 # Quiz: Civil Engineering 17 Oct 2020

Quiz: Civil Engineering
Exam: SSC-JE
Topic: Soil Mechanics

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 10 Minutes

Q1. Which one of the following correctly represents the dry unit weight of soil sample which has bulk weight of γ_t at a moisture content of w%?
(a) (wγ_t)/100
(b) γ_t (1+W/100)
(c) γt(100/(100+W))
(d) (γ_t (100-W))/100

Q2. A soil has a liquid limit of 40% and plasticity index of 20%. The plastic limit of the soil will be
(a) 20%
(b) 30%
(c) 40%
(d) 60%

Q3. A soil has a discharge velocity of 6 × 10-7 m/s and void ratio of 0.5. Its seepage velocity is
(a) 18 × 10-7 m/s
(b) 12 × 10-7 m/s
(c) 6 × 10-7 m/s
(d) 3 × 10-7 m/s

Q4. Taylor’s stability number is equal to
(a) C/(F_C H)
(b) c/(F_C γH_C )
(c) C/γH
(d) C/(γH_C )

Q5. In tri-axial compression test, the deviator stress is given by
(a) σ_1+σ¬¬_3
(b) σ_1-σ_3
(c) 1/2 (σ_1+σ_3)
(d) 1/2 (σ_1-σ_3)

Q6. Unconfined compressive strength test is :
(a) Undrained test
(b) Drained test
(c) Consolidated undrained test
(d) Consolidated drained test

Q7. A line load of infinite length has an intensity q per unit length. What is the vertical stress σ_z at a depth z below the earth at the centre of the load ?
(a) σ_z=2qz/π
(b) σ_z=2q/πz
(c) σ_z=2qz²/π
(d) σ_z=2q/πz²

Q8. A soil has plasticity index of 50, it can be described as:
(a) low plastic
(b) medium plastic
(c) Very highly plastic
(d) None of these

Q9. On analysis of particle size distribution of a soil it is found that D10 = 0.1 mm, D30 = 0.3 mm & D60 = 0.8 mm. What will be the uniformity coefficient for the given soil:
(a) 10
(b) 12
(c) 8
(d) 6

Q10. In Casagrande’s liquid limit device, the material of the test specimen is harder than the standard rubber. This hardness indicates that the liquid limit, plasticity index, flow index and toughness index, respectively, of the specimen, are
(a) more, less more and same
(b) same, less, same and more
(c) less, less, same and less
(d) less, same, less and more

SOLUTION

S1. Ans.(c)
Sol. the relation between dry unit weight ‘γ_d’ and bulk unit weight ‘γ_t’ given by –
γ_d=γ_t/((1+w%) )
=γ_t/((1+w/100) )
▭(γ_d=((100γ_t)/(100+w)) )

S2. Ans.(a)
Sol. Given
w_L=40%
I_P = 20%
w_P= ?
I_P=w_L-w_P
20%=40%-W_P
▭(W_P=20%)

S3. Ans.(a)
Sol.
V=6×10^(-7) m\/sec
e=0.5
V_s= ?
▭(V_S=V/n)
V_S=(6×10^(-7))/(e/(1+e))
=((1+0.5)×6×10^(-7))/0.5
▭(V_S=18×10^(-7) m\/sec.)

S4. Ans.(d)
Sol. Taylor’s stability No. →
S_n=C_m/yH
S_n=C/(F_c.yH) (∵F_C=C/C_m )
S_n=C/(H_C.yH) (∵F_C=H_C/H)
▭(S_n=C/(yH_C ))

S5. Ans.(b)
Sol. in tri-axial compression test →
Minor principal stress = confining pressure
σ_C=σ_3
Major Principal stress σ_1=σ_2+σ_d
σ_d is deviator stress
∵ ▭(σ_d=σ_1-σ_3 )

S6. Ans.(a)
Sol. The unconfined compression test is a special form of a tri-axial test in which the confining pressure is zero. The test can be conducted only on clayey soil which can stand without confinement. The load is applied rapidly hence it is an undrained test.

S7. Ans.(b)
Sol. A line load of infinite length has an intensity q per unit length then vertical stress (σ_z) at a depth Z below the earth is given by –
▭(σ_z=2q/πz)

S8. Ans.(c)

Sol.

 Plasticity index Soil description 0 Non – plastic 1 – 5 Slightly plastic 5 – 10 Low plastic 10 – 20 Medium plastic 20 – 40 High plastic > 40 Very high plastic

S9. Ans.(c)
Sol. Given,
D_10=0.1 mm
D_30=0.3 mm
D_60=0.8 mm
C_u= ?
Co-efficient of uniformity (Cu) = D_60/D_10
=0.8/0.1
▭(Cu=8 )

S10. Ans.(c)
Sol. If the test specimen is harder than the standard rubber, less number of blows will fill the groove for a given water content. In other words 25 number of blows will be required at less water content, so liquid limit and thereby plasticity index will be less. There will be no effect on flow index (I_F ), Toughness index =I_P/I_F will also be less.

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