Quiz: CIVIL ENGINEERING
Exam: NPCIL-SCINTIFIC ASSISTANT(CIVIL)
Topic: MISCELLANEOUS
Each question carries 1 mark
Negative marking: No negative marking
Time: 10 Minutes
Q1. In surveyor’s compass, the graduated ring is attached to the:
(a) Box
(b) Needle
(c) Sigh vane
(d) Jewel bearing
Q2. The imaginary line passing through the intersection of cross hairs and the optical center of the objective is known as ……….
(a) Line of sight
(b) Line of collimation geoid
(c) Axis of the telescope
(d) None of these
Q3. Which of the following is an expansive soil?
(a) Dune sand
(b) Black Cotton soil
(c) Loess
(d) Lateritic soil
Q4. The uniformity coefficient of soil with usual notation is defined as:
(a) D_30/D_40
(b) D_40/D_50
(c)D_50/D_60
(d) D_60/D_10
Q5. If a transition curve is introduced between a straight and a curve such that a super elevation of 15 cm is introduced over a curve and if the rate of super elevation is 1 IN 500, then the length (m) of the transition curve will be
(a) 50.00
(b) 75.00
(c) 100.00
(d) 150.00
Q6. The length of long chord and tangent of a circular curve if radius R will be equal, if the angle of deflection is
(a) 30°
(b) 90°
(c) 60°
(d) 120°
Q7. A soil engineer collects a sample of soil which having moist unit weight of a soil is 25 kN/m³, water content available is 50% and the specific gravity of the solid of the soil is 2.55. then, what is the value of porosity of soil mass? (Assume unit weight of water = 9.81 kN/m³)
(a) 66 %
(b) 33 %
(c) 30 %
(d) 40 %
Q8. If pressure due to submerged weight of the soil is equal to the seepage pressure, the effective pressure:
(a) Remains the same
(b) Becomes zero
(c) Becomes unity
(d) Becomes maximum
Q9. Coefficient of consolidation is measured in
(a) cm²/g
(b) cm²/sec
(c) g/cm²/sec
(d) g-cm/sec
Q10. The ratio of settlement at any time ‘t’ to the final settlement is known as
(a) co-efficient of consolidation
(b) degree of consolidation
(c) consolidation index
(d) consolidation of undisturbed soil
SOLUTION
S1. Ans.(a)
Sol. In surveyor compass the graduated ring fixed at the box & moves with the box while the magnetic needle remain’s stationary along the north south line.
S2. Ans.(b)
Sol. line of sight. The line passes through optical centre of objective & intersection of cross hair is called as line of sight.
Line of collimation – When line of sight become horizontal. It will be known as line of collimation
Axis of telescope – This is an imagionary line joining point’s of optical centre of eyepiece to the optical centre of objective glass.
Axis of bubble tube – This is an imagionary line which is tangential on longitudenal curve of bubble tube.
S3. Ans.(b)
Sol. Black cotton soil – it is residual soil containing high percentage & suitable for growing cotton. High swelling & shrinkage. And have low shear strength also known as expansive soil & regur soil.
S4. Ans.(d)
Sol. uniformity coefficient (Cu) = D_60/D_10
If C_u≃1→ uniformly graded soil
C_u>6→ well graded sand
C_u>4→ well graded gravel
S5. Ans.(b)
Sol. super elevation (e) = 15 cm.
Rate of super elevation = 1 in 500
Length of transition curve = 500 ×0.15
= 75 metre.
S6. Ans.(d)
Sol. length of long chord = tangent length
2R sin ∆/2=R tan ∆/2
cos ∆/2 =1/2
▭(∆/2=60) ▭(∆ =120°)
S7. Ans.(b)
Sol. unit weight of soil (r_b ) = 25 kN/m³
Water content (w) = 50%
Specific gravity (G) = 2.55
Porosity (n) = ?
r_d= r_b/(1+w)=Grw/(1+e)
=25/(1+0.5 )=(2.55×9.81 )/(1+e)
▭(e=0.50)
n=e/(1+e)=0.5/1.5
▭(n=0.33)
S8. Ans.(b)
Sol. if pressure due to submerged weight of the soil is equal to the seepage pressure, the effective pressure becomes zero.
S9. Ans.(b)
Sol.
Coefficient Unit
Coefficient of compressibility
Coefficient of consolidation
Coefficient of volume compressibility m^2/kN
m^2/sec
m^2/kN
S10. Ans.(b)
Sol. degree of consolidation
(a) when settlement at any stage is known.
▭(%U=∆h/∆H×100 )
∆h → settlement after time ‘t’
∆H → ultimate consolidation settlement
(b) when pore pressure is known
▭(%U=(ui-u)/(ui-0)×100)
u → pore water pressure after time ‘t’
ui → pore water pressure at the beginning of consolidation
(c) when void ratio is known –
▭(% U=(e_0-e)/(e_0-e_100 )×100)
e_o→ void ratio at the beginning
e → void ratio after time ‘t’
e_100 → final void ratio at the end.
