Seepage Analysis MCQ Quiz - Objective Question with Answer for Seepage Analysis - Download Free PDF

Last updated on Jun 19, 2025

Latest Seepage Analysis MCQ Objective Questions

Seepage Analysis Question 1:

Which of the following zone in zoned type embankment prevents piping through cracks? 

  1. Central core
  2. Core wall
  3. Outer zone
  4. Transition zone

Answer (Detailed Solution Below)

Option 4 : Transition zone

Seepage Analysis Question 1 Detailed Solution

Explanation:

Zoned Dam

  • Zoned darns usually have a central zone of selected soil material, to form a relatively impermeable core, a transition zone along both faces of the core to prevent piping through cracks which may form in the core, and outer zones of more pervious material for stability.
  • This construction is widely used in earth dams and is selected whenever suitable materials are available.
  • Clay, even though highly impermeable, may not make the best core if it shrinks and swells too much.
  • The most satisfactory cores are of clay mixed with sand and fine gravel

Seepage Analysis Question 2:

If Av is the area of voids in cross-section and A is the total area of cross-section, the relation between seepage velocity (vs ) and superficial velocity (v) is correctly represented as:

  1. vs = AvA
  2. vs = Avv/A
  3. v = AvsAv
  4. v = Avvs/A

Answer (Detailed Solution Below)

Option 4 : v = Avvs/A

Seepage Analysis Question 2 Detailed Solution

Explanation:

  • Seepage velocity (vs): This is the velocity at which the fluid actually moves through the voids in the material.

  • Superficial velocity (v): This is the velocity of the fluid if it were to move through the entire cross-sectional area (including both the solid and void portions of the material).

The relationship between seepage velocity and superficial velocity is given by the equation: v = Avvs/A

Where:

  •   ​ is the area of voids in the cross-section.

  •    is the total area of the cross-section (including both solids and voids).

  •    is the seepage velocity.

  •    is the superficial velocity.

Additional InformationSeepage Velocity

  • Definition: Seepage velocity refers to the actual velocity of water or fluid that moves through the void spaces (pores) of a material, such as soil or a porous medium.

  • Importance: It is used to understand how quickly fluid travels through the porous medium, which is crucial in applications like groundwater flow, soil permeability tests, and drainage.

  • Relation to Porosity: Seepage velocity is dependent on the porosity and the permeability of the material. Higher porosity generally results in higher seepage velocity for the same hydraulic gradient.

Seepage Analysis Question 3:

Which type of soil is commonly used for the core of an embankment to minimise seepage?

  1. Clay
  2. Sand
  3. Silty sand
  4. Gravel

Answer (Detailed Solution Below)

Option 1 : Clay

Seepage Analysis Question 3 Detailed Solution

Explanation:

Clay:

  • Clay is commonly used for the core of embankments because it has very low permeability, which helps minimize seepage.

  • The fine particles in clay make it highly effective at reducing the movement of water, making it ideal for controlling seepage in embankment dams.

Additional Information Sand:

  • Sand has relatively high permeability, which makes it unsuitable for the core of an embankment where seepage control is critical.

  • It would allow water to pass through more easily, undermining the stability of the embankment.

Silty Sand:

  • Silty sand has a higher permeability than clay, though not as much as pure sand.

  • It may still allow some seepage, making it less ideal for minimizing water movement compared to clay.

Gravel:

  • Gravel has very high permeability, allowing significant water flow.

  • It is typically used in the outer layers or drainage zones of embankments but not in the core, where seepage control is essential.

Seepage Analysis Question 4:

For piping phenomenon to occur in soils, the most important condition to be satisfied is that the 

  1. Specific gravity of soil solids is more than 2.8
  2. void ratio is more than 2.0
  3. hydraulic gradient is nearly unity
  4. soil is fine grained
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : hydraulic gradient is nearly unity

Seepage Analysis Question 4 Detailed Solution

Explanation:

  • When flow takes place in an upward direction, seepage pressure (ps) also acts in an upward direction and effective stress is reduced.
  • If seepage pressure equals the submerged weight of soil mass then effective stress reduces to zero. In such a case, cohesionless soil loses all its shear strength and has the tendency to flow along with water. This phenomenon is known as the quicksand condition.

For quicksand / Piping condition:

Effective stress = γ'z - Ps = 0

γ'z = Ps = i × z × γw

Where Ps = Seepage pressure = i z γw

γ = i × γw

i = (γγw)=G11+e.γw

ic = G11+e

Note:

The hydraulic gradient at which quicksand condition occurs is termed as the critical hydraulic gradient piping gradient, floating gradient, or bursting gradient.

Important Points

  • For piping, the hydraulic gradient should be more than or equals to the critical hydraulic gradient.  
  • For most of the soils, the critical hydraulic gradient is close to unity. So when the hydraulic gradient is nearly unity piping will occur.

Seepage Analysis Question 5:

The process by which a mass of saturated soil is caused by external forces to suddenly lose its sheer strength and to behave as a fluid is called

  1. piping
  2. slide
  3. quick condition
  4. liquefaction

Answer (Detailed Solution Below)

Option 4 : liquefaction

Seepage Analysis Question 5 Detailed Solution

Explanation:

Liquefaction – Process and Explanation

  • Liquefaction is the process by which a mass of saturated soil loses its shear strength due to external forces.

  • This phenomenon occurs when vibrations, such as those from an earthquake, cause soil particles to rearrange.

  • The rearrangement of soil particles leads to an increase in pore water pressure, reducing the soil's ability to support loads.

  • As a result, the soil behaves as a fluid-like material, losing its solid properties temporarily.

  • This behavior of soil during liquefaction can result in ground failures, like landslides or uneven settlement.

  • Liquefaction is most common in sandy soils or soils with poor drainage characteristics.

  • It is a critical consideration in geotechnical engineering, especially in earthquake-prone areas.

  • Liquefaction can cause structures to tilt or collapse if built on affected soil.

  • Mitigation methods include soil compaction, grouting, or designing structures to withstand liquefaction effects.

Top Seepage Analysis MCQ Objective Questions

A soil has a discharge velocity of 6 × 10-7 m/s and a void ratio of 0.5. What is its seepage velocity?

  1. 3 × 10-7 m/s
  2. 6 × 10-7 m/s
  3. 12 × 10-7 m/s
  4. 18 × 10-7 m/s

Answer (Detailed Solution Below)

Option 4 : 18 × 10-7 m/s

Seepage Analysis Question 6 Detailed Solution

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Given,

Discharge velocity (V) = 6 × 10-7 m/s

Void ratio = 0.5

Then porosity (n)

n=e1+e

n=0.51+0.5=0.51.5=13

seepage velocity (Vs) = ?

seepage velocity Vs =Vn=6×1071/3

Vs = 18 × 10-7 m/s

The pore water pressure in the soil sample of consolidometer test is __________.

  1. maximum at bottom
  2. maximum at centre
  3. maximum at top
  4. minimum at centre

Answer (Detailed Solution Below)

Option 2 : maximum at centre

Seepage Analysis Question 7 Detailed Solution

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Explanation:

In the process of loading, initially, as we increase the load the pore water pressure increases, due to which pore water escapes out from the voids of the soil.

Hence, the pore water pressure in the soil sample of the consolidometer test is maximum at the center and when the primary consolidation is completed, which leads to the expulsion of pore water from the voids of the soil stops then the excess pore water process developed due to the application of load reduced to zero.

Which of the following is the main cause of occurrence of quick sand condition?

  1. when the void ratio of the soil becomes 1.0
  2. when the upward seepage pressure in soil becomes zero
  3. when the upward seepage pressure in soil becomes equal to the saturated unit weight of the soil
  4. when the upward seepage pressure in soil becomes equal to submerged unit weight of the soil

Answer (Detailed Solution Below)

Option 4 : when the upward seepage pressure in soil becomes equal to submerged unit weight of the soil

Seepage Analysis Question 8 Detailed Solution

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Quicksand condition:

(i) This process in which soil particles are lifted over the soil mass is called quicksand condition. It is also known as 'boiling of sand' as the surface of sand looks it is boiling.

(ii) At quick sand condition, net effective stress is reduce to zero, i.e.

σ̅ = 0

∴ γ'z = ps = 0     {ps = Seepage pressure}

upward seepage pressure in the soil becomes equal to submerged unit weight of the soil

or, γ'z - izγw = 0

i = icr = γ'/γw

(iii) The hydraulic gradient under which quick sand condition occurs is termed as a critical hydraulic gradient. If void ratio and specific gravity of soil is known, then

i=icr=γγw=(G1)γw1+eγw  

⇒ i=(G1)1+e

(v) quicksand condition can be prevented by lowering the water table at the site before excavation.

  • It can also be prevented by increasing the upward flow length by providing a sheet pile wall.
  • Another method adopted at sites to avoid quicksand Conditions is the addition of surcharge (i.e additional weight) on the excavation side to increase the net weight of the soil mass.

 

(i) Quick sand condition is generally observed when excavation is done below the GWT and water is pumped out to keep the excavated area free from it or it is observed when sand is under artesian pressure and overlain by a clay layer. 

The hydrostatic pressure on the Phreatic line within a dam section is

  1. Greater than atmospheric pressure
  2. Equal to atmospheric pressure
  3. Less than atmospheric pressure
  4. None of the above

Answer (Detailed Solution Below)

Option 2 : Equal to atmospheric pressure

Seepage Analysis Question 9 Detailed Solution

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Phreatic Line:

The top flow line of a saturated soil mass below which seepage takes place, is called the Phreatic line.

  • The hydrostatic pressure on the Phreatic line within a dam section is equal to atmospheric pressure
  • This line separates a saturated soil mass from an unsaturated soil mass
  • It is not an equipotential line, but a flow line

During seepage through an earth mass, the direction of seepage is

  1. parallel to the equipotential lines
  2. perpendicular to the streamlines
  3. perpendicular to the equipotential lines
  4. along the direction of gravity

Answer (Detailed Solution Below)

Option 3 : perpendicular to the equipotential lines

Seepage Analysis Question 10 Detailed Solution

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Concept:

The velocity potential function may be defined as a scalar function of space and time such that its derivative with respect to any direction gives the fluid velocity in that direction.

Velocity potential function (φ) satisfies the Laplace equation.

So 2ϕx2+2ϕy2=0 

The solution of the above equation gives two sets of curves known as equipotential lines and stream lines (or flow lines), mutually orthogonal to each other as shown in figure.

The equipotential lines represent contours of equal head (potential).

The direction of seepage is always perpendicular to the equipotential lines.

The path along which the individual particles of water seep through the soil are called streamlines or flow lines.

The hydraulic head that would produce a quick sand condition in a sand stratum of thickness 1.8 m, specific gravity 2.65 and void ratio 0.65 is equal-to

  1. 1.0 m
  2. 1.2 m
  3. 1.6 m
  4. 1.8 m

Answer (Detailed Solution Below)

Option 4 : 1.8 m

Seepage Analysis Question 11 Detailed Solution

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Concept:

Critical hydraulic gradient (ic): 

The quick condition occurs at a critical upward hydraulic gradient ic, when the seepage force just balances the buoyant weight of an element of soil. The critical hydraulic gradient is typically around 1.0 for many soils.

At the critical conditions, the effective stress is equal to zero.

The critical hydraulic gradient (ic) is given by,

ic=G11+e

where,

G - Specific gravity of soil

e - Void ratio of soil

Calculation:

Given: G = 2.65, e = 0.65, h = 1.8 m

Critical hydraulic gradient,

ic=G11+e=2.6511+0.65=1

Head required

= ic × thickness of sand stratum

= 1 × 1.8 = 1.8 m

Which is not a method of obtaining flow nets?

  1. Electrical flow analogy
  2. Capillary flow analogy
  3. Sand model
  4. Flow model

Answer (Detailed Solution Below)

Option 4 : Flow model

Seepage Analysis Question 12 Detailed Solution

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Concept:

Flow net: It is a graphical representation of two dimensional steady state flow consisting of two sets of lines – the flow and the equipotential lines.

In case of isotropic soil these lines are orthogonal to each other but need not be orthogonal in anisotropic soil.

Various methods of obtaining flow nets are:

1) Graphical method

2) Electrical flow Analogy method

3) Sand Model

4) Capillary flow Model

5) Solution of Laplace equation/Analytical method.

Flow model is not used to obtain flow nets.

Maximum permissible upward gradient in a previous sand of porosity n = 45%, specific gravity Gs = 2.65 with a factor of safety 4 will be

  1. 0.425
  2. 0.225
  3. 0.375
  4. 0.275

Answer (Detailed Solution Below)

Option 2 : 0.225

Seepage Analysis Question 13 Detailed Solution

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Concept:

Critical hydraulic gradient (ic): The quicksand / boiling condition occurs at a critical upward hydraulic gradient typically around 1.0 for many soils, when the seepage force just balances the buoyant weight of an element of soil.

At the critical conditions, the effective stress is equal to zero.

ic=G11+eOR(G1)×(1n)

Calculations:

Critical hydraulic gradient

ic=(G1)×(1n)=1.65×(10.45)=0.9075

is = ic/4 = 0.9075/4 = 0.225

For a two dimensional flow through a constant head permeameter, the flow net will alter if:

  1. Soil in the permeameter is not altered
  2. Head causing flow is altered 
  3. Flow direction is reversed
  4. Permeameter length is changed

Answer (Detailed Solution Below)

Option 4 : Permeameter length is changed

Seepage Analysis Question 14 Detailed Solution

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Explanation:

Flow net:

(i) The entire pattern of flow lines and equipotential lines is referred as a flow net. It is the solution of Laplace's equation for relevant boundary conditions. Thus, a flow net is a graphical representation of the head and direction of seepage at every point.

(ii) Flow net is a function of only boundary condition and is independent of permeability, head loss, or number of flow lines and equipotential lines. Hence the flow length will alter if permeameter length is changed.

Additional Information

Properties of flow net:

(i) Flow lines and equipotential lines meet each other orthogonally.

(ii) There can be no flow across the flow line and velocity of flow is always perpendicular to equipotential lines.

(iii) Area bounded between two adjacent flow lines is called flow channel or flow path and quantity of water flowing through each channel is same.

(iv) Area bounded between two adjacent equipotential lines and adjacent flow lines are referred to as a flow field.

(v) For isotropic medium, flow field is rectangular which may either be linear or curvilinear.

(vi) Loss of head between two equipotential lines i.e. equipotential drop remain constant for all the equipotential lines. 

The seepage loss in cm2/sec for a hydraulic structure, if the flow net contains 5 flow lines and 9 equipotential lines and the head causing flow is 20 m, k= 2.6 x 10-6 cm/sec is:

  1. 1.88 × 10-3
  2. 1.68 × 10-3
  3. 2.88 × 10-3
  4. 2.60 × 10-3

Answer (Detailed Solution Below)

Option 4 : 2.60 × 10-3

Seepage Analysis Question 15 Detailed Solution

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Concept:

Determination of seepage discharge when the medium is isotropic:

Seepage discharge is given by,

 q=khNfNd

Where h = Hydraulic head or head difference between upstream and downstream level or head loss through the soil

Nf = Total number of flow channels = Floe lines - 1

N= Total number of equipotential drops = Equipotential lines -1

k = Coefficient of permeability

Calculations:

Given data

Nf = 5 - 1 = 4, Nd = 9 - 1 = 8, h = 20 m = 2000 cm

k = 2.6 × 10-6 cm/sec

∵ seepage discharge(q),

 q=khNfNd

q=2.6×106×2000×48

q = 2.60 × 10-3  cm2/sec

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