Surveying miscellaneous

Surveying miscellaneous

Easy Questions

Surveying

Surveying miscellaneous
  1. In the figure given below, lengths PQ(WCB : 30°) and QR (WCB : 45°) respectively up to three places of decimal are









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    Let PQ = a, QR = b
    Line PQ
    Latitude = 100 + x1 cos 30°
    Departure = 200 + x1 sin 30°
    Line QR
    Latitude = 1000 – x2 cos 45°
    Departure = 1000 – x2 sin 45°
    For closed traverse, 100 + x1 cos 30° = 1000 – x2 cos 45°
    ⇒ x1 cos 30° + x2 cos 45° = 900 200 + x1 sin 30° = 1000 – x2 sin 45°
    ⇒ x1 sin 30° + x2 sin 45° = 800
    ∴ x1 = 273.205 m , x2 = 938.186 cm.

    Correct Option: A

    Let PQ = a, QR = b
    Line PQ
    Latitude = 100 + x1 cos 30°
    Departure = 200 + x1 sin 30°
    Line QR
    Latitude = 1000 – x2 cos 45°
    Departure = 1000 – x2 sin 45°
    For closed traverse, 100 + x1 cos 30° = 1000 – x2 cos 45°
    ⇒ x1 cos 30° + x2 cos 45° = 900 200 + x1 sin 30° = 1000 – x2 sin 45°
    ⇒ x1 sin 30° + x2 sin 45° = 800
    ∴ x1 = 273.205 m , x2 = 938.186 cm.

  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 plan of a map was photo copied to a reduced size such that a line originally 100 mm, measures 90 mm. The original scale of the plan was 1 : 1000. The revised scale is








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    Shrinkage factor =
    90
    		 = 0.9
    100

    Revised side = Original scale × reduction factor/ Shrinkage factor
    =
    1
    		 × 0.9 =
    1
    10001111.11

    Correct Option: B

    Shrinkage factor =
    90
    		 = 0.9
    100

    Revised side = Original scale × reduction factor/ Shrinkage factor
    =
    1
    		 × 0.9 =
    1
    10001111.11

  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. Consider the following figure, which is an extract from a contour map (scale = 1 : 20,000) of an area. An alignment of a road at a ruling gradient of 4% is to be fixed from the point O and beyond. What should be the radius of the arc with O as the centre to get the point of a alignment on the next contour on the map.









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    Interval of contour = 20 m.

    ∴ length of arc =
    20
    		 = 500 m (4% gradient)
    0.04

    For 1 : 20,000 side, 500 m will be
    500
    		 on map i.e., 2.5 cm
    20,000

    Correct Option: C

    Interval of contour = 20 m.

    ∴ length of arc =
    20
    		 = 500 m (4% gradient)
    0.04

    For 1 : 20,000 side, 500 m will be
    500
    		 on map i.e., 2.5 cm
    20,000