Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
In plane traverse computation, which statement correctly defines the signed LATITUDE and DEPARTURE of a survey leg, taking north and east as positive?
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Explanation:
Latitude is the signed north-south component of a traverse line. Northward components are positive and southward components are negative under the stated convention. Departure is its signed east-west component. Eastward departures are positive and westward departures are negative. The pair determines the change in northing and easting between consecutive stations. These components must be computed from horizontal rather than unreduced slope distances.
వివరణ:
Question 2
ప్రశ్న 2
For a horizontal traverse leg of length L and whole-circle bearing theta measured clockwise from north, which expressions give its signed latitude and departure?
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Explanation:
Whole-circle bearing is measured clockwise from the north reference direction. Resolving a horizontal leg along the northing axis gives latitude equal to L cos(theta). Resolving it along the easting axis gives departure equal to L sin(theta). The algebraic signs follow directly from the quadrant of the bearing. Thus these formulas automatically distinguish north, south, east and west components. The length used must be the horizontal distance of the survey leg.
వివరణ:
Question 3
ప్రశ్న 3
A horizontal traverse leg is 150 m long with a whole-circle bearing of 120 degrees. Taking north and east as positive, what are its latitude and departure to three decimal places?
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Explanation:
Latitude is calculated as 150 cos(120 degrees). Because cos(120 degrees) = -0.5, the latitude is -75.000 m. Departure is 150 sin(120 degrees), which is approximately +129.904 m. The bearing lies in the southeast quadrant. Accordingly the leg has a southing and an easting. These signed components satisfy the horizontal-length relationship to the stated rounding accuracy.
వివరణ:
Question 4
ప్రశ్న 4
A horizontal traverse leg has length 100 m and whole-circle bearing 225 degrees. Which signed latitude and departure are correct to three decimal places?
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Explanation:
A bearing of 225 degrees is in the southwest quadrant. Both the north-south and east-west components are therefore negative. The latitude is 100 cos(225 degrees) = -70.710678 m. The departure is 100 sin(225 degrees) = -70.710678 m. Rounding each value to three decimals gives -70.711 m. Equal magnitudes occur because the line lies at 45 degrees to the south and west axes.
వివరణ:
Question 5
ప్రశ్న 5
A 120 m horizontal traverse line has reduced bearing N 30 degrees W. What are its signed latitude and departure, to three decimal places, with north and east positive?
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Explanation:
The notation N 30 degrees W specifies a northwest direction. Its northward latitude is 120 cos(30 degrees), or approximately +103.923 m. Its westward departure is the negative of 120 sin(30 degrees). Thus departure is -60.000 m. The signs are consistent with the northwest quadrant. The squared components also reproduce the 120 m horizontal length within rounding tolerance.
వివరణ:
Question 6
ప్రశ్న 6
Station A has known coordinates Northing = 1000.000 m and Easting = 2000.000 m. Traverse leg AB has latitude -72.000 m and departure +96.000 m. What are the coordinates of B?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Latitude is the signed change in northing along the traverse leg. Subtracting 72.000 m from 1000.000 m gives the northing of B as 928.000 m. Departure is the signed change in easting. Adding 96.000 m to 2000.000 m gives the easting of B as 2096.000 m. A negative latitude and positive departure place B southeast of A. The station coordinates are obtained by algebraically accumulating components from known control.
వివరణ:
Question 7
ప్రశ్న 7
For a closed traverse, the separately totalled north latitudes are 205.57 m, south latitudes 205.30 m, east departures 300.75 m and west departures 300.95 m. If north and east are positive, what are the unadjusted coordinate misclosures?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Total northings and southings are compared algebraically to obtain latitude misclosure. The latitude residual is 205.57 - 205.30 = +0.27 m. Total eastings and westings are similarly compared. The departure residual is 300.75 - 300.95 = -0.20 m. Thus the unadjusted computed endpoint is 0.27 m north and 0.20 m west of the required endpoint. A perfect closed traverse would have zero residual for each coordinate component.
వివరణ:
Question 8
ప్రశ్న 8
A closed traverse has an algebraic sum of latitudes of +0.300 m and an algebraic sum of departures of -0.400 m. What is the magnitude of its LINEAR MISCLOSURE?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The latitude and departure residuals are perpendicular coordinate components of the closing-error vector. The linear misclosure is the square root of the sum of their squares. Here it is sqrt[(0.300)^2 + (-0.400)^2] m. This equals sqrt(0.09 + 0.16) = sqrt(0.25) m. The magnitude is therefore 0.500 m. The negative departure sign influences the direction of the closing-error vector but not its magnitude.
వివరణ:
Question 9
ప్రశ్న 9
A closed traverse has a total horizontal perimeter of 2500 m and an unadjusted linear misclosure of 0.500 m. What is the relative precision in the conventional 1:N notation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Relative precision compares the linear misclosure with total horizontal traverse length. Its numerical denominator is perimeter divided by linear misclosure. Substituting gives 2500/0.500 = 5000. Therefore the ratio is stated as 1:5000. A larger denominator corresponds to smaller proportional closure error. The ratio must be assessed against the project survey requirements rather than treated as a universal acceptance standard.
వివరణ:
Question 10
ప్రశ్న 10
A project requires a CLOSED traverse to achieve a relative precision of at least 1:5000. If its total horizontal length is 1800 m, what is the MAXIMUM permissible linear misclosure under this requirement?
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Explanation:
A minimum relative precision of 1:5000 means the linear misclosure must not exceed one part in 5000 of the traverse length. The total traverse length is 1800 m. Dividing 1800 by 5000 gives 0.360 m. Therefore 0.360 m is the maximum closing-error magnitude that meets the stated ratio. A greater misclosure would yield a smaller precision denominator and fail this specified criterion. Other survey quality conditions may still apply even when the ratio is satisfied.
వివరణ:
Question 11
ప్రశ్న 11
Which statement correctly describes the BOWDITCH or COMPASS RULE for balancing a closed traverse?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Bowditch adjustment distributes the unadjusted latitude and departure residuals among traverse legs. For each coordinate component, a leg receives a correction proportional to its length relative to total perimeter. The correction sign is opposite to the corresponding total coordinate misclosure. This rule is conventionally chosen when angular and distance measurements have broadly comparable precision. The adjusted coordinate components must sum to zero for a closed loop. It does not make an unacceptable field survey accurate merely by forcing mathematical closure.
వివరణ:
Question 12
ప్రశ్న 12
A four-leg closed traverse has the following unadjusted data (line: horizontal length, latitude, departure; metres): AB: 100, +80.000, +60.000; BC: 150, -90.000, +120.000; CD: 100, -80.000, -60.000; DA: 150, +90.120, -119.910. Use Bowditch correction with total perimeter 500 m, taking north and east as positive. What corrections must be applied to the latitude and departure of AB?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The sum of the four unadjusted latitudes is +0.120 m. The sum of their departures is +0.090 m. The AB length fraction is 100/500 = 0.20. Under Bowditch, its latitude correction is -0.120 times 0.20 = -0.024 m. Its departure correction is -0.090 times 0.20 = -0.018 m. Both corrections oppose the positive coordinate misclosures.
వివరణ:
Question 13
ప్రశ్న 13
A four-leg closed traverse has the following unadjusted data (line: horizontal length, latitude, departure; metres): AB: 100, +80.000, +60.000; BC: 150, -90.000, +120.000; CD: 100, -80.000, -60.000; DA: 150, +90.120, -119.910. Use Bowditch correction with total perimeter 500 m, taking north and east as positive. What are the ADJUSTED latitude and departure of BC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The unadjusted closure is +0.120 m in latitude and +0.090 m in departure. The BC length fraction is 150/500 = 0.30. The Bowditch latitude correction for BC is -0.120 times 0.30 = -0.036 m. Its departure correction is -0.090 times 0.30 = -0.027 m. Adding these to the original BC components gives -90.036 m latitude and +119.973 m departure. The negative latitude is a southing and the positive departure remains an easting.
వివరణ:
Question 14
ప్రశ్న 14
A four-leg closed traverse has the following unadjusted data (line: horizontal length, latitude, departure; metres): AB: 100, +80.000, +60.000; BC: 150, -90.000, +120.000; CD: 100, -80.000, -60.000; DA: 150, +90.120, -119.910. Use Bowditch correction with total perimeter 500 m, taking north and east as positive. If station A has Northing = 1000.000 m and Easting = 2000.000 m, what are the ADJUSTED coordinates of station C, after following AB and BC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The unadjusted coordinate residuals are +0.120 m latitude and +0.090 m departure. Bowditch gives adjusted AB components of +79.976 m latitude and +59.982 m departure. The adjusted BC components are -90.036 m latitude and +119.973 m departure. Thus northing at C is 1000.000 + 79.976 - 90.036 = 989.940 m. Easting at C is 2000.000 + 59.982 + 119.973 = 2179.955 m. The result accumulates the adjusted components from the known coordinates of A.
వివరణ:
Question 15
ప్రశ్న 15
A four-leg closed traverse has the following unadjusted data (line: horizontal length, latitude, departure; metres): AB: 100, +80.000, +60.000; BC: 150, -90.000, +120.000; CD: 100, -80.000, -60.000; DA: 150, +90.120, -119.910. Use Bowditch correction with total perimeter 500 m, taking north and east as positive. Starting from A at Northing = 1000.000 m, Easting = 2000.000 m, what are the ADJUSTED coordinates of D after traversing AB, BC and CD?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Bowditch-adjusted components of AB are latitude +79.976 m and departure +59.982 m. Those of BC are latitude -90.036 m and departure +119.973 m. CD has length fraction 100/500 and therefore corrected components -80.024 m latitude and -60.018 m departure. The accumulated northing at D is 1000.000 + 79.976 - 90.036 - 80.024 = 909.916 m. Its easting is 2000.000 + 59.982 + 119.973 - 60.018 = 2119.937 m. The remaining adjusted DA components return exactly to the initial coordinates of A.
వివరణ:
Question 16
ప్రశ్న 16
A four-leg closed traverse has the following unadjusted data (line: horizontal length, latitude, departure; metres): AB: 100, +80.000, +60.000; BC: 150, -90.000, +120.000; CD: 100, -80.000, -60.000; DA: 150, +90.120, -119.910. Use Bowditch correction with total perimeter 500 m, taking north and east as positive. What are the ADJUSTED latitude and departure of DA after Bowditch balancing?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The algebraic latitude misclosure is +0.120 m and departure misclosure is +0.090 m. The DA length fraction is 150/500 = 0.30. Its latitude correction is therefore -0.036 m, and its departure correction is -0.027 m. Adding the corrections to the original DA latitude +90.120 m gives +90.084 m. Adding the departure correction to -119.910 m gives -119.937 m. Summing all four adjusted latitudes and departures separately gives zero, confirming the balanced loop.
వివరణ:
Question 17
ప్రశ్న 17
A closed traverse has latitude misclosure -0.300 m and departure misclosure +0.500 m. Its perimeter is 1200 m. Under Bowditch, what latitude and departure corrections are applied to a 240 m leg?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The line length fraction for this leg is 240/1200 = 0.20. Bowditch corrections use the negatives of the total component misclosures multiplied by this fraction. The latitude correction is -(-0.300) times 0.20 = +0.060 m. The departure correction is -(+0.500) times 0.20 = -0.100 m. These signs oppose the respective original closing errors. Corrections to all the legs must sum to +0.300 m in latitude and -0.500 m in departure.
వివరణ:
Question 18
ప్రశ్న 18
Which statement correctly distinguishes the classical TRANSIT RULE from the Bowditch rule?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The transit rule uses coordinate-component magnitudes rather than line lengths as correction weights. A line receives a latitude correction in proportion to its absolute preliminary latitude. Its departure correction is separately weighted by its absolute preliminary departure. This classical approach is conventionally associated with angular measurements of higher precision than the corresponding length measurements. The Bowditch rule instead uses the same leg-length fraction for both components. Both methods should be applied only after observational blunders and unacceptable closures have been considered.
వివరణ:
Question 19
ప్రశ్న 19
A four-leg closed traverse has preliminary signed components (latitude, departure; metres): AB (+60.000, +60.000), BC (-40.000, +89.850), CD (-59.900, -60.000), DA (+40.100, -90.150). Use the classical transit rule with corrections based on ABSOLUTE component values. The absolute-latitude sum is 200.000 m and the absolute-departure sum is 300.000 m. What are the transit-rule corrections to the latitude and departure of AB?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The unadjusted latitude residual is 60 - 40 - 59.9 + 40.1 = +0.200 m. The departure residual is 60 + 89.85 - 60 - 90.15 = -0.300 m. For AB, the absolute-latitude fraction is 60/200 = 0.30. Its latitude correction is -0.200 times 0.30 = -0.060 m. The absolute-departure fraction is 60/300 = 0.20, so its departure correction is -(-0.300) times 0.20 = +0.060 m. The corrections respectively oppose the two signed misclosures.
వివరణ:
Question 20
ప్రశ్న 20
A four-leg closed traverse has preliminary signed components (latitude, departure; metres): AB (+60.000, +60.000), BC (-40.000, +89.850), CD (-59.900, -60.000), DA (+40.100, -90.150). Use the classical transit rule with corrections based on ABSOLUTE component values. The absolute-latitude sum is 200.000 m and the absolute-departure sum is 300.000 m. What are the TRANSIT-ADJUSTED latitude and departure of BC?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The algebraic closure components are +0.200 m latitude and -0.300 m departure. The transit latitude correction for BC is -0.200 times (40/200) = -0.04000 m. The transit departure correction is +0.300 times (89.850/300) = +0.08985 m. Adding these to the preliminary BC components gives -40.04000 m in latitude. The adjusted departure is 89.85000 + 0.08985 = +89.93985 m. The latitude remains southward while the departure remains eastward.
వివరణ:
Question 21
ప్రశ్న 21
A four-leg closed traverse has preliminary signed components (latitude, departure; metres): AB (+60.000, +60.000), BC (-40.000, +89.850), CD (-59.900, -60.000), DA (+40.100, -90.150). Use the classical transit rule with corrections based on ABSOLUTE component values. The absolute-latitude sum is 200.000 m and the absolute-departure sum is 300.000 m. If A has Northing = 500.00000 m and Easting = 1000.00000 m, what are the TRANSIT-ADJUSTED coordinates of D after AB, BC and CD?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The transit-adjusted AB components are (+59.94000, +60.06000) m. The adjusted BC components are (-40.04000, +89.93985) m. The adjusted CD components are (-59.95990, -59.94000) m, using latitude correction -0.05990 m and departure correction +0.06000 m. Accumulating latitudes from A gives Northing(D) = 500 + 59.94 - 40.04 - 59.95990 = 459.94010 m. Accumulating departures gives Easting(D) = 1000 + 60.06 + 89.93985 - 59.94 = 1090.05985 m. The adjusted DA components (+40.05990, -90.05985) m return to A exactly.
వివరణ:
Question 22
ప్రశ్న 22
A closed traverse has latitude misclosure +0.400 m and departure misclosure -0.600 m. Its perimeter is 1000 m, and a specified leg is 200 m long. For TRANSIT weights, that leg has absolute latitude 80 m out of total absolute latitudes 800 m and absolute departure 120 m out of total absolute departures 600 m. Which correctly compares its Bowditch and transit CORRECTIONS?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Bowditch leg-length fraction is 200/1000 = 0.20. Thus Bowditch gives latitude correction -0.400 times 0.20 = -0.080 m and departure correction +0.600 times 0.20 = +0.120 m. The transit latitude weight is 80/800 = 0.10, giving -0.400 times 0.10 = -0.040 m. The transit departure weight is 120/600 = 0.20, giving +0.600 times 0.20 = +0.120 m. The two rules yield different latitude corrections in this example but the same departure correction. The distinction comes from their different weighting systems.
వివరణ:
Question 23
ప్రశ్న 23
In a classical transit-rule adjustment where the absolute-latitude and absolute-departure denominators are nonzero, a traverse line has ZERO preliminary latitude but NONZERO preliminary departure. Which statement is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Under the transit rule, a leg latitude correction is weighted by the absolute value of that leg latitude. A zero preliminary latitude therefore has a zero weight in the latitude correction distribution. The departure is weighted independently by its own absolute magnitude. A nonzero preliminary departure can consequently receive a departure correction if total departure misclosure is nonzero. This contrasts with Bowditch, under which a line of nonzero length generally receives a share of both component residuals. The transit correction denominators must be defined and nonzero for the stated computation.
వివరణ:
Question 24
ప్రశ్న 24
The adjusted plane coordinates of four consecutive corners of a plot are A (E=0 m, N=0 m), B (E=80 m, N=0 m), C (E=60 m, N=50 m), and D (E=10 m, N=50 m). What is the enclosed horizontal area computed from the coordinates?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The polygon can be evaluated with the coordinate or shoelace area formula. The forward coordinate-product sum is 0(0) + 80(50) + 60(50) + 10(0) = 7000 m^2. The reverse product sum is 0(80) + 0(60) + 50(10) + 50(0) = 500 m^2. One-half of their absolute difference is (7000 - 500)/2 = 3250 m^2. The same result follows from the trapezoid formula using parallel sides 80 m and 50 m separated by 50 m. The coordinates describe a plane horizontal area, not the actual sloping surface area.
వివరణ:
Question 25
ప్రశ్న 25
A closed traverse has been forced to exact coordinate closure using Bowditch corrections, although its ORIGINAL unadjusted linear misclosure exceeded the survey specification. Which conclusion is most technically justified?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Bowditch adjustment algebraically redistributes an observed coordinate misclosure to force the adjusted components to close. That mathematical balance does not prove that all original angular and distance measurements were accurate. An excessive original misclosure may indicate blunders, unsuitable field conditions, or inadequate observing precision. Original relative precision should be evaluated against the stated survey requirement before treating the adjustment as satisfactory. Problematic observations may require correction or remeasurement. The accepted adjusted coordinates must also remain consistent with independent control and the required survey accuracy.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Latitude is the north-south component and departure is the east-west component. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. Latitude = L cos(theta); departure = L sin(theta). సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: A. Latitude = -75.000 m; departure = +129.904 m. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: D. Latitude = -70.711 m; departure = -70.711 m. సమాధానం: D.
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Question 5 ప్రశ్న 5Answer: B. Latitude = +103.923 m; departure = -60.000 m. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. Northing = 928.000 m; Easting = 2096.000 m. సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: D. Latitude = +0.27 m; departure = -0.20 m. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: A. 0.500 m. సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: B. 1:5000. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: C. 0.360 m. సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: A. Both latitude and departure corrections to a leg are proportional to its horizontal length divided by the total traverse perimeter. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: D. Latitude -0.024 m; departure -0.018 m. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: C. Latitude -90.036 m; departure +119.973 m. సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: B. Northing 989.940 m; Easting 2179.955 m. సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: A. Northing 909.916 m; Easting 2119.937 m. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: D. Latitude +90.084 m; departure -119.937 m. సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: B. Latitude +0.060 m; departure -0.100 m. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: C. The transit rule distributes latitude corrections by absolute individual latitudes and departure corrections by absolute individual departures, and is traditionally applied when angular observations are more precise than distances. సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: A. Latitude -0.060 m; departure +0.060 m. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: D. Latitude -40.04000 m; departure +89.93985 m. సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: B. Northing 459.94010 m; Easting 1090.05985 m. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. Bowditch latitude -0.080 m; transit latitude -0.040 m; both departure corrections +0.120 m. సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: A. It receives zero latitude correction, while its departure correction can be nonzero. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: D. 3250 m^2. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: C. Adjusted coordinate sums can close exactly, but the original excessive misclosure still requires investigation and possibly re-observation before the traverse is accepted. సమాధానం: C.