Surveying miscellaneous

Surveying miscellaneous

Easy Questions

Surveying

Surveying miscellaneous
  1. During a levelling work along a falling gradient using a Dumpy level and a Staff of 3m length, following successive readings were taken: 1.785, 2.935, 0.360, 1.320. What will be the correct order of booking these four readings in a level book? (BS : Back Sight IS : Intermediate Sight, FS : Fore Sight)








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    For first reading it is BS and then FS. Sudden change in value represent shifting of instrument. Therefore again first reading BS and then FS.

    Correct Option: A

    For first reading it is BS and then FS. Sudden change in value represent shifting of instrument. Therefore again first reading BS and then FS.

  1. Following bearings are observed while traversing with a compass.
    Line Fore Bearing Back Bearing
    AB 126°45' 308°00'
    BC 49°15' 227°30'
    CD 340°30' 161°45'
    DE 258°30' 78°30'
    EA 212°30' 31°45'

    After applying the correction due to local attraction, the corrected fore bearing of line BC will be








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    The difference between back bearing and fore bearing of line DE = 258°30' – 78° 30' = 180°
    ∴ Station D and E are free from local attraction
    Fore bearing of line EA = 212° 30'
    Correct back bearing of line EA = 216° 30' – 180° = 36° 30'
    ∴ Error at A = 31° 45' – 36° 30' = – 4° 45'
    Hence correction at A = + 4° 45'
    ∴ Correct fore bearing of line AB = 126°45' + 4°45'= 131° 30'
    Correct back bearing of AB = 131° 30 + 180° = 311° 30'
    ∴ Error at B = – 3° 30'
    Correction at B = + 3° 30'
    Hence correct bearing of BC = 45° 15' + 3° 30' = 48° 45'

    Correct Option: D

    The difference between back bearing and fore bearing of line DE = 258°30' – 78° 30' = 180°
    ∴ Station D and E are free from local attraction
    Fore bearing of line EA = 212° 30'
    Correct back bearing of line EA = 216° 30' – 180° = 36° 30'
    ∴ Error at A = 31° 45' – 36° 30' = – 4° 45'
    Hence correction at A = + 4° 45'
    ∴ Correct fore bearing of line AB = 126°45' + 4°45'= 131° 30'
    Correct back bearing of AB = 131° 30 + 180° = 311° 30'
    ∴ Error at B = – 3° 30'
    Correction at B = + 3° 30'
    Hence correct bearing of BC = 45° 15' + 3° 30' = 48° 45'

  1. 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 m

    Correct 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

  1. 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

  1. 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 m

    Correct 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