Quiz: Civil Engineering
Exam: OPSC
Topic: MISCELLANEOUS
Each question carries 1 mark
Negative marking: 1/3 mark
Time: 10 Minutes
Q1. The ordinate of the instantaneous Unit Hydrograph (IUH) of a catchment at any time t, is
(a) the slope of the 1 hour unit hydrograph at that time
(b) the slope of the direct runoff unit hydrograph at that time
(c) difference in the slope of the S curve and 1 hour unit hydrograph
(d) the slope of the S curve with effective rainfall intensity of 1 cm/hr
Q2. 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
Q3. Dickens formula predicts maximum flood discharge, Q, in terms of the area, A, and the coefficient, c, as Q = cA^n. The value of n is
(a) 0.25
(b) 0.50
(c) 0.67
(d) 0.75
Q4. Energy loss in a hydraulic jump in a rectangular channel is given by: (Symbols have their usual meanings)
(a) ∆E =(y_2-y_1 )³/(4y_1 y_2 )
(b) ∆E =(V_1-V_2 )³/(2g√(V_1+V_2 ))
(c) Both (a) & (b)
(d) None of the above
Q5. The following bearing were observed while traversing with a compass, which stations are affected by local attraction?
Line F.B. B.B.
AB 104°30’ 284°30’
BC 48°15’ 226°0’
CD 290°30’ 115°15’
DA 180°15’ 357°15’
(a) C and D
(b) A and D
(c) A and B
(d) B and C
Q6. The difference between face left and face right observation of a theodolite is 3’. The error is
(a) 3’
(b) 1°30”
(c) 45”
(d) 0’
Q7. Consider the following statements:
Reciprocal levelling eliminates the effect of
1. errors due to atmospheric refraction
2. errors due to earth’s curvature
3. errors due to line of collimation
4. mistake in taking levelling staff readings
Which of these statements are correct?
(a) 2, 3 and 4
(b) 1, 3 and 4
(c) 1, 2 and 3
(d) 1, 2 and 4
Q8. Chlorine usage in the treatment of 25000 m³/day of water has been 9 kg/day. The residual chlorine after 10 minutes contact is 0.2 mg/l. The chlorine demand of water would be nearly
(a) 0.12 mg/l
(b) 0.16 mg/l
(c) 0.22 mg/l
(d) 0.28 mg/l
Q9. The correct relationship between theoretical oxygen demand (TOD), biochemical oxygen demand (BOD) and, chemical oxygen demand (COD) is given by.
(a) TOD > BOD > COD
(b) BOD > COD > TOD
(c) COD > BOD > TOD
(d) TOD > COD > BOD
Q10. The unit in which both sedimentation and digestion processes of sludge take place simultaneously is called:
(a) Skimming tank
(b) Detritus tank
(c) Imhoff tank
(d) Digestion tank
SOLUTION
S1. Ans.(d)
Sol. The ordinate of one Instantaneous unit hydrograph at any time ‘t’ is the slope of s – curve of intensity 1 cm/hr.
S2. 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
S3. Ans.(d)
Sol. Dickens formula
▭(Q=CA^(3/4) )
▭(n=3/4)
S4. Ans. (a)
Sol. Head loss (Energy loss) (h_L) = ((y_2-y_1)³)/(4y_1 y_2 )
S5. Ans.(a)
Sol. Since difference of F.B and B.B of line AB differs by 180°. Hence both stations A and B are unaffected by local attraction and station C and D are affected by local attraction.
S6. Ans.(b)
Sol.
▭(Error=(Difference between face left and face right observation)/2)
Error = (3’)/2
▭(Error=1°30″)
S7. Ans.(c)
Sol. Reciprocal levelling is used to determine the correct difference of elevations of two points. It eliminates the error due to curvature of earth, atmospheric refraction, and collimation error.
S8. Ans.(b)
Sol. Given,
Quantity of water treated = 25000 m³/day
Chlorine use = 9 kg/day
Chlorine dose = (Chlorine used)/(Quantity of water treated)
=(9 kg/day)/(25000 m^3/day)
=(9×10^6 mg/day)/(25000×10^3 l/day)
= 0.36 mg/l
Chlorine demand = Chlorine dose – Residual chlorine
= 0.36 – 0.2
= 0.16 mg/l
S9. Ans.(d)
Sol. The correct relationship between TOD, COD and BOD is
▭(TOD>COD>BOD)
S10. Ans.(c)
Sol. An Imhoff tank is an improvement over septic tank in which both sedimentation and digestion process of sludge takes place.
