Surveying miscellaneous

Surveying miscellaneous

Easy Questions

Surveying

Surveying miscellaneous
  1. The observation from a closed loop traverse around an obstacle are

    What is the value of the missing measurement (rounded off to the nearest 10 mm)?








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    Latitude of each segment = L × cos θ.
    PQ = L cos 33.75°
    QR = 300 × cos 86.3847 = 18.917
    RS = 354.524 × cos 169.3819 = – 348.453
    ST = 450 × cos 243.9003° = – 197.97
    TP = 268 × cos 317.5° = 197.590
    Since it is a closed loop traverse ∑ latitude = 0
    ∴ L cos 33.75° + 18.917 – 348.45 – 197.97 + 197.59.
    ∴ L = 369.89 m

    Correct Option: A

    Latitude of each segment = L × cos θ.
    PQ = L cos 33.75°
    QR = 300 × cos 86.3847 = 18.917
    RS = 354.524 × cos 169.3819 = – 348.453
    ST = 450 × cos 243.9003° = – 197.97
    TP = 268 × cos 317.5° = 197.590
    Since it is a closed loop traverse ∑ latitude = 0
    ∴ L cos 33.75° + 18.917 – 348.45 – 197.97 + 197.59.
    ∴ L = 369.89 m

  1. A Bench Mark has been established at the soffit of an ornamental arch at the known elevation of 100.0 m above sea level. The back sight used to establish height of instrument is an inverted staff reading of 2.105 m. A forward sight reading with normally held staff of 1.105 m is taken on a recently constructed plinth. The elevation of the plinth is








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    HI = BM + BS = 100 – 2.105 (inverted reading) = 97.895
    FS = HI – level. = 97.895 – 1.105 = 96.79 m

    Correct Option: D

    HI = BM + BS = 100 – 2.105 (inverted reading) = 97.895
    FS = HI – level. = 97.895 – 1.105 = 96.79 m

  1. The local mean time at a place located in longitude 90° 40' E when the standard time is 6 hours and 30 minutes and the standard meridian is 82° 30' E is








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    Difference between standard meridian and longitude of place = 90° 40' E – 82° 30'. = 8° 10'.
    1° means 4 minutes


    ∴ 8° 10' means
    49
    		 °
    6

    ∴ Time =
    49°
    		 × 4 = 32 min 40 seconds
    6

    ∴ Total time = 6 hours + 30 min + 32 min 40 sec = 7h 2 m 40s.

    Correct Option: D

    Difference between standard meridian and longitude of place = 90° 40' E – 82° 30'. = 8° 10'.
    1° means 4 minutes


    ∴ 8° 10' means
    49
    		 °
    6

    ∴ Time =
    49°
    		 × 4 = 32 min 40 seconds
    6

    ∴ Total time = 6 hours + 30 min + 32 min 40 sec = 7h 2 m 40s.

  1. The magnetic bearing of a line AB was N 59° 30' W in the year 1967, when the declination was 4° 10' E. If the present declination is 3°W, the whole circle bearing of the line is








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    True bearing = magnetic bearing ± declination
    = N 59° 30' W – 4° 10' E
    = N 55° 20' W
    WCB = 360° – 55° 20'. = 304° 40'
    W.C. magnetic bearing = 304° 40' – 3° = 301° 40'

    Correct Option: D

    True bearing = magnetic bearing ± declination
    = N 59° 30' W – 4° 10' E
    = N 55° 20' W
    WCB = 360° – 55° 20'. = 304° 40'
    W.C. magnetic bearing = 304° 40' – 3° = 301° 40'

  1. In quadrantal bearing system, bearing of a line varies from








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    NA

    Correct Option: C

    NA