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GATE’22 CE:- Daily Practice Quiz 27-Aug-2021

IIT Kharagpur has released the GATE 2022 exam dates. As per the GATE 2022 exam date, the exam will be conducted on February 5, 6, 12 & 13. Candidates preparing for GATE 2022 must attempt this Civil Engineering Quiz to boost your GATE exam preparation.

Quiz: Civil Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 marks
Negative marking: 1/3 mark
Time: 12 Minutes

Q1. A 4-hour rainfall in a catchment of 250 sq.km, produces rainfall depth of 6.2 cm and 5.0 cm in successive 2-hour unit period. Assuming the ϕ – index of the soil to be 1.2 cm/hour, the runoff volume in hectare meter will be?
(a) 2200
(b) 1600
(c) 22
(d) 16

Q2. If the principal stresses at a point in a stressed body are 150 kN/m² tensile and 50 kN/m² compressive, then maximum shear stress at this point will be
(a) 250 kN/m²
(b) 150 kN/m²
(c) 100 kN/m²
(d) 50 kN/m²

Q3. The batching tolerance for cement as per IS 456 is
(a) ±1%
(b) ±1.5%
(c) ±2%
(d) ±3%

Q4. The optimistic, most likely, and pessimistic estimates of time for an activity are 4 days, 11 days and 12 days respectively. the expected completion time of this activity is
(a) 8 days
(b) 9 days
(c) 10 days
(d) 11 days

Q5. The concrete mix design is conducted as per
(a) IS: 383
(b) IS: 456
(c) IS: 10262
(d) IS: 13920

Solutions

S1. Ans.(b)
Sol.

Time(hour) 0-2 2-4
Rainfall (Cm) 6.2 5.0

ϕ-index=1.2 cm\/hour
A=250 km^2=250×10^(6 ) m^2
We know,
ϕ-index=(P-Q)/t
Where,
P=Precipitation\Rainfall
t=Duration of rainfall
Q = Runoff
1.2=((6.2+5)-Q)/4
Q=11.2-4.8
▭(Q=6.4 cm)
Runoff volume = total runoff × Area of catchment
=6.4×10^(-2)×250×10^6
=1600×10^4 m^3
=1600 ha-m (∵1hecatare=10^(4 ) m^2 )

S2. Ans.(c)
Sol, Given,
σ_1= +150 kN\/m^2 (tensile)
σ_2= -50 kN\/m^2 (compressive)
τ=0
We know →
Maximum shear stress = Radius of Mohr circle
√(((σ_1-σ_2)/2)^2+τ^2 )
√([150-(-50)]^2/4+(0)^2 )
√((100)^2 )
=100 kN\/m^2

S3. Ans.(c)
Sol.
Material and their Batching tolerance limit –
Aggregate, Water, admixture = ±3%
Cement = ± 2%

S4. Ans.(c)
Sol. Given, Optimistic time (t_0 )=4
Most likely time (t_m )=11
Pessimistic time (t_P )=12
Expected time (t_E )= ?
We know,
t_E=(t_0+4 t_m+t_P)/6
=(4+(4×11)+12)/6
=60/6
▭(t_E=10 days)

S5. Ans.(c)
Sol. IS : 10262 → Concrete mix design
IS : 456 → Reinforced concrete
IS : 1343 → Prestress concrete

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