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JNTU B.Tech II SURVEYING AND OFFSHORE STRUCTURES Previous Question Paper (March - 2021 - R18/R16)

B.Tech | II | I | R18/R16 | March - 2021

Answer PaperPaper Code: 153BW/133BVMaximum Marks: 75Time: 3 hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

Explain the different obstacles encountered in chain surveying.

7 marks

Answer

Obstacles in chain surveying prevent either the direct measurement of distance or the direct ranging of a line. They are classified into three categories:

  1. Obstacles to Ranging: These occur when the ground is high or there is a structure (like a building or hill) between the two endpoints, making it impossible to see one end from the other. This is solved by reciprocal ranging.
  2. Obstacles to Chaining: These occur when the chain cannot be laid on the ground due to physical barriers like ponds, rivers, or buildings. If the obstacle can be chained around (e.g., a building), it is solved by setting up perpendicular offsets. If it cannot be chained through (e.g., a river), it is solved by using similar triangles or geometric constructions.
  3. Obstacles to both Ranging and Chaining: These are the most difficult, such as a large lake or a dense forest, where neither the line can be seen nor the chain can be laid. These are solved using advanced trigonometric methods or by establishing a parallel line to bypass the obstacle.
b

A 20 m chain was found to be 15 cm too long after chaining a distance of 1600 m. It was found to be 30 cm too long at the end of day’s work after chaining a total distance of 3200 m. Determine the correct distance if the chain was correct before the commencement of the work.

8 marks

Answer

The total distance of 3200 m is divided into two parts: the first 1600 m and the remaining 1600 m.

For the first 1600 m, the chain length changed from 20 m to 20.15 m. The average length of the chain is:

Lavg1 = (20 + 20.15) / 2 = 20.075 m

Correct distance for the first part:

D1 = (Lavg1 / L) × Measured Distance = (20.075 / 20) × 1600 = 1606 m

For the second 1600 m, the chain length changed from 20.15 m to 20.30 m. The average length of the chain is:

Lavg2 = (20.15 + 20.30) / 2 = 20.225 m

Correct distance for the second part:

D2 = (Lavg2 / L) × Measured Distance = (20.225 / 20) × 1600 = 1618 m

The total correct distance is:

Total Correct Distance = D1 + D2 = 1606 + 1618 = 3224 m
215 marks
a

What are fundamental lines of Theodolite? What should be the relationship between them?

8 marks

Answer

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b

Explain in detail how permanent adjustments are carried using theodolite instrument.

7 marks

Answer

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315 marks
a

The following consecutive readings were taken with a level and a 4m levelling staff on a continuously sloping ground at common interval of 30m. 0.855 (on A), 1.545, 2.335, 3.115, 3.825, 0.455, 1.380, 2.055, 2.855, 3.455, 0.585, 1.015, 1.850, 2.755, 3.845 (on B). The R.L. of A was 380.500 make a level book and apply usual checks. Determine the gradient of the line AB.

7 marks

Answer

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b

Explain briefly the following:

  • Indirect methods of contouring
  • Characteristics of contours.
8 marks

Answer

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415 marks
a

What do you understand by Global positioning system? Give an overview of GPS.

8 marks

Answer

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b

Explain the application of total station in surveying.

7 marks

Answer

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5

What are the types of gauges? Explain atleast three types of tide gauges.

15 marks

Answer

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6

How bathymetry survey is done by sounding methods, explain in detail.

15 marks

Answer

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7

Write a note on:

15 marks
a

Seismic survey

7 marks

Answer

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b

Bathymetry survey

8 marks

Answer

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8

A wooden cylinder of length L and diameter L/2 is floating on water with its axis vertical. Find the metacentric height if the specific gravity of wood is 0.6.

15 marks

Answer

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