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Question 1
ప్రశ్న 1
A nominal 20 m chain has become 20.08 m long because of wear at its joints. It is used to measure a horizontal line without any other measurement errors. What is the correct effect of this defect on the recorded distance and its length correction?
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Explanation:
The actual chain length exceeds its nominal length. Every nominal twenty-metre chain reading therefore spans more than twenty metres on the ground. The accumulated recorded distance is smaller than the actual ground distance. The length-standardization correction must consequently be positive. It is calculated from the ratio of actual to nominal chain length. This is a systematic instrumental error rather than ordinary random scatter.
వివరణ:
Question 2
ప్రశ్న 2
A horizontal survey line is recorded as 320.00 m using a chain whose nominal length is 20.00 m but whose calibrated actual length is 20.05 m. Assume uniform chain length and no other errors. What is the TRUE horizontal length?
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Explanation:
The chain is longer than its nominal twenty metres. The correct distance equals the recorded distance multiplied by actual chain length divided by nominal length. Thus the multiplier is 20.05/20.00 = 1.0025. Applying it to 320.00 metres gives 320.80 metres. The correction is positive 0.80 metre. The result assumes the calibrated chain length remained unchanged throughout measurement.
వివరణ:
Question 3
ప్రశ్న 3
A chain is nominally 20.00 m long but is found to be only 19.92 m long during measurement. A level chain line is recorded as 250.00 m using that chain. Neglect other errors. What is the corrected line length?
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Explanation:
A chain shorter than its nominal length produces an overstated recorded distance. The correction factor is actual length divided by nominal length. For the given chain, this factor is 19.92/20.00 = 0.996. Multiplying 250.00 metres by 0.996 gives 249.00 metres. The required correction is therefore negative 1.00 metre. No additional correction is needed under the question's stated assumptions.
వివరణ:
Question 4
ప్రశ్న 4
An area was plotted and computed as 12,000 m² using chain measurements with a nominal 20.00 m chain. The chain was actually 20.10 m long throughout the survey, and BOTH coordinate directions used the same uncorrected chain scale. Neglect all other errors. What is the corrected area?
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Explanation:
The actual-to-nominal chain ratio is 20.10/20.00 = 1.005. All measured horizontal coordinates have been uniformly scaled by this factor. An area computed from both coordinate directions must therefore be multiplied by the square of the linear factor. The corrected area is 12,000 × (1.005)² = 12,120.30 square metres. Merely applying a single linear factor would not fully correct a uniformly scaled area. This square-law correction applies because both dimensions were obtained with the same incorrect chain scale.
వివరణ:
Question 5
ప్రశ్న 5
A survey line was measured in two separate portions using a nominal 20 m chain. The first portion was recorded as 120 m while the chain's actual length was 20.05 m. After repair, the second portion was recorded as 80 m while the actual chain length was 19.95 m. No other corrections apply. What is the corrected TOTAL distance?
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Explanation:
The two recorded portions were measured under different chain calibrations. The first true portion equals 120 × 20.05/20 = 120.30 metres. The second true portion equals 80 × 19.95/20 = 79.80 metres. Their sum is 200.10 metres. Applying only one chain correction to the full 200 metres would ignore the change in calibration. Segment-wise correction is therefore essential when the instrument length changes during the survey.
వివరణ:
Question 6
ప్రశ్న 6
A nominal 20.00 m chain was exactly 20.00 m long at the start of a survey and 20.10 m long at the end. Its actual length increased linearly with accumulated recorded distance, and 400.00 m was recorded evenly over that change. Using the mean-length approximation specified by these conditions, what is the corrected distance?
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Explanation:
The starting actual chain length is 20.00 metres and the final length is 20.10 metres. Under the stated uniform-change assumption, the mean actual length is 20.05 metres. The mean standardization factor is consequently 20.05/20.00 = 1.0025. Applying it to 400.00 recorded metres gives 401.00 metres. The average-length approach is appropriate because the question specifies a linear change distributed over the measured work. A single final-length correction would overstate the correction for earlier measurements.
వివరణ:
Question 7
ప్రశ్న 7
During chain surveying, an incorrectly calibrated chain causes almost the same proportional distance bias on every line. Small independent reading fluctuations occur above and below their expected values. Which error interpretation is correct?
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Explanation:
An incorrect chain length introduces a repeatable scale bias into successive measurements. Such a systematic effect tends to accumulate rather than cancel when comparable lengths are measured. Small independent observational errors may take positive or negative signs. These random or compensating errors can partially offset one another. Their presence does not remove the systematic error from incorrect chain length. Instrument calibration and repeated checked measurements address the two error categories differently.
వివరణ:
Question 8
ప్రశ్న 8
A slope distance of 50.00 m is measured accurately between two points with a vertical difference of 14.00 m. What HORIZONTAL distance should be used in the chain-survey plan? Assume a straight slope and right-angle vertical-horizontal geometry.
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Explanation:
The measured fifty metres is the sloping hypotenuse between the points. The vertical difference forms the perpendicular side of a right triangle. The square of horizontal distance equals 50² - 14². This is 2500 - 196 = 2304 square metres. Taking the square root gives a horizontal distance of forty-eight metres. The slope measurement must therefore be reduced by two metres for horizontal plotting.
వివరణ:
Question 9
ప్రశ్న 9
Two end stations A and B are exactly 100 m apart on a straight horizontal line. Instead of properly ranging, the crew measures via an intermediate point P located 50 m along AB and 1 m perpendicular to it, recording AP + PB. Approximately how large is the error in the recorded distance?
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Explanation:
The true straight distance AB is one hundred metres. Point P divides the horizontal alignment into two fifty-metre projections but lies one metre off line. Each measured segment is sqrt(50² + 1²) metres long. The broken route therefore measures 2sqrt(2501), approximately 100.020 metres. Its recorded length exceeds the direct separation by about 0.020 metre. Poor ranging creates a positive alignment error even when the individual segment lengths are measured accurately.
వివరణ:
Question 10
ప్రశ్న 10
A steel tape is standard at 20°C and is used at a uniform tape temperature of 40°C to record a level line as 200.000 m. Its coefficient of linear thermal expansion is 12 × 10^-6 per °C. Using Ct = alpha(T - T0)L and neglecting other corrections, what corrected distance is obtained?
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Explanation:
The tape temperature exceeds its standardization temperature by twenty degrees Celsius. The thermal correction is alpha times temperature difference times recorded distance. Substitution gives 12 × 10^-6 × 20 × 200 = 0.048 metre. A warmer steel tape becomes longer and consequently records too small a distance for a given ground length. The correction is positive under the stated model. Adding 0.048 metre to 200.000 metres gives 200.048 metres.
వివరణ:
Question 11
ప్రశ్న 11
A steel tape records a horizontal distance of 100.000 m under a pull of 150 N, although its standardization pull was 50 N. The tape has cross-sectional area 5 mm² and Young's modulus 200,000 N/mm². Using Cp = (P - P0)L/(AE), and neglecting all other corrections, what is the corrected distance?
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Explanation:
The field pull exceeds the standard pull by one hundred newtons. Greater tension extends the tape and makes the marked reading understate the actual measured distance. The denominator AE is 5 × 200,000 = 1,000,000 newtons when consistent millimetre-based units are used. The pull correction is 100 × 100/1,000,000 = 0.010 metre. Its sign is positive because the tape is stretched more than at standardization. Thus the corrected length is 100.010 metres.
వివరణ:
Question 12
ప్రశ్న 12
An intervening ridge prevents the two terminal rods of a chain line from being intervisible, but the ground route can be physically chained after intermediate points are established. Suitable intermediate assistants can see the rods needed for reciprocal adjustment. How should the obstruction principally be classified and handled?
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Explanation:
The ridge prevents direct visibility between the terminal stations. The problem is therefore one of establishing a straight alignment by ranging. The question states that physical chaining remains possible once suitable intermediate points are determined. Reciprocal ranging uses successive adjustments of intermediate rods when the endpoints are not intervisible. After alignment, horizontal measurements can be taken along the required line. This is classified as an obstacle to ranging rather than an obstacle to direct distance measurement.
వివరణ:
Question 13
ప్రశ్న 13
Both terminal stations of a survey line are visible across a river, but the crew cannot lay a chain through the water. Accessible ground on one bank is available for constructing measured auxiliary triangles. What is the appropriate classification and response?
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Explanation:
The terminal stations are visible, so their intended alignment can be identified. Water prevents a chain from being laid directly along that line. The obstruction therefore concerns chaining rather than terminal ranging. Suitable triangles or perpendicular constructions on accessible ground can be measured instead. The unknown line length may then be derived from the verified geometric relationships. Visual estimation of river width would not provide an equivalent measured survey result.
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Question 14
ప్రశ్న 14
A substantial warehouse occupies a section of the proposed straight chain line and completely blocks both intervisibility and direct access between stations. Which response best preserves the intended control?
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Explanation:
The warehouse blocks the line of sight through the intended survey line. It also prevents placing the measuring chain along the obstructed segment. The obstacle therefore affects both ranging and chaining. An accessible parallel detour may be constructed by checked perpendicular offsets. The original alignment and distance can then be transferred geometrically beyond the building. The transferred control should be checked independently wherever practicable.
వివరణ:
Question 15
ప్రశ్న 15
For a line AB obstructed by a warehouse, the surveyor constructs perpendiculars AA' and BB', each 6 m long on the SAME side of AB. The points A' and B' are joined by a measured straight line of 54 m, and the perpendiculars are accurately established so that AA'B'B is a rectangle. What is the inaccessible length AB?
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Explanation:
The four stations are explicitly stated to form a rectangle. Opposite sides of a rectangle are equal in length. Therefore the inaccessible side AB equals the parallel measured side A'B'. The two six-metre perpendicular offsets transfer the line laterally without changing the length between corresponding endpoints. Since A'B' measures fifty-four metres, AB also measures fifty-four metres. The offsets should not be added to the parallel line when determining AB.
వివరణ:
Question 16
ప్రశ్న 16
Accessible stations A and C on one bank of a river are 40 m apart. An inaccessible station B on the opposite bank is visible from both, and the measured horizontal triangle has angle CAB = 60° and angle ACB = 30°. Neglect curvature and assume the triangle lies in a plane. What is the inaccessible distance AB?
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Explanation:
The two given triangle angles add to ninety degrees. The third angle at B is therefore ninety degrees. The accessible baseline AC is opposite the right angle and is the hypotenuse of length forty metres. The inaccessible side AB is opposite the thirty-degree angle at C. Thus AB equals 40 sin30° = 20 metres. This illustrates indirect determination of an inaccessible length from a measured baseline and observed geometry.
వివరణ:
Question 17
ప్రశ్న 17
In a chain-survey field book, a lamp post is recorded at chainage 60 m with a 8 m perpendicular offset to the RIGHT of line AB. During plotting, the drafter places the post 8 m to the LEFT at the SAME chainage. How far apart are the plotted and correctly positioned points?
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Explanation:
Both positions have the same chainage and thus the same perpendicular foot on AB. The correct point lies eight metres on the right side of the line. The incorrectly plotted point lies eight metres on the left side. Their separation is therefore 8 + 8 = 16 metres. This is a field-book interpretation or plotting blunder rather than an error in the recorded offset length. Properly identifying the direction of chaining is essential when interpreting right and left entries.
వివరణ:
Question 18
ప్రశ్న 18
A straight chain line has perpendicular offsets to an irregular boundary at chainages 0, 10, 20 and 30 m. The corresponding offsets on the SAME side are 2, 5, 7 and 3 m. Using the TRAPEZOIDAL rule between successive offsets, what is the enclosed strip area?
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Explanation:
The three adjacent chainage intervals are each ten metres long. The trapezoidal rule averages the two bounding offsets in every interval. Their individual areas are 10(2 + 5)/2 = 35, 10(5 + 7)/2 = 60 and 10(7 + 3)/2 = 50 square metres. Adding these gives 145 square metres. Equivalently, the formula is 10[(2 + 3)/2 + 5 + 7]. This result uses straight-line interpolation of the boundary between successive offset positions.
వివరణ:
Question 19
ప్రశ్న 19
Five perpendicular offsets to a curved boundary, all on the SAME side of a straight chain line, are 0, 6, 9, 6 and 0 m at equal 10 m intervals. Assuming Simpson's one-third rule is appropriate for this four-interval dataset, what is the estimated area?
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Explanation:
There are five offsets and four equal intervals, satisfying the basic even-interval requirement of Simpson's one-third rule. The endpoint offsets are zero and zero. The two interior odd-position offsets of six metres receive coefficient four, and the central nine-metre offset receives coefficient two. The area equals (10/3)[0 + 0 + 4(6 + 6) + 2(9)]. The bracket is sixty-six, giving an area of 220 square metres. This differs from the trapezoidal estimate because Simpson's rule represents the curved boundary by parabolic segments.
వివరణ:
Question 20
ప్రశ్న 20
A rectangular plot was measured as 100.00 m long along a nominal 20.00 m chain and exactly 6.00 m wide using a separately calibrated correct tape. The chain's actual length was 20.10 m throughout. What is the corrected plot area, assuming both dimensions are horizontal?
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Explanation:
Only the one-hundred-metre length was measured with the incorrectly calibrated chain. Its corrected length is 100 × 20.10/20.00 = 100.50 metres. The width of six metres was measured independently with a correct tape and must not be scaled. The corrected area is therefore 100.50 × 6.00 = 603.00 square metres. Squaring the chain correction factor would be inappropriate here because only one dimension contains that error. This illustrates the importance of identifying which observations share a systematic scale bias.
వివరణ:
Question 21
ప్రశ్న 21
At chainages 0, 10 and 20 m on a straight survey line, perpendicular boundary offsets to the LEFT are 2, 4 and 2 m, while offsets to the RIGHT are 3, 5 and 3 m. Using trapezoids between successive cross-sections, what is the combined enclosed area on BOTH sides of the line?
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Explanation:
The strip width between left and right boundaries equals the sum of their respective offsets. The widths at the three chainages are 5, 9 and 5 metres. Two intervals of ten metres separate these cross-sections. The trapezoidal area is 10[(5 + 5)/2 + 9]. This evaluates to 10(5 + 9) = 140 square metres. The left and right parts could also be integrated separately and then added, giving the same result.
వివరణ:
Question 22
ప్రశ్న 22
A surveyed check line has verified ground length 80.0 m. On a drawing at scale 1:1000, the plotted separation of its endpoints is 7.7 cm. A plotting discrepancy greater than 0.5 m on the ground is unacceptable. What should the surveyor conclude?
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Explanation:
A scale of 1:1000 represents one centimetre on the plan as ten metres on the ground. A plotted length of 7.7 centimetres therefore corresponds to seventy-seven metres. The independently verified check length is eighty metres. The plotting discrepancy is three metres on the ground. This is greater than the permitted half-metre difference. The plotted geometry, source chainages and station identification should be checked before the plan is accepted.
వివరణ:
Question 23
ప్రశ్న 23
At a plotting scale of 1:500, the position of a perpendicular-offset endpoint is displaced by 0.4 mm on the drawing. What ground-position error does that plotting displacement represent?
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Explanation:
The representative fraction 1:500 means one drawing unit equals five hundred identical ground units. A displacement of 0.4 millimetre on the drawing corresponds to 0.4 × 500 = 200 millimetres on the ground. Two hundred millimetres equals 0.20 metre. Thus the plotting displacement represents twenty centimetres of ground-position error. Larger-scale drawings reduce the ground distance represented by the same small plotting error. The calculation concerns plotting alone and does not include field-observation errors.
వివరణ:
Question 24
ప్రశ్న 24
A line's independent check establishes that it contains 14 complete lengths of a nominal 20 m chain, but the field book counts only 13 complete lengths with no fractional remainder. The chain length itself is correct. What is the principal error if one complete chain count was omitted?
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Explanation:
Fourteen complete twenty-metre chain lengths represent 280 metres. Thirteen complete lengths represent only 260 metres. The field-book value is therefore twenty metres too small. An omitted full chain count is a discrete counting or booking blunder. It is not the small proportional correction associated with thermal expansion or calibration. The original tally and independent measurement should be reconciled rather than hiding the discrepancy in statistical adjustment.
వివరణ:
Question 25
ప్రశ్న 25
A chain-survey framework is to be plotted for a pumping-station site. One main line was measured with a calibrated chain of incorrect nominal length, some offsets were taken with an independently verified correct tape, and a check-line disagreement remains unresolved. Which workflow is technically MOST appropriate?
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Explanation:
An incorrect chain length creates a systematic scale effect in measurements made with that chain. The correction should be applied to the affected observations, not indiscriminately to all survey data. Offsets measured with a separately verified correct tape do not inherit the chain's calibration error. An unresolved check-line discrepancy may reveal further measurement, identification or plotting mistakes. Such discrepancies should be investigated before finalizing the survey framework. Only verified and consistently reduced observations should be transferred to the plotted plan.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. The recorded distance is too small, and a positive correction is required. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. 320.80 m. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. 249.00 m. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 12,120.30 m². సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 200.10 m. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: D. 401.00 m. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: A. The chain-length bias is cumulative and systematic; the small independent fluctuations are compensating or random errors. సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. 48.00 m. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 0.020 m too long. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: D. 200.048 m. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: C. 100.010 m. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: A. Obstacle to ranging only; apply reciprocal ranging before measuring along the line. సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: D. Chaining is obstructed but ranging is possible; determine the inaccessible length from checked auxiliary geometry. సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: B. Treat it as an obstacle to both ranging and chaining, transferring the alignment around it with checked perpendicular and parallel lines. సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: C. 54 m. సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: A. 20 m. సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: B. 16 m. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. 145 m². సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. 220 m². సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. 603.00 m². సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 140 m². సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. The plotted line represents 77.0 m, giving a 3.0 m discrepancy that exceeds tolerance. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. 0.20 m. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. A gross counting or booking blunder causing an understatement of 20 m. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: B. Apply the applicable correction only to observations affected by the incorrect chain, verify the check-line discrepancy, and plot accepted observations at a consistent scale. సమాధానం: B.