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Question 1
ప్రశ్న 1
Which combination of components most clearly distinguishes a conventional TOTAL STATION from a purely optical theodolite?
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Explanation:
A total station integrates electronic measurement of horizontal and vertical angles with electronic distance measurement. An onboard processor can combine these observations with station data to calculate coordinates. A display and control interface allow the observer to choose survey and stakeout operations. Internal memory or a connected data collector can retain measured points and descriptive codes. These functions distinguish the instrument from a purely optical theodolite. A total station still requires correct setup and observation procedures for reliable results.
వివరణ:
Question 2
ప్రశ్న 2
What is the principal function of the ONBOARD MICROPROCESSOR in an ordinary surveying total station?
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Explanation:
The microprocessor handles measured data collected by the electronic angle encoders and EDM. It can calculate horizontal distance, height difference and coordinates using the appropriate conventions. It may also support programs for stakeout, area and other survey tasks. Results can be stored with point identifiers and later transferred to office software. The computations depend on correct input coordinates, target heights and instrument settings. Computing capability does not automatically eliminate systematic or observational errors.
వివరణ:
Question 3
ప్రశ్న 3
Before computing ABSOLUTE grid coordinates of detail points with a total station set over a known survey control point, which setup is essential?
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Explanation:
An occupied control station gives the starting easting and northing for coordinate computation. The horizontal circle must also be oriented to a known reference direction. A backsight to a suitably controlled point can establish that orientation. The measured horizontal angle to a new point is then converted into a direction in the adopted coordinate system. EDM observations supply the distance component needed for the coordinate differences. If orientation is wrong, calculated details may be rotated even when the station coordinates are correct.
వివరణ:
Question 4
ప్రశ్న 4
A total station reports corrected slope distance S and ZENITH ANGLE Z, where Z is measured from the upward vertical. Under plane-survey geometry, which formulas give horizontal distance H and signed vertical difference v from instrument centre to target centre?
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Explanation:
A zenith angle is measured from the upward vertical rather than from the horizontal plane. The horizontal component of slope distance is therefore S sin Z. The signed vertical component from instrument centre to target centre is S cos Z. When Z is below 90 degrees, the vertical component is positive. When Z exceeds 90 degrees, it is negative. These geometric components must be distinguished from ground-station elevation differences, which also require instrument and target heights.
వివరణ:
Question 5
ప్రశ్న 5
A total station measures a CORRECTED slope distance of 200.000 m and a zenith angle of 60 degrees. Neglect curvature and refraction. What is the horizontal distance to three decimals?
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Explanation:
The zenith angle is measured downward from the upward vertical. The horizontal distance is H = S sin Z. Substituting S = 200.000 m and Z = 60 degrees gives H = 200 sin 60 degrees. Evaluating this expression gives approximately 173.205081 m. The requested horizontal distance is 173.205 m after rounding. The corresponding vertical component between optical centres is +100.000 m, not the required horizontal component.
వివరణ:
Question 6
ప్రశ్న 6
A total station occupies A at Northing 1000.000 m and Easting 2000.000 m. A target P has corrected slope distance 200.000 m, zenith angle 60 degrees and correctly oriented whole-circle azimuth 60 degrees clockwise from grid north. Neglect curvature and refraction. What are the plane coordinates of P to three decimals?
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Explanation:
First reduce the slope distance to horizontal distance using H = 200 sin 60 degrees = 173.205081 m. A grid azimuth measured clockwise from north gives delta northing = H cos azimuth. Therefore delta northing = 173.205081 cos 60 degrees = +86.602540 m. The easting difference is H sin 60 degrees = +150.000000 m. Adding the components to A yields Northing 1086.602540 m and Easting 2150.000000 m. The rounded answer is N = 1086.603 m and E = 2150.000 m.
వివరణ:
Question 7
ప్రశ్న 7
At a total-station setup, the known AZIMUTH from the occupied station toward the backsight is 70 degrees clockwise from north. The instrument measures a clockwise horizontal angle of 115 degrees FROM THAT BACKSIGHT DIRECTION to a new target. What is the target azimuth?
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Explanation:
The azimuth to the backsight is the reference direction at the occupied station. The question specifies a clockwise measured angle from that reference toward the new point. Clockwise azimuths increase by the measured clockwise angle. Adding 70 degrees and 115 degrees gives 185 degrees. The result is already in the standard whole-circle range of 0 to 360 degrees. There is no need to reverse the backsight azimuth because the given reference is explicitly from the occupied station toward the backsight.
వివరణ:
Question 8
ప్రశ్న 8
A total station occupies A at Northing 500.000 m and Easting 1000.000 m. The azimuth FROM A to its backsight is 30 degrees. A clockwise horizontal angle of 90 degrees is observed from the backsight to point P, and the corrected horizontal distance AP is 100.000 m. What are the coordinates of P to three decimals?
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Explanation:
The target azimuth is 30 + 90 = 120 degrees clockwise from north. The northing difference is 100 cos 120 degrees = -50.000 m. The easting difference is 100 sin 120 degrees = +86.602540 m. Adding the latitude to the occupied station gives N = 500 - 50 = 450.000 m. Adding the departure gives E = 1000 + 86.602540 = 1086.603 m after rounding. The negative northing and positive easting increments agree with the southeast direction of the target.
వివరణ:
Question 9
ప్రశ్న 9
Ground station A has RL 120.000 m. A total station is set up at A with instrument height 1.500 m and measures slope distance 100.000 m at elevation angle +30 degrees to a prism whose centre is 2.000 m above ground point B. Neglect curvature and refraction. What is RL of B?
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Explanation:
The instrument-centre elevation is 120.000 + 1.500 = 121.500 m. The sight line rises 100 sin 30 degrees = 50.000 m to the prism centre. The prism-centre RL is therefore 121.500 + 50.000 = 171.500 m. The prism is 2.000 m above the ground point B. Subtracting the prism height gives RL(B) = 171.500 - 2.000 = 169.500 m. Both instrument height and prism height are necessary to convert the optical-centre difference to a ground-station difference.
వివరణ:
Question 10
ప్రశ్న 10
Ground station A has RL 150.000 m. A total station has instrument height 1.600 m, a corrected slope distance of 80.000 m and a zenith angle of 120 degrees to a prism of height 1.800 m at B. Neglect curvature and refraction. What is RL of B?
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Explanation:
The zenith angle exceeds 90 degrees, so the observed target is below the instrument optical centre. The vertical component is 80 cos 120 degrees = -40.000 m. The instrument-centre RL is 150.000 + 1.600 = 151.600 m. The prism-centre RL is 151.600 - 40.000 = 111.600 m. Subtracting prism height 1.800 m gives RL(B) = 109.800 m. Using the cosine of the zenith angle correctly preserves the negative vertical sign.
వివరణ:
Question 11
ప్రశ్న 11
A total station observes a corrected slope distance of 80.000 m at zenith angle 120 degrees. What is the HORIZONTAL separation between instrument and target centres to three decimals?
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Explanation:
The horizontal component of a slope distance is H = S sin Z when Z is measured from the upward vertical. The slope distance is 80 m and the zenith angle is 120 degrees. Therefore H = 80 sin 120 degrees. Since sin 120 degrees is approximately 0.8660254, H is approximately 69.282032 m. Rounded to three decimal places, the horizontal separation is 69.282 m. The signed vertical component is -40.000 m and should not be substituted for the horizontal result.
వివరణ:
Question 12
ప్రశ్న 12
For a REMOTE ELEVATION observation, a base reflector and the inaccessible upper target lie on the same vertical line. Their common horizontal distance from the total station is 100.000 m. The elevation angle to the BASE REFLECTOR CENTRE is +10 degrees, and the elevation angle to the UPPER TARGET is +30 degrees, both measured from the same instrument centre. What is the height of the upper target ABOVE the base reflector centre to three decimals?
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Explanation:
The remote elevation method can use a measured base point to establish the horizontal distance to an inaccessible upper point. The question states that the base reflector centre and the upper target are vertically aligned. Their vertical positions above the same instrument centre are H tan 10 degrees and H tan 30 degrees. The vertical separation is therefore 100(tan 30 degrees - tan 10 degrees). This equals approximately 40.102329 m. Thus the upper target is 40.102 m above the base reflector centre, without requiring either ground-point RL.
వివరణ:
Question 13
ప్రశ్న 13
During coordinate observations, the ACTUAL prism-centre height is 1.800 m above the target ground mark, but the operator enters 1.500 m in the total station. Assume all angular and distance observations and the instrument height are otherwise correct. How does this error affect the COMPUTED RL of that ground mark?
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Explanation:
The instrument determines an elevation for the observed prism centre from the occupied-station RL, instrument height and vertical line-of-sight component. To obtain the ground-point RL, it subtracts the prism height entered by the operator. The entered height is 1.500 m rather than the true 1.800 m. Subtracting 0.300 m too little makes the calculated ground-point RL 0.300 m too high. Under the stated conventional geometry, changing the entered target height does not alter the measured horizontal angle and distance. Checking prism-height entries is therefore essential in height surveys.
వివరణ:
Question 14
ప్రశ్న 14
A design point has coordinate differences from the occupied station of +300.000 m NORTH and +400.000 m EAST. What are the required horizontal distance and grid azimuth measured clockwise from north, to the nearest second?
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Explanation:
The horizontal distance is the hypotenuse of the northing and easting differences. Therefore H = sqrt(300 squared + 400 squared) = 500.000 m. Because both increments are positive, the design direction lies in the northeast quadrant. Its clockwise azimuth from north is arctan(400/300). This equals approximately 53.130102 degrees or 53 degrees 07 minutes 48 seconds to the nearest second. The resulting design observations are 500.000 m and 53°07'48".
వివరణ:
Question 15
ప్రశ్న 15
Relative to a total station at a known point, a proposed stake has coordinate differences of -60.000 m in NORTHING and +80.000 m in EASTING. What are the required horizontal distance and whole-circle grid azimuth clockwise from north, to the nearest second?
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Explanation:
The horizontal distance is sqrt[(-60) squared + 80 squared] = 100.000 m. The negative northing and positive easting increments place the point southeast of the occupied station. The acute reference angle from the south direction is arctan(80/60) = 53.130102 degrees. Hence the azimuth clockwise from north is 180 - 53.130102 = 126.869898 degrees. This converts to 126°52'12" to the nearest second. The southeast quadrant must be considered when converting an inverse-tangent result into a whole-circle azimuth.
వివరణ:
Question 16
ప్రశ్న 16
A total station occupies (N = 1000.000 m, E = 1000.000 m), with its horizontal direction correctly oriented to GRID NORTH. A design stake has coordinates (N = 1080.000 m, E = 1060.000 m). Which horizontal distance and clockwise azimuth should be used for its coordinate stakeout, to the nearest second?
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Explanation:
Subtracting occupied-station coordinates from the design coordinates gives delta northing +80.000 m and delta easting +60.000 m. The horizontal range is sqrt(80 squared + 60 squared) = 100.000 m. Since the target is northeast, its azimuth is arctan(60/80). The azimuth is approximately 36.869898 degrees. Converted to degrees, minutes and seconds, it is 36°52'12" to the nearest second. This is the required polar distance and direction when the instrument is already oriented to grid north.
వివరణ:
Question 17
ప్రశ్న 17
A total-station stakeout screen is specifically configured to show HORIZONTAL DISTANCE RESIDUAL = measured distance MINUS design distance. The prism is correctly aligned on the design azimuth, the measured horizontal distance is 100.300 m, and the design horizontal distance is 100.000 m. What field movement is required to reach the design location?
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Explanation:
The specified display convention is measured horizontal distance minus design horizontal distance. Here the displayed distance residual is 100.300 - 100.000 = +0.300 m. The positive result means that the reflector is 0.300 m farther from the instrument than the design position. Since the prism is already on the correct design azimuth, a lateral correction is not required. Moving the prism 0.300 m toward the total station reduces the distance residual to zero. Some instruments use the opposite display sign, which is why the convention is explicitly stated in the question.
వివరణ:
Question 18
ప్రశ్న 18
For a coordinate stakeout, the design point has N = 2150.000 m and E = 3100.000 m, while the CURRENT measured prism ground-point coordinates are N = 2149.960 m and E = 3100.080 m. Which coordinate movements are required to reach the design point, and what is their horizontal magnitude, to three decimals?
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Explanation:
Stakeout coordinate movements are calculated as design coordinates minus the current observed coordinates. The required northing movement is 2150.000 - 2149.960 = +0.040 m. The required easting movement is 3100.000 - 3100.080 = -0.080 m. Hence the prism must move 0.040 m north and 0.080 m west. The horizontal displacement magnitude is sqrt(0.040 squared + 0.080 squared) = approximately 0.089443 m. Rounding to three decimals gives 0.089 m.
వివరణ:
Question 19
ప్రశ్న 19
Which statement most accurately contrasts PRISM and REFLECTORLESS total-station EDM observations?
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Explanation:
A prism is designed to return incident EDM light approximately toward the instrument. The instrument must be configured for the selected reflector type and its applicable constant. Reflectorless observation instead uses light returned directly from the surface being sighted. Surface reflectance, incidence angle and competing background targets can affect the result. Both modes can be combined with electronic angular observations to determine survey-point coordinates. Neither measurement mode removes the need for suitable quality checks or safe use of the instrument.
వివరణ:
Question 20
ప్రశ్న 20
Which statement correctly describes an ELECTRONIC COMPENSATOR and proper levelling of a total station?
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Explanation:
Modern total stations often contain tilt sensors or compensators to account for small residual inclinations. These systems function only within their designed correction range. Proper centring over the occupied station is still necessary. The instrument must also be levelled sufficiently for the compensator to operate as intended. A compensator cannot correct incorrect station coordinates, target identification or a wrong backsight orientation. Setup and check observations remain essential parts of reliable surveying.
వివరణ:
Question 21
ప్రశ్న 21
After setting a total station over a known station and orienting it using a known backsight, which observation offers an important INDEPENDENT CHECK before collecting numerous detail points?
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Explanation:
A setup can contain centring error, an incorrect backsight direction or wrong control-point inputs. Observing a second independently known point can expose some of these problems. The measured bearing or coordinate residual can be compared with the established control information. Distance and elevation differences can also provide useful checks when available. Such control verification should be completed before an extensive detail survey or important stakeout. A successful computer computation does not independently validate the data entered into the instrument.
వివరణ:
Question 22
ప్రశ్న 22
A surveyor sets up a total station on an UNKNOWN ground point and measures suitable angles and distances to MULTIPLE control points with known coordinates, so that the unknown station coordinates and orientation can be computed. What is this operation generally called?
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Explanation:
Resection determines an occupied position using observations to points of known coordinates. In total-station work it is often implemented as a free-station setup. Measurements can include directions and distances to multiple control points. The available observations and control geometry must be adequate to solve for position and orientation. Redundant observations allow residual checks and may improve the reliability of the result. Poorly arranged control points or inaccurate inputs can make the solution weak or unreliable.
వివరణ:
Question 23
ప్రశ్న 23
What is the main surveying benefit of assigning POINT NUMBERS and FEATURE CODES while storing total-station observations for a topographic or utility survey?
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Explanation:
Total stations can store observations and calculated coordinates under unique point identifiers. Feature codes describe the type of object or survey feature observed. For example, codes can distinguish kerbs, drains, manholes and building corners. When data are transferred to survey or CAD software, these codes help organize or connect the points appropriately. Accurate coding reduces the risk of confusing different surveyed objects. Nevertheless, recorded coordinates still require suitable control, field verification and quality assurance.
వివరణ:
Question 24
ప్రశ్న 24
A total station has surveyed the boundary of a horizontal plot as consecutive coordinates (E,N) in metres: A(100,200), B(160,200), C(140,250), D(110,250), returning to A. Using the coordinate (shoelace) method, what is the enclosed PLANE AREA?
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Explanation:
The supplied points form a four-sided horizontal polygon and are listed in boundary order. The lower parallel side AB is 160 - 100 = 60 m long. The upper side DC is 140 - 110 = 30 m long. Their perpendicular separation in northing is 250 - 200 = 50 m. Therefore the trapezoid area is (60 + 30)/2 multiplied by 50 = 2250 m squared. A shoelace calculation from the ordered coordinates gives the same result.
వివరణ:
Question 25
ప్రశ్న 25
A construction surveyor is about to use a total station to SET OUT design points of a water-supply structure from site control. Which procedure is most technically defensible?
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Explanation:
Construction stakeout requires a consistent design coordinate reference and reliable site control. The occupied station and orientation must be established from suitable known points. EDM settings, reflector constants, heights and angular conventions should be checked before measurements are accepted. The total station can calculate the direction and distance needed to mark each design point. The marked points should be observed again and compared against the design coordinates and project tolerances. Independent control checks help detect setup mistakes that onboard computations cannot automatically identify.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Electronic angle measurement, an electronic distance meter and an onboard processor for computation and data handling. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. To process measured angles and distances with entered station information and display or store derived survey results. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. Enter the occupied station coordinates and establish the instrument orientation using suitable known direction or backsight control. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. H = S sin Z; v = S cos Z. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: A. 173.205 m. సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: D. N = 1086.603 m; E = 2150.000 m. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: B. 185 degrees. సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: C. N = 450.000 m; E = 1086.603 m. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: A. 169.500 m. సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: C. 109.800 m. సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: B. 69.282 m. సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 40.102 m. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: A. The computed ground RL is 0.300 m too high. సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: D. 500.000 m and 53°07'48". సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: B. 100.000 m and 126°52'12". సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: C. 100.000 m and 36°52'12". సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: A. Move the prism 0.300 m toward the total station along the sight line. సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: D. Move north 0.040 m and west 0.080 m; magnitude 0.089 m. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: B. A prism provides a controlled retroreflecting target, while reflectorless measurement uses return signals from a surface and depends on surface properties and measurement conditions. సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: C. A compensator senses limited instrument tilt for electronic angular correction, but accurate setup and operation within its specified range remain necessary. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: A. Sight another independently known control point and compare its measured direction, distance or calculated coordinates with the control values. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: D. Free-station setup or resection. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: B. They preserve point identity and feature meaning for data transfer, drafting and later interpretation. సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: C. 2250 m^2. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: D. Verify control coordinates and orientation, check instrument and target heights and EDM settings, compute design directions/distances, stake the targets, and independently check the final positions. సమాధానం: D.