Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
A simple rectangular building has external plan dimensions 14.60 m by 9.40 m and uniform external wall thickness 0.30 m. There are no internal walls. What is the centreline perimeter of the closed wall circuit?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For opposite walls of uniform thickness, the centreline dimensions are the external dimensions less one wall thickness. The centreline length dimension is 14.60 - 0.30 = 14.30 m. The centreline width dimension is 9.40 - 0.30 = 9.10 m. Their closed perimeter is 2(14.30 + 9.10) = 46.80 m. The four ordinary corners require no separate junction deductions for this uniform closed circuit.
వివరణ:
Question 2
ప్రశ్న 2
A rectangular building has external dimensions 11.23 m by 7.23 m, uniform external masonry thickness 0.23 m and masonry height 3.20 m. There are no internal walls, openings or projections. What masonry volume is obtained by the centreline method before rounding?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The opposite wall centrelines are separated by 11.23 - 0.23 = 11.00 m and 7.23 - 0.23 = 7.00 m. Hence the closed centreline perimeter is 2(11.00 + 7.00) = 36.00 m. The wall has uniform thickness of 0.23 m and height of 3.20 m. Its masonry volume is 36.00 x 0.23 x 3.20 = 26.496 m3. Since no openings or branches occur, no further corrections are required.
వివరణ:
Question 3
ప్రశ్న 3
A building has one closed wall circuit of centreline perimeter 38.40 m and no internal walls. A foundation trench 0.90 m wide and 1.25 m deep is symmetrically centred on that circuit. With no additional excavation or deductions, what in-situ excavation quantity is required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The closed foundation trench follows the given 38.40 m centreline perimeter. Its width is 0.90 m and its excavation depth is 1.25 m. For a uniform centred strip without branches, volume equals centreline length times width times depth. The volume is therefore 38.40 x 0.90 x 1.25 = 43.20 m3. No separate correction is needed for the corners of this uniform closed circuit.
వివరణ:
Question 4
ప్రశ్న 4
An external closed foundation circuit has centreline perimeter 42.00 m. One internal foundation strip is booked for 5.60 m BETWEEN the centrelines of two opposite external strips and creates TWO T-junctions. Every strip at excavation level is 0.80 m wide and 1.10 m deep. What is the net excavation quantity, assuming no other intersections?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The uncorrected sum of centreline lengths is 42.00 + 5.60 = 47.60 m. Each T-junction requires a correction of half the excavation strip width. The total correction is 2(0.80/2) = 0.80 m. The effective centreline length is 46.80 m, and volume equals 46.80 x 0.80 x 1.10. The net excavation quantity is 41.184 m3.
వివరణ:
Question 5
ప్రశ్న 5
A closed external foundation circuit has a centreline perimeter of 42.00 m, and one internal strip is booked for 5.60 m BETWEEN opposite external strip centrelines, creating TWO T-junctions. A PCC bed of width 0.60 m and thickness 0.15 m is placed under all these strips, without other overlaps. What is the total PCC volume?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The uncorrected centreline sum is 42.00 + 5.60 = 47.60 m. At PCC level each T-junction requires deduction of half the PCC width, or 0.30 m. Two junctions give a combined deduction of 0.60 m. The effective PCC length is therefore 47.00 m. Multiplying 47.00 x 0.60 x 0.15 gives 4.230 m3.
వివరణ:
Question 6
ప్రశ్న 6
Two vertically distinct masonry footing courses share a wall-axis layout whose UNCORRECTED total centreline length is 47.60 m, with exactly TWO T-junctions and no other intersections. The lower course is 0.70 m wide and 0.30 m high; the upper course is 0.50 m wide and 0.25 m high. Both widths are centred on the axes. What is their combined masonry volume before rounding?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The two footing courses have different widths and therefore require separate junction corrections. For the lower course, effective length is 47.60 - 0.70 = 46.90 m and volume is 46.90 x 0.70 x 0.30 = 9.849 m3. For the upper course, effective length is 47.60 - 0.50 = 47.10 m and volume is 47.10 x 0.50 x 0.25 = 5.8875 m3. The footing courses occupy different vertical levels, so their volumes are added. The combined volume is 9.849 + 5.8875 = 15.7365 m3.
వివరణ:
Question 7
ప్రశ్న 7
A simple rectangular building has external dimensions 10.50 m by 8.50 m, uniform external wall thickness 0.25 m, and superstructure height 2.80 m. It contains no partitions or openings. What superstructure masonry volume follows from the centreline method?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The centreline dimensions are 10.50 - 0.25 = 10.25 m and 8.50 - 0.25 = 8.25 m. Their closed centreline perimeter is 2(10.25 + 8.25) = 37.00 m. The masonry thickness is 0.25 m. Multiplying the perimeter by thickness and 2.80 m height gives 37.00 x 0.25 x 2.80 = 25.90 m3. There are no opening or branch corrections under the stated geometry.
వివరణ:
Question 8
ప్రశ్న 8
A uniform plinth-wall layout has an external closed-circuit centreline perimeter of 36.80 m and a cross wall booked for 4.20 m BETWEEN opposite external wall centrelines, creating TWO T-junctions. All plinth walls are 0.23 m wide and 0.60 m high. What is their combined masonry volume before rounding?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The raw total centreline length is 36.80 + 4.20 = 41.00 m. Two T-junctions require a combined deduction equal to one 0.23 m wall breadth. The effective plinth-wall length is 41.00 - 0.23 = 40.77 m. Its cross-sectional area is 0.23 x 0.60 = 0.138 m2. The masonry volume is 40.77 x 0.138 = 5.62626 m3.
వివరణ:
Question 9
ప్రశ్న 9
The centreline perimeter of a closed external superstructure wall is 42.00 m. The wall is 0.23 m thick and 3.20 m high, and it has one door 1.00 m by 2.10 m and two windows each 1.20 m by 1.50 m. For this problem, deduct the FULL through-wall volumes of all openings and make no other adjustments. What is the net masonry volume?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The gross wall volume is 42.00 x 0.23 x 3.20 = 30.912 m3. The total opening area is 1.00 x 2.10 + 2(1.20 x 1.50) = 5.70 m2. Each opening passes through the complete 0.23 m wall thickness. Their specified total volume deduction is 5.70 x 0.23 = 1.311 m3. The net wall quantity is 30.912 - 1.311 = 29.601 m3.
వివరణ:
Question 10
ప్రశ్న 10
For a uniform masonry layer, the UNCORRECTED sum of wall-centreline lengths is 53.20 m. Its plan has THREE independent T-junctions and ONE independent full plus-shaped crossing; no junction regions overlap. All strips are 0.40 m wide and 0.80 m high. What masonry volume results?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Each T-junction requires a half-breadth correction of 0.20 m. The three T-junctions therefore account for 0.60 m of length deduction. The full crossing requires one additional full-breadth deduction of 0.40 m. The corrected centreline length is 53.20 - 0.60 - 0.40 = 52.20 m. The masonry volume is 52.20 x 0.40 x 0.80 = 16.704 m3.
వివరణ:
Question 11
ప్రశ్న 11
A 0.12 m thick internal partition meets a continuous 0.30 m thick main wall at right angles. The partition centreline is booked for 5.00 m UP TO the main-wall centreline, and the main wall has already been measured completely. Both walls are 3.00 m high with no openings. What ADDITIONAL partition volume should be booked?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The main wall has already been included in the estimate. The branch partition overlaps that wall by half of the main-wall thickness. The overlap length is 0.30/2 = 0.15 m, giving a net branch length of 4.85 m. The partition cross-sectional area is 0.12 x 3.00 = 0.36 m2. Its additional volume is 4.85 x 0.36 = 1.746 m3.
వివరణ:
Question 12
ప్రశ్న 12
Two parallel continuous main walls are each 0.35 m thick and have already been measured completely. A perpendicular internal partition, 0.15 m thick and 3.00 m high, is booked for 7.00 m FROM ONE main-wall centreline TO THE OTHER. What ADDITIONAL net partition masonry volume is required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The internal partition reaches the centreline of each already measured main wall. At each end its overlap length is half of 0.35 m, or 0.175 m. The two overlaps total 0.35 m, leaving an additional length of 7.00 - 0.35 = 6.65 m. The partition has a cross-sectional area of 0.15 x 3.00 = 0.45 m2. Its additional masonry volume is 6.65 x 0.45 = 2.9925 m3.
వివరణ:
Question 13
ప్రశ్న 13
At a footing-course level, a continuous main strip 0.80 m wide has already been fully measured. A perpendicular branch footing strip is 0.50 m wide, 0.25 m high and booked for 4.80 m UP TO the main-strip centreline. Both strips are centred on their axes, and there are no other intersections. What ADDITIONAL branch-footing volume is required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Because the continuous footing has already been measured, its overlapping portion must be excluded from the additional branch volume. The booked branch terminates at the main-strip centreline, so the overlap length is 0.80/2 = 0.40 m. The additional branch length is 4.80 - 0.40 = 4.40 m. Its cross-sectional area is 0.50 x 0.25 = 0.125 m2. The additional volume is 4.40 x 0.125 = 0.550 m3.
వివరణ:
Question 14
ప్రశ్న 14
A damp-proof course is payable in square metres over a uniform 0.25 m wide plinth-wall layout. The external closed circuit has centreline perimeter 37.50 m, and an internal wall booked for 5.50 m BETWEEN opposite external wall centrelines forms TWO T-junctions. All DPC strips are 0.25 m wide, with no openings or other deductions. What DPC area should be measured?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The uncorrected sum of centrelines is 37.50 + 5.50 = 43.00 m. The two T-junctions require a combined correction equal to one full DPC strip breadth of 0.25 m. The effective length is consequently 43.00 - 0.25 = 42.75 m. DPC quantity under the specified item is a plan area obtained from length multiplied by breadth. The measured area is 42.75 x 0.25 = 10.6875 m2.
వివరణ:
Question 15
ప్రశ్న 15
A closed uniform foundation circuit has centreline perimeter 44.00 m and NO internal branches. Vertically separate courses comprise a PCC bed 0.90 m wide and 0.15 m thick, a lower masonry footing 0.70 m wide and 0.30 m high, and an upper masonry footing 0.50 m wide and 0.25 m high. What are the PCC quantity and the TOTAL footing-masonry quantity respectively?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The closed centreline circuit has no T-junctions, so its 44.00 m perimeter applies to the centred courses. PCC volume is 44.00 x 0.90 x 0.15 = 5.940 m3. Lower masonry footing volume is 44.00 x 0.70 x 0.30 = 9.240 m3. Upper masonry footing volume is 44.00 x 0.50 x 0.25 = 5.500 m3. Total footing masonry is therefore 9.240 + 5.500 = 14.740 m3.
వివరణ:
Question 16
ప్రశ్న 16
A closed rectangular external wall circuit has OUTSIDE dimensions 12.40 m by 8.40 m, uniform wall thickness 0.40 m and no internal partitions. A symmetrically centred foundation excavation strip is 1.10 m wide and 0.90 m deep along the same axes. What is the in-situ excavation volume?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The centreline dimensions of the external walls are 12.40 - 0.40 = 12.00 m and 8.40 - 0.40 = 8.00 m. The closed centreline perimeter is 2(12.00 + 8.00) = 40.00 m. The wider trench remains centred on the same axis because the excavation is symmetrical. Its volume is 40.00 x 1.10 x 0.90 = 39.600 m3. No deductions at ordinary L-corners are needed for the uniform closed strip.
వివరణ:
Question 17
ప్రశ్న 17
A circular masonry ring has CENTRELINE radius 3.50 m, uniform wall thickness 0.30 m and height 2.20 m, without openings. Taking pi = 3.1416, what is the wall volume to two decimal places?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The ring centreline circumference is 2 times pi times its centreline radius. It equals 2 x 3.1416 x 3.50 = 21.9912 m. The uniform ring has cross-sectional area 0.30 x 2.20 = 0.66 m2. Its volume is therefore 21.9912 x 0.66 = 14.514192 m3. Reporting this value to two decimal places gives 14.51 m3.
వివరణ:
Question 18
ప్రశ్న 18
A rectangular closed-wall building has OUTSIDE dimensions 10.23 m by 7.23 m and uniform wall thickness 0.23 m, with no internal partitions. A centred trench 1.20 m wide and 1.00 m deep contains a separate centred PCC bed 0.80 m wide and 0.20 m thick along the same closed axis. What is the NUMERICAL DIFFERENCE between the trench excavation volume and PCC volume, without assuming it is all backfill?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The closed centreline dimensions are 10.23 - 0.23 = 10.00 m and 7.23 - 0.23 = 7.00 m. The centreline perimeter is 2(10.00 + 7.00) = 34.00 m. Trench excavation is 34.00 x 1.20 x 1.00 = 40.80 m3. PCC quantity is 34.00 x 0.80 x 0.20 = 5.44 m3. The numerical difference is 40.80 - 5.44 = 35.36 m3 and is not automatically the backfilling quantity.
వివరణ:
Question 19
ప్రశ్న 19
A perpendicular branch joins an already measured continuous main strip at ONE T-junction. Its axis is booked for 6.00 m UP TO the main-strip centreline at every level. At trench level both strips are 0.80 m wide and the excavation is 1.00 m deep; at PCC level both strips are 0.60 m wide and the PCC is 0.20 m thick. What ADDITIONAL branch excavation and PCC quantities are required respectively?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The main strip is already measured, so a half-width overlap must be excluded from the branch at each level. At trench level the net branch length is 6.00 - 0.80/2 = 5.60 m. Its excavation is 5.60 x 0.80 x 1.00 = 4.480 m3. At PCC level the net branch length is 6.00 - 0.60/2 = 5.70 m. The additional PCC volume is 5.70 x 0.60 x 0.20 = 0.684 m3.
వివరణ:
Question 20
ప్రశ్న 20
Two equal-width masonry wall runs cross at ONE complete plus-shaped intersection. Their UNCORRECTED sum of centreline lengths is 49.50 m, and no other intersections occur. Both have thickness 0.45 m and height 0.60 m. What is their combined net masonry volume before rounding?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At a full crossing of equal-width strips, the common square area must be counted only once. This gives an effective-length correction equal to one full strip breadth. The net centreline length is 49.50 - 0.45 = 49.05 m. The wall cross-sectional area is 0.45 x 0.60 = 0.27 m2. The net masonry quantity is 49.05 x 0.27 = 13.2435 m3.
వివరణ:
Question 21
ప్రశ్న 21
An external closed superstructure circuit has centreline perimeter 35.50 m. One internal wall is booked for 5.00 m BETWEEN opposite external wall centrelines, forming TWO T-junctions. Every wall is 0.23 m thick and 3.00 m high. The external walls contain one door 1.00 m by 2.10 m and two windows 1.20 m by 1.50 m each, all through-wall; deduct ALL opening volumes in full and make no other adjustments. What is the net masonry volume?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The uncorrected centreline sum is 35.50 + 5.00 = 40.50 m. The two T-junctions require a 0.23 m total correction, leaving 40.27 m. Gross masonry is 40.27 x 0.23 x 3.00 = 27.7863 m3. The stated openings total 2.10 + 2(1.20 x 1.50) = 5.70 m2 and displace 5.70 x 0.23 = 1.311 m3. Hence the net masonry is 27.7863 - 1.311 = 26.4753 m3.
వివరణ:
Question 22
ప్రశ్న 22
A continuous RCC ring beam of rectangular section 0.23 m wide by 0.30 m deep follows a CLOSED uniform wall circuit with centreline perimeter 43.60 m. It contains no intersecting beam branches, openings or overlapping concrete with other items. What is its concrete quantity before rounding?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The ring beam follows a closed geometric centreline whose perimeter is given as 43.60 m. Its rectangular cross-sectional area is 0.23 x 0.30 = 0.069 m2. A uniform closed strip is measured using the centreline perimeter without separate ordinary-corner corrections. Multiplying 43.60 by 0.069 gives 3.0084 m3. No other concrete volume is subtracted because the question excludes intersections and overlaps.
వివరణ:
Question 23
ప్రశ్న 23
For a closed wall circuit WITHOUT internal branches, the common geometric centreline perimeter is 34.00 m. Excavation is 0.90 m wide and 1.00 m deep at Rs 900/m3; PCC is a separate 0.90 m wide and 0.15 m thick layer at Rs 6,500/m3; superstructure masonry is 0.23 m thick and 3.00 m high at Rs 6,000/m3. With no openings or other allowances, what is the combined estimated cost?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The excavation quantity is 34.00 x 0.90 x 1.00 = 30.60 m3, costing Rs 27,540. The PCC quantity is 34.00 x 0.90 x 0.15 = 4.59 m3, costing Rs 29,835. The superstructure masonry quantity is 34.00 x 0.23 x 3.00 = 23.46 m3, costing Rs 1,40,760. The three layers are separately measured and priced with compatible units. Their combined estimated cost is Rs 27,540 + Rs 29,835 + Rs 1,40,760 = Rs 1,98,135.
వివరణ:
Question 24
ప్రశ్న 24
A rectangular building has OUTSIDE dimensions 13.60 m by 9.60 m and uniform 0.30 m thick external walls. One internal wall, also 0.30 m thick, spans in the 9.60 m direction BETWEEN the two corresponding external-wall centrelines, creating TWO T-junctions. All walls are 2.70 m high and have no openings. What is the combined superstructure masonry volume?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The external wall centreline dimensions are 13.60 - 0.30 = 13.30 m and 9.60 - 0.30 = 9.30 m. Hence the external circuit has centreline perimeter 2(13.30 + 9.30) = 45.20 m. The internal wall is booked for 9.30 m, so the uncorrected total is 54.50 m. The two T-junctions require a 0.30 m correction, giving effective length 54.20 m. The masonry quantity is 54.20 x 0.30 x 2.70 = 43.902 m3.
వివరణ:
Question 25
ప్రశ్న 25
A wall-axis layout comprises one external CLOSED circuit of centreline perimeter 40.00 m and one internal wall booked for 6.00 m BETWEEN opposite external-wall centrelines, forming exactly TWO T-junctions. At each layer all strips have the same stated width and are symmetrically centred on these axes. Separate vertical masonry layers are: lower foundation course 0.60 m wide by 0.40 m high; plinth course 0.30 m wide by 0.70 m high; superstructure 0.30 m thick by 3.00 m high. ONLY the superstructure has a full-volume deduction for one 1.00 m by 2.10 m door and two 1.50 m by 1.20 m windows, all through the 0.30 m wall. What is the COMBINED net masonry volume of the three courses before rounding?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The uncorrected centreline sum is 40.00 + 6.00 = 46.00 m. At foundation level, correcting two T-junctions gives length 45.40 m and masonry volume 45.40 x 0.60 x 0.40 = 10.896 m3. The plinth length is 45.70 m and its volume is 45.70 x 0.30 x 0.70 = 9.597 m3. The superstructure gross volume is 45.70 x 0.30 x 3.00 = 41.130 m3, and the fully deducted opening volume is [1.00 x 2.10 + 2(1.50 x 1.20)] x 0.30 = 1.710 m3. The combined net masonry is 10.896 + 9.597 + (41.130 - 1.710) = 59.913 m3.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. 46.80 m. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: A. 26.496 m3. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: D. 43.20 m3. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: B. 41.184 m3. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: C. 4.230 m3. సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: A. 15.7365 m3. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: D. 25.90 m3. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: B. 5.62626 m3. సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: A. 29.601 m3. సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: C. 16.704 m3. సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: D. 1.746 m3. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: B. 2.9925 m3. సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: A. 0.550 m3. సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: D. 10.6875 m2. సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: C. PCC 5.940 m3; masonry 14.740 m3. సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: B. 39.600 m3. సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: A. 14.51 m3. సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: D. 35.36 m3. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: C. Excavation 4.480 m3; PCC 0.684 m3. సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: B. 13.2435 m3. సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: D. 26.4753 m3. సమాధానం: D.
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Question 22 ప్రశ్న 22Answer: A. 3.0084 m3. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: C. Rs 1,98,135. సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: B. 43.902 m3. సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: A. 59.913 m3. సమాధానం: A.