QUIZ : CIVIL ENGINEERING(20-01-2020)

Quiz: CIVIL ENGINEERING
Exam: NPCIL-SCINTIFIC ASSISTANT(CIVIL)
Topic: MISCELLANEOUS

Each question carries 1 mark
Negative marking: No negative marking
Time: 10 Minutes

Q1. Consumptive use of water for a crop represents
(a) The transpiration needs of the crop
(b) evaporation needs of the cropped area
(c) evapotranspiration needs of the cropped area plus the minor quantity required in plant metabolism
(d) None of the above

Q2. The efficiency of water application does not depend upon
(a) climatic conditions
(b) type of the soil
(c) method of application
(d) geometry of the conveyance system
Q3. Given that for a triangulation survey
D= distance in km
H=the visible horizon from a station of known elevation above the datum (in metres)
If there is no obstruction due to intervening ground, then H is equal to
(a) 0.6735 D²
(b) 6.735 D²
(c) 0.06735 D²
(d) 0.006735 D²

Q4. A line of levels was run from Bench Mark A (770.815) to a Bench Mark B (772.940). The sum of back sights was 32.665 m and sum of fore sights was 30.44 m. The closing error is (in m)
(a) 0.105
(b) 0.100
(c) 0.205
(d) 0.200

Q5. The distance of visible horizon from a height of 36 m above mean sea level is given by
(a) √(36/0.6728)=km
(b) 36√(1/0.06728) km
(c) √(36/0.06728) km
(d) 36√0.06728 km
Q6. If arithmetic sum of latitudes of a closed traverse is Σ Lat and closing error in latitude is dx, the correction for a side whose latitude is l, as given by Transit Rule, is
(a) l×dx/(Σ Lat)
(b) l×ΣLat/dx
(c) ΣLat×dx/1
(d) None of these
Q7. If the reduced bearing of a line AB is N60° W and length is 100 m, then the latitude and departure respectively of the line AB will be
(a) +50 m, + 86.6 m
(b) +86.6 m, –50m
(c) +50 m, –86.6 m
(d) + 70.7 m, –50m

Q8. The error due to eccentricity of inner and outer axes can be eliminated by
(a) reading both verniers and taking the mean of the two
(b) taking both face observations and taking the mean of the two
(c) double sighting
(d) taking mean of several readings distributed over different portions of the graduated circle.
Q9. The length of a line measured with a 20 metre chain was found to be 250 metres. What is the true length of line if the chain was 10 cm too long
(a) 251.25 m
(b) 248.75 m
(c) 250.1 m
(d) 249.9 m
Q10. Which of the following trees yields hard wood?
(a) deodar
(b) chir
(c) sheesham
(d) pine

Solutions

S1. Ans.(c)
Sol. Consumptive use of water for a crop represents evapotranspiration needs of the cropped area plus the minor quantity required in plant metabolism.

S2. Ans.(d)
Sol. Efficiency of water depends on
→ Climate condition
→ Type of soil
→ Method of application.

S3. Ans.(c)
Sol. ▭(H=0.06735 D^2 )
D → Distance in km.
H → distance of visible Horizon (m)

S4. Ans.(b)
Sol. RL_B = RL_A + (ΣBS – ΣFS)
RL_B = 770.815 + 32.665 – 30.44
= 773.035
Closing error = 773.035 – 772.940
= 0.10

S5. Ans.(c)
Sol. Height (H) = 36 m
Distance of visible Horizon = √(H/0.0673) = √(36/0.0673)

S6. Ans.(a)
Sol. Transit method → Correction of latitude and departure of any side
= Total error in latitude or departure ×[(latitude or departure )/(sum of latitude or departure )]

S7. Ans.(c)
Sol. Reduced bearing = N 60° W
Length = 100 m
Latitude = 100 cos 60°
= 50
Departure = – 100 sin 60°
= – 86.6

S8. Ans.(a)
Sol. Error due to eccentricity of inner and outer axis means that the centre of graduated horizontal circle does not coinside with centre of vernier plate.

S9. Ans.(a)
Sol. True length of line given by
l = l^’ (L^’/L)
= 250 (20.1/20)
= 251.25 m.

S10. Ans.(c)
Sol. The deciduous or broad leaf trees have flat broad levees. These tree yield hard woods which are generally close grained, strong, heavy dark coloured. The example of such tree as teak, mahagony, sheesham etc.

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