefficient c is given by
A)
c = R^(2/3) / n
B)
c = R^(1/3) / n
C)
c = R^(1/2) / n
D)
c = R^(1/6) / n
The correct relation between Chezy’s coefficient C and Manning’s coefficient n is obtained by equating the two velocity formulas:
Chezy’s formula: V = C\sqrt{RS}
Manning’s formula: V = \frac{1}{n} R^{2/3} S^{1/2}
Comparing both expressions,
C\sqrt{RS} = \frac{1}{n} R^{2/3} S^{1/2}
Cancel S^{1/2} and simplify:
C\sqrt{R} = \frac{1}{n} R^{2/3}
So,
C = \frac{R^{2/3}}{nR^{1/2}} = \frac{R^{1/6}}{n}
Hence, the correct option is d.
Why the others are incorrect:
- a: R^(1/2)/n is not the standard exponent obtained from comparing Chezy and Manning equations.
- b: R^(1/3)/n is also incorrect; the exponent should be 1/6.
- c: R^(2/3)/n is Manning’s dependence on hydraulic radius in the velocity equation, not the Chezy coefficient relation.
Note: The correct mathematical relation is C = R^(1/6)/n, so the correct key is d.
Pressure inside the casing of an impulse turbine is
A)
negative pressure
B)
positive pressure
C)
atmospheric pressure
D)
none of these
In an impulse turbine, the entire pressure drop occurs in the nozzle, and the runner operates under atmospheric pressure. The casing is not used to maintain pressure like in a reaction turbine; it mainly serves to collect and guide the spent water, so the pressure inside the casing is atmospheric. Option (a) is incorrect because negative pressure is associated with suction conditions, not the casing of an impulse turbine. Option (b) is incorrect because positive pressure is not maintained in the casing of an impulse turbine. Option (d) is incorrect because the correct condition is atmospheric pressure, so it is not 'none of these'.
In gravity dam, main overturning force is
A)
self-weight of the dam
B)
wind pressure
C)
water pressure
D)
uplift pressure
In a gravity dam, the principal overturning force is the hydrostatic water pressure acting horizontally on the upstream face. This pressure increases with depth and creates a large overturning moment about the toe of the dam, so option (c) is correct. The self-weight of the dam acts downward and provides the resisting moment against overturning, so option (a) is incorrect. Uplift pressure acts upward at the base and reduces the effective weight and stability, but it is not the main overturning force, so option (d) is incorrect. Wind pressure is generally negligible compared with water pressure in dam stability analysis, so option (b) is incorrect.
Rainfall mass curve shows the variations of
A)
rainfall intensity with time
B)
rainfall intensity with cumulative rainfall
C)
rainfall excess with time
D)
cumulative rainfall with time
A rainfall mass curve is a plot of cumulative rainfall depth against time. It shows how total rainfall accumulates over the duration of the storm, so option (d) is correct. Option (a) is incorrect because rainfall intensity vs time is an intensity hyetograph, not a mass curve. Option (b) is incorrect because cumulative rainfall is not plotted against rainfall intensity. Option (c) is incorrect because rainfall excess vs time refers to effective rainfall, which is a different hydrologic concept.
The Muskingham method of flood routing is a
A)
form of hydraulic routing of a flood
B)
form of reservoir routing
C)
complete numerical solution of St. Venant equations
D)
hydrological channel routing method
Muskingum method is a hydrological channel routing method. It routes flood hydrographs through a river/channel using a storage relationship based on inflow, outflow, and storage, without solving the full hydraulic equations.
Why others are incorrect:
- a) Hydraulic routing of a flood: Incorrect. Hydraulic routing uses the full unsteady flow equations (Saint-Venant equations) and is more detailed than Muskingum.
- b) Form of reservoir routing: Incorrect. Reservoir routing is a different routing concept; Muskingum is specifically used for channel routing.
- c) Complete numerical solution of St. Venant equations: Incorrect. Muskingum is an approximate hydrologic method, not a complete hydraulic solution.
The correct sequence of processes in a water treatment plant for rural water supply is
A)
chlorination, aeration, sedimentation, rapid sand filter
B)
coagulation, sedimentation, slow sand filter, chlorination
C)
coagulation, flocculation, clarification, pressure filter
D)
aeration, plain sedimentation, slow sand filter, chlorination
For rural water supply, the usual treatment train is: aeration (if needed to remove dissolved gases/odour/iron), plain sedimentation to settle suspended matter, slow sand filtration for biological and fine particulate removal, and finally chlorination for disinfection. Hence option d is correct.
Why others are wrong:
- a: Chlorination is placed first, but disinfection is done at the end, not before treatment; the sequence is incorrect.
- b: Coagulation is generally followed by flocculation and then sedimentation/clarification, not directly by slow sand filtration; also coagulation is not the typical first step for simple rural supply schemes.
- c: Coagulation, flocculation, clarification, and pressure filter is a conventional rapid-treatment sequence, more suited to larger/urban plants, not the standard rural water-supply sequence asked here.
Two primary air pollutants are
A)
sulphur oxide and ozone
B)
nitrogen oxide and peroxyacetylnitrate
C)
sulphur oxide and hydrocarbon
D)
ozone and peroxyacetylnitrate
Primary air pollutants are those emitted directly from a source into the atmosphere. Sulphur oxides and hydrocarbons are directly released from combustion and industrial processes, so option (c) is correct. Ozone and peroxyacetylnitrate (PAN) are secondary pollutants formed in the atmosphere by photochemical reactions, so option (d) is incorrect. Nitrogen oxides are primary pollutants, but PAN is secondary, so option (b) is incorrect. Sulphur oxides are primary, but ozone is secondary, so option (a) is incorrect.
The unit of moment of inertia of an area is
A)
kg-m2
B)
kg-m-s2
C)
kg/m2
D)
m4
The moment of inertia of an area (second moment of area) is defined as \(I=\int y^2 \, dA\). Since area has units of \(m^2\) and distance squared contributes \(m^2\), the unit becomes \(m^4\). Hence option d is correct.
Why the others are incorrect:
- a) \(kg\!\cdot\!m^2\) is the unit of mass moment of inertia, not area moment of inertia.
- b) \(kg\!\cdot\!m\!\cdot\!s^{-2}\) is the unit of force (newton).
- c) \(kg/m^2\) is a mass-per-area type unit, not a moment of inertia unit.
The number of bricks required per cubic metre of brick masonry is
A)
400
B)
450
C)
500
D)
550
For standard brick masonry, the number of bricks required per 1 m³ is about 500. This comes from the usual brick size with mortar joints, where the volume of one brick in masonry works out to roughly 0.002 m³, so 1/0.002 ≈ 500 bricks. Hence option (c) 500 is correct. The other options are incorrect because 400, 450, and 550 do not match the standard approximate quantity used in estimation for brick masonry.
For the truss shown in the Figure, the force in the member QR is
A)
zero
B)
P/ 2
C)
P
D)
2P
Based on established principles in Civil Engineering, the correct and most appropriate choice is 'zero' (Option A).
A simply supported beam of span l carries over its full span a load varying linearly from zero at both ends to w/unit length at midspan, then maximum bending moment is equal to
A)
wl2 / 8
B)
wl2 / 4
C)
wl2 / 12
D)
wl2 / 10
The load is symmetric and varies linearly from 0 at each support to w at midspan, so the load diagram is a triangle on each half of the beam. For the left half (0 to l/2), the load intensity is q(x)=2wx/l. The total load on the beam is the area of the two triangles: 2 × (1/2 × l/2 × w) = wl/2. By symmetry, each support reaction is wl/4.
Maximum bending moment occurs at midspan due to symmetry. Consider the left half beam. The moment at midspan is:
M_mid = Reaction at A × (l/2) − moment of left triangular load about midspan.
The left triangular load has resultant = (1/2)(l/2)w = wl/4, acting at a distance l/6 from the larger end (midspan) or l/3 from the left support. About midspan, its lever arm is l/6. Hence its moment about midspan is (wl/4)(l/6)=wl^2/24.
So,
M_max = (wl/4)(l/2) − wl^2/24 = wl^2/8 − wl^2/24 = wl^2/12.
Therefore, the correct option is c.
Why others are incorrect:
- a (wl^2/8): this is only the moment due to the support reaction at midspan, ignoring the opposing load moment.
- b (wl^2/4): too large; it would correspond to an incorrect load/resultant or lever arm.
- d (wl^2/10): not the correct value from the symmetric triangular loading calculation.