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# Quiz: Civil Engineering 28 May 2021

Quiz: Civil Engineering
Exam: DFCCIL
Topic: Surveying

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 10 Minutes

Q1. Survey of India was publishing toposheets using a scale of :
(a) 1 : 5000
(b) 1 : 1000
(c) 1 : 10000
(d) 1 : 50000

Q2. A rectangular plot of 16 km² in area is shown on a map by a similar rectangular area of 1 cm², R.F. of the scale to measure a distance of 40 km will be :
(a) 1/1600
(b) 1/400000
(c) 1/400
(d) 1/16000

Q3. A 30 m, metric chain is found to be 0.1 m too short throughout the measurement. If the distance measured is recorded as 300 m, then the actual distance measured will be :
(a) 300.1 m
(b) 299.0 m
(c) 301.0 m
(d) 310.0 m

Q4. Magnetic bearing of line 48°20’ and magnetic declination is 5°38’ E, then the true bearing is ………
(a) 44°02’
(b) 34°02’
(c) 64°02’
(d) 53°58’

Q5. If the forbearing of a line AB is 35° and that of line BC is 15°, then the included angle between the line is
(a) 20°
(b) 50°
(c) 160°
(d) 230°

Q6. The most accurate method of orientation by the three point method of plane tabling is the
(a) Tracing paper method
(b) Graphical method
(c) Trail-and-error method
(d) Both (a) & (b)

Q7. If the focus length of the object glass is 25cm and the distance from object glass to the trunnion axis is 15 cm, the additive constant is
(a) 0.1
(b) 0.4
(c) 0.6
(d) 1.33

Q8. The tangent distance of a curve of radius ‘R’ deflected through an angle of 60 degrees will be:
(a) R
(b) R√3
(c) R/√3
(d) infinite (∞)

Q9. The staff reading at a distance of 80m from a level with the bubble at centre is 1.31 m. when the bubble is moved 5 divisions out of the centre, the reading is 1.39 m. then the angular value of one division of the bubble is
(a) 28.78 sec
(b) 41.25 Sec
(c) 14.25 Sec
(d) 25.05 Sec

Q10. In leveling between two points A and B on opposite banks of a river, the following reading were taken:

A          B

A                                              1.500   1.00

B                                              1.350   0.850
If R.L. of A is 100.0 m, the R. L. of B
(a) is less than 100.0m
(b) is more than 100.0m
(c) is 100.0m
(d)Cannot be determined by given data

Solutions

S1. Ans.(d)
Sol. survey of India published toposheets using scale. 1/25000 to 1/250000
Geographical map 1/500000 to 1/15000000 &
Land survey 1/500 to 1/5000

S2. Ans.(b)
Sol. Area of plot =16 km^2
Area of plot in map = 1 cm²
1 cm² = 16 km²
1 cm = 4 km
Representative fraction (R.F.) = 1cm/(4×100000 cm)
▭(R.F.1/400000)

S3. Ans.(b)
Sol. True length of chain (L) = 30 m.
False length of chain (L^1) = 29.9 m.
False length of chain (l^1) = 300 m.
True length of chain (l) = ?
L×l=L^1×l^1
30×l=29.9×300
l=(29.9×300)/30
▭(l=299m)

S4. Ans.(d)
Sol. Magnetic declination = S°38’ E
Magnetic bearing = 48°20’
True bearing = ?

True bearing of AB= 48°20’ +5°38’
= 53°58’

S5. Ans.(c)
Sol. fore bearing of AB = 35°
Fore bearing of BC = 15°
Included angle = (90-35) + 15 + 90°
→160°

S6. Ans.(c)
Sol. Three point method of plane tabling is done by following
(i) Lehman’s rule
(ii) Tracing paper method
(iii) Bessel’s rule (graphical method)
(iv) trial & error method.
Trial & error method is most accurate method.

S7. Ans.(b)
Sol. f = 25 cm.
d = 15 cm.
In tocheometre. ▭(D=kS+c)
K ⇒ Multiplying constant
▭(K=f/i)
▭(C=f+d)
C = 25+15
▭(C=40 cm)=0.40 m.

S8. Ans.(c)
Sol. ∆ =60°
Tagent distance = R tan ∆/2
=R tan 60/2
=R tan⁡30°
=R/√3

S9. Ans.(b)
Sol.

Angular value of one division =s/nd
=((1.39-1.31))/(5×80)
= 2×10^(-4)×206265 sec.
=(41.253 sec) ̲

S10. Ans.(b)
Sol. h_A = 1.5, h_B = 1
h_A’ = 1.35, h_B’ = 0.850
Difference in level (H) = ((h_B-h_A )+(h_B’-h_A’ ))/2
= ((1-1.5)+(0.850-1.35))/2
=(-1)/2
=-0.5
R_L of A=100 m
R_L of B=100+0.5
=100.5m

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