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This mock test of Surveying - 1 for GATE helps you for every GATE entrance exam.
This contains 10 Multiple Choice Questions for GATE Surveying - 1 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The sensitiveness of a bubble tube in a level would decrease if

Solution:

QUESTION: 2

Hydrographic surveys deal with the mapping of

Solution:

QUESTION: 3

If "L" is the length of the chain , "W" is the weight of the chain and 'T' is the tension , the Sag Correction for the chain line is

Solution:

QUESTION: 4

If the whole circle bearing is 315°20′, its quadrantal bearing would be

Solution:

*Answer can only contain numeric values

QUESTION: 5

The radius of a 1 degree curve is __________ m

Solution:

QUESTION: 6

A level when set up 25 m from peg A and 50 m from peg B reads 2.847 on a staff held on A and 3.462 on a staff held on B, keeping bubble at its centre while reading. If the reduced levels of A and B are 283.665 m and 284.295 m respectively, the collimation error per 100 m is

Solution:

Actual height difference between A and B = 284.295 - 283.665 = 0.63.

Height difference between A and B based on staff readings = 3.462 - 2.847 = 0.615.

Distance between A and B = 50-25 = 25 m.

Therefore collimation error per 100 m.

= (0.63 - 0.615)*(100/25).

= 0.060 m.

QUESTION: 7

An invar tape, 50 m in length, standardized at 20oC temperature and 10kg pull, is used to measure a baseline. The Correction per tape length, if at the time of measurement the temperature was 30o and the coefficient of linear expansion of the tape was 1x10-6 per oC , will be

Solution:

Correction for temperature

Ct = L ∝ ( Tm - To)

∝ = Co-efficient of thermal expansion

Tm = Mean temperature in the field during measurement

To = the standard temperature in the field during

L = measured length

Ct = 1 x 10^{-6} (30 - 20) x 50

= 0.0005 m

QUESTION: 8

A Rectangular plot of 16km2 in area is shown on a map by a similar rectangular area of 1 cm2 , R.F of the scale to measure a distance of 40km will be

Solution:

R.F is ratio of two measurements with common unit

QUESTION: 9

Following observations were taken with a transit fitted with stadia wires.

The line of sight was horizontal and the staff was held vertical.

The tacheometric constants k and C are 100 and 0.4 m, respectively. The horizontal distance between staff and instrument is

Solution:

D = KS + C

S = 2.830 - 1.726 = 1.104

D = 100 x 1.104 + 0.4 = 110.8m

QUESTION: 10

Following observations were taken during a reciprocal leveling:

If reduced level of P is 140.815 m, the reduced level of Q is

Solution:

True difference = 1.824-0.928/ 2.748-1.606 = 0.7845

Reduced level of Q is = 140.815-0.7845 = 140.030m

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