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Quiz: Civil Engineering 7 June 2021

Quiz: Civil Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/3 mark
Time: 20 Minutes

Q1. The distance between the points measured along a slope is 428 m. if the difference in level is 62 m, then the horizontal distance between them is
(a) 643.48 m
(b) 423.48 m
(c) 523.48 m
(d) 323. 48 m

Q2. The BOD_5 of a surface water sample is 200 mg/litre at 20°C. the value of the reaction constant is K = 0.2 day^(-1)with base ‘e’. the ultimate BOD of the sample is
(a) 126 mg/litre
(b) 544 mg/litre
(c) 146 mg/litre
(d) 316 mg/litre

Q3. The extra widening required for a two-lane national highway at a horizontal curve of 300 m radius, considering a wheel base of 8m and a design speed of 100 kmph is
(a) 0.42 m
(b) 0.62 m
(c) 0.82 m
(d) 0.92 m

Q4. A layer of normally consolidated, saturated silty clay of 1m thickness is subjected to one dimensional consolidation under a pressure increment of 20 kPa. The properties of the soil are: specific gravity = 2.7, natural moisture content = 45% compression index = 0.45% and recompression index = 0.05. the initial average effective stress within the layer is 100 kPa. Assuming Terzaghi’s theory to be applicable, the primary consolidation settlement (rounded off to the nearest mm) is
(a) 2 mm
(b) 9 mm
(c) 14 mm
(d) 16 mm

Q5. Critical depth at a section of a rectangular channel is 1.5 m. the specific energy at that section is
(a) 0.75 m
(b) 1.0 m
(c) 1.5 m
(d) 2.25 m

Q6. A solid circular shaft of diameter d and length L is fixed at one end and free at the other end. A torque T is applied at the free end. The shear modulus of the material is G. the angle of twist at the free end is
(a) 16TL/(πd^4 G)
(b) 32TL/(πd^4 G)
(c) 64TL/(πd^4 G)
(d) 128TL/(πd^4 G)

Q7. What is the total degree of indeterminacy, both internal and external of the plane frame shown below?

Quiz: Civil Engineering 7 June 2021_30.1

(a) 10
(b) 11
(c) 12
(d) 14

Q8. An earthen channel has been designed on Lacey formulae to carry a full supply discharge of 30 m³/s. the mean velocity of flow at this discharge is :
(a) 0.98 m/s
(b) 0.77 m/s
(c) 2.2 m/s
(d) 1.36 m/s

Q9. On a section of a highway the speed-density relationship is linear and is given by V =
[80 – (2/3)k]; where v is in km/h and k is in veh/km. the capacity (in veh/h) of this section of the highway would be
(a) 1200
(b) 2400
(c) 4800
(d) 9600

Q10. Using 2 legged vertical stirrups of 6 mm diameter mild steel (Fe 250) find spacing to resists 70.0 kN shear force. Size of the beam is 300 mm×1000 (effective).
(a) 175 mm
(b) 117 mm
(c) 300 mm
(d) 250 mm


S1. Ans.(b)

Quiz: Civil Engineering 7 June 2021_40.1

AB=62 m.
BC=428 m.p
AC= ?
From Pythagoras theorem
AC= √((428)^2-(62)^2 )
= √((428+62)(428-62) )
▭(AC=423.48 m)

S2. Ans.(d)
Sol. Given,
BOD_5=200 mg\/l.
K=0.2\/day (at base e)
=0.2×0.434 (at base 10)
Ultimate BOD (L)= ?
▭(BOD_5=L(1-10^(-K_D t) ) )
200=L(1-10^(-0.0868×5) )
L=200/((1-10^(-0.0868×5) ) )
▭(L=316.52 mg\/l)

S3. Ans.(c)
Sol. Given,
R=300 m.
l=8 m.
V=100 kmph.
E_W= ?
▭(E_w=0.82 m.)

S4. Ans.(d)
Sol. Given,
∆σ ̅=20 kPa
C_C =0.45
(σ_0 ) ̅=100 kPa
∆H= ?
▭(∆H=(C_C H_O)/(1+e_0 ) log_10 [((σ_0 ) ̅+(∆σ) ̅)/(σ_0 ) ̅ ] )
We know-
∆H=(0.45×1)/(1+1.215) log_10 [(100+20)/100]
=0.45/2.215 log_10 (120/100)
=0.016 m

S5. Ans.(d)
Sol. given,
y_c=1.5 m
E_c= ?
E_c=3/2 y_c
▭(E_C=2.25 m.)

S6. Ans.(b)
Sol. Angle of Twist (θ_P )-
For solid circular shaft = (J) = π/32 d^4
▭(θ_P=32TL/(πd^4 G))

S7. Ans.(a)
No.of members (m)=10
No.of joints(j) = 10
No.of external reactions (r_e )=12
Internal hinged reactions (r_r )=2
Degree of static indeterminacy(D_s)=(3m+r_e )-3j-r_r
= [(3×10) + 12] –(3×10) – 2
= 42 – 32
▭(D_S=10 )

S8. Ans.(b)
Sol. Note → Silt factor range varies from = 0.8 – 1.2
Let silt factor be (f) = 1.0
Discharge (Q) = 30 m³/sec.
Velocity of flow (V) = ?
According to lacey,
▭(V=0.77 m\/sec.)

S9. Ans.(b)
Sol. Given, V=[80-(2/3)k]
We know,
capacity (veh\/h)=velocity (km\/h)×Traffic density (Veh\/km)
q=80 k-2/3 k^2—P(1)
For q to be maximum →
▭( k=60)
Now, maximum capacity
▭(q_max=2400 veh\/h)

S10. Ans.(a)
Sol. given, d=6mm
Asv=π/4 d^2×2
=56.54 mm^2
V_s=70kN=70×10^3 N
Spacing (S) = ?
S=(0.87fy×Asv×Effective depth)/V_s
S=(0.87×250×56.54×1000)/(70×10^3 )
▭(S=175.67≃175 mm.)

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