# 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?

(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

SOLUTIONS

S1. Ans.(b)
Sol.

AB=62 m.
BC=428 m.p
AC= ?
From Pythagoras theorem
BC^2=AB^2+AC^2
AC^2=(428)^2-(62)^2
AC= √((428)^2-(62)^2 )
= √((428+62)(428-62) )
=√(490×366)
▭(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)
=0.0868
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,
n=2
R=300 m.
l=8 m.
V=100 kmph.
E_W= ?
E_W=(nl^2)/2R+V/(9.5√R)
=(2×(8)^2)/(2×300)+100/(9.5√300)
=0.213+0.608
▭(E_w=0.82 m.)

S4. Ans.(d)
Sol. Given,
H_0=1m.
∆σ ̅=20 kPa
G=2.7
w=0.45
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-
Se=wG
1×e=0.45×2.7
▭(e=1.215)
∆H=(0.45×1)/(1+1.215) log_10 [(100+20)/100]
=0.45/2.215 log_10 (120/100)
=0.016 m
▭(∆H=16mm.)

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

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

S7. Ans.(a)
Sol.
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=((Qf^2)/140)^(1\/6)
=[(30×(1)^2)/140]^(1\/6)
▭(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-(2/3)k]×k
q=80 k-2/3 k^2—P(1)
For q to be maximum →
dq/dk=0
80=(-4k)/3=0
▭( k=60)
Now, maximum capacity
q_max=(80×60)-2/3×(60)^2
▭(q_max=2400 veh\/h)

S10. Ans.(a)
Sol. given, d=6mm
Asv=π/4 d^2×2
=π/4×(6)^2×2
=56.54 mm^2
V_s=70kN=70×10^3 N
f_y=250
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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