**Quiz: Civil EngineeringExam: All AE/JE ExamsTopic: Miscellaneous**

**Each question carries 1 markNegative marking: 1/4 markTime: 10 Minutes**

Q1. The reduction coefficient of a reinforced concrete column with an effective length of 4.8 m and size 250 mm × 300 mm is

(a) 0.80

(b) 0.85

(c) 0.90

(d) 0.95

Q2. What will be the dry density of soil if a soil has a bulk density of 22 kN/m^3 and water content 10%

(a) 18.6 kN\/m^3

(b) 22.0 kN\/m^2

(c) 23.2 kN\/m^3

(d) 20.0 kN\/m^3

Q3. Secondary consolidation is caused due to

(a) Triaxial stress

(b) Creep

(c) Hydrostatic pressure

(d) Steel gradient

Q4. The natural bedding plane of stones and the direction of pressure in stone masonry is –

(a) At 45°

(b) Parallel

(c) At 30°

(d) Normal

Q5. The process of mixing water with quick lime is called:

(a) Cooling of lime

(b) Slaking of lime

(c) Curing of lime

(d) Calcination of lime

Q6. A channel bed slope 0.0009 carries a discharge of 30 m³/s when the depth of flow is 1.0m. what is the discharge carried by an exactly similar channel at the same depth of flow if the slope is decreased to 0.0001?

(a) 10 m³/s

(b) 15 m³/s

(c) 60 m³/s

(d) 90 m³/s

Q7. What will be the effect on Euler’s crippling load for a column with both ends hinged, if the length of the column is halved?

(a) 0.25

(b) 0.5

(c) 2

(d) 4

Q8. Find the number of sleepers required for 1280 m long B.G. track. Take sleeper density (M +5) and length of rail as 12.8 m. (terms have their usual meaning)

(a) 1800

(b) 1000

(c) 1200

(d) 1500

Q9. Influence line diagrams can be drawn for

(a) Only beams

(b) Beams arches, trusses etc.

(c) Only arches

(d) Only trusses

Q10. The quantity of cement required in 1 cum of 1:2:4 cement concrete is approximately

(a) 0.11 cu m

(b) 0.22 cu m

(c) 0.33 cu m

(d) 0.44 cu m

Solutions

S1. Ans.(b)

Sol. Leff = 4.8 m.

Size = 250 mm × 350 mm

Reduction coefficient (Cr) = 1.25 – left/48D

=1.25-4800/(48×300)

=0.92

(And)

C_r=1.25-left/48B

=1.25-4800/(48×250)

=0.85

Cr=[0.92,0.85]_(whichever is leaser )

▭(Cr=0.85)

S2. Ans.(d)

Sol. Bulk density (Υ_b )=22 kN\/m^3

Water content (w)=10%

Dry density (Υ_d )=Υ_b/(1+w)

=22/(1+0.1)

=20.0 kN\/m^3

S3. Ans.(b)

Sol. Secondary consolidation is caused due to creep by the plastic rearrangement of soil solids.

S4. Ans.(d)

Sol. In stone masonry, the stones are placed in position such that the natural bedding plane is normal to the direction of pressure they carry. The stone used in stone masonry should be hard, strong and durable.

S5. Ans.(b)

Sol. Slaked lime is obtained when quick lime reacts with water, this reaction is highly exothermic. This process is known as slaking of lime. During slaking process, the volume of lime increases.

CaO+H_2 O→Ca(OH)_2+Heat

S6. Ans.(a)

Sol.

Q_1= 30 m³/sec

S_1 = 0.009

Q_2 = ?

S_2 = 0.0001

→ Similar channel means that area and wetted perimeter are same.

Q_2/Q_1 =√(S_2/S_1 )

Q_2/30=√(0.0001/0.0009)

▭(Q_2=10 m^3 \/sec)

S7. Ans.(d)

Sol. Euler’s Buckling load formula-

P_C=(π^2 EI)/(le^2 )

For both End hinged (le = l)

P_c=(π^2 EI)/l^2

If length column is halved [le=l\/2]

Pc^’=(π^2 EI)/(le)^2 =(4π^2 EI)/l^2

Pc^’=4(Pc)

S8. Ans.(a)

Sol. Number of sleepers per rail length generally varies from (M+2) to (M+7).

M = Length of a rail (in m)

for 12.8m rail length, the sleeper density is given = (M+5)

= (12.8 + 5)

= 17.8 ≃ 18

sleeper No. of sleeper required for 1280 m. long B.G. track = (1280×18)/128

= 1800 sleeper.

S9. Ans.(b)

Sol. The influence line diagram is the graphical representation for reaction, shear force and bending moment due to moving load. It can be drawn for beams, arches, trusses etc.

S10. Ans.(b)

Sol. The volume of dry concrete is 1.52 times of quantity of quantity of wet concrete.

Hence.

Quantity of dry cement for 1 m³ wet cement concrete = 1 × 1.52 = 1.52 m³.

Quantity of cement in 1:2:4 proportion of cement = 1/7×1.52=0.217 m^3 ≃ 0.22 m³.