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Question 1
ప్రశ్న 1
Which statement most accurately describes the basic purpose of ELECTRONIC DISTANCE MEASUREMENT (EDM) in surveying?
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Explanation:
Electronic distance measurement uses electromagnetic radiation to determine the separation between an instrument and a target. Depending on the equipment, a travel-time or phase-comparison technique is used. The radiation may be optical or microwave in different classes of instruments. A measurement ordinarily refers to the line of sight and therefore gives a slope distance unless further reduced. Instrument, reflector and atmospheric effects may require correction before the distance is used in survey computations. EDM is a distance-measurement technique and does not by itself replace the angular observations needed for traverse orientation.
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Question 2
ప్రశ్న 2
In a phase-comparison EDM, what is the principal role of MODULATING a transmitted optical or microwave carrier?
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Explanation:
Phase-comparison instruments impose a known modulation on an electromagnetic carrier. The returned modulation is delayed because the signal travels to the target and back. Comparing its phase with the transmitted reference provides the fractional part of the round-trip modulation cycle. The unknown whole-cycle count must also be resolved before the full distance is known. Practical instruments may use several modulation frequencies to establish the correct distance range. The modulation principle does not require the surveyor to count optical carrier wavelengths directly.
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Question 3
ప్రశ్న 3
Which of the following statements about the SIGNALS used by surveying EDM instruments is technically correct?
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Explanation:
Optical EDM instruments commonly use infrared or laser light. Microwave-based distance-measuring systems have also been used in surveying. These signals are forms of electromagnetic radiation rather than ultrasonic mechanical vibrations. Electromagnetic propagation speed through air depends on its refractive properties. Temperature, pressure and other atmospheric conditions therefore matter for precise EDM work. The selected signal type also influences operating range and target requirements.
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Question 4
ప్రశ్న 4
An EDM measures the ROUND-TRIP travel time t of a pulse sent from the instrument to a reflector and returned to the instrument. If the wave speed along the path is v, which formula gives the one-way distance D?
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Explanation:
The transmitted pulse travels from the EDM instrument to the target. It then returns along the measurement path to the receiver. The measured interval t corresponds to approximately twice the one-way instrument-to-target distance. The total signal path length is therefore 2D = vt. Rearrangement gives D = vt/2. The value of v must correspond to the effective wave speed in the measurement medium or incorporate the relevant correction.
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Question 5
ప్రశ్న 5
In a simplified pulse EDM calculation, the measured round-trip travel time is 4.00 microseconds and the electromagnetic-wave speed is assumed to be 3.00 × 10^8 m/s. Neglect all other corrections. What is the ONE-WAY distance?
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Explanation:
A time interval of 4.00 microseconds equals 4.00 × 10^-6 seconds. The total round-trip path is the propagation speed multiplied by this interval. This gives 3.00 × 10^8 multiplied by 4.00 × 10^-6 = 1200 m. The measured one-way distance is half the round-trip path length. Therefore D = 1200/2 = 600 m. The calculation assumes the stated propagation speed is applicable and neglects instrument delays and other corrections.
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Question 6
ప్రశ్న 6
A pulse EDM is aimed at a reflector 750 m away, measured along the signal path. Assuming a propagation speed of exactly 3.00 × 10^8 m/s and no instrument delay, what ROUND-TRIP travel time should be recorded?
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Explanation:
The one-way distance from instrument to reflector is 750 m. The signal must travel to the reflector and return, so its total path length is 1500 m. The round-trip time is t = 2D/v. Substitution gives t = 1500/(3.00 × 10^8) seconds. This equals 5.0 × 10^-6 seconds, or 5.0 microseconds. Dividing the one-way distance by the wave speed without accounting for the return path would be incorrect.
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Question 7
ప్రశ్న 7
In a phase-comparison EDM, let lambda be the MODULATION wavelength, N the known whole number of complete modulation cycles in the TOTAL ROUND-TRIP signal path (instrument to reflector and back), and phi the measured phase remainder of that round-trip path in degrees, with 0 <= phi < 360. If D is the one-way distance, which idealised equation is correct?
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Explanation:
The transmitted and reflected signals together traverse the complete round-trip distance of 2D. This total path contains N complete modulation wavelengths and a remaining fractional wavelength. A phase remainder of phi degrees represents phi/360 of one modulation cycle. Therefore the complete round-trip distance is 2D = lambda(N + phi/360). Dividing this expression by two gives the required one-way distance. The formula assumes that the integer number of complete round-trip cycles has already been correctly resolved and that the effective modulation wavelength is known.
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Question 8
ప్రశ్న 8
In an ideal phase-comparison EDM, the modulation wavelength is 20.0 m, the ROUND-TRIP path contains 15 complete wavelengths, and the measured remainder is 90 degrees. What is the ONE-WAY distance?
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Explanation:
The 90-degree phase remainder is one-quarter of a complete modulation cycle. The round-trip path therefore contains 15 + 90/360 = 15.25 wavelengths. Multiplying by the 20.0 m wavelength gives 2D = 305.0 m. Halving the round-trip path gives D = 152.5 m. The calculation assumes that the integer number of whole wavelengths has already been resolved. Omitting the factor of one-half would confuse the return path with the surveyed distance.
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Question 9
ప్రశ్న 9
Why is a SINGLE phase measurement at ONE modulation frequency generally insufficient to determine a long EDM distance unambiguously?
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Explanation:
A phase comparison measures the fractional part of the modulation cycle remaining in the return signal. The phase reading repeats whenever the round-trip path gains a whole wavelength. Consequently several different distances can produce the same fractional phase reading. The unknown integer number of completed cycles is called an ambiguity. Multiple modulation frequencies or equivalent ambiguity-resolution techniques help determine that integer. Once the ambiguity is resolved, the measured fractional phase can be converted into the complete distance.
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Question 10
ప్రశ్న 10
A phase-comparison EDM has modulation wavelength 12.0 m. Its resolved ROUND-TRIP path contains N = 8 complete wavelengths and a further phase remainder of 180 degrees. Neglect other corrections. What is the one-way distance?
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Explanation:
A phase remainder of 180 degrees corresponds to half a modulation wavelength. The double-path length therefore contains 8.5 wavelengths. Multiplying by lambda = 12.0 m gives 2D = 8.5 × 12.0 = 102.0 m. The one-way distance is D = 102.0/2. Hence the measured distance is 51.0 m. This calculation explicitly assumes the whole-cycle integer has already been correctly determined.
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Question 11
ప్రశ్న 11
What is the principal function of a CORNER-CUBE PRISM reflector used with a reflector-type optical EDM?
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Explanation:
A corner-cube reflector is designed to send incident light approximately back toward its source. This retroreflection helps the EDM receive the returning optical signal. The prism must still be correctly aimed and positioned according to the equipment requirements. Its effective optical path can produce an additive reflector constant that depends on instrument and reflector conventions. The prism itself does not perform the geometric reduction from slope distance to horizontal or grid distance. Target centering and the correct reflector setting are important for precise observations.
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Question 12
ప్రశ్న 12
A surveyor switches a total station from prism mode to REFLECTORLESS EDM mode to measure a wall. Which statement is most appropriate?
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Explanation:
Reflectorless EDM uses a signal returned from the observed object rather than a dedicated prism. The amount and quality of returning light vary with the target surface. Dark, highly oblique, transparent or specular surfaces can create measurement difficulties. The actual performance range and accuracy depend on the instrument and field conditions. Reflectorless mode does not eliminate the need for correct pointing and appropriate corrections. A surveyor should confirm that the returned signal corresponds to the intended surface and not an unintended background.
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Question 13
ప్రశ్న 13
Which statement correctly distinguishes an ADDITIVE EDM reflector constant from a DISTANCE-PROPORTIONAL scale correction?
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Explanation:
An additive or zero constant is expressed in units of length, such as millimetres. Its effect is approximately independent of the length of the measured line for a given setup and reflector combination. A scale correction is proportional to the distance and is often expressed in parts per million. Changes in modulation frequency and atmospheric velocity modelling are possible sources of scale error. Prism conventions may differ among manufacturers, so the signed constant must follow the actual instrument settings. Both types should be handled without accidentally applying a correction twice.
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Question 14
ప్రశ్న 14
An EDM displays a slope distance of 1200.000 m BEFORE a required reflector correction is applied. For this question, the signed ADDITIVE correction to be added to the displayed distance is -0.030 m, and all other corrections are zero. What is the corrected slope distance?
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Explanation:
The measured slope distance is explicitly stated to exclude the reflector correction. The correction to be added is -0.030 m. Therefore the corrected distance is 1200.000 + (-0.030) m. This gives 1199.970 m. The negative sign means that the specified correction reduces the displayed value. Real instruments may apply prism constants internally, so the user must check the instrument settings before making an additional manual correction.
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Question 15
ప్రశ్న 15
Why are AIR TEMPERATURE and ATMOSPHERIC PRESSURE relevant to accurate EDM observations?
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Explanation:
The propagation velocity of electromagnetic radiation in air depends on the refractive index. Air temperature and pressure influence that refractive index. Humidity may also matter, with its influence depending on signal wavelength and required precision. If the instrument uses an unsuitable atmospheric correction, a distance-dependent error can remain. Appropriate meteorological measurements and manufacturer procedures are therefore important for precise EDM work. Atmospheric effects are distinct from an additive prism constant or geometric slope reduction.
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Question 16
ప్రశ్న 16
An EDM slope distance of 1500.000 m has NOT yet received its atmospheric correction. The signed atmospheric SCALE correction to be applied is +12 ppm, where corrected distance = measured distance × (1 + 12 × 10^-6). What is the corrected slope distance?
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Explanation:
One part per million represents a scale change of one unit in one million units of distance. A correction of +12 ppm corresponds to the multiplier 1 + 12 × 10^-6. For 1500.000 m, the correction magnitude is 1500 × 12 × 10^-6 = 0.018 m. The positive sign means the correction is added to the distance. Hence the corrected slope distance is 1500.018 m. This question assumes that the stated atmospheric correction has not already been applied internally by the EDM.
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Question 17
ప్రశ్న 17
A surveying EDM is specified to have distance accuracy of the form ±(a mm + b ppm of measured distance). What do the TWO terms represent?
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Explanation:
EDM distance specifications commonly combine a fixed-length term with a distance-dependent term. The fixed term a is expressed in millimetres and does not scale with line length in the stated model. The proportional term b ppm corresponds to b millimetres per kilometre. Therefore the ppm contribution becomes larger on longer measured lines. The stated plus sign in such a specification gives a conventional combined limiting accuracy expression rather than a guarantee that all field errors are absent. Target centering, atmospheric conditions and calibration can further affect the actual quality of a survey measurement.
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Question 18
ప్రశ్న 18
An EDM has a stated distance accuracy of ±(2 mm + 2 ppm of measured distance). According to THIS specified accuracy expression, what magnitude applies to a measured distance of 2.000 km?
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Explanation:
The fixed component of the stated accuracy expression is 2 mm. The distance-proportional component is 2 ppm. Since 1 ppm corresponds to 1 mm per kilometre, 2 ppm over 2 km equals 4 mm. The model combines the fixed and proportional terms by addition. Therefore the stated magnitude is 2 + 4 = 6 mm. It is not appropriate to interpret 2 ppm as 2 mm regardless of the line length.
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Question 19
ప్రశ్న 19
An EDM gives a corrected SLOPE DISTANCE S between the instrument centre and a prism centre. If alpha is the signed vertical angle measured from the horizontal, which plane-survey formula gives the corresponding HORIZONTAL distance H?
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Explanation:
The slope distance is the line-of-sight hypotenuse of the vertical-plane right triangle. The horizontal distance is the side adjacent to the angle measured from the horizontal. Consequently H = S cos(alpha). The formula works for both upward and downward vertical angles because cosine is even. For a zenith angle Z measured downward from the upward vertical, the equivalent formula is H = S sin(Z). These basic reductions assume ordinary plane-survey geometry and do not themselves convert a ground distance to a map-projection grid distance.
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Question 20
ప్రశ్న 20
A total station measures a corrected SLOPE DISTANCE of 500.000 m and a ZENITH ANGLE of 60 degrees, where 0 degrees is vertically upward and 90 degrees is horizontal. Neglect curvature and refraction. What is the horizontal distance to three decimals?
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Explanation:
A zenith angle is measured from the upward vertical. Therefore the horizontal reduction is H = S sin(Z), not S cos(Z). Substituting S = 500.000 m and Z = 60 degrees gives H = 500 sin(60 degrees). Since sin(60 degrees) is approximately 0.8660254, H is approximately 433.012702 m. Rounding to three decimals gives 433.013 m. The corresponding signed upward component between the instrument and prism centres is 500 cos(60 degrees) = 250.000 m.
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Question 21
ప్రశ్న 21
The optical centres of an EDM and its target have a measured SLOPE separation of 130 m and a known vertical separation of 50 m. Assuming plane geometry, what is their HORIZONTAL separation?
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Explanation:
The slope distance, horizontal separation and vertical separation form a right triangle. By the Pythagorean relationship, H squared equals S squared minus the square of the vertical separation. Here H = sqrt(130^2 - 50^2). This is sqrt(16900 - 2500), or sqrt(14400). Therefore the horizontal separation is 120 m. The given vertical separation is between the optical centres, so no additional instrument-height or reflector-height adjustment is required for this horizontal calculation.
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Question 22
ప్రశ్న 22
Station A has RL = 100.000 m. A total station at A has instrument height 1.500 m and sights a prism whose centre is 1.800 m above ground station B. The corrected slope distance from instrument centre to prism centre is 100.000 m at a vertical ELEVATION angle of +30 degrees. Neglect curvature and refraction. What is RL of B?
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Explanation:
The instrument optical centre has RL = 100.000 + 1.500 = 101.500 m. The line-of-sight elevation difference from instrument centre to prism centre is S sin(alpha). With S = 100.000 m and alpha = +30 degrees, this vertical difference is +50.000 m. Therefore the prism centre has RL = 151.500 m. The prism centre is 1.800 m above the ground station B, so RL(B) = 151.500 - 1.800. The required RL of B is 149.700 m.
వివరణ:
Question 23
ప్రశ్న 23
An EDM displays an UNCORRECTED SLOPE distance of 1000.000 m. Apply a signed ADDITIVE correction of -0.020 m, then a signed SCALE correction of +10 ppm to that result, using S_corrected = (1000.000 - 0.020)(1 + 10 × 10^-6). The zenith angle is 60 degrees. Neglect curvature and refraction. What HORIZONTAL distance results to three decimals?
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Explanation:
The signed additive correction first changes the slope distance to 999.980 m. Applying the specified +10 ppm factor gives 999.980 × 1.000010 = 999.9899998 m. A zenith angle of 60 degrees requires horizontal reduction by sin(60 degrees). The horizontal distance is therefore 999.9899998 × sin(60 degrees). This evaluates to approximately 866.016743 m. Rounding to three decimal places gives 866.017 m. The correction order and the fact that both corrections were previously unapplied are explicitly stated in the question.
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Question 24
ప్రశ్న 24
A precise survey EDM is checked periodically on a CONTROLLED CALIBRATION BASELINE with suitable meteorological observations. Which statement correctly describes an important purpose of this exercise?
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Explanation:
EDM calibration compares instrument distance observations with reliable reference lengths. These comparisons help reveal fixed additive offsets and distance-proportional scale errors. Suitable observations may also reveal cyclic or periodic measuring errors. Atmospheric and geometric effects must be treated consistently when comparing the measurements. Calibration does not remove the need for correct instrument centering, reflector selection or field observations. A properly maintained calibration history improves confidence in the systematic behaviour of the measuring system.
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Question 25
ప్రశ్న 25
A survey team is using a total station with EDM to compute coordinates of a closed traverse. Which procedure is most technically defensible before using the distances for coordinate calculations?
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Explanation:
Reliable EDM traverse work starts with correct instrument and target setup. The selected prism or reflectorless mode and any reflector constants must match the actual target. Atmospheric corrections should reflect the measurement conditions and must not be applied a second time if already included by the instrument. Slope observations are reduced to horizontal distances before ordinary plan-coordinate calculations, with any further required reference-surface or grid factors handled separately. Independent checks and traverse closure help identify inconsistencies that a single distance display may not reveal. A strong return signal is helpful but does not by itself establish complete positional accuracy.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. To determine distance using the propagation and return of electromagnetic signals, with appropriate calibration and corrections. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. To provide a measurable periodic reference whose returned phase can be compared with the transmitted reference to infer travel distance. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. EDM may use infrared or laser light, and some EDM systems use microwaves; all are electromagnetic radiation. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. D = vt/2. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: D. 600 m. సమాధానం: D.
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Question 6 ప్రశ్న 6Answer: B. 5.0 microseconds. సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: A. 2D = lambda(N + phi/360). సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. 152.5 m. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. Because phase repeats every complete modulation cycle, leaving the integer number of whole cycles unknown. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: D. 51.0 m. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: C. To return incident light approximately toward its source over an appropriate acceptance range. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: A. The instrument may measure from light returned by the wall surface, but achievable range and reliability depend on surface reflectivity, incidence angle and operating conditions. సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: B. An additive constant is a fixed distance offset for the specified instrument-target combination, whereas a scale correction changes proportionally with distance. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: D. 1199.970 m. సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: A. They influence the refractive index of air and therefore the propagation speed of the electromagnetic signal used to calculate distance. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. 1500.018 m. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. The a term is a fixed-length component, and the b ppm term increases in magnitude with measured distance. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: A. 6 mm. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: C. H = S cos(alpha). సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: D. 433.013 m. సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: A. 120 m. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: B. 149.700 m. సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: C. 866.017 m. సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: D. To assess systematic EDM effects such as additive (zero), scale and potentially cyclic errors against known reference distances. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: A. Verify centering and target identification, use the correct reflector/EDM mode and atmospheric settings, apply any outstanding distance corrections, reduce slope distances to the required horizontal reference, and check traverse closure. సమాధానం: A.