Surveying miscellaneous
- Two Pegs A and B were fixed on opposite banks of a 50 m wide river. The level was set up at A and the staff readings on Pegs A and B were observed as 1.350 m and 1.550 m, respectively. Thereafter the instrument was shifted and set up at B. The staff readings on Pegs B and A were observed as 0.750 m and 0.550 m, respectively. If the R.L. of Peg A is 100.200 m, the R.L. (in m) of Peg B is ______ .
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Thus is reciprocal levelling
∆h = (b1 - a1) + (b2 - a2) 2 ∆h = (1.55 - 1.35) + (10.75 - 0.55) = 0.20 2
RL at B = RL at A – ∆h = 100.200 – 0.20 = 100Correct Option: C
Thus is reciprocal levelling
∆h = (b1 - a1) + (b2 - a2) 2 ∆h = (1.55 - 1.35) + (10.75 - 0.55) = 0.20 2
RL at B = RL at A – ∆h = 100.200 – 0.20 = 100
- In a leveling work, sum of the Back Sight (B.S.) and Fore Sight (F.S.) have been found to be 3.085 m and 5.645 m respectively. If the Reduced Level (R.L.) of the starting station is 100.000 m, the R.L. (in m) of the last station is ______ .
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∑BS = 3.085
∑F.S = 5.645 m
Fall = ∑FS – ∑BS
= 5.645 – 3.085 = 2.560
R.L of last station = R.L of first station – Fall
= 100 – 2.560 = 97.44 mCorrect Option: A
∑BS = 3.085
∑F.S = 5.645 m
Fall = ∑FS – ∑BS
= 5.645 – 3.085 = 2.560
R.L of last station = R.L of first station – Fall
= 100 – 2.560 = 97.44 m
- A tacheometer was placed at point P to estimate the horizontal distances PQ and PR. The corresponding stadia intercepts with the telescope kept horizontal, are 0.320 m and 0.210 m, respectively. The ∠QPR is measured to be 61° 30' 30''. If the stadia multiplication constant = 100 and stadia addition constant = 0.10 m, the horizontal distance (in m) between the points Q and R is ___________
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NA
Correct Option: A
NA
- A theodolite is set up at station A and a 3 m long staff is held vertically at station B. The depression angle reading at 2.5 m marking on the staff is 6°10'. The horizontal distance between A and B is 2200 m. Height of instrument at station A is 1.1 m and R.L. of A is 880.88 m. Apply the curvature and refraction correction, and determine the R.L. of B (in m).
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R.L. of A = 880.88 m
R.L. of Plane of Collimation = 880.88 + 1.2 = 882.08m
True staff reading at station B = 2.5 – 0.0673 × 2.22 = 2.174 m
Now, B'B'' = (tan6º 10') × 2200 = 237.701 m
R.L. of station B = R.L. of Plane of Collimation – true staff reading – B'B''
= 882.08 – 2.174 – 237.701 = 642.205 mCorrect Option: B
R.L. of A = 880.88 m
R.L. of Plane of Collimation = 880.88 + 1.2 = 882.08m
True staff reading at station B = 2.5 – 0.0673 × 2.22 = 2.174 m
Now, B'B'' = (tan6º 10') × 2200 = 237.701 m
R.L. of station B = R.L. of Plane of Collimation – true staff reading – B'B''
= 882.08 – 2.174 – 237.701 = 642.205 m
- The bearings of two inaccessible stations, S1 (Easting 500 m, Northing 500 m) and S2 (Easting 600 m, Northing 450 m) from a station S3 were observed as 225° and 153° 26' respectively. The independent Easting (in m) of station S3 is :
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Northing of S3 = 500 + L1 cos 45°
= 450 + L2 cos 26° 34'
⇒ L1 cos 45° – L2 cos26° 34' = –50
Easting of S3
= 500 + L1 sin 45° = 600 – L2 sin 26° 34'
L1 sin 45° + L2 sin 26°34' = 100
∴ L1 = 70.71, L2 = 111.80
Easting of S3 = 500 + 70.71 × sin 45° ≈ 550mCorrect Option: C
Northing of S3 = 500 + L1 cos 45°
= 450 + L2 cos 26° 34'
⇒ L1 cos 45° – L2 cos26° 34' = –50
Easting of S3
= 500 + L1 sin 45° = 600 – L2 sin 26° 34'
L1 sin 45° + L2 sin 26°34' = 100
∴ L1 = 70.71, L2 = 111.80
Easting of S3 = 500 + 70.71 × sin 45° ≈ 550m