Surveying miscellaneous

Surveying miscellaneous

Easy Questions

Surveying

Surveying miscellaneous
  1. The following table gives data of consecutive co-ordinates in respect of a closed theodolite traverse PQRSP.

    The magnitude and direction of error of closure in whole circle bearing are








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    ∑L = 400.75 + 100.25 – 199 – 300 = 2
    ∑D = – 300.5 + 199.25 + 399.75 – 200.5 = – 2
    Error of closure = √(∑C)² + (∑D)² = √4 + 4 = 2√2

    tan θ =
    ∑D
    		 = - 1
    ∑L

    ∴ θ = 315°

    Correct Option: C

    ∑L = 400.75 + 100.25 – 199 – 300 = 2
    ∑D = – 300.5 + 199.25 + 399.75 – 200.5 = – 2
    Error of closure = √(∑C)² + (∑D)² = √4 + 4 = 2√2

    tan θ =
    ∑D
    		 = - 1
    ∑L

    ∴ θ = 315°

  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 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

  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 observed magnetic bearing of a line OE was found to be 185°. It was later discovered that station O had a local attraction of + 1.5°. The true bearing of the line OE, considering a magnetic declination of 3.5° E shall be








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    Corrected magnetic bearing = measured value – Local attraction
    = 185 – 1.5 = 183.5°
    True bearing = magnetic bearing + magnetic declination
    = 183.5 + 3.5 = 187°

    Correct Option: B

    Corrected magnetic bearing = measured value – Local attraction
    = 185 – 1.5 = 183.5°
    True bearing = magnetic bearing + magnetic declination
    = 183.5 + 3.5 = 187°