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Question 1
ప్రశ్న 1
For M35 concrete, take the assumed standard deviation S = 5.0 N/mm² and the IS 10262:2019 grade factor X = 6.5 N/mm². What target mean compressive strength should be adopted?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 10262:2019 requires the target mean strength to be the higher of fck + 1.65S and fck + X. For M35 concrete, the statistical expression gives 35 + 1.65 × 5 = 43.25 N/mm². The grade-factor expression gives 35 + 6.5 = 41.50 N/mm². The higher value is therefore 43.25 N/mm². Hence Option C is the correct target mean strength for proportioning.
వివరణ:
Question 2
ప్రశ్న 2
For M50 concrete, the assumed standard deviation for good site control is 5.0 N/mm². If site control is fair, increase S by 1.0 N/mm² as prescribed, and take X = 6.5 N/mm². What target mean strength governs?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For fair site control, the assumed standard deviation is increased from 5.0 to 6.0 N/mm². The statistical target is 50 + 1.65 × 6 = 59.90 N/mm². The alternative grade-factor target is 50 + 6.5 = 56.50 N/mm². IS 10262:2019 requires the higher of these two results to be adopted. Therefore the governing target mean strength is 59.90 N/mm² and Option C is correct.
వివరణ:
Question 3
ప్రశ్న 3
For M20 concrete, an adequate production record justifies an established standard deviation of 3.0 N/mm². If X = 5.5 N/mm², what target mean strength is required by the IS 10262:2019 dual-check rule?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Using the established standard deviation, fck + 1.65S equals 20 + 1.65 × 3 = 24.95 N/mm². The second expression gives fck + X = 20 + 5.5 = 25.50 N/mm². The larger value must govern even though it does not arise from the statistical expression. Thus the required target mean strength is 25.50 N/mm². Option B is therefore the unique correct answer.
వివరణ:
Question 4
ప్రశ్న 4
For 20 mm nominal maximum size angular aggregate, take the reference water content as 186 kg/m³ at 50 mm slump. Using a 3 percent increase for each additional 25 mm slump, what water content is estimated for a 75 mm slump before admixture correction?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The slump is increased from 50 mm to 75 mm, which is one increment of 25 mm. The corresponding water-content increase is 3 percent of 186 kg/m³. This increase equals 5.58 kg/m³. The estimated water content is therefore 186 + 5.58 = 191.58 kg/m³. Hence Option B is correct before any water reduction due to an admixture is considered.
వివరణ:
Question 5
ప్రశ్న 5
The estimated water content for a mix is 191.58 kg/m³ before using a superplasticizer. Trial data justify a 22 percent water reduction at the selected dosage. What revised water content should be used?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A 22 percent reduction means that 78 percent of the original estimated water is retained. The revised water content is therefore 191.58 × 0.78. This gives 149.4324 kg/m³. Rounded to two decimal places, the value is 149.43 kg/m³. Option B is consequently the correct revised water content.
వివరణ:
Question 6
ప్రశ్న 6
A concrete mix requires 168 kg/m³ of free water and the adopted free water-cement ratio is 0.42. What cement content is required before minimum and maximum cement-content checks?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The free water-cement ratio is the mass of free water divided by the mass of cement. Rearranging gives cement content = water content ÷ water-cement ratio. Substitution gives 168 ÷ 0.42 = 400 kg/m³. This is the calculated strength-based cement content before durability and maximum-content checks are applied. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
A strength relationship suggests a free water-cement ratio of 0.47, while the applicable durability requirement limits it to 0.45. If the selected free water content is 171 kg/m³, which adopted ratio and corresponding cement content are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The strength-based ratio must be checked against the durability limit, and the lower value governs. Therefore the adopted free water-cement ratio is 0.45 rather than 0.47. The required cement content is 171 ÷ 0.45 = 380 kg/m³. Using 0.47 would violate the stated durability limit even though it came from the strength relationship. Hence Option B correctly combines the governing ratio and cement quantity.
వివరణ:
Question 8
ప్రశ్న 8
A mix has 160 kg/m³ of free water and an adopted water-cement ratio of 0.50, giving a calculated cement content of 320 kg/m³. If the applicable durability requirement specifies a minimum cement content of 340 kg/m³, what cement content and actual water-cement ratio should finally be used, assuming the water content is unchanged?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The strength-based calculation gives 160 ÷ 0.50 = 320 kg/m³ of cement. The stated durability requirement is more stringent because it requires at least 340 kg/m³. Therefore 340 kg/m³ must be adopted while the free water remains 160 kg/m³. The resulting actual water-cement ratio is 160 ÷ 340 = 0.4706, or approximately 0.471. Option B is thus the correct final combination.
వివరణ:
Question 9
ప్రశ్న 9
For 1 m³ of concrete, take cement = 380 kg with specific gravity 3.15, free water = 171 kg, chemical admixture = 3.8 kg with specific gravity 1.20, and entrapped air = 1.0 percent by volume. Approximately what absolute volume remains for total aggregate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The absolute volume of cement is 380/(3.15 × 1000) = about 0.1206 m³. The water occupies 0.1710 m³ and the admixture occupies 3.8/(1.20 × 1000) = about 0.00317 m³. Entrapped air occupies 0.0100 m³ of the one-cubic-metre volume. The aggregate volume is therefore 1 − 0.1206 − 0.1710 − 0.00317 − 0.0100 = about 0.6952 m³. The nearest option is 0.695 m³, so Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
A mix has a total aggregate absolute volume of 0.6952 m³ per cubic metre of concrete. If the coarse-aggregate fraction is 0.64, with specific gravities 2.72 for coarse aggregate and 2.65 for fine aggregate, what are the approximate aggregate masses?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The coarse-aggregate absolute volume is 0.6952 × 0.64 = 0.4449 m³ approximately. Its mass is therefore about 0.4449 × 2.72 × 1000 = 1,210 kg/m³. The fine-aggregate absolute volume is 0.6952 × 0.36 = 0.2503 m³ approximately. Its mass is about 0.2503 × 2.65 × 1000 = 663 kg/m³. Thus Option A gives the correct pair of aggregate masses to the stated precision.
వివరణ:
Question 11
ప్రశ్న 11
A chemical admixture is dosed at 1.0 percent by mass of a 420 kg/m³ cementitious-material content. If the admixture specific gravity is 1.15, what are its mass and absolute volume per cubic metre of concrete?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A dosage of 1.0 percent of 420 kg/m³ gives an admixture mass of 4.20 kg/m³. Absolute volume equals mass divided by specific gravity times the density of water. Therefore the volume is 4.20/(1.15 × 1000) = about 0.00365 m³. This volume must be included in the absolute-volume balance when calculating aggregate quantities. Hence Option A correctly states both the mass and volume.
వివరణ:
Question 12
ప్రశ్న 12
A concrete mix is to contain 450 kg/m³ of total cementitious material, of which fly ash is 20 percent by mass. If the selected water-cementitious materials ratio is 0.38, what are the OPC, fly ash and free-water quantities?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Twenty percent of 450 kg/m³ is 90 kg/m³, so that is the fly-ash content. The remaining 80 percent is OPC, equal to 360 kg/m³. At a water-cementitious materials ratio of 0.38, the free-water content is 0.38 × 450 = 171 kg/m³. The three required quantities are therefore 360 kg/m³ OPC, 90 kg/m³ fly ash and 171 kg/m³ water. Option C is the only combination satisfying both the replacement percentage and the selected ratio.
వివరణ:
Question 13
ప్రశ్న 13
A preliminary mix uses 141 kg/m³ water at w/cm = 0.29, giving the initial cementitious-material content. Based on trial experience, the designer increases that content by 10 percent and then uses 15 percent fly ash and 5 percent silica fume by mass of the revised total. Approximately how much OPC remains?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The initial cementitious-material content is 141 ÷ 0.29 = about 486.21 kg/m³. A trial-based 10 percent increase gives 486.21 × 1.10 = about 534.83 kg/m³. Fly ash and silica fume together constitute 20 percent of this revised total, leaving 80 percent as OPC. The OPC content is therefore 0.80 × 534.83 = about 427.86 kg/m³. Rounded suitably, this is 428 kg/m³, so Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
For 20 mm aggregate and Zone II fine aggregate, the reference coarse-aggregate fraction is 0.62 at w/c = 0.50. If the actual w/c is 0.38, and the fraction is increased by 0.01 for every 0.05 decrease in w/c, what corrected coarse-aggregate fraction results?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The water-cement ratio has decreased by 0.50 − 0.38 = 0.12. The specified adjustment rate is 0.01 in coarse-aggregate fraction for every 0.05 decrease in the ratio. Therefore the required increase is (0.12/0.05) × 0.01 = 0.024. Adding this to the reference value gives 0.62 + 0.024 = 0.644. Hence Option C is the correct corrected coarse-aggregate fraction.
వివరణ:
Question 15
ప్రశ్న 15
A corrected coarse-aggregate fraction is 0.644. For a pumpable mix, the designer adopts a full 10 percent reduction in this estimated coarse-aggregate fraction. What coarse- and fine-aggregate fractions should then be used?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A 10 percent reduction means retaining 90 percent of the corrected coarse-aggregate fraction. Thus the new coarse-aggregate fraction is 0.644 × 0.90 = 0.5796. The fine-aggregate fraction is the balance of the total aggregate volume, equal to 1 − 0.5796 = 0.4204. This adjustment increases the mortar fraction and can assist pumpability when it is technically justified. Therefore Option A gives the correct pair of fractions.
వివరణ:
Question 16
ప్రశ్న 16
For 1 m³ of concrete, take OPC = 360 kg with specific gravity 3.15, fly ash = 90 kg with specific gravity 2.20, water = 171 kg, liquid admixture = 4.5 kg with specific gravity 1.10, and entrapped air = 1 percent. Approximately what volume remains for total aggregate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The OPC volume is 360/(3.15 × 1000) = about 0.1143 m³. The fly-ash volume is 90/(2.20 × 1000) = about 0.0409 m³, while water occupies 0.1710 m³ and admixture about 0.00409 m³. Entrapped air occupies 0.0100 m³. The balance is 1 − 0.1143 − 0.0409 − 0.1710 − 0.00409 − 0.0100 = about 0.6597 m³. The nearest option is therefore 0.660 m³, making Option C correct.
వివరణ:
Question 17
ప్రశ్న 17
The total aggregate absolute volume in a mix is 0.670 m³, of which 62 percent is coarse aggregate. If the specific gravities of coarse and fine aggregate are 2.70 and 2.65 respectively, what are the approximate SSD aggregate masses?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The coarse-aggregate volume is 0.670 × 0.62 = 0.4154 m³. Multiplying by 2.70 × 1000 gives a coarse-aggregate mass of about 1,121.6 kg/m³. The fine-aggregate volume is 0.670 × 0.38 = 0.2546 m³. Multiplying by 2.65 × 1000 gives about 674.7 kg/m³ of fine aggregate. Rounded to whole kilograms, Option A gives the correct SSD masses.
వివరణ:
Question 18
ప్రశ్న 18
A mix design requires 1,200 kg/m³ of coarse aggregate in SSD condition. The aggregate absorption is 0.5 percent by mass of oven-dry aggregate, and the stock aggregate is completely dry. What dry aggregate mass should be batched and how much additional water is required for absorption?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
An SSD mass includes the water held within the aggregate at its absorption capacity. The oven-dry mass is therefore 1200/(1 + 0.005) = about 1,194.03 kg. The difference between the SSD and oven-dry masses is about 5.97 kg. Because the stock aggregate is completely dry, that amount of water will be absorbed and must be supplied in addition to the design free water. Hence Option A correctly gives both the batch mass and absorption-water correction.
వివరణ:
Question 19
ప్రశ్న 19
A mix requires 650 kg/m³ of fine aggregate at SSD condition. Its water absorption is 1.0 percent and its total moisture content is 4.0 percent. Following the IS 10262 illustrative adjustment approach, take free surface moisture as total moisture minus absorption. What wet fine-aggregate mass should be batched and how much water does it contribute above SSD condition?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The free surface moisture is 4.0 − 1.0 = 3.0 percent. Following the IS 10262 illustrative field-adjustment method, the wet batch mass is 650 × 1.03 = 669.5 kg. The water contributed above SSD condition is the difference between wet mass and SSD mass. That contribution is 669.5 − 650 = 19.5 kg. Therefore Option C correctly gives both the wet fine-aggregate mass and the water contribution.
వివరణ:
Question 20
ప్రశ్న 20
A mix is designed on an SSD basis with 1,200 kg/m³ coarse aggregate and 650 kg/m³ fine aggregate, and design free water of 180 kg/m³. Coarse aggregate has absorption 0.5 percent and total moisture 2.0 percent, while fine aggregate has absorption 1.0 percent and total moisture 5.0 percent. Using the IS 10262 illustrative adjustment method, approximately how much water should be added at the mixer?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The coarse aggregate has free surface moisture of 2.0 − 0.5 = 1.5 percent, so its wet mass is 1200 × 1.015 = 1218 kg and it contributes 18 kg of water above SSD. The fine aggregate has free surface moisture of 5.0 − 1.0 = 4.0 percent, so its wet mass is 650 × 1.04 = 676 kg and it contributes 26 kg. The total water contributed by the wet aggregates is therefore 18 + 26 = 44 kg/m³. This amount must be deducted from the design free water of 180 kg/m³. The separate mixer water is thus 180 − 44 = 136 kg/m³, making Option A correct.
వివరణ:
Question 21
ప్రశ్న 21
A mix requires SSD masses of 1,250 kg/m³ coarse aggregate and 700 kg/m³ fine aggregate. Both stock aggregates are completely dry, and their water absorptions are 0.8 percent and 1.2 percent respectively. Approximately how much extra water is required to bring the aggregates to SSD condition?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For completely dry aggregate, the dry batch mass is obtained from SSD mass divided by one plus the absorption fraction. The coarse-aggregate dry mass is 1250/1.008 = about 1240.08 kg, so it requires about 9.92 kg of absorption water. The fine-aggregate dry mass is 700/1.012 = about 691.70 kg, so it requires about 8.30 kg of absorption water. The total extra water required is therefore about 9.92 + 8.30 = 18.22 kg/m³. Hence Option C is correct.
వివరణ:
Question 22
ప్రశ్న 22
A mix contains 400 kg/m³ cement and is to remain exactly at a maximum effective water-cement ratio of 0.45. A liquid admixture dosage is 5 kg/m³ and the admixture contains 60 percent water by mass. Ignoring other water sources, how much separate mixing water may be added?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At a water-cement ratio of 0.45 with 400 kg/m³ cement, the maximum effective water is 0.45 × 400 = 180 kg/m³. The liquid admixture contributes 5 × 0.60 = 3 kg/m³ of water. Therefore only 180 − 3 = 177 kg/m³ should be supplied as separate mixing water. If 180 kg/m³ were added separately, the effective water would become 183 kg/m³ and the ratio would rise to 0.4575. Option B is therefore the correct separate-water quantity.
వివరణ:
Question 23
ప్రశ్న 23
A trial mix is proportioned per cubic metre as cement 400 kg, water 180 kg, fine aggregate 650 kg, coarse aggregate 1,200 kg and admixture 4 kg. Which set gives the corresponding quantities for a 0.05 m³ laboratory trial batch?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A 0.05 m³ batch is exactly 5 percent of the one-cubic-metre proportions. Cement required is 400 × 0.05 = 20 kg and water is 180 × 0.05 = 9 kg. Fine aggregate is 650 × 0.05 = 32.5 kg, while coarse aggregate is 1200 × 0.05 = 60 kg. The admixture quantity is 4 × 0.05 = 0.20 kg. Only Option D contains all five correctly scaled quantities.
వివరణ:
Question 24
ప్రశ్న 24
Trial Mix No. 1 contains 160 kg/m³ water and 400 kg/m³ cement, giving w/c = 0.40. To achieve the required workability, the water content must be increased by 8 kg/m³ while the preselected w/c ratio is maintained. How much additional cement is required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
After the workability adjustment, the total water content becomes 160 + 8 = 168 kg/m³. To maintain the preselected water-cement ratio of 0.40, the cement content must become 168/0.40 = 420 kg/m³. The original cement content was 400 kg/m³. Therefore the additional cement required is 420 − 400 = 20 kg/m³. Option C is the correct adjustment.
వివరణ:
Question 25
ప్రశ్న 25
Trial Mix No. 2 uses 168 kg/m³ water at a selected w/c ratio of 0.40. Trial Mixes No. 3 and No. 4 are prepared at approximately 10 percent below and 10 percent above this ratio while keeping the same water content. What cement contents correspond to w/c ratios of 0.36 and 0.44 respectively?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A 10 percent reduction from 0.40 gives a trial water-cement ratio of 0.36, while a 10 percent increase gives 0.44. With the water content fixed at 168 kg/m³, the cement content at 0.36 is 168/0.36 = 466.67 kg/m³. At a ratio of 0.44, the cement content is 168/0.44 = about 381.82 kg/m³. These two trial points bracket the selected ratio and help establish the strength relationship around the working mix. Therefore Option C gives the correct pair of cement contents.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. 43.25 N/mm² సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: C. 59.90 N/mm² సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: B. 25.50 N/mm² సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: B. 191.58 kg/m³ సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: B. 149.43 kg/m³ సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. 400 kg/m³ సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: B. 0.45 and 380.00 kg/m³ సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: B. 340 kg/m³ and approximately 0.471 సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: C. 0.695 m³ సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: A. Coarse aggregate 1,210 kg/m³ and fine aggregate 663 kg/m³ సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: A. 4.20 kg and approximately 0.00365 m³ సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. OPC 360 kg/m³, fly ash 90 kg/m³, water 171 kg/m³ సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: C. 428 kg/m³ సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: C. 0.644 సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: A. Coarse 0.5796 and fine 0.4204 సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. 0.660 m³ సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: A. Coarse 1,122 kg/m³ and fine 675 kg/m³ సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: A. 1,194.03 kg and 5.97 kg water సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: C. 669.5 kg and 19.5 kg సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: A. 136 kg/m³ సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: C. 18.22 kg/m³ సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: B. 177 kg/m³ సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: D. Cement 20 kg, water 9 kg, fine aggregate 32.5 kg, coarse aggregate 60 kg, admixture 0.20 kg సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: C. 20 kg/m³ సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: C. 466.67 kg/m³ and 381.82 kg/m³ సమాధానం: C.