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APSC AE CE Paper 1 Mock Test - 5 - Civil Engineering (CE) MCQ


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30 Questions MCQ Test APSC AE CE Mock Test Series 2024 - APSC AE CE Paper 1 Mock Test - 5

APSC AE CE Paper 1 Mock Test - 5 for Civil Engineering (CE) 2024 is part of APSC AE CE Mock Test Series 2024 preparation. The APSC AE CE Paper 1 Mock Test - 5 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The APSC AE CE Paper 1 Mock Test - 5 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for APSC AE CE Paper 1 Mock Test - 5 below.
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APSC AE CE Paper 1 Mock Test - 5 - Question 1

The saturated unit weight of the soil sample having specific gravity = 2.66, void ratio = 0.63 and unit weight of water = 9.81 kN/m3 is:

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 1

The relation between the degree of saturation (S), voids ratio (e), moisture content (W), and specific gravity (G) is given by,

e × S = W × G

The relation between the unit weight of saturated soil (γ­Sat), specific gravity (G), unit weight of water (γ­W), and voids ratio (e) is given by

APSC AE CE Paper 1 Mock Test - 5 - Question 2

If the area of cross-section of a single angle discontinuous strut is 30 cm2 and allowable working stress corresponding to its slenderness ratio is 625 kg/cm2 , the safe load carrying capacity of the member is 

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 2

As per IS : 800 - 1984 , When a single angle strut is connected to a gusset plate with one rivet, then allowable working stress corresponding to the slenderness ratio of the member, is reduced to 80%. So, Safe load carrying capacity of the member = (0.80*625) * (30) = 15 tonnes.

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APSC AE CE Paper 1 Mock Test - 5 - Question 3

Velocity distribution in a turbulent boundary layer follows:

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 3
In turbulent boundary layer velocity distribution follows ((1/7)th) power-law i.e., logarithmic law.
APSC AE CE Paper 1 Mock Test - 5 - Question 4

The plan of a building is in the form of a rectangular with centre dimensions of the outer walls as 10.3 and 17.3m. The thickness of the walls in the superstructure is 0.3m. Then its carpet area is

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 4

Calculation of Carpet Area:


  • Length of the building = 17.3m

  • Breadth of the building = 10.3m

  • Thickness of walls = 0.3m (applicable to all sides)

  •  

Calculating the Inner Dimensions:


  • Length of inner rectangle = Length of building - 2 * Thickness = 17.3m - 2 * 0.3m = 16.7m

  • Breadth of inner rectangle = Breadth of building - 2 * Thickness = 10.3m - 2 * 0.3m = 9.7m

  •  

Calculating Carpet Area:


  • Carpet Area = Length * Breadth = 16.7m * 9.7m = 162.19m2

Therefore, the carpet area of the building is 162.19 square meters, which is closest to option D: 170m2.

APSC AE CE Paper 1 Mock Test - 5 - Question 5

The variation in declination due to magnetic storms is called

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 5
Diurnal: It is a daily variation which changes with locality, season (more in summertime, more in day time)

Annual Variation: Yearly swing 1' or 2' in amplitude, it varies from place to place

Secular Variation: Has periodic character, (Pd. 250 years)

APSC AE CE Paper 1 Mock Test - 5 - Question 6

The relationship between tropical ground

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 6
Tropical year = 365.2422 days

Sidereal year = 365.2564 days

APSC AE CE Paper 1 Mock Test - 5 - Question 7

The Pitot tube is used to measure.

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 7
Pitot tube: It is a simple device used for measuring the velocity of flow. The basic principle used in this device is that if the speed of flow at a particular pit is reduced to zero, which is known as stagnation point, the pressure there is increased due to the conversion of the kinetic energy into pressure energy, and by measuring the increase in the the pressure energy at this point, the velocity of flow may be determined.
APSC AE CE Paper 1 Mock Test - 5 - Question 8

Venturimeter is advantageous because:

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 8
Venturimeter is advantageous because

(→) It has a much smaller head loss

(→) Its coefficient of discharge ((Cd=0.98)) is more

(→) Its accuracy is quite good.

APSC AE CE Paper 1 Mock Test - 5 - Question 9

The detention period for oxidation ponds is usually kept is -

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 9
Oxidation ponds have a long detention period between 10 to 15 days.The BOD removal efficiency of an oxidation pond lies between 80% and 90%, so the maximum BOD removal efficiency is 90%.
APSC AE CE Paper 1 Mock Test - 5 - Question 10

Centre of buoyancy always

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 10

Coincides with the centroid of the volume of displaced fluid

APSC AE CE Paper 1 Mock Test - 5 - Question 11

Directions: A pair of number is given. Select the pair which shows the similar relationship that is shown by the given pair. 1984 : 4891

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 11

The pattern is as follows,
1984 is the reverse of 4891
Similarly, 2874 is the reverse of 4782.
Hence the required pair is 2874 : 4782.

APSC AE CE Paper 1 Mock Test - 5 - Question 12

Irrigation efficiency of an irrigation system is the ratio of

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 12

It is defined as the ratio between the amount of water utilized (i.e., used to meet the consumptive use requirement of crop plus that necessary to maintain a favorable salt balance in the crop root zone) to the total volume of water diverted, stored or pumped for irrigation.

APSC AE CE Paper 1 Mock Test - 5 - Question 13

The length of the straight portion of a bar beyond the end of the hook should be at least-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 13

The length of the straight portion of a bar beyond the end of the hook curve should be at least four times the diameter.

Hence the option C is correct.

APSC AE CE Paper 1 Mock Test - 5 - Question 14

Slump test is used for-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 14
Slump test, Compacting factor test, and Vee-Bee consistometer method are used to measure workability.

Hence the option C is correct.

APSC AE CE Paper 1 Mock Test - 5 - Question 15

The purpose of lateral ties in short concrete columns is-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 15
The purpose of lateral ties in short concrete columns is to avoid buckling of longitudinal bars.

Hence the option A is correct.

APSC AE CE Paper 1 Mock Test - 5 - Question 16

Minimum thickness of main steel members, not exposed to the weather is-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 16
Where steelwork is directly exposed to weather, the steel is directly exposed to weather and is fully accessible for cleaning and repainting. The thickness shall be not less than 6 mm.

The steel is directly exposed to weather and is not accessible for cleaning and repainting. The thickness shall be not less than 8 mm. These provisions do not apply to the web of Indian standard rolled steel joists and channels to packings.

Steelwork not Directly Exposed to Weather. The thickness of steel in main members not directly exposed to weather shall be not less than 6 mm.

The thickness of steel in secondary members not directly exposed to weather shall be not less than 4-5 mm.

Hence the option B is correct.

APSC AE CE Paper 1 Mock Test - 5 - Question 17

If the LMT is 8h12m165 AM at 3845′ W longitude, the GMT will be

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 17
1 degree =4 minute time difference

Difference in longitude

∴3845′W=38.75

∴ Time difference =38.75×4=155 minutes

∵ GMT will be at east

∴ time will be 155 minutes ahead.

∴8hr12 minutes 16sec+155 minutes =10hr47 minutes 16secAM

Hence the option B is correct.

APSC AE CE Paper 1 Mock Test - 5 - Question 18

The modular ratio M is given by:

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 18
The modular ratio m has the value

(m=280/3σcbc)

Hence the option C is correct.

APSC AE CE Paper 1 Mock Test - 5 - Question 19

Shear reinforcement is provided in the form of-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 19
In practice, shear reinforcement is provided in three forms; stirrups, inclined bent-up bars, and a combination system of stirrups and bent-up bars. In reinforced concrete building construction, stirrups are most commonly used as shear reinforcement for their simplicity in fabrication and installation.

Hence the option D is correct.

APSC AE CE Paper 1 Mock Test - 5 - Question 20

IS code for designing pile foundation is :

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 20
Following are the Indian Standard Codes on Pile Foundations: IS 2911: Part 1: Sec 1: 1979 Driven cast-in-situ concrete piles. IS 2911: Part 1: Sec 2: 1979 Bored cast-in-situ piles. IS 2911: Part 1: Sec 3: 1979 Driven precast concrete pile.

Hence the correct answer is option A.

APSC AE CE Paper 1 Mock Test - 5 - Question 21

Chemical grouting is generally used for-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 21
The method is suitable for medium and fine sands. However, the effect of chemical grouting is not permanent.

Hence the correct answer is option C.

APSC AE CE Paper 1 Mock Test - 5 - Question 22

The property of the stones to withstand the adverse action of weather is known as

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 22
The stone used in pavement construction should be durable and should resist disintegration due to the action of the weather. The aggregates are subjected to the physical and chemical activity of rain and groundwater and also the atmosphere. Hence the road stones used in the construction should be sound enough to withstand the weathering action.

Hence the correct answer is option B.

APSC AE CE Paper 1 Mock Test - 5 - Question 23

For the determination of water content, the soil sample is heated for a period of 24 hrs. at a temperature of-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 23
Water Content Determination - Civil Engineering Notes. 110 ± 5°C. The period of 24 hours is sufficient for all normal soils to cause the evaporation of water from them. If the soil contains organic matter, then the oven temperature should be between 60 to 80 degrees Centigrade.
APSC AE CE Paper 1 Mock Test - 5 - Question 24

The effective length of a fillet weld should not be less than-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 24
The effective length of a fillet weld may be taken as the overall length of the full-size fillet less one leg length, s, for each end which does not continue around a corner. However, a fillet weld with an effective length less than 4s or less than 40 mm should not be used to carry the load.
APSC AE CE Paper 1 Mock Test - 5 - Question 25

Factor of safety is the ratio of-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 25
Factor of safety =Yield stress/ working stress
APSC AE CE Paper 1 Mock Test - 5 - Question 26

If the maximum bending moment of a simply supported slab is M kg-cm, the effective depth of the slab is-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 26
If the maximum bending moment of a simply supported slab is M kg-cm, the adequate depth of the slab is; hence correct answer is option D.
APSC AE CE Paper 1 Mock Test - 5 - Question 27

For a number of columns constructed in a row, the type of foundation provided is-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 27
For a number of columns constructed in a row, the type of foundation provided is the strip.
APSC AE CE Paper 1 Mock Test - 5 - Question 28

According to IS: 875 Part 3, desiring wind speed is obtained by multiplying the basic wind speed by factors (k1,k2) and (k3), where (k3) is-

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 28
According to IS: 875 Part 3, desiring wind speed is obtained by multiplying the basic wind speed by factors (k1,k2) and (k3), where (k3) is the topography factor.

Hence the correct answer is option C.

APSC AE CE Paper 1 Mock Test - 5 - Question 29

Air valves in a distribution system are provided at:

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 29
Air Valve Location Basically, air valves for exhausting air should be located at points on the pipeline to which air tends to be drawn or where air tends to collect. Air valves for air intake should be located at points on the pipeline that are most susceptible to sub-atmospheric pressures.
APSC AE CE Paper 1 Mock Test - 5 - Question 30

When a coolie walks on a horizontal platform with a load on his head, the work done by the coolie on the load is zero.

Detailed Solution for APSC AE CE Paper 1 Mock Test - 5 - Question 30
When a coolie walks on a horizontal platform with a load on his head, he applies force in an upward direction equal to its weight. The displacement of the load is along the horizontal direction. Thus the work done by the coolie on the load is zero.

Hence, the correct option is (A)

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