DFCCIL'21 CE: Daily Practice Quiz. 21-September-2021

DFCCIL (Dedicated Freight Corridor Corporation of India Limited) Exam Date 2021 on September and October 2021 so boost your preparation by attempt this Civil Engineering Geo Tech Engineering quiz for DFCCIL recruitment 2021.

Quiz: Civil Engineering
Exam: DFCCIL-JE
Topic: Geo Tech Engineering

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes

Q1. Arrange the following lists of soils in increasing order of surface areas?
(a) Silt, sand, colloids, clay
(b) Sand, silt, colloids, clay
(c) Sand, silt, clay, colloids
(d) Clay, silt, sand, colloids

Q2. Void ratio of an undisturbed sample of soil is 0.6. the values of maximum and minimum possible void ratio are found as 0.8 and 0.4, respectively. The relative density in percentage, for this soil sample will be
(a) 25
(b) 50
(c) 75
(d) 90

Q3. Relationship between dry density γ_d’ percentage air voids n_a’ water content w and specific gravity G of any soil is
(a) γ_d=((1+b_a )Gγ_w)/(1+wG)
(b) γ_d=((1+n_a )Gγ_w)/(1-wG)
(c) γ_d=((1-n_a )Gγ_w)/(1+wG)
(d) γ_d=((1-n_a )Gγ_w)/(1-wG)

Q4. A soil has percentage air voids of the order of 30%. It has a porosity of 0.4 the air content of that soil shall be:
(a) 0.75
(b) 0.12
(c) 1.33
(d) 0.70

Q5. Two specimens of clay A and B are tested in a consolidation apparatus.
If (m_v )_A=3.6×10^(-4) m^2/kW
(m_v )_B=1.8×10^(-4) m^2/ kW
(C_v )_B=1.9×10^(-4) cm^2/s
(C_v )_A=3.8×10^(-4) cm^2/s
Then the ratio K_A / K_B is equal to (where m_v is coefficient of volume change and C_v is coefficient of consolidation):
(a) 0.0625
(b) 0.25
(c) 1.0
(d) 4.0

Q6. A loose uniform sand with rounded grains has effective grains size of 0.05 cm. Co-efficient of permeability of the sand is…………….
(a) 0.25 cm/sec
(b) 0.5 cm/sec
(c) 1 cm/sec
(d) 1.25 cm/sec

SOLUTION

S1. Ans.(c)
Sol. Smaller Size of soil have higher surface area.
→ soil in increasing order of surface
Gravel → sand → silt → clay → colloids

S2. Ans.(b)
Sol. e_max=0.8
e_min=0.4
e=0.6
Relative density = (e_(max )-e)/(e_max-e_min )×100
(0.8-0.6)/(0.8-0.4)×100
0.2/0.4×100
=50%

S3. Ans.(c)
Sol. r_d=((1-η_a )Gr_ω)/((1+e) )
r_d→ dry density
η_a → percentage air void
G → Specific gravity
e → void ratio
▭(Se=W.G)
▭(r_d=((1-η_a )Gr_ω)/(1+WG))

S4. Ans.(a)
Sol. percentage air void (η_a ) = 30%
Porosity (n) = 0.4
Air content (a_c) = ?
η_a = na_c
0.30 = 0.4 × a_c
▭(a_c=0.75)

S5. Ans.(d)
Sol.
Clay A Clay B
(m_v )_A = 3.6 × 10^(-4) m²/kW (m_v )_B = 1.8 × 10^(-4) m²/kW
(C_v )_A = 3.8 × 10^(-4) cm²/kW (C_v )_B = 1.9 × 10^(-4) cm²/kW
▭(k=C_v m_v r_w )
K_A/K_B =((C_V )_A (m_v )_B r_w)/((C_V )_B (m_v )_B r_w )
=(3.8×10^(-4)×3.6× 10^(-4))/(1.9×10^(-4)×1.8×10^(-4) )
▭(K_A/K_B =4)
S6. Ans.(a)
Sol. effective size (D_10 ) = 0.05 cm.
Co-efficient of permeability (k) =?
According to Hazen eq^n –
K = 100 (D_10 )^2
K = 100 × (0.05)²
K = 0.25 cm/sec