Surveying 2
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1) Question.
A lighthouse is visible just above the horizon at a certain station at the sea level. The distance between the station and the lighthouse is 40km. The height of the lighthouse is approximately
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2) Question.
Which one of the following methods estimates best the area of an irregular and curved boundary?
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3) Question.
An observer standing on the deck of a ship just sees the top of the a lighthouse which is 30m above the sea level. If the height of the observer's eye is 10m above the sea level, then the distance if the observer from the lighthouse will be nearly
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4) Question.
The true length of a line is known to be 200 m . When this is measured with a 20m tape, the length is 200.80 m. The correct length of the 20m tape is
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5) Question.
For a chord of 60 m, the mid ordinate for a circular curve of 50 m radius will be
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6) Question.
In an external focussing tacheometer, the fixed interval between stadia hairs is 5mm; the focal length of the objective is 25cm and the distance of the vertical axis of the instrument from the optical centre of the objective is 15cm. Which one of the following is the set of constants of the tacheometer?
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7) Question.
In a closed traverse, the sum of south latitudes exceeds the sum of north latitudes and the sum of east departures exceeds the sum of west departures. The closing line will lie in the
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8) Question.
A lemniscate curve between the tangent is transitional throughout, if the polar deflection angle of its apex is equal to (¢ is the deflection angle between the initial and final tangents)
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9) Question.
The combined correction for curvature and refraction for a distance of 1400 m
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10) Question.
To find the R.L of a roof slab of a building, staff readings were taken from a particular set-up of the levelling instrument. The readings were 1.050 m with staff on the benchmark and 2.300 m with staff below the roof slab and held inverted. Taking the R.L of the B.M. as 135.150 m, the R.L. of the roof slab will be
Answer.HI of instrument = RL + BS = 135.15 + 1.050 = 136.2 m
FS = -2.300 m (as inverted)
RL of roof slab = HI – FS = 136.2 – (-2.300) = 138.5 m