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UPSSSC-JE’21 CE: Daily Practice Quiz. 19-October-2021

Know your strengths and practice your concepts with this quiz on UPSSSC-JE Recruitment 2021. This quiz for UPSSSC-JE Recruitment 2021 is designed specially according to UPSSSC-JE Syllabus 2021.

Quiz: Civil Engineering

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes

Q1. Calculate the length (m.) of the longer chord of 260 m radius curve having deflection angle of 90 degree.
(a) 353.6 m
(b) 392.7 m
(c) 367.7 m
(d) 500 m

Q2. What is the purpose of conducting the resection method in the plane table surveying?
(a) To survey hilly region
(b) To survey plain region
(c) To plot the details
(d) To determine the location of the instrument

Q3. The depth of water below the spillway and after hydraulic jump are 1 m and 6m respectively. The head loss will be ……………
(a) 174 m
(b) 5.2 m
(c) 1.7 m
(d) None of these

Q4. While designing a hydraulic structure, the piezometric head at bottom of the floor is computed as 10m. The datum is 3 m below floor bottom. The assured standing water depth above the floor is 2 m. The specific gravity of the floor material is 2.5 The floor thickness should be
(a) 2.00 m
(b) 3.33 m
(c) 4.40 m
(d) 6.00 m

Q5. The average rainfall for a 3 hour duration storm is 2.7 cm and the loss rate is 0.3 cm/hr. The flood hydrograph has a base flow of 20 m³/s and produces a peak flow of 210 m³/s. The peak of a 3-h unit hydrograph is
(a) 125.50 m³/s
(b) 105.50 m³/s
(c) 77.77 m³/s
(d) 70.37 m³/s

Q6. In a river, discharge is 173 m³/s; water surface slope is 1 in 6000; and stage at the gauge station is 10.0 m. If during a flood, the stage at the gauge station is same and the water surface slope is 1 in 2000, the flood discharge in m³/s, is approximately
(a) 371
(b) 100
(c) 519
(d) 300


S1. Ans.(c)
Sol. Given, Radius of curve (R) = 260 m.
Deflection angle (∆) = 90°
Length of Long chord is given by–
T₁DT₂ = 2R sin ∆/2
= 2 × 260 × sin 90/2
▭(T_1 DT_2=367.7 m)

S2. Ans.(d)
Sol. The purpose of conducting the resection method in the plane table surveying is to determine the location of the instrument point.

S3. Ans. (b)
Sol. y_1 = 1 m.
y_2 = 6 m.
Head loss (h_L) = ((y_2-y_1)³)/(4y_1 y_2 )
= ((6-1)³)/(4 × 6 ×1)

h_L = 5.20 m

S4. Ans.(b)
Sol. Floor thickness,
h = 10 – (3 + 2) = 5 m
t=5/(2.5-1)=3.33 m.

S5. Ans.(b)
Sol. Rainfall excess = 2.7 – (0.3 × 3)
= 1.8 cm.
Peak discharge of direct runoff hydrograph
= 210 – 20
= 190 m³/sec
Peak of 3-hour unit hydrograph =
=190/1.8=105.55 m^3/sec

S6. Ans.(d)
Sol. Q₁ = 173 m³/sec
S₁ = 1 in 6000
S₂ = 1 in 2000
Q₂ = ?
(By manning^’ eq^n)/(Q=1/n AR^(2/3) S^(1/2) )
Q ∝√S
Q_1/Q_2 =√((S₁)/(S₂))
173/Q_2 =√((1/6000)/(1/2000))
▭(Q_2=299.64≈300 m^3/sec)

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