Seepage Analysis MCQ Quiz in मराठी - Objective Question with Answer for Seepage Analysis - मोफत PDF डाउनलोड करा

Last updated on May 8, 2024

पाईये Seepage Analysis उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Seepage Analysis एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Top Seepage Analysis MCQ Objective Questions

Seepage Analysis Question 1:

The constant of proportionality between seepage velocity and hydraulic gradient is called

1. Seepage coefficient
2. Coefficient of transmissibility
3. Coefficient of percolation
4. Modified coefficient of permeability

Option 3 : Coefficient of percolation

Seepage Analysis Question 1 Detailed Solution

Concept-

Coefficient of transmissibility (T)-

The rate of flow of water through the entire medium of unit width under unit hydraulic gradient is termed as coefficient of transmissibility

Darcy's law states that there is a linear relationship between flow velocity (v) and hydraulic gradient (i) for any given saturated soil under steady laminar flow conditions.

If the rate of flow is q (volume/time) through cross-sectional area (A) of the soil mass, Darcy's Law can be expressed as

v = q/A = k.i

where k = permeability of the soil
i = Dh/L
Dh = difference in total heads
L = length of the soil mass

The flow velocity (v) is also called the Darcian velocity or the superficial velocity.

It is different from the actual velocity inside the soil pores, which is known as the seepage velocity, vS.

At the particulate level, the water follows a tortuous path through the pores.

Seepage velocity is always greater than the superficial velocity, and it is expressed as:

$${V_s} = \frac{q}{{{A_v}}} = \frac{q}{{{A_v}}} \times \frac{A}{A} \approx \frac{v}{n}$$

where AV = Area of voids on a cross section normal to the direction of flow
n = porosity of the soil

Seepage velocity can be written as,

v = kp i

kp = Coefficient of percolation

The constant of proportionality between seepage velocity and hydraulic gradient is called coefficient of percolation.

Seepage Analysis Question 2:

What will be the capillary uplift when a glass tube of diameter 4 mm is immersed vertically in water?

(Given, Surface tension = 1 N/m, = 1000 kg/m3, g = 10 m/s2)

1. 1 cm
2. 0.1 cm
3. 1 m
4. 0.1 m

Option 4 : 0.1 m

Seepage Analysis Question 2 Detailed Solution

Concept:

Capillary rise (hc) $$= \frac{{4{\rm{T}}}}{{{{\rm{\gamma }}_{\rm{w}}}{\rm{d}}}}$$

Where

T = Surface tension of water, and d = effective pore diameter

Calculation:

Given: T 1 N/m, d = 4 mm = 4 × 10-3 m

γw = 1000kg/m3 * 10 m/s2 = 10 kN/m3

$${\rm{Capillary\;rise\;}}\left( {{{\rm{h}}_{\rm{c}}}} \right) = \frac{{4 × 1}}{{10 × {{10}^3} × 4 × {{10}^{ - 3}}}}$$

hc = 0.1 m

Seepage Analysis Question 3:

The rate of discharge of percolating water per unit cross-sectional area of voids perpendicular to the direction of flow is called:

1. normal velocity
2. discharged velocity
3. pore velocity
4. seepage velocity

Option 4 : seepage velocity

Seepage Analysis Question 3 Detailed Solution

Concept:

Apparent velocity (V):

• It is called apparent velocity because the actual flow is through pores in the cross-section and not through the entire cross-sectional area.
• It is also called discharge velocity (V)

Seepage velocity (Vs):

• The rate of discharge of percolating water per unit cross-sectional area of voids perpendicular to the direction of flow is called Seepage velocity.
• The velocity of water through pores is called seepage velocity (VS). As the flow is continuous, discharge Q must be the same throughout the system

Q = A × V = AV × Vs

Where,

A = total cross-sectional area,

AV = area of voids in the total cross-sectional area

$${\rm{V}} = \frac{{{{\rm{A}}_{\rm{V}}}}}{{\rm{A}}} \times {{\rm{V}}_{\rm{s}}}$$

The porosity of soil is given by,

$${\rm{n}} = \frac{{{\rm{volume\;of\;void}}}}{{{\rm{total\;volume}}}} = \frac{{{{\rm{A}}_{\rm{V}}} \times {\rm{L}}}}{{{\rm{A}} \times {\rm{L}}}} = \frac{{{{\rm{A}}_{\rm{V}}}}}{{\rm{A}}}{\rm{\;}}$$

V = n × Vs

$$\therefore {{\bf{V}}_{\bf{s}}} = \frac{{\bf{V}}}{{\bf{n}}}$$

As porosity (n) ranges between 0 < n < 1

$$\frac{V}{V_S} <1$$

Hence,  $$V_s>V$$

Seepage velocity (Vs) is always greater than or equal to discharge velocity.

Seepage Analysis Question 4:

Darcy’s law is not applicable for flow in:

1. Fine sands
2. Clays
3. Silts
4. Gravels

Option 4 : Gravels

Seepage Analysis Question 4 Detailed Solution

Concepts:

Darcy’s Law states that velocity of flow through porous media is directly proportional to hydraulic gradient for any given saturated soil under steady laminar flow conditions.

If the rate of flow is Q through cross-sectional area (A) of the soil mass, Darcy's Law can be expressed as
V = Q/A = k × i
Where,

k = permeability of the soil
i = dh/dl

dh = difference in total heads
dl = length of small soil element

Darcy’s law assumes the flow must be laminar i.e. flow of water through soil voids should have very less velocity. This only possible in case of soils having small but large no of voids. Therefore, it is applicable for soils such as fine saturated sands, clays, silts etc. It is not applicable for soils having large size voids like gravel, boulder etc because in these soils water is drained immediately due to action of gravity and hence, not possible to maintain laminar flow in these soils.

Seepage Analysis Question 5:

What is the representative height of capillary rise for fine sand?

1. 0.10 to 0.15 m
2. 1.0 to 10.0 m
3. 0.30 to 1.00 m
4. 0.02 to 0.10 m

Option 3 : 0.30 to 1.00 m

Seepage Analysis Question 5 Detailed Solution

Concepts:

The some portion of water in voids of soil particles is held by surface tension and this water cannot move under the influence of gravitational force.  This held water sometimes travel in upward direction in the soil even against the gravitational forces. It happens because of a phenomenon called the capillarity or capillary effect.  This risen water is called capillary water.

The approximate height of Capillary in different soils is given as:

 Soil Height of Capillary rise (m) Coarse Sand 0.03-0.15 Medium Sand 0.12-1.1 Fine Sand 0.3-3.5 Silt 1.5-2 Clay > 10

Additional InformationThe height of capillary in soils is determined by an approximate formula suggested by Terzaghi, which is given as:

$$G = C/eD_10$$

C is constant in Sq.cm ranges from 0.1 to 0.5 Sq.cm.

e is the Void ratio

D10 is effective grain size in cm.

Seepage Analysis Question 6:

In a soil specimen, the total stress, effective stress, hydraulic gradient, and critical hydraulic gradient are σ, σ’, i and ic, respectively. For initiation of quicksand condition, which one of the following statements is TRUE?

1. σ’ = 0 and i = iC
2. σ = 0 and i = iC
3. σ’ ≠ 0 and i = iC
4. σ’ ≠ 0 and i ≠ iC

Option 1 : σ’ = 0 and i = iC

Seepage Analysis Question 6 Detailed Solution

Concept:

For a structure constructed on cohesionless soil, and in contact with the water body. Due to the difference in the water level across the structure, the hydraulic gradient would be set up resulting in upward seepage. Water movement in upward direction would exert upward thrust on the soil and thereby reducing the effective stress.

If the head causing upward flow is increased, a stage is reached when the effective stress is reduced to zero. The condition so developed is known as quicksand condition. Quicksand is not any special type of sand, but it is a condition of zero effective stress in cohesionless soils.

For Quick sand condition, effective stress in the soil becomes zero (σ = 0)

At the bottom of the soil column,

$$\sigma = \gamma \times L$$

$$u = {\gamma _w}\;\left( {L + {\rm{\Delta }}H} \right)$$

During the quick sand condition, the effective stress is reduced to zero.

$$\gamma \times L = {\gamma _w} \times \left( {L + {\rm{\Delta }}H} \right)$$

$$L\;\left( {\gamma - {\gamma _w}} \right) = {\gamma _w} \times {\rm{\Delta }}H$$

$$L \times {\gamma _b} = {\gamma _w} \times {\rm{\Delta }}H$$

$$\frac{{{\rm{\Delta }}H}}{L} = \frac{{{\gamma _b}}}{{{\gamma _w}}} = {i_c}\;$$

$${\rm{where\;}}$$ic = critical hydraulic gradient

Seepage Analysis Question 7:

Sand drains are used in saturated clay soil to

1. Reduce settlement
2. Accelerate the consolidation
3. Increase the permeability
4. Increase pore pressure

Option 2 : Accelerate the consolidation

Seepage Analysis Question 7 Detailed Solution

Explanation:

Sand drains:

• It is a process of radial consolidation that increases the rate of drainage by providing multiple drainage path in the radial direction in addition to vertical drainage
• These holes are backfilled with a suitable grade of sand.

Need of sand drains:

• In highly compressible soil, like soft marine clay deposits show large differential settlement, which may even cause the failure of the structure.
• Also, due to this high compressibility and settlement, the movement to heavy equipment for construction becomes difficult.
• So, sand drains are used here to accelerate the process of consolidation by providing multiple radial paths for drainage.

1. The provision of sand drains allows the drainage of pore water in the radial direction in addition to the drainage in the vertical direction.
2. Since the permeability of soil is much larger in the horizontal direction than vertical direction, the rate of consolidation becomes much larger
3. With faster consolidation, the soil gains strength rapidly, allowing a faster pace of work and reducing the cost of the project.

Some important specification  regarding soil drains:

1) Spacing of sand drains =  2 - 5 m

2) Depth of sand drain = 3 - 35 m

3) Radius of sand drain well = 0.2 - 0.3 m

4) Thickness of sand blanket = 0.6 - 1 m

Seepage Analysis Question 8:

If the void ratio and discharge velocity for soil is 0.5 and 6 × 10-7 m/s respectively, what is the value of seepage velocity (m/s)?

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

Option 4 : 18 × 10-7

Seepage Analysis Question 8 Detailed Solution

Given:

Void ratio, e = 0.5

Discharge velocity = 6 × 10-7 m/s

Also, Porosity, $$\eta = \frac{{\rm{e}}}{{1 + {\rm{e}}}} = \frac{{0.5}}{{1 + 0.5}} = 0.333$$

Seepage velocity, $$Vs = \frac{V}{\eta } = \frac{{6 \times {{10}^{ - 7}}}}{{0.333}} = 18 \times {10^{ - 7}}m/s$$

Seepage Analysis Question 9:

Quick sand is a -

1. Type of sand
2. Cohesionless soil
3. Flow condition occuring in cohesive soils
4. Flow condition occuring in cohesionless soil

Option 4 : Flow condition occuring in cohesionless soil

Seepage Analysis Question 9 Detailed Solution

Explanation:

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.

Generally occurs in silts and sand, which are cohesioless, in cohesive soil like clay, quick sand is hard to occur as very is high cohesion in case of clay

(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

Seepage Analysis Question 10:

A flownet of a cofferdam foundation has 6 flow channels and 18 equipotential drops. The head loss during seepage is 6 m. If the coefficient of permeability of soil is 4 x 10-5 m/min, then the seepage loss (m3/day) is

1. 72
2. 8
3. 0.115
4. 1.037

Option 3 : 0.115

Seepage Analysis Question 10 Detailed Solution

Concept:

Determination of seepage discharge when the medium is isotropic:

Seepage discharge is given by, $$q = kh\frac{{{N_f}}}{{{N_d}}}$$

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

Nf = Total number of flow channels

Nd = Total number of equipotential drops

k = Coefficient of permeability

Calculations:

Given,

Nf = 6, Nd = 18, h = 6 m

k = 4 × 10-5 m/min

∵ seepage discharge, $$q = kh\frac{{{N_f}}}{{{N_d}}}$$

$$q = 4 × {10^{ - 5}} × 6 × \frac{6}{{18}} = 8 × {10^{ - 5}}\ {m^2}/\min /m$$

q = 8 × 10-5 × 24 × 60 = 0.115 m3/day
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