GATE'22 CE: Daily Practices Quiz. 16-July-2021 |_00.1
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GATE’22 CE: Daily Practices Quiz. 16-July-2021

Quiz: Civil Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 marks
Negative marking: 1/3 mark
Time: 12 Minutes

Q1. The back sight at a station A is 0.865 m and the height of instrument is 561.365 m. the reduced level at A will be (in unit m)
(a) 558.705
(b) 559.260
(c) 560.500
(d) 560.550

Q2. A trapezoidal channel with base of 6 m and side slope of two horizontal to one vertical conveys water at 17 m³/sec with a depth of 1.5 m. The flow situation in the channel is:
(a) Critical
(b) Supercritical
(c) Subcritical
(d) None of the above

Q3. The area between the two isohyets 45 cm and 55 cm is 100 km², and that between 55 cm and 65 cm is 150 km². what is the average depth of annual precipitation over the basin of 250 km²?
(a) 50 cm
(b) 52 cm
(c) 56 cm
(d) 60 cm

Q4. The following data were obtained when a sample of medium sand was tested in a constant head permeameter:
Cross-section area of sample : 100 cm²
Hydraulic gradient : 10
Discharge collected : 10 cc/s
The coefficient of permeability of the sand is
(a) 0.1 m/s
(b) 0.01 m/s
(c) 1 × 10^(-4) m/s
(d) 1 × 10^(-8) m/s

Q5. A crest vertical curve joins two gradients of +3% and -2% for a design speed of 80 km/h and the corresponding stopping sight distance of 120 m. the height of driver’s eye and the object above the road surface are 1.20 m and 0.15 m respectively. The curve length (which is less than stopping sight distance) to be provided is
(a) 120 m
(b) 152 m
(c) 163 m
(d) 240 m

Solutions

S1. Ans.(c)
Sol.

GATE'22 CE: Daily Practices Quiz. 16-July-2021 |_40.1

HI=RL_A+BS
RL_A=HI-BS
RL_A=561.365-0.865
▭(RL_A=560.500 m.)

S2. Ans.(c)
Sol. Given,
B = 6m.
Q = 17 m³/s
y = 1.5 m.
2H:1V = mH:1V ⇒ m = 2
T = B + 2my
T = 6 + (2 × 2 × 1.5)
▭(T=12m)
A = (B + my) y
= [6 + (2 × 1.5)] × 1.5
= 13.5 m²
We know,
Froude no.(F)=(Q^2 T)/(gA^3 )
=((17)^2×12)/(9.81×(13.5)^3 )
▭(F=0.14368<1) Hence,flow is Subcritical

S3. Ans.(c)
Sol.
P_1=45 cm.
P_2=55 cm
P_3=65 cm
A_1=100 km^2
A_2=150 km^2
P_avg= ?
Average depth of annual precipitation by Iso-hyetal method is given by →
P_avg=(((P_1+P_2)/2).A_1+((P_2+P_3)/2) A_2)/(A_1+A_2 )
=(((45+55)/2)×100+((55+65)/2)×150)/(100+150)
=(5000+9000)/250
▭(P_avg=56 cm)

S4. Ans.(c)
Sol. Given,
Q = 10 cc/sec.
i = 10
A = 100 cm²
K = ?
We know,
Q = KiA
10 = K × 10 × 100
K=1/100 cm\/sec⁡
▭(K=1×10^(-4) m\/sec.)

S5. Ans.(b)
Sol. Given,
N_1= +3%=0.03
N_2= -2%= -0.02
N=|N_1-N_2 |=|0.03-(-0.02)|=0.05
S=120m
Note→ When, ▭(SSD<Curve length), then curve length is given by –
▭(L=2S-4.4/N)
=(2×120)-4.4/0.05
=240-88
▭(L=152m)

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