Surveying miscellaneous

Surveying miscellaneous

Easy Questions

Surveying

Surveying miscellaneous
  1. The following measurements were made during testing a levelling instrument.

    P1 is close to P and Q1 is close to Q. If the reduced level of station P is 100.000 m, the reduced level of station Q is








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    Apparant diff. in elevation (from P) = 1.700 – 2.800 = – 1.1 m
    Apparant diff. in elevation (from Q) = 1.800 – 2.700 = – 0.90 m

    ∴ True difference =
    1.1 + 0.9
    		 = 1 m
    2

    ∴ True elevation of Q = 100 + 1 = 101 m

    Correct Option: C

    Apparant diff. in elevation (from P) = 1.700 – 2.800 = – 1.1 m
    Apparant diff. in elevation (from Q) = 1.800 – 2.700 = – 0.90 m

    ∴ True difference =
    1.1 + 0.9
    		 = 1 m
    2

    ∴ True elevation of Q = 100 + 1 = 101 m

  1. Two straight lines intersect at an angle of 60°. The radius of a curve joining the two straight lines is 600 m. The length of long chord and mid-ordinates in metres of the curve are








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    Length of long chord = 2R sin
    θ
    		 = 2 × 600 × sin 50 = 600 m
    2


    Length of mid ordinate = L1 - cos
    θ
    2

    = 600 (1 – cos 30°) = 80.38 m

    Correct Option: B

    Length of long chord = 2R sin
    θ
    		 = 2 × 600 × sin 50 = 600 m
    2


    Length of mid ordinate = L1 - cos
    θ
    2

    = 600 (1 – cos 30°) = 80.38 m

  1. The magnetic bearing of a line AB is S 45° E and the declination is 5° West. The true bearing of the line AB is








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    SE – therefore 2nd quadrant
    True bearing = S 40° E.

    Correct Option: B

    SE – therefore 2nd quadrant
    True bearing = S 40° E.

  1. The plan of a survey plotted to a scale of 10 m to 1 cm is reduced in such a way that a line originally 10 cm long now measures 9 cm. The area of the reduced plan is measured as 81 cm2. The actual (m2) of the survey is








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    Actual area =
    Actual scale
    		2 × measured area × scale
    Reduced scale

    =
    100
    		 × 81 × 100 = 10000 m2
    81

    Correct Option: A


    Actual area =
    Actual scale
    		2 × measured area × scale
    Reduced scale

    =
    100
    		 × 81 × 100 = 10000 m2
    81

  1. The length and bearings of a closed traverse PQRSP are given below.

    The missing length and bearing, respectively of the line SP are








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    For closed traverse in WCB,
    Sum of latitude, ∑L cos θ = 0 and sum of departure ∑L sin θ = 0.
    ∑L cos θ = 200 cos 0° + 1000 cos 45° + 907 cos 180° + x cos θ = 0
    ∴ x cos θ = – 0.1068°
    ∑L sin θ = 0 = 200 sin 0° + 1000 sin 45° + 907 sin 180° + x sin θ = 0
    ∴ x sin θ = – 707.11
    ∴ from both equation, θ = 270°
    x = 707.1068 m

    Correct Option: B

    For closed traverse in WCB,
    Sum of latitude, ∑L cos θ = 0 and sum of departure ∑L sin θ = 0.
    ∑L cos θ = 200 cos 0° + 1000 cos 45° + 907 cos 180° + x cos θ = 0
    ∴ x cos θ = – 0.1068°
    ∑L sin θ = 0 = 200 sin 0° + 1000 sin 45° + 907 sin 180° + x sin θ = 0
    ∴ x sin θ = – 707.11
    ∴ from both equation, θ = 270°
    x = 707.1068 m