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Question 1
ప్రశ్న 1
In reducing staff readings by the HEIGHT-OF-INSTRUMENT (HI) method, which operation establishes the line-of-collimation RL for a new setup on a benchmark of known RL?
Explanation:
The height of instrument is the reduced level of the horizontal line of collimation. A backsight on a point of known RL gives the vertical distance from that point to the line of sight. Therefore HI = RL of the reference point + BS. Intermediate sights and foresights are subsequently subtracted from that HI to obtain point RLs. An instrument move requires a newly established HI based on the backsight at the new setup. The reference datum remains unchanged.
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Question 2
ప్రశ్న 2
A benchmark has RL 104.265 m, and the backsight from a freshly set level is 1.385 m. What is the HI for this setup?
Explanation:
The level sight is above the benchmark by the backsight reading. HI is obtained as 104.265 + 1.385 = 105.650 m. Subtracting the reading would incorrectly place the sight below the benchmark. The benchmark retains its RL of 104.265 m. The computed 105.650 m is an elevation relative to the adopted datum. It can be used for all valid staff observations made without moving this setup.
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Question 3
ప్రశ్న 3
A valve-chamber floor is observed by an intermediate sight of 2.375 m from a setup whose HI is 106.800 m. What is the floor RL?
Explanation:
An upright staff reading measures the vertical interval between the point and the horizontal sight line. For an intermediate point, RL = HI - IS. Substitution gives 106.800 - 2.375 = 104.425 m. Adding the reading would reverse the physical relationship. This is an intermediate sight, so the instrument setup remains the same. Its value is not added to the backsight total for the endpoint arithmetic check.
వివరణ:
Question 4
ప్రశ్న 4
From a single undisturbed level setup, successive readings on two different points are 2.485 m and 1.760 m. What is the RISE or FALL from the first point to the second?
Explanation:
When two readings share one horizontal line of sight, the smaller reading belongs to the higher point. The difference is 2.485 - 1.760 = +0.725 m. A positive difference of previous minus current reading denotes a rise. Hence the second point is 0.725 m above the first. The readings must be from the same setup for this direct comparison. Absolute RLs are unnecessary to determine the rise.
వివరణ:
Question 5
ప్రశ్న 5
Two consecutive readings from the SAME instrument setup are 0.955 m at A and 2.230 m at B. What entry belongs in the rise-and-fall columns for B?
Explanation:
The reading increases from 0.955 m to 2.230 m. A larger staff reading from an unchanged horizontal sight line signifies lower ground. The magnitude is 2.230 - 0.955 = 1.275 m. Therefore the fall column at B receives 1.275 m. The rise column is left blank or treated as zero for this interval. The RL of B is the RL of A minus 1.275 m.
వివరణ:
Question 6
ప్రశ్న 6
A one-setup level book contains BM (RL 110.000 m; BS 1.620 m), point P (IS 2.335 m) and CP (FS 1.475 m). Which pair correctly gives the RLs of P and CP?
Explanation:
Compute the setup HI as 110.000 + 1.620 = 111.620 m. The intermediate point has RL(P) = 111.620 - 2.335 = 109.285 m. The change point has RL(CP) = 111.620 - 1.475 = 110.145 m. The rise from P to CP is 2.335 - 1.475 = 0.860 m. The initial fall from BM to P is 2.335 - 1.620 = 0.715 m. The arithmetic check is 1.620 - 1.475 = 0.860 - 0.715 = 110.145 - 110.000 = +0.145 m.
వివరణ:
Question 7
ప్రశ్న 7
The HI of the old level setup is 121.550 m. Its foresight on a change point is 2.175 m; after shifting, the backsight on the SAME change point is 0.940 m. What is the new HI?
Explanation:
The old setup first establishes the change-point RL. RL(CP) = 121.550 - 2.175 = 119.375 m. The new backsight of 0.940 m is added to this unchanged point RL. Consequently the new HI is 119.375 + 0.940 = 120.315 m. Directly adding the new backsight to the old HI would ignore the old foresight. The two readings at the same change point transfer vertical control between the two instrument setups.
వివరణ:
Question 8
ప్రశ్న 8
Reduce this TWO-SETUP level book by the HI method. Start: BM RL 80.250 m. Setup 1: BS on BM 1.465 m, IS on A 2.250 m, FS on CP 1.980 m. Setup 2: BS on the same CP 0.845 m, IS on B 1.105 m, FS on endpoint E 1.755 m. What is RL(E)?
Explanation:
For setup 1, HI1 = 80.250 + 1.465 = 81.715 m. Hence RL(A) = 81.715 - 2.250 = 79.465 m and RL(CP) = 81.715 - 1.980 = 79.735 m. For setup 2, HI2 = 79.735 + 0.845 = 80.580 m. Thus RL(B) = 80.580 - 1.105 = 79.475 m and RL(E) = 80.580 - 1.755 = 78.825 m. As an independent endpoint check, sum BS = 2.310 m and sum FS = 3.735 m. Their difference, -1.425 m, equals 78.825 - 80.250 = -1.425 m.
వివరణ:
Question 9
ప్రశ్న 9
For a levelling run, the consecutive ground changes are: fall 0.785 m, rise 0.270 m, fall 0.260 m and fall 0.650 m. Which rise-and-fall totals and net elevation change are correct?
Explanation:
The only upward change is 0.270 m, giving a total rise of 0.270 m. The falls sum to 0.785 + 0.260 + 0.650 = 1.695 m. The net elevation change is sum rise minus sum fall. Thus the net change is 0.270 - 1.695 = -1.425 m. This indicates the final point is 1.425 m below the first point. In a correctly reduced complete run, this must also equal sum BS minus sum FS.
వివరణ:
Question 10
ప్రశ్న 10
A change point is observed as FS = 2.450 m just before the instrument is moved and as BS = 0.780 m just after the move. From the second setup the next point P has IS = 1.340 m. What is the rise or fall from the change point to P?
Explanation:
The FS and following BS on a change point belong to different horizontal sight lines. They cannot be directly compared to compute ground rise or fall. The valid comparison uses the new BS of 0.780 m and the next IS of 1.340 m from the same setup. Their difference is 0.780 - 1.340 = -0.560 m. Therefore P is 0.560 m below the change point. The change point itself has only one RL even though two observations are booked against it.
వివరణ:
Question 11
ప్రశ్న 11
A THREE-SETUP level book starts at BM RL 150.000 m. Setup 1: BS BM 1.240, IS A 1.850, FS CP1 0.955. Setup 2: BS CP1 1.335, IS P 0.765, FS CP2 2.105. Setup 3: BS CP2 0.925, IS Q 1.415, FS endpoint E 1.085 (all readings in m). What is RL(Q)?
Explanation:
The first HI is 150.000 + 1.240 = 151.240 m, giving RL(A) = 149.390 m and RL(CP1) = 150.285 m. The second HI is 150.285 + 1.335 = 151.620 m, giving RL(P) = 150.855 m and RL(CP2) = 149.515 m. The third HI is 149.515 + 0.925 = 150.440 m. Therefore RL(Q) = 150.440 - 1.415 = 149.025 m. The endpoint RL is separately 150.440 - 1.085 = 149.355 m. The independent endpoint check gives sum BS 3.500 m minus sum FS 4.145 m = -0.645 m, equal to 149.355 - 150.000 m.
వివరణ:
Question 12
ప్రశ్న 12
In a correctly booked rise-and-fall reduction, the successive changes between distinct points are: fall 0.610 m, rise 0.895 m, rise 0.570 m, fall 1.340 m, fall 0.490 m and rise 0.330 m. What are the sums of rises and falls?
Explanation:
The rises are 0.895 m, 0.570 m and 0.330 m. Their total is 1.795 m. The falls are 0.610 m, 1.340 m and 0.490 m. Their total is 2.440 m. The rise-minus-fall check therefore gives 1.795 - 2.440 = -0.645 m. This negative result means that the last ground point is 0.645 m below the first.
వివరణ:
Question 13
ప్రశ్న 13
A three-setup run starts at RL 150.000 m and has sum BS = 3.500 m, sum FS = 4.145 m, sum rises = 1.795 m and sum falls = 2.440 m. Which statement correctly completes BOTH arithmetic checks?
Explanation:
The BS-FS difference is 3.500 - 4.145 = -0.645 m. The rise-fall difference is 1.795 - 2.440 = -0.645 m. Both independently calculated differences agree. The final RL must be 150.000 - 0.645 = 149.355 m. Hence the third equality, final RL minus initial RL, is also -0.645 m. This establishes the standard arithmetic agreement of the completed level book, not proof against field-observation errors.
వివరణ:
Question 14
ప్రశ్న 14
A completed levelling run begins at RL 92.650 m and ends at RL 91.965 m. If the sum of backsights is 4.125 m, what must be the sum of foresights for an arithmetically consistent book?
Explanation:
The change in RL is 91.965 - 92.650 = -0.685 m. The level-book relation is sum BS - sum FS = final RL - initial RL. Therefore 4.125 - sum FS = -0.685 m. Rearrangement gives sum FS = 4.125 + 0.685 = 4.810 m. A larger foresight total is consistent with an overall fall in elevation. Intermediate sight readings do not enter this endpoint sum.
వివరణ:
Question 15
ప్రశ్న 15
In a levelling book the starting RL is 125.250 m, final RL is 126.025 m and sum FS is 2.980 m. What is the required sum BS?
Explanation:
The final point is higher than the initial point by 126.025 - 125.250 = 0.775 m. The levelling arithmetic relation requires sum BS - sum FS = +0.775 m. With sum FS = 2.980 m, sum BS must be 2.980 + 0.775 m. This gives sum BS = 3.755 m. Checking the result, 3.755 - 2.980 = 0.775 m. The calculation does not require the intervening intermediate sights.
వివరణ:
Question 16
ప్రశ్న 16
For the HEIGHT-OF-INSTRUMENT method, which standard arithmetic relation checks the endpoint reduction of a properly booked levelling run?
Explanation:
Every new backsight increases the new line-of-sight elevation relative to its reference point. Every terminating foresight determines the RL of the change point or final point by subtraction. On linking setups, the intervening change-point terms cancel algebraically. Thus sum BS minus sum FS equals the difference between final and initial RL. Intermediate sights do not appear in this telescoping endpoint relation. Agreement does not, by itself, verify every intermediate-point reduction.
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Question 17
ప్రశ్న 17
Which identity is the COMPLETE standard arithmetic check used in a consistently computed RISE-AND-FALL level book?
Explanation:
Each rise contributes a positive height difference between consecutive ground points. Each fall contributes a negative height difference. Their signed sum telescopes to last RL minus first RL. When the same point is used across instrument changes, the signed height difference also equals the total backsights minus the total foresights. Therefore all three quantities must agree for a correctly reduced book. The check supports the arithmetic but cannot certify the actual quality of observations.
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Question 18
ప్రశ్న 18
For a stable change point used between two levelling setups, what is the correct treatment in a rise-and-fall level book?
Explanation:
A change point is one physical ground point observed from two instrument positions. The old setup ends with a foresight on it and the new setup begins with a backsight on the same point. Its reduced level does not change merely because the instrument is moved. The two sightings therefore do not represent a rise or fall between distinct points. A rise or fall may be computed by comparing readings within each individual setup. Keeping one RL for the change point prevents double-counting and invalid differences.
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Question 19
ప్రశ్న 19
A level-book arithmetic check balances perfectly, but the staff was tilted during one observation. What conclusion is justified?
Explanation:
An endpoint or rise-and-fall equality is an arithmetic consistency test. It does not measure whether an observed staff value was accurate. A tilted staff can give an erroneous reading even when the recorded numbers are reduced flawlessly. An inaccurate observation may propagate through the book while all arithmetic identities remain satisfied. Independent control, good field procedure and checks of the observations are still necessary. The balance alone cannot certify survey accuracy.
వివరణ:
Question 20
ప్రశ్న 20
Which statement MOST ACCURATELY compares the internal checking capability of the two standard RL-reduction methods?
Explanation:
In the HI method, each intermediate RL is found directly from one HI and its staff reading. The usual BS-minus-FS endpoint check does not include intermediate sights. The rise-and-fall method forms consecutive height differences and propagates the RL through each point, offering additional checks on intermediate reductions. Its rise-minus-fall total must agree with the endpoint change and the BS-minus-FS total. These checks are about correct booking and calculation rather than guaranteed correctness of field readings. Neither method substitutes for independent verification when measurement errors are suspected.
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Question 21
ప్రశ్న 21
For a water-main longitudinal profile with MANY intermediate ground points observed from each instrument position, which statement best describes the usual choice of reduction method?
Explanation:
Once HI is established, each intermediate RL follows from one subtraction. This makes the HI method efficient for many profile or cross-section points. The rise-and-fall method instead compares successive readings and records each height change. That method is more laborious but makes the intervening RL progression explicit. For correctly observed and reduced data, both methods yield the same elevations relative to the same datum. The choice concerns workflow and internal checking, not a difference in the underlying elevations.
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Question 22
ప్రశ్న 22
One intermediate-sight reading was MISREAD in the field, but that wrong numerical value was booked and every RL, rise and fall was recomputed consistently from it. What can be said about the standard arithmetic checks?
Explanation:
The two methods operate on the values actually entered in the level book. If an intermediate observation is wrong but all reductions use it consistently, the arithmetic can remain self-consistent. In the rise-and-fall method, that intermediate value affects adjacent differences, often with offsetting effects in the total. The BS-minus-FS endpoint sum does not contain intermediate sights. Therefore even the triple equality may hold despite the misreading. A repeat observation or independent check is needed to establish observational correctness.
వివరణ:
Question 23
ప్రశ్న 23
A run closes on a benchmark whose established RL is 98.500 m. The reduced RL calculated from the new observations is 98.482 m, and the level-book arithmetic check balances. What is the closing misclosure defined as CALCULATED minus ESTABLISHED RL?
Explanation:
The independently established benchmark RL is 98.500 m. The levelling result for the same benchmark is 98.482 m. Misclosure on the stated sign convention is 98.482 - 98.500 = -0.018 m. A perfect arithmetic balance only means that the booked readings were reduced consistently. It does not force agreement with independent benchmark control. The significance of an 18 mm misclosure depends on the applicable accuracy or closure tolerance.
వివరణ:
Question 24
ప్రశ్న 24
From one setup, point P has RL 132.640 m and an IS reading of 1.920 m. Point Q, observed without moving the instrument, has an IS of 0.875 m. What is RL(Q)?
Explanation:
The common HI can be reconstructed from P as 132.640 + 1.920 = 134.560 m. Q has RL equal to HI minus its staff reading. Therefore RL(Q) = 134.560 - 0.875 = 133.685 m. An independent rise-and-fall calculation gives 1.920 - 0.875 = 1.045 m rise from P to Q. Adding 1.045 m to 132.640 m gives the same result, 133.685 m. Both calculations require the setup to remain undisturbed between the two readings.
వివరణ:
Question 25
ప్రశ్న 25
A two-setup book starts at BM RL 95.000 m. Setup 1: BS BM 1.280, IS P 1.870, FS CP 1.540. Setup 2: BS CP 1.050, IS Q 0.820, FS E 1.690 (m). A clerk recorded RLs P = 94.510 m, CP = 94.740 m, Q = 94.970 m and E = 94.100 m. Which recorded RL is INCORRECT even though the BS-FS endpoint check balances?
Explanation:
The first HI is 95.000 + 1.280 = 96.280 m. Thus the correct intermediate RL(P) is 96.280 - 1.870 = 94.410 m, not the recorded 94.510 m. The first CP RL is 96.280 - 1.540 = 94.740 m, while HI2 is 94.740 + 1.050 = 95.790 m. Therefore RL(Q) = 95.790 - 0.820 = 94.970 m and RL(E) = 95.790 - 1.690 = 94.100 m, both correctly recorded. The endpoint check gives sum BS 2.330 minus sum FS 3.230 = -0.900 m, equal to 94.100 - 95.000 m, so it misses the clerical intermediate RL error. Checking the BM-to-P fall of 1.870 - 1.280 = 0.590 m independently exposes the incorrect P entry.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. Add the backsight to the benchmark RL. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: B. 105.650 m. సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: D. 104.425 m. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: A. Rise of 0.725 m. సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: C. Fall = 1.275 m. సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: B. P = 109.285 m; CP = 110.145 m. సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: D. 120.315 m. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: C. 78.825 m. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. Rise 0.270 m, fall 1.695 m, net -1.425 m. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. Fall of 0.560 m. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: A. 149.025 m. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: D. Rise 1.795 m; fall 2.440 m. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: B. End RL = 149.355 m; both differences are -0.645 m. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. 4.810 m. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: D. 3.755 m. సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: A. Sum BS - sum FS = last RL - first RL. సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: C. Sum BS - sum FS = sum rise - sum fall = last RL - first RL. సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: B. Give the change point one RL, book the old FS and new BS against it, and compute no rise or fall between those two sightings. సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: A. The arithmetic is internally consistent, but the balance alone does not establish observational accuracy. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. Rise-and-fall makes consecutive intermediate RL computations more directly checkable; the usual HI endpoint check does not independently check each intermediate RL. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. The HI method is often faster for numerous intermediate points, whereas rise-and-fall is more detailed for checking successive RL reductions. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. The checks can still balance because algebraic consistency does not establish that the observed staff value was true. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: B. -0.018 m. సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: D. 133.685 m. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: A. RL(P) = 94.510 m. సమాధానం: A.
